Class 12 physics, NCERT chapter 3. Attempt each question, check your answer, then open the step-by-step solution.
Three $6\ \Omega$ resistors are connected in parallel across an ideal $12\ \text{V}$ battery. The current drawn from the battery is
Answer: (D) $6\ \text{A}$
$$R_{eq} = \frac{6}{3} = 2\ \Omega, \qquad I = \frac{V}{R_{eq}} = \frac{12}{2} = 6\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics
The equivalent resistance between points A and B in the circuit shown is
Answer: (B) $6\ \Omega$
The $3\ \Omega$ and $6\ \Omega$ resistors are in parallel:
$$R_p = \frac{3\times 6}{3+6} = 2\ \Omega$$
This is in series with $4\ \Omega$: $R_{AB} = 4 + 2 = 6\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
Consider two circuits, (A) and (B), each having two resistors. One of them has a positive temperature coefficient of resistance, $+\alpha$, while the other one has a negative temperature of coefficient, $-\alpha$, as shown in the figure. The current through these circuits are denoted by $I_A$ and $I_B$. At initial temperature, the resistance of the two resistors is $R_0$. As the temperature is increased, the correct option that describes the variation of current in these circuits is :
Answer: (A) $I_A$ remains constant while $I_B$ increases
Circuit (A), series:
$$R_A = R_0(1 - \alpha\Delta T) + R_0(1 + \alpha\Delta T) = 2R_0$$
This does not change, so $I_A$ remains constant.
Circuit (B), parallel:
$$R_B = \frac{R_0(1 - \alpha\Delta T)\,R_0(1 + \alpha\Delta T)}{2R_0} = \frac{R_0}{2}\left(1 - \alpha^2\Delta T^2\right)$$
This decreases as $\Delta T$ increases, so $I_B$ increases.
Solution by Sreeraj P, M.Sc Physics
The current passing through the battery in the given circuit, is:
Answer: (B) $0.5$ A
A and F are joined by a plain wire, and so are C and D. The network between B and E therefore has:
$R_{BA} = 5\ \Omega$, $R_{BD} = 2.5\ \Omega$ (through C), $R_{AE} = 3\ \Omega$ (through F), $R_{DE} = 1.5\ \Omega$, and $R_{AD} = 6\ \Omega$ as the bridge.
Check the Wheatstone condition:
$$\frac{R_{BA}}{R_{BD}} = \frac{5}{2.5} = 2, \qquad \frac{R_{AE}}{R_{DE}} = \frac{3}{1.5} = 2$$
The bridge is balanced, so no current flows through the $6\ \Omega$ resistor.
$$R_{BE} = (5 + 3) \parallel (2.5 + 1.5) = \frac{8 \times 4}{8 + 4} = \frac{8}{3}\ \Omega$$
Total resistance in the battery loop:
$$R = \frac{8}{3} + 1.5 + 5.5 + \frac{1}{3} = 3 + 7 = 10\ \Omega$$
$$I = \frac{5}{10} = 0.5\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics
A wire of resistance R is cut into $8$ equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is :
Answer: (C) $\dfrac{R}{16}$
Each piece has resistance $\dfrac{R}{8}$.
Four in parallel: $\dfrac{R/8}{4} = \dfrac{R}{32}$.
Two such sets in series: $2 \times \dfrac{R}{32} = \dfrac{R}{16}$.
Solution by Sreeraj P, M.Sc Physics
A constant voltage of $50$ V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is :
Answer: (B) $2.0$ A
C and D are joined by a wire, so they are at the same potential. The circuit is ($1\ \Omega \parallel 3\ \Omega$) in series with ($2\ \Omega \parallel 4\ \Omega$):
$$R = \frac{1 \times 3}{4} + \frac{2 \times 4}{6} = \frac{3}{4} + \frac{4}{3} = \frac{25}{12}\ \Omega$$
$$I = \frac{50}{25/12} = 24\ \text{A}$$
Current through $1\ \Omega$ (into C): $24 \times \dfrac{3}{4} = 18$ A.
Current through $2\ \Omega$ (out of C): $24 \times \dfrac{4}{6} = 16$ A.
Current in CD $= 18 - 16 = 2.0$ A (from C to D).
Solution by Sreeraj P, M.Sc Physics
A wire of length '$l$' and resistance $100\ \Omega$ is divided into $10$ equal parts. The first $5$ parts are connected in series while the next $5$ parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:
Answer: (D) $52\ \Omega$
Each part has resistance $\dfrac{100}{10} = 10\ \Omega$.
Five in series: $5 \times 10 = 50\ \Omega$.
Five in parallel: $\dfrac{10}{5} = 2\ \Omega$.
Total: $50 + 2 = 52\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
The terminal voltage of the battery, whose emf is $10$ V and internal resistance $1\ \Omega$, when connected through an external resistance of $4\ \Omega$ as shown in the figure is :
Answer: (A) $8$ V
$$I = \frac{E}{R + r} = \frac{10}{4 + 1} = 2\ \text{A}$$
$$V = E - Ir = 10 - 2 \times 1 = 8\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics
Two heaters A and B have power rating of $1$ kW and $2$ kW, respectively. Those two are first connected in series and then in parallel to a fixed power source. The ratio of power outputs for these two cases is:
Answer: (D) $2:9$
With the same rated voltage $V$: $R_A = \dfrac{V^2}{1000}$ and $R_B = \dfrac{V^2}{2000}$.
Series: $R_s = R_A + R_B = \dfrac{3V^2}{2000}$, so
$$P_s = \frac{V^2}{R_s} = \frac{2000}{3}\ \text{W}$$
Parallel: the powers add, so
$$P_p = 1000 + 2000 = 3000\ \text{W}$$
$$\frac{P_s}{P_p} = \frac{2000/3}{3000} = \frac{2}{9}$$
Solution by Sreeraj P, M.Sc Physics
If the galvanometer $G$ does not show any deflection in the circuit shown, the value of $R$ is given by :
Answer: (C) $100\ \Omega$
No current flows through $G$, so the $2$ V cell carries no current. The potential difference across $R$ must then equal the emf of the $2$ V cell.
The $10$ V battery drives a current through $400\ \Omega$ and $R$ in series:
$$I = \frac{10}{400 + R}$$
$$IR = 2 \;\Rightarrow\; \frac{10R}{400 + R} = 2 \;\Rightarrow\; 10R = 800 + 2R \;\Rightarrow\; R = 100\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
Resistance of a carbon resistor determined from colour codes is $(22000 \pm 5\%)\ \Omega$. The colour of third band must be :
Answer: (C) Orange
$22000 = 22 \times 10^3$.
The first two bands give the digits $2, 2$ (red, red). The third band is the multiplier $10^3$, which is orange.
(The $\pm 5\%$ tolerance is the gold fourth band.)
Solution by Sreeraj P, M.Sc Physics
The magnitude and direction of the current in the following circuit is
Answer: (B) $0.5$ A from $A$ to $B$ through $E$
The two cells have their positive terminals facing each other at E, so they oppose. The $10$ V cell wins, and the net emf is $10 - 5 = 5$ V.
Total resistance: $2 + 1 + 7 = 10\ \Omega$.
$$I = \frac{5}{10} = 0.5\ \text{A}$$
The $10$ V cell drives current out of its positive terminal towards E, i.e. from A to B through E.
Solution by Sreeraj P, M.Sc Physics
$10$ resistors, each of resistance $R$ are connected in series to a battery of emf $E$ and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased $n$ times. The value of $n$ is :
Answer: (B) $100$
Series: $R_s = 10R$, so $I_s = \dfrac{E}{10R}$.
Parallel: $R_p = \dfrac{R}{10}$, so $I_p = \dfrac{10E}{R}$.
$$n = \frac{I_p}{I_s} = 100$$
Solution by Sreeraj P, M.Sc Physics
The resistance of platinum wire at $0°$C is $2\ \Omega$ and $6.8\ \Omega$ at $80°$C. The temperature coefficient of resistance of the wire is :
Answer: (C) $3 \times 10^{-2}\ °\text{C}^{-1}$
$R_t = R_0(1 + \alpha t)$:
$$\alpha = \frac{R_t - R_0}{R_0\,t} = \frac{6.8 - 2}{2 \times 80} = \frac{4.8}{160} = 3 \times 10^{-2}\ °\text{C}^{-1}$$
Solution by Sreeraj P, M.Sc Physics
In a potentiometer circuit a cell of EMF $1.5$ V gives balance point at $36$ cm length of wire. If another cell of EMF $2.5$ V replaces the first cell, then at what length of the wire, the balance point occurs ?
Answer: (B) $60$ cm
The balancing length is proportional to the emf: $E \propto l$.
$$l_2 = 36 \times \frac{2.5}{1.5} = 60\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics
The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is $0.25\ \Omega$. What will be the effective resistance if they are connected in series ?
Answer: (A) $4\ \Omega$
Parallel: $\dfrac{R}{4} = 0.25$, so each wire has $R = 1\ \Omega$.
Series: $4R = 4\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
Column - I gives certain physical terms associated with flow of current through a metallic conductor. Column - II gives some mathematical relations involving electrical quantities. Match Column - I and Column - II with appropriate relations.
Column - I:
(A) Drift Velocity
(B) Electrical Resistivity
(C) Relaxation Period
(D) Current Density
Column - II:
(P) $\dfrac{m}{ne^2\rho}$
(Q) $nev_d$
(R) $\dfrac{eE}{m}\tau$
(S) $\dfrac{E}{J}$
Answer: (B) (A)-(R), (B)-(S), (C)-(P), (D)-(Q)
(A) Drift velocity: $v_d = \dfrac{eE}{m}\tau$ (R).
(B) Resistivity: $\rho = \dfrac{E}{J}$ (from $J = \sigma E$) (S).
(C) Relaxation time: from $\rho = \dfrac{m}{ne^2\tau}$, $\tau = \dfrac{m}{ne^2\rho}$ (P).
(D) Current density: $J = nev_d$ (Q).
(A)-(R), (B)-(S), (C)-(P), (D)-(Q).
Solution by Sreeraj P, M.Sc Physics
Three resistors having resistances $r_1$, $r_2$ and $r_3$ are connected as shown in the given circuit. The ratio $\dfrac{i_3}{i_1}$ of currents in terms of resistances used in the circuit is :
Answer: (C) $\dfrac{r_2}{r_2 + r_3}$
$i_1$ splits between $r_2$ and $r_3$, which are in parallel. The current divides in inverse ratio of resistance:
$$i_3 = i_1\frac{r_2}{r_2 + r_3} \;\Rightarrow\; \frac{i_3}{i_1} = \frac{r_2}{r_2 + r_3}$$
Solution by Sreeraj P, M.Sc Physics
The solids which have the negative temperature coefficient of resistance are :
Answer: (D) insulators and semiconductors
In semiconductors and insulators, heating frees more charge carriers across the band gap, so the resistance falls as the temperature rises: a negative temperature coefficient.
In metals, the carrier number is fixed and more collisions at higher temperature raise the resistance: a positive coefficient.
Solution by Sreeraj P, M.Sc Physics
Which of the following graph represents the variation of resistivity ($\rho$) with temperature (T) for copper ?
Answer: (C) see figure
For a metal such as copper, resistivity rises with temperature. It has a small non-zero value at low temperature and increases more steeply as the temperature rises (it is not a straight line over a wide range).
Graph (3) shows this rising, upward-curving shape.
Solution by Sreeraj P, M.Sc Physics
The color code of a resistance is given below : The values of resistance and tolerance, respectively, are :
Answer: (D) $470\ \Omega$, $5\%$
Yellow $= 4$, Violet $= 7$, Brown $=$ multiplier $10^1$, Gold $=$ tolerance $\pm 5\%$.
$$R = 47 \times 10 = 470\ \Omega,\ \pm 5\%$$
Solution by Sreeraj P, M.Sc Physics
A charged particle having drift velocity of $7.5 \times 10^{-4}\ \text{m s}^{-1}$ in an electric field of $3 \times 10^{-10}\ \text{V m}^{-1}$, has a mobility in $\text{m}^2\text{V}^{-1}\text{s}^{-1}$ of :
Answer: (B) $2.5 \times 10^6$
$$\mu = \frac{v_d}{E} = \frac{7.5 \times 10^{-4}}{3 \times 10^{-10}} = 2.5 \times 10^6\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics
A resistance wire connected in the left gap of a metre bridge balances a $10\ \Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3 : 2$. If the length of the resistance wire is $1.5$ m, then the length of $1\ \Omega$ of the resistance wire is :
Answer: (B) $1.0 \times 10^{-1}$ m
Balance condition: $\dfrac{R}{10} = \dfrac{3}{2}$, so $R = 15\ \Omega$.
$15\ \Omega$ over $1.5$ m, so $1\ \Omega$ corresponds to
$$\frac{1.5}{15} = 0.1\ \text{m} = 1.0 \times 10^{-1}\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics
A current of $3.2$ A flows through a wire. The number of electrons crossing any cross-section per second is ($e = 1.6 \times 10^{-19}$ C)
Answer: (A) $2 \times 10^{19}$
$n = \dfrac{I}{e} = \dfrac{3.2}{1.6 \times 10^{-19}} = 2 \times 10^{19}$ per second.
Solution by Sreeraj P, M.Sc Physics
A copper wire of cross-section $1\ \text{mm}^2$ carries $1.6$ A. If the number density of free electrons is $10^{29}\ \text{m}^{-3}$, the drift speed is ($e = 1.6 \times 10^{-19}$ C)
Answer: (B) $0.1$ mm/s
$v_d = \dfrac{I}{neA} = \dfrac{1.6}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}} = 10^{-4}$ m/s $= 0.1$ mm/s.
Solution by Sreeraj P, M.Sc Physics
A wire is stretched uniformly so that its length doubles. Its resistance becomes
Answer: (C) $4$ times
Volume is constant, so doubling $L$ halves $A$. $R = \dfrac{\rho L}{A}$ becomes $\dfrac{\rho(2L)}{A/2} = 4R$.
Solution by Sreeraj P, M.Sc Physics
A wire of resistance $R$ is stretched until its radius becomes half. Its new resistance is
Answer: (D) $16R$
Halving the radius makes $A$ one-fourth; constant volume makes $L$ four times. $R' = R \times 4 \times 4 = 16R$.
Solution by Sreeraj P, M.Sc Physics
A $12\ \Omega$ wire is bent into a circle. The resistance between two points at the ends of a diameter is
Answer: (A) $3\ \Omega$
The two semicircles ($6\ \Omega$ each) are in parallel: $\dfrac{6 \times 6}{12} = 3\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
A $12\ \Omega$ wire is bent into a circle. The resistance between two points that subtend $90°$ at the centre is
Answer: (B) $2.25\ \Omega$
The arcs are $3\ \Omega$ and $9\ \Omega$ in parallel: $\dfrac{3 \times 9}{12} = 2.25\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
A cell of emf $2$ V and internal resistance $0.5\ \Omega$ is connected to a $3.5\ \Omega$ resistor. The current and terminal voltage are
Answer: (C) $0.5$ A and $1.75$ V
$I = \dfrac{2}{3.5 + 0.5} = 0.5$ A. $V = E - Ir = 2 - 0.25 = 1.75$ V.
Solution by Sreeraj P, M.Sc Physics
A cell gives $0.5$ A through a $4\ \Omega$ resistor and $0.9$ A through a $2\ \Omega$ resistor. Its internal resistance is
Answer: (D) $0.5\ \Omega$
$E = 0.5(4 + r) = 0.9(2 + r) \Rightarrow 2 + 0.5r = 1.8 + 0.9r \Rightarrow r = 0.5\ \Omega$ (and $E = 2.25$ V).
Solution by Sreeraj P, M.Sc Physics
A battery of emf $E$ and internal resistance $r$ delivers maximum power to an external resistor $R$ when
Answer: (A) $R = r$
$P = \dfrac{E^2R}{(R + r)^2}$ is maximum when $\dfrac{dP}{dR} = 0$, which gives $R = r$. Then $P_{max} = \dfrac{E^2}{4r}$.
Solution by Sreeraj P, M.Sc Physics
Four identical cells, each of emf $1.5$ V and internal resistance $0.5\ \Omega$, are connected in series across a $4\ \Omega$ resistor. The current through the resistor is
Answer: (B) $1$ A
$E = 6$ V, internal resistance $= 2\ \Omega$. $I = \dfrac{6}{4 + 2} = 1$ A.
Solution by Sreeraj P, M.Sc Physics
Two cells of emf $2$ V and $4$ V, with internal resistances $1\ \Omega$ and $2\ \Omega$, are connected in parallel (like terminals together). The emf of the combination is
Answer: (C) $\dfrac{8}{3}$ V
$E_{eq} = \dfrac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \dfrac{2 + 2}{1 + 0.5} = \dfrac{8}{3}$ V.
Solution by Sreeraj P, M.Sc Physics
A $6$ V battery of negligible internal resistance is connected across $3\ \Omega$ in series with a parallel combination of $6\ \Omega$ and $3\ \Omega$. The current through the $6\ \Omega$ resistor is
Answer: (D) $0.4$ A
$6\ \Omega \parallel 3\ \Omega = 2\ \Omega$. Total $= 5\ \Omega$, so $I = \dfrac{6}{5} = 1.2$ A.
Voltage across the parallel pair $= 1.2 \times 2 = 2.4$ V, so the $6\ \Omega$ resistor carries $\dfrac{2.4}{6} = 0.4$ A.
Solution by Sreeraj P, M.Sc Physics
In a Wheatstone bridge, $P = 10\ \Omega$, $Q = 20\ \Omega$ and $R = 15\ \Omega$. For balance, the fourth resistance $S$ must be
Answer: (A) $30\ \Omega$
Balance: $\dfrac{P}{Q} = \dfrac{R}{S} \Rightarrow S = \dfrac{QR}{P} = \dfrac{20 \times 15}{10} = 30\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
In a metre bridge, the null point is at $40$ cm from the left end when a $6\ \Omega$ resistor is in the left gap. The resistance in the right gap is
Answer: (B) $9\ \Omega$
$\dfrac{6}{S} = \dfrac{40}{60} \Rightarrow S = 9\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
The emfs of two cells are compared with a potentiometer. The balancing lengths are $300$ cm and $200$ cm when used singly, and the cells are then connected in series so as to aid each other. The balancing length for the combination is
Answer: (C) $500$ cm
$E \propto l$, so aiding emfs add: $300 + 200 = 500$ cm.
Solution by Sreeraj P, M.Sc Physics
A potentiometer wire $4$ m long has a potential difference of $2$ V across it. A cell balances at $2.5$ m. Its emf is
Answer: (D) $1.25$ V
Potential gradient $= \dfrac{2}{4} = 0.5$ V/m. $E = 0.5 \times 2.5 = 1.25$ V.
Solution by Sreeraj P, M.Sc Physics
A cell balances at $60$ cm on a potentiometer when on open circuit, and at $50$ cm when a $5\ \Omega$ resistor is connected across it. Its internal resistance is
Answer: (A) $1\ \Omega$
$r = R\left(\dfrac{l_1}{l_2} - 1\right) = 5\left(\dfrac{60}{50} - 1\right) = 1\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
A $100$ W, $220$ V bulb is operated on a $110$ V supply. The power it consumes is
Answer: (B) $25$ W
$R = \dfrac{V^2}{P}$ is fixed, so $P \propto V^2$: $100 \times \left(\dfrac{1}{2}\right)^2 = 25$ W.
Solution by Sreeraj P, M.Sc Physics
A $60$ W bulb and a $100$ W bulb, both rated for $220$ V, are connected in series to a $220$ V supply. Which bulb glows brighter?
Answer: (C) the $60$ W bulb
$R = \dfrac{V^2}{P}$, so the $60$ W bulb has the larger resistance. In series the current is the same, and $P = I^2R$ is larger for the larger resistance: the $60$ W bulb glows brighter.
Solution by Sreeraj P, M.Sc Physics
An electric heater of $1$ kW is used for $2$ hours a day for $30$ days. At ₹$6$ per unit, the cost is
Answer: (D) ₹$360$
Energy $= 1 \times 2 \times 30 = 60$ kWh (units). Cost $= 60 \times 6 = $ ₹$360$.
Solution by Sreeraj P, M.Sc Physics
A heater coil boils a certain amount of water in $10$ min. If the coil is cut in half and one half is used on the same supply, the same water boils in
Answer: (A) $5$ min
Half the length gives half the resistance, so $P = \dfrac{V^2}{R}$ doubles and the time halves: $5$ min.
Solution by Sreeraj P, M.Sc Physics
Using Kirchhoff's laws, find the current, in amperes, through a $2\ \Omega$ resistor connected between the positive terminals of two cells whose negative terminals are joined, with emfs $10$ V and $4$ V and negligible internal resistance.
Numerical value type. Enter your answer.
Answer: 3
The loop has net emf $10 - 4 = 6$ V (the cells oppose each other around the loop) and resistance $2\ \Omega$: $I = 3$ A.
Solution by Sreeraj P, M.Sc Physics
The resistivity of a wire is $1.7 \times 10^{-8}\ \Omega$ m. Find the resistance, in ohm, of $100$ m of it with cross-section $1\ \text{mm}^2$.
Numerical value type. Enter your answer.
Answer: 1.7
$R = \dfrac{\rho L}{A} = \dfrac{1.7 \times 10^{-8} \times 100}{10^{-6}} = 1.7\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
The stored charge in the capacitor in steady state of the following circuit is ______ $\mu$C.
Numerical value type. Enter your answer.
Answer: 200
In steady state no current flows through the capacitor branch, so the $10\ \Omega$ in series with it carries no current and the capacitor voltage equals the voltage across the $4\ \Omega$ vertical resistor.
Reduce the ladder from the right:
Voltage across the $10\ \Omega$ vertical $= 12 - 1 \times 7 = 5$ V. Current in the right branch $= \dfrac{5}{10} = 0.5$ A, so the $4\ \Omega$ vertical has $2$ V.
$Q = CV = 100\ \mu\text{F} \times 2\ \text{V} = 200\ \mu$C.
Solution by Sreeraj P, M.Sc Physics
Refer to the circuit diagram given below. The heat generated across the $6\ \Omega$ resistance in $100$ second is $\dfrac{\alpha}{100}$ J. The value of $\alpha$ is ______. (Nearest integer)
Numerical value type. Enter your answer.
Answer: 3477
Label the node joining the $3\ \Omega$, $4\ \Omega$ and $6\ \Omega$ as $X$, and take the right-hand wire (joined to the negative terminals of both cells) as $0$ V. The left wire is then at $+2$ V, and the $6\ \Omega$ branch ends at the $+$ terminal of the $3$ V cell, i.e. at $3$ V.
KCL at $X$:
$$\frac{V_X - 2}{3} + \frac{V_X}{4} + \frac{V_X - 3}{6} = 0 \;\Rightarrow\; 9V_X = 14 \;\Rightarrow\; V_X = \frac{14}{9}\ \text{V}$$
Current in $6\ \Omega$: $I = \dfrac{3 - 14/9}{6} = \dfrac{13}{54}$ A.
Heat in $100$ s: $I^2Rt = \dfrac{169}{2916} \times 6 \times 100 \approx 34.77$ J $= \dfrac{3477}{100}$ J, so $\alpha = 3477$.
Solution by Sreeraj P, M.Sc Physics
Two cells of emfs $1$ V and $2$ V and internal resistance $2\ \Omega$ and $1\ \Omega$, respectively connected in parallel, gave a current of $1$ A through an external resistance. If the polarity of one cell is reversed, then value of current through the external resistance will be $\dfrac{\alpha}{5}$ A. The value of $\alpha$ is ______.
Numerical value type. Enter your answer.
Answer: 3
Concept. Cells in parallel act as one cell: $E_{\text{eq}}=\dfrac{E_1/r_1+E_2/r_2}{1/r_1+1/r_2}$, $r_{\text{eq}}=\dfrac{r_1r_2}{r_1+r_2}$. Reversing a cell changes the sign of its emf.
Step 1. Original. $E_{\text{eq}}=\dfrac{0.5+2}{1.5}=\dfrac53$ V, $r_{\text{eq}}=\dfrac23\ \Omega$; $1=\dfrac{5/3}{2/3+R}\Rightarrow R=1\ \Omega$.
Step 2. One reversed. $E_{\text{eq}}=\dfrac{-0.5+2}{1.5}=1$ V.
Step 3. Current. $\dfrac{1}{2/3+1}=\dfrac35$ A.
Answer. $\alpha=3$.
Solution by Sreeraj P, M.Sc Physics
Refer to the figure given below. The values of $I_1, I_2$ and $I_3$ are ______.
Answer: (A) $I_1 = 2.5$ A, $I_2 = 1.875$ A, $I_3 = 1.875$ A
Concept. Use node potentials with Kirchhoff's junction law. The 5 V cell fixes the difference between the top and bottom corners, so the unknowns reduce to two potentials. Take the left corner as $0$ V; call the top $T$, the bottom $D$ and the centre $C$.
Step 1. Fix the cells. The 5 V cell (positive at the top) gives $V_T=V_D+5$. The 10 V cell (positive towards the left corner) with $1\ \Omega$: the current $I_1$ from $C$ to the left corner is $I_1=V_C+10$.
Step 2. Junction law at $C$. Current leaving $C$ through the two $2\ \Omega$ resistors and the cell branch is zero: $\dfrac{V_C-V_T}{2}+\dfrac{V_C-V_D}{2}+(V_C+10)=0$.
Step 3. Junction law for $T$ and $D$ together (the 5 V cell's current stays inside): $\dfrac{0-V_T}{4}+\dfrac{0-V_D}{4}+\dfrac{V_C-V_T}{2}+\dfrac{V_C-V_D}{2}=0\Rightarrow4V_C=3(V_T+V_D)$.
Step 4. Solve. Step 2 is $2V_C-\dfrac{V_T+V_D}{2}+10=0$; with $V_T+V_D=\dfrac{4V_C}{3}$ from Step 3, $2V_C-\dfrac{2V_C}{3}+10=0\Rightarrow V_C=-7.5$ V. Then $V_T+V_D=-10$ and $V_T-V_D=5$: $V_T=-2.5$ V, $V_D=-7.5$ V.
Step 5. Currents. $I_1=-7.5+10=2.5$ A. In the lower $4\ \Omega$: $I_3=\dfrac{0-(-7.5)}{4}=1.875$ A. At $T$: $\dfrac{V_T-V_C}{2}=2.5$ A leaves towards $C$ and $\dfrac{0-V_T}{4}=0.625$ A arrives from the left, so the cell branch supplies $I_2=2.5-0.625=1.875$ A.
Answer. $I_1=2.5$ A, $I_2=1.875$ A, $I_3=1.875$ A (Option A).
Solution by Sreeraj P, M.Sc Physics
Refer to the figure given below, current between terminals $A$ and $B$ is ______ A.
Answer: (B) $1.25$
Each of the top three rows has three $5$ V cells (all the same way round) and three $3\ \Omega$ resistors: emf $15$ V, resistance $9\ \Omega$. The three rows are in parallel between the two side rails, so together they act as a $15$ V source with $3\ \Omega$ internal resistance.
The bottom row from $A$ to $B$ ($3 \times 3 = 9\ \Omega$) is the load:
$$I = \frac{15}{3 + 9} = 1.25\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics
Under steady state condition the potential difference across the capacitor in the circuit is ______ V.
Answer: (A) $0.5$
In steady state no current flows in the capacitor branch, so there is no drop across its $4\ \Omega$ resistor.
Current flows round the loop of the $2$ V cell, the $6\ \Omega$ and the $2\ \Omega$: $I = \dfrac{2}{8} = 0.25$ A.
The capacitor branch is in parallel with the $2\ \Omega$, so $V_C = 0.25 \times 2 = 0.5$ V.
Solution by Sreeraj P, M.Sc Physics
When an external resistance of $5\ \Omega$ is connected across terminals of a cell, a current of $0.25$ A flows through it. When the $5\ \Omega$ resistor is replaced by a $2\ \Omega$ resistor, a current of $0.5$ A flows through it. The internal resistance of the cell is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 1
Concept. $E=I(R+r)$ with the same $E$ and $r$ for both loads.
Step 1. Equate. $0.25(5+r)=0.5(2+r)\Rightarrow1.25+0.25r=1+0.5r$.
Answer. $r=1\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
The voltage and the current between $A$ and $B$ points shown in the circuit are ______.
Answer: (B) $24$ V, $4$ A
Concept. Identical branches (same emf, same internal resistance) in parallel behave like one cell with the same emf and the internal resistance divided by the number of branches. The third branch's 14 V and 13 V cells add to 27 V, so all four branches are identical.
Step 1. Equivalent source. $E=27$ V, $r=\dfrac34=0.75\ \Omega$.
Step 2. Load. The two $3\ \Omega$ resistors from $A$ to $B$ are in series: $6\ \Omega$.
Step 3. Current. $I=\dfrac{27}{6+0.75}=4$ A.
Step 4. Voltage. $V_{AB}=4\times6=24$ V.
Answer. $24$ V, $4$ A (Option B).
Solution by Sreeraj P, M.Sc Physics
Two resistors of $200\ \Omega$ and $400\ \Omega$ are connected in series with a battery of $100$ V. A bulb rated at $200$ V, $100$ W is connected across the $400\ \Omega$ resistance. The potential drop across the bulb is ______ V.
Answer: (B) $50$
Concept. The bulb is a resistor, $R=\dfrac{V^2}{P}=400\ \Omega$, in parallel with the $400\ \Omega$ resistor.
Step 1. Parallel. $200\ \Omega$, in series with $200\ \Omega$.
Step 2. Voltage. Half of $100$ V $=50$ V.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
A Wheatstone bridge is initially at room temperature and all arms of the bridge have same value of resistances ($R_1 = R_2 = R_3 = R_4$). When $R_3$ resistance is heated to some temperature, its resistance value has gone up by 10%. The potential difference $(V_a - V_b)$ (after $R_3$ is heated) is ______ V.
Answer: (B) 0.95
Concept. In an unbalanced bridge each side is a potential divider across the cell. Find the potential at each midpoint and subtract.
Step 1. Side through $a$. $R_1=R_2$: $V_a=\dfrac{40}{2}=20$ V (taking the bottom as 0).
Step 2. Side through $b$. $R_3=1.1R$, $R_4=R$: $V_b=40\times\dfrac{R}{1.1R+R}=\dfrac{40}{2.1}\approx19.05$ V.
Step 3. Difference. $V_a-V_b\approx20-19.05=0.95$ V.
Answer. $0.95$ V (Option B).
Solution by Sreeraj P, M.Sc Physics
For the two cells having same EMF $E$ and internal resistance $r$, the current passing through the external resistor $6\Omega$ is same when both the cells are connected either in parallel or in series. The value of internal resistance $r$ is ______ $\Omega$.
Answer: (B) 6
Concept. Two identical cells: series gives $\dfrac{2E}{R+2r}$, parallel gives $\dfrac{E}{R+r/2}$.
Step 1. Equate. $2\left(6+\dfrac r2\right)=6+2r\Rightarrow r=6\ \Omega$.
Answer. Option B. (In general they are equal when $r=R$.)
Solution by Sreeraj P, M.Sc Physics
In the potentiometer, when the cell in the secondary circuit is shunted with $4\Omega$ resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with $12\Omega$ resistance, the balance is shifted to a length of 180 cm . The internal resistance of cell is ______ $\Omega$
Answer: (A) 4
Concept. A shunted cell supplies a current, so its terminal voltage $V=\dfrac{ER}{R+r}$ is what the potentiometer balances: $V\propto l$.
Step 1. Ratio. $\dfrac{120}{180}=\dfrac{4/(4+r)}{12/(12+r)}=\dfrac{4(12+r)}{12(4+r)}$.
Step 2. Solve. $2\cdot12(4+r)=3\cdot4(12+r)\Rightarrow96+24r=144+12r\Rightarrow r=4\ \Omega$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
The equivalent resistance between the points $A$ and $B$ in the following circuit is $\dfrac x5\Omega$. The value of $x$ is ______ .
Numerical value type. Enter your answer.
Answer: 21
Concept. This bridge is unbalanced ($\frac63\neq\frac36$) and has no symmetry that removes the middle resistor, so use Kirchhoff's junction law with node potentials. Apply 1 V across $A$ and $B$, find the current, and $R=\dfrac{1}{I}$.
Step 1. Set up. $V_A=1$, $V_B=0$. Node $M$: $6\ \Omega$ to $A$, $3\ \Omega$ to $B$, $3\ \Omega$ to $N$. Node $N$: $3\ \Omega$ to $A$, $6\ \Omega$ to $B$.
Step 2. Junction law at $M$ (×6). $(1-V_M)+2(0-V_M)+2(V_N-V_M)=0\Rightarrow1-5V_M+2V_N=0$.
Step 3. Junction law at $N$ (×6). $2(1-V_N)+(0-V_N)+2(V_M-V_N)=0\Rightarrow2-5V_N+2V_M=0$.
Step 4. Solve. $V_M=\dfrac37$ V, $V_N=\dfrac47$ V.
Step 5. Current from $A$. $\dfrac{1-3/7}{6}+\dfrac{1-4/7}{3}=\dfrac{4}{42}+\dfrac{6}{42}=\dfrac{5}{21}$ A, so $R=\dfrac{21}{5}\ \Omega$.
Answer. $x=21$.
Solution by Sreeraj P, M.Sc Physics
The reading of the ammeter $(A)$ in steady state in the following circuit (assuming negligible internal resistance of the ammeter) is ______ A.
Answer: (B) 1
In steady state no current flows through the capacitor, so the diagonal branch ($10\ \mu$F in series with $8\ \Omega$) carries no current.
Let $P$ be the junction after the $1\ \Omega$ resistor. From $P$ the current has two paths back to the cell:
These two $8\ \Omega$ paths are in parallel: $4\ \Omega$. Total resistance $= 1 + 4 = 5\ \Omega$.
$$I = \frac{10}{5} = 2\ \text{A}$$
It divides equally between the two equal paths, so the ammeter reads $1$ A.
Solution by Sreeraj P, M.Sc Physics
A regular hexagon is formed by six wires each of resistance $r\Omega$ and the corners are joined to the centre by wires of same resistance. If the current enters at one corner and leaves at the opposite corner, the equivalent resistance of the hexagon between the two opposite corners will be
Answer: (B) $\dfrac45r$
Concept. Use symmetry. With $V$ at the entry corner A and $0$ at the opposite corner D, the centre O is at $V/2$; neighbours B, F of A are equipotential at $x$, and C, E are at $V-x$.
Step 1. Node B. Currents out of B sum to zero: $(x-V)+(x-\tfrac V2)+(x-(V-x))=0\Rightarrow x=\tfrac58V$.
Step 2. Current from A. To B and F: $2\cdot\dfrac{3V/8}{r}$; to O: $\dfrac{V/2}{r}$; total $\dfrac{5V}{4r}$.
Step 3. Resistance. $\dfrac{V}{I}=\dfrac45r$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
In a meter bridge experiment to determine the value of unknown resistance, first the resistances $2\Omega$ and $3\Omega$ are connected in the left and right gaps of the bridge and the null point is obtained at a distance $l$ cm from the left. Now when an unknown resistance $x\Omega$ is connected in parallel to $3\Omega$ resistance, the null point is shifted by 10 cm to the right of wire. The value of unknown resistance $x$ is
______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 6
First balance: $\dfrac23 = \dfrac{l}{100 - l}\Rightarrow l = 40$ cm.
New null point at $50$ cm, so the right gap now equals the left gap:
$$\frac{2}{R_{\text{right}}} = \frac{50}{50}\Rightarrow R_{\text{right}} = 2\ \Omega$$
$$\frac{3x}{3 + x} = 2\Rightarrow 3x = 6 + 2x\Rightarrow x = 6\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
Two resistors of $100\Omega$ each are connected in series with a 9 V battery. A voltmeter of $400\Omega$ resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is ______ V.
Answer: (C) 4
Concept. A real voltmeter is a resistor in parallel with what it measures, so it lowers the reading.
Step 1. Parallel. $100\parallel400=80\ \Omega$.
Step 2. Divider. $9\times\dfrac{80}{180}=4$ V.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Two resistors $2\Omega$ and $3\Omega$ are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire $XY$. When an unknown resistor is connected in parallel with $3\Omega$ resistor, the null point is shifted by 22.5 cm toward $Y$. The resistance of unknown resistor is ______ $\Omega$.
Answer: (D) 2
Initially: $\dfrac23 = \dfrac{l}{100 - l}\Rightarrow l = 40$ cm.
New null point: $40 + 22.5 = 62.5$ cm.
$$\frac{2}{R'} = \frac{62.5}{37.5} = \frac53\Rightarrow R' = 1.2\ \Omega$$
$R'$ is $3\ \Omega$ in parallel with $x$:
$$\frac{3x}{3 + x} = 1.2\Rightarrow 3x = 3.6 + 1.2x\Rightarrow x = 2\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
Given below are two statements:
Statement-I: The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs.
Statement-II: The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries.
In the light of the above statements, choose the correct answer from the options given below.
Answer: (B) Statement-I is false but Statement-II is true
Concept. Two unequal cells in parallel: $E_{eq}=\dfrac{E_1r_2+E_2r_1}{r_1+r_2}$, a weighted average of the two emfs, and $r_{eq}=\dfrac{r_1r_2}{r_1+r_2}$.
Step 1. Statement I. A weighted average lies between $E_1$ and $E_2$, so it is not smaller than both. Statement I is false.
Step 2. Statement II. $\dfrac{r_1r_2}{r_1+r_2}$ is smaller than either $r_1$ or $r_2$ (like resistors in parallel). Statement II is true.
Answer. Statement I is false but Statement II is true (Option B).
Solution by Sreeraj P, M.Sc Physics
The sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance $R_p = 1\ \Omega$ as shown in the figure. An external resistance of $R_e = 2\ \Omega$ is connected via the sliding contact. The electric current in the circuit is
Answer: (D) $1.0\ \text{A}$
The slider divides the wire into two halves of $0.5\ \Omega$ each.
$R_e = 2\ \Omega$ is connected between the left end of the wire and the slider, so it is in parallel with the left half:
$$R_{\parallel} = \frac{0.5\times2}{0.5+2} = 0.4\ \Omega$$
This is in series with the right half:
$$R_{total} = 0.4 + 0.5 = 0.9\ \Omega$$
$$I = \frac{0.9\ \text{V}}{0.9\ \Omega} = 1.0\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics
Which of the following resistivity ($\rho$) vs temperature ($T$) curves is most suitable to be used in wire bound standard resistors?
Answer: (D) see figure
A standard resistor must keep its resistance almost constant when its temperature changes. It is made of an alloy such as manganin or constantan, which has a fairly high resistivity and a very small temperature coefficient of resistivity.
So the suitable curve is a nearly flat line at a non-zero resistivity: option (4).
Solution by Sreeraj P, M.Sc Physics
The net current flowing in the given circuit is ______ A.
Numerical value type. Enter your answer.
Answer: 1
In steady state no current flows through the $1\ \mu\text{F}$ capacitor, so the branch $2.5\ \Omega$–$1\ \mu\text{F}$ and everything to its right ($1\ \Omega$, $5\ \Omega$, $8\ \Omega$, $4\ \Omega$) carry no current.
The $3\ \Omega$ and $6\ \Omega$ resistors both connect the second top junction to the bottom wire, so they are in parallel:
$$\frac{3\times6}{3+6} = 2\ \Omega$$
This is in series with the $2\ \Omega$: $2 + 2 = 4\ \Omega$. That branch is in parallel with the diagonal $4\ \Omega$ across the cell:
$$R = \frac{4\times4}{4+4} = 2\ \Omega$$
$$I = \frac{2\ \text{V}}{2\ \Omega} = 1\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics
At steady state the charge on the capacitor, as shown in the circuit below, is ______ $\mu\text{C}$.
Numerical value type. Enter your answer.
Answer: 16
In steady state no current flows into the capacitor, so the $10\ \Omega$ and $15\ \Omega$ resistors carry the same current:
$$I = \frac{5}{10 + 15} = 0.2\ \text{A}$$
The capacitor is connected across the $10\ \Omega$ resistor:
$$V_C = 0.2\times10 = 2\ \text{V}, \qquad Q = CV = 8\ \mu\text{F}\times2\ \text{V} = 16\ \mu\text{C}$$
Solution by Sreeraj P, M.Sc Physics
A wire of resistance $9\ \Omega$ is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be ______ ohm.
Numerical value type. Enter your answer.
Answer: 2
Concept. Between two vertices of a triangle, one side is in parallel with the other two sides in series.
Step 1. Sides. $3\ \Omega$ each.
Step 2. Combine. $\dfrac{3\times6}{9}=2\ \Omega$.
Answer. $2$.
Solution by Sreeraj P, M.Sc Physics
A wire of resistance $R$ is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is
Answer: (C) $32/27$
Concept. Across one edge of a polygon: that edge in parallel with the rest. For $n$ sides, $R_{\text{edge}}=\dfrac{(n-1)R}{n^2}$.
Step 1. Triangle. $\dfrac{2R}{9}$. Square. $\dfrac{3R}{16}$.
Step 2. Ratio. $\dfrac{2/9}{3/16}=\dfrac{32}{27}$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Find the equivalent resistance between the two ends of the following circuit.
Answer: (A) $\dfrac{r}{9}$
Label the junctions along the line: left end $L$, then $P$ (between the first and second resistors), then $Q$ (between the second and third), then the right end $R'$.
Each $r/3$ resistor now connects the node $\{L, Q\}$ to the node $\{P, R'\}$: all three are in parallel between the two ends.
$$R_{eq} = \frac{r/3}{3} = \frac{r}{9}$$
Solution by Sreeraj P, M.Sc Physics
The value of current $I$ in the electrical circuit given below, when the potential at A is equal to the potential at B, will be ______ A.
Numerical value type. Enter your answer.
Answer: 2
With $V_A = V_B$ no current flows through the $30\ \Omega$ resistor, and the bridge is balanced:
$$\frac{10}{20} = \frac{R}{40} \Rightarrow R = 20\ \Omega$$
The $30\ \Omega$ can be removed: $(10 + 20) = 30\ \Omega$ in parallel with $(20 + 40) = 60\ \Omega$ gives $20\ \Omega$.
$$I = \frac{40}{20} = 2\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics
The reading of the ideal voltmeter $V$ in the circuit shown is (each cell has emf $5\ \text{V}$ and internal resistance $0.2\ \Omega$)
Answer: (C) $0\ \text{V}$
Concept. In a closed ring of identical cells all aiding each other, the current is $I=\dfrac{nE}{nr}=\dfrac Er$, and the terminal voltage of each cell is $E-Ir=0$.
Step 1. Current. Eight cells: $I=\dfrac{8\times5}{8\times0.2}=25$ A.
Step 2. Terminal voltage of one cell. $V=5-25\times0.2=0$.
Step 3. Reading. The ideal voltmeter across one cell reads its terminal voltage.
Answer. $0$ V (Option C).
Solution by Sreeraj P, M.Sc Physics
The current in a conductor is expressed as $I = 3t^2 + 4t^3$, where $I$ is in ampere and $t$ is in second. The amount of electric charge that flows through a section of the conductor from $t = 1\ \text{s}$ to $t = 2\ \text{s}$ is ______ C.
Numerical value type. Enter your answer.
Answer: 22
Concept. Current is the rate of flow of charge, so the charge is the area under the $I$–$t$ graph: $q=\int I\,dt$.
Step 1. Integrate. $\displaystyle\int_1^2(3t^2+4t^3)\,dt=\left[t^3+t^4\right]_1^2=(8+16)-(1+1)$.
Answer. $22$ C.
Solution by Sreeraj P, M.Sc Physics
In a metre bridge, when a resistance in the left gap is $2\ \Omega$ and an unknown resistance is in the right gap, the balance length is found to be $40\ \text{cm}$. On shunting the unknown resistance with $2\ \Omega$, the balance length changes by
Answer: (A) $22.5\ \text{cm}$
Concept. Meter bridge balance: $\dfrac{R_{\text{left}}}{R_{\text{right}}}=\dfrac{l}{100-l}$.
Step 1. Unknown. $\dfrac{2}{X}=\dfrac{40}{60}\Rightarrow X=3\ \Omega$.
Step 2. Shunted. $3\parallel2=1.2\ \Omega$; $\dfrac{2}{1.2}=\dfrac{l}{100-l}\Rightarrow l=62.5$ cm.
Step 3. Shift. $62.5-40=22.5$ cm.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are $8\ \Omega$ and $10\ \Omega$ respectively. After inserting it in a hot bath of temperature $400^\circ\text{C}$, the resistance of the platinum wire is
Answer: (C) $16\ \Omega$
Concept. $R_t=R_0(1+\alpha t)$. Two readings give $\alpha$; then predict.
Step 1. $\alpha$. $10=8(1+100\alpha)\Rightarrow\alpha=2.5\times10^{-3}\ ^\circ$C$^{-1}$.
Step 2. At $400^\circ$C. $8(1+1)=16\ \Omega$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Twelve wires each having resistance $2\ \Omega$ are joined to form a cube. A battery of $6\ \text{V}$ emf is joined across points $a$ and $c$. The voltage difference between $e$ and $f$ is ______ V.
Numerical value type. Enter your answer.
Answer: 1
Concept. Use symmetry to fix the potentials of as many corners as possible, then apply Kirchhoff's junction law at one remaining corner.
Step 1. Potentials of the symmetric corners. Let $V_a=6$ V, $V_c=0$. The plane midway between $a$ and $c$ contains $b$, $d$ and the corners below them, $e$ and $g$, so $V_b=V_d=V_e=V_g=3$ V.
Step 2. Corner $h$ (below $a$). Its neighbours are $a$ (6 V), $e$ (3 V) and $g$ (3 V), all through equal resistors: $(6-V_h)+(3-V_h)+(3-V_h)=0\Rightarrow V_h=4$ V.
Step 3. Corner $f$ (below $c$). By the same symmetry (reversing the battery), $V_f=6-V_h=2$ V.
Step 4. Difference. $V_e-V_f=3-2=1$ V.
Answer. $1$.
Solution by Sreeraj P, M.Sc Physics
An electric bulb rated $50\ \text{W}$ – $200\ \text{V}$ is connected across a $100\ \text{V}$ supply. The power dissipated by the bulb is
Answer: (B) $12.5\ \text{W}$
The resistance is fixed, so $P \propto V^2$:
$$P = 50\left(\frac{100}{200}\right)^2 = 12.5\ \text{W}$$
Solution by Sreeraj P, M.Sc Physics
Two wires A and B are made of the same material and have the same mass. Wire A has a radius of $2.0\ \text{mm}$ and wire B has a radius of $4.0\ \text{mm}$. The resistance of wire B is $2\ \Omega$. The resistance of wire A is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 32
Concept. Same material and same mass means the same volume, so $l\propto\dfrac1A$ and $R=\dfrac{\rho l}{A}\propto\dfrac{1}{A^2}\propto\dfrac{1}{r^4}$.
Step 1. Ratio. $\dfrac{R_A}{R_B}=\left(\dfrac{4}{2}\right)^4=16$.
Step 2. Value. $16\times2=32\ \Omega$.
Answer. $32$.
Solution by Sreeraj P, M.Sc Physics
In the given figure $R_1 = 10\ \Omega$, $R_2 = 8\ \Omega$, $R_3 = 4\ \Omega$ and $R_4 = 8\ \Omega$. The battery is ideal with emf $12\ \text{V}$. The equivalent resistance of the circuit and the current supplied by the battery are respectively
Answer: (B) $12\ \Omega$ and $1\ \text{A}$
Concept. Follow each resistor's two ends to the nodes. $R_2$, $R_3$ and the diagonal $R_4$ all run between the top wire and the bottom wire, so they are in parallel, however they are drawn.
Step 1. Parallel group. $\dfrac1{R_p}=\dfrac18+\dfrac14+\dfrac18=\dfrac12$, so $R_p=2\ \Omega$.
Step 2. Total. $R=R_1+R_p=10+2=12\ \Omega$.
Step 3. Current. $I=\dfrac{12}{12}=1$ A.
Answer. $12\ \Omega$ and $1$ A (Option B).
Solution by Sreeraj P, M.Sc Physics
A $20\ \Omega$ resistor is connected in series with a parallel combination of $5\ \Omega$ and $10\ \Omega$ resistors across a $10\ \text{V}$ battery. The ratio of heat dissipated per second through the $5\ \Omega$ and $10\ \Omega$ resistances is
Answer: (B) $2:1$
The $5\ \Omega$ and $10\ \Omega$ resistors are in parallel, so they have the same voltage. With $P = \dfrac{V^2}{R}$:
$$\frac{P_5}{P_{10}} = \frac{10}{5} = 2:1$$
Solution by Sreeraj P, M.Sc Physics
A wire of resistance $20\ \Omega$ is divided into 10 equal parts. Pairs of these parts are connected in parallel, and the resulting parallel pairs are then connected in series. The equivalent resistance of the final combination is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 5
Each part: $2\ \Omega$. Each parallel pair: $1\ \Omega$. Five pairs in series: $5\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
A wire of resistance $R$ and radius $r$ is stretched till its radius becomes $r/2$. If the new resistance of the stretched wire is $xR$, then the value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 16
Concept. Stretching keeps volume constant. Radius halves, so area becomes $\tfrac14$ and length $4$ times: $R\times16$.
Answer. $x=16$.
Solution by Sreeraj P, M.Sc Physics
A wire of resistance $R$ and length $L$ is cut into 5 equal parts. If these parts are joined parallelly, then resultant resistance will be:
Answer: (A) $\dfrac{R}{25}$
Each part has resistance $R/5$. Five equal resistors in parallel give
$$R_{eq} = \frac{R/5}{5} = \frac{R}{25}$$
Solution by Sreeraj P, M.Sc Physics
A wire of length $10\ \text{cm}$ and radius $\sqrt7\times10^{-4}\ \text{m}$ is connected across the right gap of a meter bridge. When a resistance of $4.5\ \Omega$ is connected on the left gap by using a resistance box, the balance length is found to be at $60\ \text{cm}$ from the left end. If the resistivity of the wire is $R\times10^{-7}\ \Omega\,\text{m}$, then value of $R$ is:
Answer: (C) 66
Concept. Find the wire's resistance from the balance, then $\rho=\dfrac{RA}{l}$.
Step 1. Resistance. $\dfrac{4.5}{X}=\dfrac{60}{40}\Rightarrow X=3\ \Omega$.
Step 2. Resistivity. $\rho=\dfrac{3\times\frac{22}{7}\times7\times10^{-8}}{0.1}=66\times10^{-7}\ \Omega$ m.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Wheatstone bridge principle is used to measure the specific resistance $S_1$ of a given wire, having length $L$ and radius $r$. If $X$ is the resistance of the wire, then specific resistance is $S_1 = X\dfrac{\pi r^2}{L}$. If the length of the wire gets doubled, then the value of specific resistance will be:
Answer: (D) $S_1$
Specific resistance (resistivity) is a property of the material only. Doubling the length doubles $X$, and $X\pi r^2/L$ remains unchanged, so it stays $S_1$.
Solution by Sreeraj P, M.Sc Physics
Three voltmeters, all having different internal resistances, are joined as shown in figure. When some potential difference is applied across A and B, their readings are $V_1$, $V_2$ and $V_3$. Choose the correct option.
Answer: (D) $V_1 + V_2 = V_3$
Concept. A voltmeter reads the potential difference across its own terminals, whatever its resistance. Going round any closed loop, the potential differences add up to zero (Kirchhoff's loop law).
Step 1. Upper branch. $V_1$ and $V_2$ are in series between $A$ and $B$, so $V_1+V_2=V_A-V_B$.
Step 2. Lower branch. $V_3$ is directly across $A$ and $B$: $V_3=V_A-V_B$.
Step 3. Compare. $V_1+V_2=V_3$, regardless of the internal resistances. ($V_1=V_2$ only if their resistances were equal.)
Answer. $V_1+V_2=V_3$ (Option D).
Solution by Sreeraj P, M.Sc Physics
The electric current through a wire varies with time as $I = I_0 + \beta t$, where $I_0 = 20\ \text{A}$ and $\beta = 3\ \text{A s}^{-1}$. The amount of electric charge crossed through a section of the wire in $20\ \text{s}$ is:
Answer: (B) $1000\ \text{C}$
$$q = \int_0^{20}(20 + 3t)\,dt = 20\times20 + \frac32(20)^2 = 400 + 600 = 1000\ \text{C}$$
Solution by Sreeraj P, M.Sc Physics
A $16\ \Omega$ wire is bent to form a square loop. A $9\ \text{V}$ battery with internal resistance $1\ \Omega$ is connected across one of its sides. If a $4\ \mu\text{F}$ capacitor is connected across one of its diagonals, the energy stored by the capacitor will be $\dfrac{x}{2}\ \mu\text{J}$, where $x =$ ______.
Numerical value type. Enter your answer.
Answer: 81
Concept. In steady state no current flows into the capacitor, so it just measures the voltage between two corners of the loop.
Step 1. Circuit. Sides $4\ \Omega$; the side across the battery is parallel with the other three ($12\ \Omega$): $3\ \Omega$. Current $\dfrac{9}{3+1}=2.25$ A; voltage $6.75$ V.
Step 2. Diagonal. Along the $12\ \Omega$ path the current is $0.5625$ A; two sides take $0.5625\times8=4.5$ V.
Step 3. Energy. $\tfrac12\times4\times10^{-6}\times4.5^2=40.5\ \mu$J $=\dfrac{81}{2}\ \mu$J.
Answer. $x=81$.
Solution by Sreeraj P, M.Sc Physics
In the given circuit, the current in resistance $R_3$ is:
Answer: (A) $1\ \text{A}$
$R_2 \parallel R_3 = 2\ \Omega$, so the total resistance is $2 + 2 + 1 = 5\ \Omega$.
$$I = \frac{10}{5} = 2\ \text{A}$$
This divides equally between the equal resistors $R_2$ and $R_3$, so the current in $R_3$ is $1\ \text{A}$.
Solution by Sreeraj P, M.Sc Physics
In the given circuit, the current flowing through the resistance $20\ \Omega$ is $0.3\ \text{A}$, while the ammeter reads $0.9\ \text{A}$. The value of $R_1$ is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 30
Voltage across the parallel combination: $V = 0.3\times20 = 6\ \text{V}$.
Current in $15\ \Omega$: $\dfrac{6}{15} = 0.4\ \text{A}$. Current in $R_1$: $0.9 - 0.3 - 0.4 = 0.2\ \text{A}$.
$$R_1 = \frac{6}{0.2} = 30\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
In the given circuit, the terminal potential difference of the cell is:
Answer: (A) $2\ \text{V}$
The two $4\ \Omega$ resistors in parallel give $2\ \Omega$.
$$I = \frac{3}{2 + 1} = 1\ \text{A},\qquad V = \varepsilon - Ir = 3 - 1 = 2\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics
Resistance of a wire at $0\,{}^\circ\text{C}$, $100\,{}^\circ\text{C}$ and $t\,{}^\circ\text{C}$ is found to be $10\ \Omega$, $10.2\ \Omega$ and $10.95\ \Omega$ respectively. The temperature $t$ in Kelvin scale is ______.
Numerical value type. Enter your answer.
Answer: 748
Concept. Same law, $R_t=R_0(1+\alpha t)$; then convert to kelvin.
Step 1. $\alpha$. $\dfrac{0.2}{10\times100}=2\times10^{-4}$.
Step 2. Temperature. $10.95=10(1+2\times10^{-4}t)\Rightarrow t=475^\circ$C $=748$ K.
Answer. $748$.
Solution by Sreeraj P, M.Sc Physics
Water boils in an electric kettle in 20 minutes after being switched on. Using the same main supply, the length of the heating element should be ______ to ______ times of its initial length if the water is to be boiled in 15 minutes.
Answer: (A) decreased, $3/4$
Concept. Same supply: $P=\dfrac{V^2}{R}$, and the heat needed is fixed, so time $\propto R\propto$ length.
Step 1. Ratio. $\dfrac{15}{20}=\dfrac34$.
Answer. Decreased to $\dfrac34$ (Option A).
Solution by Sreeraj P, M.Sc Physics
A heater is designed to operate with a power of $1000\ \text{W}$ in a $100\ \text{V}$ line. It is connected in combination with a resistance of $10\ \Omega$ and a resistance $R$ to a $100\ \text{V}$ mains as shown in the figure. For the heater to operate at $62.5\ \text{W}$, the value of $R$ should be ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 5
Concept. The heater's resistance follows from its rating. To run it at 62.5 W, find the voltage it needs, then use the series $10\ \Omega$ and the parallel $R$ to share the 100 V.
Step 1. Heater resistance. $R_h=\dfrac{100^2}{1000}=10\ \Omega$.
Step 2. Voltage for 62.5 W. $V_h=\sqrt{62.5\times10}=25$ V; heater current $\dfrac{25}{10}=2.5$ A.
Step 3. Main current. The $10\ \Omega$ takes $100-25=75$ V: $I=7.5$ A.
Step 4. $R$. It carries $7.5-2.5=5$ A at 25 V: $R=5\ \Omega$.
Answer. $5$.
Solution by Sreeraj P, M.Sc Physics
The equivalent resistance between A and B is:
Answer: (B) $19\ \Omega$
The vertical wire joins the top and bottom midpoints into one node X; the horizontal wire (which hops over it without contact) joins the left and right midpoints into another node Y.
A to X: $6\ \Omega$. Between X and Y there are three paths: $10 + 5 = 15\ \Omega$, $7 + 8 = 15\ \Omega$ and $11 + 4 = 15\ \Omega$, in parallel giving $5\ \Omega$. Y to B: $8\ \Omega$.
$$R_{AB} = 6 + 5 + 8 = 19\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
The current flowing through the $1\ \Omega$ resistor is $\dfrac{n}{10}\ \text{A}$. The value of $n$ is ______.
Numerical value type. Enter your answer.
Answer: 25
Take $V_D = 0$; the 5 V cell gives $V_B = 5\ \text{V}$. Let the potentials of A and C be $V_A$ and $V_C$. The 10 V cell raises the potential towards A, so the current in the $1\ \Omega$ branch (C to A) is $I_1 = V_C + 10 - V_A$.
KCL at A: $I_1 = \dfrac{V_A - 5}{4} + \dfrac{V_A}{4}$
KCL at C: $I_1 + \dfrac{V_C - 5}{2} + \dfrac{V_C}{2} = 0$
Solving: $V_C = 0$, $V_A = 7.5\ \text{V}$, so
$$I_1 = 0 + 10 - 7.5 = 2.5\ \text{A} = \frac{25}{10}\ \text{A} \;\Rightarrow\; n = 25$$
(Check at A: $\dfrac{2.5}{4} + \dfrac{7.5}{4} = 2.5\ \text{A}$.)
Solution by Sreeraj P, M.Sc Physics
The effective resistance between $A$ and $B$, if the resistance of each resistor is $R$, will be:
Answer: (A) $\dfrac{8R}{3}$
Concept. Inside the diamond, the top, centre and bottom points are symmetric between its left and right corners, so they are at the same potential. The vertical resistors between them carry no current.
Step 1. Remove the verticals. Left corner to right corner: three identical paths (top, centre, bottom), each $R+R=2R$.
Step 2. Diamond. Three $2R$ paths in parallel: $\dfrac{2R}{3}$.
Step 3. Add the end resistors. $R+\dfrac{2R}3+R=\dfrac{8R}{3}$.
Answer. $\dfrac{8R}{3}$ (Option A).
Solution by Sreeraj P, M.Sc Physics
At room temperature ($27\,{}^\circ\text{C}$), the resistance of a heating element is $50\ \Omega$. The temperature coefficient of the material is $2.4\times10^{-4}\ ^\circ\text{C}^{-1}$. The temperature of the element, when its resistance is $62\ \Omega$, is ______ $^\circ\text{C}$.
Numerical value type. Enter your answer.
Answer: 1027
$$R = R_{27}\left[1 + \alpha(T - 27)\right] \;\Rightarrow\; 62 = 50\left[1 + 2.4\times10^{-4}(T - 27)\right]$$
$$T - 27 = \frac{0.24}{2.4\times10^{-4}} = 1000 \;\Rightarrow\; T = 1027\,{}^\circ\text{C}$$
Solution by Sreeraj P, M.Sc Physics
To determine the resistance ($R$) of a wire, a circuit is designed below. The $V$–$I$ characteristic curve for this circuit is plotted for the voltmeter and the ammeter readings as shown in the figure. The value of $R$ is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 2500
From the graph, $\dfrac VI = \dfrac{8\ \text{V}}{4\ \text{mA}} = 2\ \text{k}\Omega$. This is the resistance of $R$ in parallel with $10\ \text{k}\Omega$:
$$\frac{10R}{10 + R} = 2 \;\Rightarrow\; 8R = 20 \;\Rightarrow\; R = 2.5\ \text{k}\Omega = 2500\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
A potential divider circuit is shown in the figure. The output voltage $V_0$ is
Answer: (C) $0.5\ \text{V}$
Concept. In a series chain, the current is the same everywhere, so the output across part of the chain is the current times the resistance of that part.
Step 1. Total resistance. $3300+7\times100=4000\ \Omega$.
Step 2. Current. $I=\dfrac{4}{4000}=1$ mA.
Step 3. Output. $V_0$ is across five $100\ \Omega$ resistors: $1\ \text{mA}\times500\ \Omega=0.5$ V.
Answer. $0.5$ V (Option C).
Solution by Sreeraj P, M.Sc Physics
An electric toaster has resistance of $60\ \Omega$ at room temperature ($27^\circ\text{C}$). The toaster is connected to a $220\ \text{V}$ supply. If the current flowing through it reaches $2.75\ \text{A}$, the temperature attained by the toaster is around: (if $\alpha = 2\times10^{-4}\ {}^\circ\text{C}^{-1}$)
Answer: (C) $1694^\circ\text{C}$
Concept. The toaster's resistance rises with temperature, $R=R_0(1+\alpha\,\Delta T)$, so the hot resistance from $V/I$ tells how much it has heated. Here the given value $60\ \Omega$ at $27^\circ$C is used as the reference.
Step 1. Hot resistance. $R=\dfrac{220}{2.75}=80\ \Omega$.
Step 2. Rise in temperature. $80=60\left(1+2\times10^{-4}\,\Delta T\right)\Rightarrow\Delta T=\dfrac{1/3}{2\times10^{-4}}\approx1667^\circ$C.
Step 3. Temperature. $T=27+1667\approx1694^\circ$C.
Answer. $1694^\circ$C (Option C).
Solution by Sreeraj P, M.Sc Physics
Two cells are connected in opposition in a single closed loop. Cell $E_1$ has $8\ \text{V}$ emf and $2\ \Omega$ internal resistance; cell $E_2$ has $2\ \text{V}$ emf and $4\ \Omega$ internal resistance. There is no other resistance in the circuit. The terminal potential difference of cell $E_2$ is ______ V.
Numerical value type. Enter your answer.
Answer: 6
Concept. Cells in opposition: net emf is the difference. The weaker cell is charged, so its terminal voltage is $E+Ir$.
Step 1. Current. $\dfrac{8-2}{2+4}=1$ A.
Step 2. Terminal voltage of $E_2$. $2+1\times4=6$ V.
Answer. $6$.
Solution by Sreeraj P, M.Sc Physics
When a potential difference $V$ is applied across a wire of resistance $R$, it dissipates energy at a rate $W$. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the energy dissipation rate will become:
Answer: (D) $4W$
Each half has resistance $R/2$; in parallel they give $R/4$.
$$W' = \frac{V^2}{R/4} = 4\frac{V^2}{R} = 4W$$
Solution by Sreeraj P, M.Sc Physics
Two resistances of $100\ \Omega$ and $200\ \Omega$ are connected in series with a battery of $4\ \text{V}$ and negligible internal resistance. A voltmeter is used to measure the voltage across the $100\ \Omega$ resistance, and it gives a reading of $1\ \text{V}$. The resistance of the voltmeter must be ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 200
Concept. Series circuit: voltages divide in proportion to resistance. Reading $1$ V means the other $3$ V is across $200\ \Omega$.
Step 1. Current. $\dfrac{3}{200}=0.015$ A.
Step 2. Parallel resistance. $\dfrac{1}{0.015}=\dfrac{200}{3}\ \Omega$.
Step 3. Voltmeter. $\dfrac1{R_V}=\dfrac{3}{200}-\dfrac1{100}=\dfrac1{200}$.
Answer. $200\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
Two conductors have the same resistance at $0^\circ\text{C}$ but their temperature coefficients of resistance are $\alpha_1$ and $\alpha_2$. The respective temperature coefficients for their series and parallel combinations are:
Answer: (B) $\dfrac{\alpha_1 + \alpha_2}{2},\ \dfrac{\alpha_1 + \alpha_2}{2}$
Concept. Write each resistance as $R_0(1+\alpha t)$, combine, and read off the coefficient of $t$ in the form $R_{eq,0}(1+\alpha_{eq}t)$.
Step 1. Series. $R_s=R_0(1+\alpha_1t)+R_0(1+\alpha_2t)=2R_0\left(1+\dfrac{\alpha_1+\alpha_2}{2}t\right)$, so $\alpha_s=\dfrac{\alpha_1+\alpha_2}{2}$.
Step 2. Parallel. For small $\alpha t$, $\dfrac1{R_p}=\dfrac{1}{R_0}\left[(1-\alpha_1t)+(1-\alpha_2t)\right]=\dfrac{2}{R_0}\left(1-\dfrac{\alpha_1+\alpha_2}{2}t\right)$, so $R_p=\dfrac{R_0}{2}\left(1+\dfrac{\alpha_1+\alpha_2}{2}t\right)$.
Step 3. Compare. Both combinations have $\alpha_{eq}=\dfrac{\alpha_1+\alpha_2}{2}$.
Answer. $\dfrac{\alpha_1+\alpha_2}{2},\ \dfrac{\alpha_1+\alpha_2}{2}$ (Option B).
Solution by Sreeraj P, M.Sc Physics
The equivalent resistance between $A$ and $B$ of the following network is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 1
Concept. In a ladder, start at the end away from the terminals and fold the network towards $A$ and $B$ one rung at a time. A plain wire across the rails shorts anything in parallel with it.
Step 1. Far end. The plain wire at the left shorts $R_1$ ($6\ \Omega$), so the left end is $0\ \Omega$.
Step 2. Next rung. $R_2$ ($2\ \Omega$) in series with the short gives $2\ \Omega$, in parallel with $R_3$ ($2\ \Omega$): $1\ \Omega$.
Step 3. Next. Add $R_4$: $1+2=3\ \Omega$, in parallel with $R_5$ ($3\ \Omega$): $1.5\ \Omega$.
Step 4. Last rung. In parallel with $R_6$ ($3\ \Omega$): $\dfrac{1.5\times3}{4.5}=1\ \Omega$.
Answer. $1$.
Solution by Sreeraj P, M.Sc Physics
By what percentage will the illumination of the lamp decrease if the current drops by $20\%$?
Answer: (C) $36\%$
Concept. The illumination is proportional to the power, $P=I^2R$, so it changes as the square of the current.
Step 1. New power. $I\to0.8I$: $P\to0.64P$.
Step 2. Decrease. $1-0.64=0.36$, i.e. $36\%$.
Answer. $36\%$ (Option C).
Solution by Sreeraj P, M.Sc Physics
The resistance per centimetre of a metre bridge wire is $r$, with $X\ \Omega$ resistance in the left gap. The balancing length from the left end is at $40\ \text{cm}$ with $25\ \Omega$ resistance in the right gap. Now the wire is replaced by another wire of $2r$ resistance per centimetre. The new balancing length for the same settings will be at
Answer: (D) $40\ \text{cm}$
Concept. In a metre bridge with a uniform wire, the two parts of the wire have resistances $rl$ and $r(100-l)$. The ratio $\dfrac{l}{100-l}$ does not depend on $r$.
Step 1. Balance. $\dfrac X{25}=\dfrac{rl}{r(100-l)}=\dfrac{l}{100-l}$.
Step 2. New wire. Doubling $r$ cancels in the ratio, so the balance length stays the same.
Answer. $40$ cm (Option D).
Solution by Sreeraj P, M.Sc Physics
In the following circuit, the battery has an emf of $2\ \text{V}$ and an internal resistance of $\dfrac23\ \Omega$. The power consumption in the entire circuit is ______ W.
Numerical value type. Enter your answer.
Answer: 3
Concept. The left and right sides of the rectangle are plain wires, so every resistor branch (the top, the bottom and each diagonal) runs between the same two nodes. The crossing diagonals are not joined.
Step 1. Branches. Top $2\ \Omega$, bottom $2\ \Omega$, each diagonal $2+2=4\ \Omega$.
Step 2. Parallel. $\dfrac1R=\dfrac12+\dfrac12+\dfrac14+\dfrac14=\dfrac32$, so $R=\dfrac23\ \Omega$.
Step 3. Current. $I=\dfrac{2}{\frac23+\frac23}=1.5$ A.
Step 4. Power of the whole circuit. $P=EI=2\times1.5=3$ W.
Answer. $3$.
Solution by Sreeraj P, M.Sc Physics
Ratio of thermal energy released in two resistors $R$ and $3R$ connected in parallel in an electric circuit is
Answer: (A) $3:1$
In parallel the voltage is common, so $H=\dfrac{V^2t}{R}\propto\dfrac1R$.
$$H_R:H_{3R}=\frac1R:\frac1{3R}=3:1$$
Solution by Sreeraj P, M.Sc Physics
In a metre bridge experiment the balance point is obtained if the gaps are closed by $2\ \Omega$ and $3\ \Omega$. A shunt of $X\ \Omega$ is added to the $3\ \Omega$ resistor to shift the balancing point by $22.5\ \text{cm}$. The value of $X$ is ______.
Numerical value type. Enter your answer.
Answer: 2
Concept. Same balance condition. Shunting the right resistor lowers it, so the null point moves right.
Step 1. First balance. $\dfrac23=\dfrac{l}{100-l}\Rightarrow l=40$ cm.
Step 2. New right resistance. At $62.5$ cm: $\dfrac{2}{R}=\dfrac{62.5}{37.5}\Rightarrow R=1.2\ \Omega$.
Step 3. Shunt. $\dfrac{3X}{3+X}=1.2\Rightarrow X=2\ \Omega$.
Answer. $2$.
Solution by Sreeraj P, M.Sc Physics
With the help of potentiometer, we can determine the value of emf of a given cell. The sensitivity of the potentiometer is
(A) directly proportional to the length of the potentiometer wire
(B) directly proportional to the potential gradient of the wire
(C) inversely proportional to the potential gradient of the wire
(D) inversely proportional to the length of the potentiometer wire
Choose the correct option for the above statements:
Answer: (B) A and C only
Concept. A potentiometer is more sensitive when a small change in voltage gives a larger change in balancing length, that is, when the potential gradient $k$ is small. For a given current, a longer wire gives a smaller $k$.
Step 1. Gradient. Sensitivity $\propto\dfrac1k$: (C) is true and (B) is false.
Step 2. Length. $k=\dfrac{V}{L}$, so a longer wire gives a smaller $k$ and higher sensitivity: (A) is true and (D) is false.
Answer. A and C only (Option B).
Solution by Sreeraj P, M.Sc Physics
A null point is found at $200\ \text{cm}$ in potentiometer when cell in secondary circuit is shunted by $5\ \Omega$. When a resistance of $15\ \Omega$ is used for shunting, null point moves to $300\ \text{cm}$. The internal resistance of the cell is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 5
The terminal voltage $\dfrac{\varepsilon R}{R+r}$ is balanced, so it is proportional to the balancing length:
$$\frac{5/(5+r)}{15/(15+r)}=\frac{200}{300}\ \Rightarrow\ 3\times5(15+r)=2\times15(5+r)$$
$$225+15r=150+30r\ \Rightarrow\ r=5\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
When two resistances $R_1$ and $R_2$ connected in series and introduced into the left gap of a meter bridge and a resistance of $10\ \Omega$ is introduced into the right gap, a null point is found at $60\ \text{cm}$ from left side. When $R_1$ and $R_2$ are connected in parallel and introduced into the left gap, a resistance of $3\ \Omega$ is introduced into the right gap to get null point at $40\ \text{cm}$ from left end. The product of $R_1R_2$ is ______ $\Omega^2$.
Numerical value type. Enter your answer.
Answer: 30
Concept. Two balances give the series sum and the parallel value; their product gives $R_1R_2$.
Step 1. Series. $\dfrac{R_1+R_2}{10}=\dfrac{60}{40}\Rightarrow R_1+R_2=15$.
Step 2. Parallel. $\dfrac{R_p}{3}=\dfrac{40}{60}\Rightarrow R_p=2$.
Step 3. Product. $R_1R_2=R_p(R_1+R_2)=30\ \Omega^2$.
Answer. $30$.
Solution by Sreeraj P, M.Sc Physics
The charge flowing in a conductor changes with time as $Q(t)=\alpha t-\beta t^2+\gamma t^3$, where $\alpha$, $\beta$ and $\gamma$ are constants. Minimum value of current is
Answer: (D) $\alpha-\dfrac{\beta^2}{3\gamma}$
Concept. $I=\dfrac{dQ}{dt}$. The minimum of $I(t)$ is where $\dfrac{dI}{dt}=0$.
Step 1. Current. $I=\alpha-2\beta t+3\gamma t^2$.
Step 2. Minimum. $\dfrac{dI}{dt}=-2\beta+6\gamma t=0\Rightarrow t=\dfrac{\beta}{3\gamma}$.
Step 3. Value. $I_{\min}=\alpha-\dfrac{2\beta^2}{3\gamma}+\dfrac{\beta^2}{3\gamma}=\alpha-\dfrac{\beta^2}{3\gamma}$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
The equivalent resistance between $A$ and $B$ is ______.
Answer: (A) $\dfrac23\ \Omega$
The left vertical wire is a single node (joined to $A$) and the right vertical wire is a single node (joined to $B$). So every path runs between $A$ and $B$:
All in parallel:
$$\frac1{R}=\frac12+\frac1{12}+\frac14+\frac16+\frac12=\frac{18}{12}\ \Rightarrow\ R=\frac23\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
If the potential difference between $B$ and $D$ is zero, the value of $x$ is $\dfrac1n\ \Omega$. The value of $n$ is ______.
Numerical value type. Enter your answer.
Answer: 2
Concept. $V_B=V_D$ means the bridge is balanced: the ratio of the two upper arms equals the ratio of the two lower arms. Reduce each arm first.
Step 1. Arms. Left to $B$: $6\parallel3=2\ \Omega$. $B$ to right: $x\parallel1=\dfrac{x}{x+1}$. Left to $D$: $1+2=3\ \Omega$. $D$ to right: $x$.
Step 2. Balance. $\dfrac{2}{x/(x+1)}=\dfrac3x\Rightarrow2(x+1)=3\Rightarrow x=\dfrac12\ \Omega$.
Step 3. Compare. $x=\dfrac1n$ gives $n=2$.
Answer. $2$.
Solution by Sreeraj P, M.Sc Physics
The drift velocity of electrons for a conductor connected in an electrical circuit is $V_d$. The conductor is now replaced by another conductor with same material and same length but double the area of cross-section. The applied voltage remains same. The new drift velocity of electrons will be
Answer: (A) $V_d$
Concept. $v_d=\dfrac{eE\tau}{m}$ and $E=\dfrac VL$. The drift speed depends on the field, not on the area.
Step 1. Same $V$ and $L$. $E$ is unchanged, so $v_d$ is unchanged (the larger area simply carries more current).
Answer. $V_d$ (Option A).
Solution by Sreeraj P, M.Sc Physics
Two identical cells, when connected either in parallel or in series give same current in an external resistance $5\ \Omega$. The internal resistance of each cell will be ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 5
Series: $I=\dfrac{2E}{5+2r}$. Parallel: $I=\dfrac{E}{5+r/2}=\dfrac{2E}{10+r}$.
Equal currents: $5+2r=10+r\Rightarrow r=5\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
$H$ amount of thermal energy is developed by a resistor in $10\ \text{s}$ when a current of $4\ \text{A}$ is passed through it. If the current is increased to $16\ \text{A}$, the thermal energy developed by the resistor in $10\ \text{s}$ will be
Answer: (B) $16H$
$H=I^2Rt\propto I^2$. The current is $4$ times larger, so the heat is $16H$.
Solution by Sreeraj P, M.Sc Physics
For the given circuit, in the steady state, $|V_B-V_D|=$ ______ V.
Numerical value type. Enter your answer.
Answer: 1
Concept. In steady state the capacitor branches carry no current, so between $B$ and $C$ only the $1\ \Omega$ resistor conducts. Then each side of the network is a simple potential divider across 6 V.
Step 1. Path $ABC$. $2+1=3\ \Omega$: $I=\dfrac63=2$ A, so the drop $A\to B$ is $2\times2=4$ V.
Step 2. Path $ADC$. $10+2=12\ \Omega$: $I=\dfrac6{12}=0.5$ A, so the drop $A\to D$ is $0.5\times10=5$ V.
Step 3. Difference. $|V_B-V_D|=|(V_A-4)-(V_A-5)|=1$ V.
Answer. $1$.
Solution by Sreeraj P, M.Sc Physics
As shown in the figure, a network of resistors is connected to a battery of $24\ \text{V}$ with an internal resistance of $3\ \Omega$. The currents through the resistors $R_4$ and $R_5$ are $I_4$ and $I_5$ respectively. The values of $I_4$ and $I_5$ are
Answer: (D) $I_4=\dfrac25\ \text{A}$ and $I_5=\dfrac85\ \text{A}$
Concept. Reduce each parallel pair, add everything in series with the internal resistance, find the main current, then split it inversely as the resistances.
Step 1. Parallel pairs. $R_1\parallel R_2=\dfrac{2\times2}{4}=1\ \Omega$; $R_4\parallel R_5=\dfrac{20\times5}{25}=4\ \Omega$.
Step 2. Total resistance. $1+2+4+2+3=12\ \Omega$ (including the $3\ \Omega$ internal resistance).
Step 3. Main current. $I=\dfrac{24}{12}=2$ A.
Step 4. Divide between $R_4$ and $R_5$. $I_4=2\times\dfrac{5}{25}=\dfrac25$ A, $I_5=2\times\dfrac{20}{25}=\dfrac85$ A.
Answer. $I_4=\dfrac25$ A and $I_5=\dfrac85$ A (Option D).
Solution by Sreeraj P, M.Sc Physics
A hollow cylindrical conductor has length of $3.14\ \text{m}$, while its inner and outer diameters are $4\ \text{mm}$ and $8\ \text{mm}$ respectively. The resistance of the conductor is $n\times10^{-3}\ \Omega$. If the resistivity of the material is $2.4\times10^{-8}\ \Omega\,\text{m}$, the value of $n$ is ______.
Numerical value type. Enter your answer.
Answer: 2
Concept. For a hollow cylinder, use the area of the ring: $A=\pi(r_2^2-r_1^2)$, with radii from the diameters.
Step 1. Area. $\pi(16-4)\times10^{-6}=12\pi\times10^{-6}$ m$^2$.
Step 2. Resistance. $\dfrac{2.4\times10^{-8}\times3.14}{12\times3.14\times10^{-6}}=2\times10^{-3}\ \Omega$.
Answer. $n=2$.
Solution by Sreeraj P, M.Sc Physics
A cell of emf $90\ \text{V}$ is connected across series combination of two resistors each of $100\ \Omega$ resistance. A voltmeter of resistance $400\ \Omega$ is used to measure the potential difference across each resistor. The reading of the voltmeter will be
Answer: (A) $40\ \text{V}$
Voltmeter across one resistor: $100\parallel400=80\ \Omega$, in series with $100\ \Omega$.
$$V=90\times\frac{80}{180}=40\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics
If a copper wire is stretched to increase its length by $20\%$, the percentage increase in resistance of the wire is ______ %.
Numerical value type. Enter your answer.
Answer: 44
Volume is constant, so $R=\dfrac{\rho l}{A}=\dfrac{\rho l^2}{V}\propto l^2$. $R'=(1.2)^2R=1.44R$: an increase of $44\%$.
Solution by Sreeraj P, M.Sc Physics
A uniform metallic wire carries a current $2\ \text{A}$, when $3.4\ \text{V}$ battery is connected across it. The mass of uniform metallic wire is $8.92\times10^{-3}\ \text{kg}$, density is $8.92\times10^3\ \text{kg m}^{-3}$ and resistivity is $1.7\times10^{-8}\ \Omega\,\text{m}$. The length of wire is
Answer: (B) $l=10\ \text{m}$
Concept. $R=\dfrac{\rho l}{A}$, and the volume $Al$ is fixed by the mass and density. Writing $A=\dfrac{\text{volume}}{l}$ gives $R=\dfrac{\rho l^2}{\text{volume}}$.
Step 1. Resistance. $R=\dfrac VI=\dfrac{3.4}{2}=1.7\ \Omega$.
Step 2. Volume. $Al=\dfrac md=\dfrac{8.92\times10^{-3}}{8.92\times10^{3}}=10^{-6}\ \text{m}^3$.
Step 3. Length. $l^2=\dfrac{R\,(Al)}{\rho}=\dfrac{1.7\times10^{-6}}{1.7\times10^{-8}}=100$, so $l=10$ m.
Answer. $l=10$ m (Option B).
Solution by Sreeraj P, M.Sc Physics
In the given circuit, the equivalent resistance between the terminals $A$ and $B$ is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 10
The plain vertical wire joins the top and bottom rails, so everything to its right (the two $4\ \Omega$ resistors and the end wire) carries no current, and the second vertical $2\ \Omega$ resistor has both ends on the same node, so it is shorted out.
What remains: $3\ \Omega$, then the top $2\ \Omega$ in parallel with the first vertical $2\ \Omega$ ($=1\ \Omega$), then $6\ \Omega$:
$$R_{AB}=3+1+6=10\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
The resistance of a wire is $5\ \Omega$. Its new resistance in ohm, if stretched to $5$ times of its original length will be
Answer: (B) 125
At constant volume $R\propto l^2$: $R'=5\times25=125\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
Two cells are connected between points $A$ and $B$ as shown. Cell 1 has emf of $12\ \text{V}$ and internal resistance of $3\ \Omega$. Cell 2 has emf of $6\ \text{V}$ and internal resistance of $6\ \Omega$. An external resistor $R$ of $4\ \Omega$ is connected across $A$ and $B$. The current flowing through $R$ will be ______ A.
Numerical value type. Enter your answer.
Answer: 1
The cells are connected with opposite polarities. Treat each as a source between $A$ and $B$ and use the parallel-cell formula with one emf negative:
$$\varepsilon_{eq}=\frac{\frac{12}{3}-\frac{6}{6}}{\frac13+\frac16}=\frac{3}{1/2}=6\ \text{V},\qquad r_{eq}=\frac{3\times6}{3+6}=2\ \Omega$$
$$I=\frac{6}{2+4}=1\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics
Given below are two statements:
Statement I: The equivalent resistance of resistors in a series combination is smaller than least resistance used in the combination.
Statement II: The resistivity of the material is independent of temperature.
In the light of the above statements, choose the correct answer from the options given below:
Answer: (A) Both Statement I and Statement II are false
In series the equivalent resistance exceeds the largest resistance, so Statement I is false. Resistivity changes with temperature, so Statement II is false.
Solution by Sreeraj P, M.Sc Physics
A network of four resistances is connected to $9\ \text{V}$ battery, as shown in figure. The magnitude of voltage difference between the points $A$ and $B$ is ______ V.
Numerical value type. Enter your answer.
Answer: 3
Each branch has $6\ \Omega$ and carries $1.5\ \text{A}$. Measuring from the left corner:
drop to $A$: $1.5\times2=3\ \text{V}$; drop to $B$: $1.5\times4=6\ \text{V}$.
$|V_A-V_B|=6-3=3\ \text{V}$.
Solution by Sreeraj P, M.Sc Physics
In a metallic conductor, under the effect of applied electric field, the free electrons of the conductor
Answer: (D) Move in the curved paths from lower potential to higher potential
Concept. Free electrons keep their fast random thermal motion. Between collisions the field accelerates them (force $-e\vec E$, opposite to the field), so each path bends; the net effect is a slow drift towards higher potential.
Step 1. Direction. The field points from higher to lower potential; electrons are pushed the other way, from lower to higher potential. Option A is wrong.
Step 2. Path. A charge accelerated sideways to its velocity moves on a curve (like a projectile), and collisions keep changing its direction. So the paths are neither straight nor uniform. Options B and C are wrong.
Answer. They move in curved paths from lower potential to higher potential (Option D).
Solution by Sreeraj P, M.Sc Physics
A rectangular parallelepiped is measured as $1\ \text{cm}\times1\ \text{cm}\times100\ \text{cm}$. If its specific resistance is $3\times10^{-7}\ \Omega\,\text{m}$, then the resistance between its two opposite rectangular faces will be ______ $\times10^{-7}\ \Omega$.
Numerical value type. Enter your answer.
Answer: 3
The rectangular faces are $1\ \text{cm}\times100\ \text{cm}$ and are $1\ \text{cm}$ apart:
$$R=\frac{\rho l}{A}=\frac{3\times10^{-7}\times0.01}{0.01\times1}=3\times10^{-7}\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
Two identical heater filaments are connected first in parallel and then in series. At the same applied voltage, the ratio of heat produced in same time for parallel to series will be
Answer: (B) $4:1$
Concept. At the same applied voltage, the heat in a given time is $H=\dfrac{V^2t}{R}$, so it is inversely proportional to the equivalent resistance.
Step 1. Resistances. Parallel: $\dfrac R2$; series: $2R$.
Step 2. Ratio. $\dfrac{H_p}{H_s}=\dfrac{2R}{R/2}=4$.
Answer. $4:1$ (Option B).
Solution by Sreeraj P, M.Sc Physics
In the circuit diagram shown in figure given below, the current flowing through resistance $3\ \Omega$ is $\dfrac x3$ A. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 1
Concept. Cells in series opposing each other: the net emf is the difference, the internal resistances add, and the current then divides in the parallel pair.
Step 1. Net emf. $8-4=4$ V.
Step 2. Total resistance. $0.5+1+4.5+(3\parallel6)=6+2=8\ \Omega$.
Step 3. Main current. $I=\dfrac48=0.5$ A.
Step 4. Current in $3\ \Omega$. $0.5\times\dfrac{6}{9}=\dfrac13$ A, so $x=1$.
Answer. $1$.
Solution by Sreeraj P, M.Sc Physics
The current flowing through $R_2$ is
Answer: (C) $\dfrac13\ \text{A}$
$R_6$ is directly across the cell and does not affect the rest.
Between $C$ and $A$: $R_3=6\ \Omega$ in parallel with $R_2+R_1=6\ \Omega$, i.e. $3\ \Omega$. Then $D\to C\to A$ is $3+3=6\ \Omega$, in parallel with $R_7=3\ \Omega$: $2\ \Omega$.
With $R_5$: $2+2=4\ \Omega$, so the current from $E$ is $\dfrac84=2\ \text{A}$ and $V_D-V_A=2\times2=4\ \text{V}$.
Current $D\to C\to A$: $\dfrac46=\dfrac23\ \text{A}$, which splits equally at $C$: $I_{R_2}=\dfrac13\ \text{A}$.
Solution by Sreeraj P, M.Sc Physics
Two identical cells each of emf $1.5\ \text{V}$ are connected in series across a $10\ \Omega$ resistance. An ideal voltmeter connected across $10\ \Omega$ resistance reads $1.5\ \text{V}$. The internal resistance of each cell is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 5
$I=\dfrac{1.5}{10}=0.15\ \text{A}=\dfrac{3}{10+2r}\Rightarrow10+2r=20\Rightarrow r=5\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
A wire of resistance $160\ \Omega$ is melted and drawn in a wire of one-fourth of its length. The new resistance of the wire will be
Answer: (B) $10\ \Omega$
Concept. Melted and redrawn: volume constant. Length $\tfrac14$ means area $\times4$: $R\times\dfrac{1}{16}$.
Step 1. Value. $\dfrac{160}{16}=10\ \Omega$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
The current flowing through a conductor connected across a source is $2\ \text{A}$ and $1.2\ \text{A}$ at $0^\circ\text{C}$ and $100^\circ\text{C}$ respectively. The current flowing through the conductor at $50^\circ\text{C}$ will be ______ $\times10^2$ mA.
Numerical value type. Enter your answer.
Answer: 15
$R\propto\dfrac1I$ and $R=R_0(1+\alpha T)$: $\dfrac{2}{1.2}=1+100\alpha\Rightarrow\alpha=\dfrac1{150}\ ^\circ\text{C}^{-1}$.
At $50^\circ\text{C}$: $R=R_0\left(1+\dfrac13\right)=\dfrac43R_0$, so $I=\dfrac{2}{4/3}=1.5\ \text{A}=15\times10^2\ \text{mA}$.
Solution by Sreeraj P, M.Sc Physics
A potential $V_0$ is applied across a uniform wire of resistance $R$. The power dissipation is $P_1$. The wire is then cut into two equal halves and a potential of $V_0$ is applied across the length of each half. The total power dissipation across two wires is $P_2$. The ratio of $P_2:P_1$ is $\sqrt x:1$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 16
Concept. At fixed voltage $P=\dfrac{V^2}{R}$. Each half has $\dfrac R2$ and gets the full $V_0$.
Step 1. Total. $2\times\dfrac{V_0^2}{R/2}=4P_1$.
Step 2. Form. $4=\sqrt{16}$.
Answer. $x=16$.
Solution by Sreeraj P, M.Sc Physics
The length of a metallic wire is increased by $20\%$ and its area of cross-section is reduced by $4\%$. The percentage change in resistance of the metallic wire is ______.
Numerical value type. Enter your answer.
Answer: 25
Concept. $R=\dfrac{\rho l}{A}$. Use the given changes directly (the volume need not stay constant here).
Step 1. Ratio. $\dfrac{1.2}{0.96}=1.25$.
Answer. $25\%$ increase.
Solution by Sreeraj P, M.Sc Physics
Figure shows a part of an electric circuit. The potentials at points $a$, $b$ and $c$ are $30\ \text{V}$, $12\ \text{V}$ and $2\ \text{V}$ respectively. The current through the $20\ \Omega$ resistor will be
Answer: (B) $0.4\ \text{A}$
Concept. One unknown node joined to points of known potential: by the junction law, the currents into the node add to zero.
Step 1. Junction law at the node $O$. $\dfrac{30-V}{10}+\dfrac{12-V}{20}+\dfrac{2-V}{30}=0$.
Step 2. Solve (×60). $180-6V+36-3V+4-2V=0\Rightarrow V=20$ V.
Step 3. Current in $20\ \Omega$. $\dfrac{20-12}{20}=0.4$ A (from $O$ to $b$).
Answer. $0.4$ A (Option B).
Solution by Sreeraj P, M.Sc Physics
As shown in the figure the voltmeter reads $2\ \text{V}$ across $5\ \Omega$ resistor. The resistance of the voltmeter is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 20
The $2\ \Omega$ resistor has $3-2=1\ \text{V}$ across it, so the total current is $0.5\ \text{A}$.
Current in $5\ \Omega$: $\dfrac25=0.4\ \text{A}$; the voltmeter takes $0.1\ \text{A}$, so $R_V=\dfrac{2}{0.1}=20\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
A current of $2$ A flows through a wire of cross-sectional area $25.0\ \text{mm}^2$. The number of free electrons in a cubic metre are $2.0\times10^{28}$. The drift velocity of the electrons is ______ $\times10^{-6}\ \text{m s}^{-1}$ (given, charge on electron $=1.6\times10^{-19}$ C).
Numerical value type. Enter your answer.
Answer: 25
Concept. $I=neAv_d$: current equals (number per volume)(charge)(area)(drift speed).
Step 1. Solve. $v_d=\dfrac{2}{2\times10^{28}\times1.6\times10^{-19}\times25\times10^{-6}}=\dfrac{2}{80}=2.5\times10^{-2}$ mm/s.
Answer. $25\times10^{-6}$ m s$^{-1}$, so $25$.
Solution by Sreeraj P, M.Sc Physics
The equivalent resistance between $A$ and $B$ as shown in figure is
Answer: (B) $5\ \text{k}\Omega$
Concept. Points joined by a plain wire are at the same potential, so they are the same node. Rename the nodes, then see which resistors share the same two nodes.
Step 1. Merge nodes. The wire from $A$ to $D$ makes $D\equiv A$; the wire from $C$ to $B$ makes $C\equiv B$.
Step 2. Place each resistor. $AC$ ($20$ k$\Omega$) is between $A$ and $B$; $CD$ ($20$ k$\Omega$) is between $B$ and $A$; $DB$ ($10$ k$\Omega$) is between $A$ and $B$.
Step 3. Parallel. All three join $A$ and $B$: $\dfrac1R=\dfrac1{20}+\dfrac1{20}+\dfrac1{10}=\dfrac{4}{20}$, so $R=5$ k$\Omega$.
Answer. $5$ k$\Omega$ (Option B).
Solution by Sreeraj P, M.Sc Physics
The number density of free electrons in copper is nearly $8\times10^{28}\ \text{m}^{-3}$. A copper wire has its area of cross-section $=2\times10^{-6}\ \text{m}^2$ and is carrying a current of $3.2$ A. The drift speed of the electrons is ______ $\times10^{-6}\ \text{m s}^{-1}$.
Numerical value type. Enter your answer.
Answer: 125
$v_d=\dfrac{I}{neA}=\dfrac{3.2}{8\times10^{28}\times1.6\times10^{-19}\times2\times10^{-6}}=1.25\times10^{-4}=125\times10^{-6}\ \text{m s}^{-1}$.
Solution by Sreeraj P, M.Sc Physics
The equivalent resistance of the circuit shown below between points $a$ and $b$ is
Answer: (B) $3.2\ \Omega$
Concept. Points symmetric about the line midway between $a$ and $b$ are at the same potential, so a resistor joining two such points carries no current.
Step 1. Symmetry. The top vertex and the inner point are both placed symmetrically between $a$ and $b$, so both are at the midway potential. The $4\ \Omega$ resistor between them carries no current.
Step 2. Remaining paths. $a\to$top$\to b$: $8\ \Omega$; $a\to$inner$\to b$: $8\ \Omega$; base: $16\ \Omega$.
Step 3. Parallel. $\dfrac1R=\dfrac18+\dfrac18+\dfrac1{16}=\dfrac{5}{16}$, so $R=3.2\ \Omega$.
Answer. $3.2\ \Omega$ (Option B).
Solution by Sreeraj P, M.Sc Physics
$10$ resistors each of resistance $10\ \Omega$ can be connected in such a way as to get maximum and minimum equivalent resistance. The ratio of maximum and minimum equivalent resistance will be ______.
Numerical value type. Enter your answer.
Answer: 100
Concept. With $n$ equal resistors $R$, the maximum is all in series ($nR$) and the minimum is all in parallel ($R/n$), so the ratio is $n^2$.
Step 1. Maximum. $10\times10=100\ \Omega$.
Step 2. Minimum. $\dfrac{10}{10}=1\ \Omega$.
Answer. $100$.
Solution by Sreeraj P, M.Sc Physics
The equivalent resistance between $A$ and $B$ of the network shown in figure is
Answer: (D) $\dfrac83R$
Concept. A bridge is balanced when the ratios of its arms match; then the middle resistor carries no current and can be removed.
Step 1. Identify the bridge. From $A$: $2R$ to node $X$ and $R$ to node $Y$. To $B$: $6R$ from $X$ and $3R$ from $Y$. $9R$ joins $X$ and $Y$.
Step 2. Balance check. $\dfrac{2R}{6R}=\dfrac{R}{3R}=\dfrac13$, so $9R$ carries no current.
Step 3. Two paths in parallel. $(2R+6R)\parallel(R+3R)=\dfrac{8R\times4R}{12R}=\dfrac{8R}{3}$.
Answer. $\dfrac83R$ (Option D).
Solution by Sreeraj P, M.Sc Physics
In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf $1.5$ V is found to be $60$ cm. If this cell is replaced by another cell of emf $E$, the length of null point increases by $40$ cm. The value of $E$ is $\dfrac x{10}$ V. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 25
Concept. With the same potential gradient, the emf balanced is proportional to the balancing length: $\dfrac{E_1}{E_2}=\dfrac{l_1}{l_2}$.
Step 1. New length. $60+40=100$ cm.
Step 2. emf. $E=1.5\times\dfrac{100}{60}=2.5$ V $=\dfrac{25}{10}$ V.
Answer. $25$.
Solution by Sreeraj P, M.Sc Physics
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: For measuring the potential difference across a resistance of $600\ \Omega$, the voltmeter with resistance $1000\ \Omega$ will be preferred over voltmeter with resistance $4000\ \Omega$.
Reason R: Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance.
In the light of the above statements, choose the most appropriate answer from the options given below.
Answer: (A) A is not correct but R is correct
Concept. A voltmeter is a resistor placed in parallel. The higher its resistance compared with the resistor being measured, the less it changes the circuit, so the reading is closer to the true value.
Step 1. Assertion. Across $600\ \Omega$, the $4000\ \Omega$ voltmeter disturbs the circuit less than the $1000\ \Omega$ one, so it is preferred. The assertion is wrong.
Step 2. Reason. At the same voltage, a higher-resistance voltmeter draws a smaller current. The reason is correct.
Answer. A is not correct but R is correct (Option A).
Solution by Sreeraj P, M.Sc Physics
Equivalent resistance between the adjacent corners of a regular $n$-sided polygon of uniform wire of resistance $R$ would be
Answer: (A) $\dfrac{(n-1)R}{n^2}$
Concept. Between adjacent corners: $\dfrac Rn$ in parallel with $\dfrac{(n-1)R}{n}$.
Step 1. Combine. $\dfrac{(R/n)(n-1)R/n}{R}=\dfrac{(n-1)R}{n^2}$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
In the given circuit the value of $\left|\dfrac{I_1+I_3}{I_2}\right|$ is ______.
Numerical value type. Enter your answer.
Answer: 2
Take the left wire at $0$ V. The positive plate of each cell faces right, so the middle node is at $10$ V and the right wire at $30$ V.
$I_1=I_2=\dfrac{10-0}{10}=1$ A (towards the left wire, as marked). $I_3$, marked from left to right, is $\dfrac{0-30}{10}=-3$ A.
$$\left|\frac{I_1+I_3}{I_2}\right|=\left|\frac{1-3}{1}\right|=2$$
Solution by Sreeraj P, M.Sc Physics
Two identical cells each of emf $1.5\ \text{V}$ are connected in parallel across a parallel combination of two resistors each of resistance $20\ \Omega$. A voltmeter connected in the circuit measures $1.2\ \text{V}$. The internal resistance of each cell is:
Answer: (C) $5\ \Omega$
Two identical cells in parallel act as one cell of emf $1.5\ \text{V}$ and internal resistance $r/2$. The external resistance is $20\parallel20 = 10\ \Omega$.
Terminal voltage:
$$V = \frac{E R}{R + r/2}\ \Rightarrow\ 1.2 = \frac{1.5\times10}{10 + r/2}$$
$$10 + \frac r2 = 12.5\ \Rightarrow\ r = 5\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
In a potentiometer arrangement, a cell gives a balancing point at $75\ \text{cm}$ length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emfs of the two cells respectively is $3 : 2$, the difference in the balancing length of the potentiometer wire in the above two cases will be ______ cm.
Numerical value type. Enter your answer.
Answer: 25
$E\propto l$: $\dfrac{E_1}{E_2} = \dfrac{75}{l_2} = \dfrac32\Rightarrow l_2 = 50\ \text{cm}$.
Difference $= 75 - 50 = 25\ \text{cm}$.
Solution by Sreeraj P, M.Sc Physics
What will be the most suitable combination of three resistors $A = 2\ \Omega$, $B = 4\ \Omega$, $C = 6\ \Omega$ so that $\left(\dfrac{22}{3}\right)\Omega$ is the equivalent resistance of the combination?
Answer: (B) Parallel combination of $A$ and $B$ connected in series with $C$
Concept. Test the options: $2\parallel4=\dfrac43$, then add $6$ in series.
Step 1. Value. $\dfrac43+6=\dfrac{22}{3}\ \Omega$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
A potentiometer wire of length $10\ \text{m}$ and resistance $20\ \Omega$ is connected in series with a $25\ \text{V}$ battery and an external resistance $30\ \Omega$. A cell of emf $E$ in the secondary circuit is balanced by $250\ \text{cm}$ long potentiometer wire. The value of $E$ (in volt) is $\dfrac{x}{10}$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 25
Current in the wire $= \dfrac{25}{20+30} = 0.5\ \text{A}$; potential drop across the wire $= 0.5\times20 = 10\ \text{V}$, so the gradient is $1\ \text{V m}^{-1}$.
$$E = 1\times2.5 = 2.5\ \text{V} = \frac{25}{10}\ \Rightarrow\ x = 25$$
Solution by Sreeraj P, M.Sc Physics
A resistor develops $300\ \text{J}$ of thermal energy in $15\ \text{s}$ when a current of $2\ \text{A}$ is passed through it. If the current increases to $3\ \text{A}$, the energy developed in $10\ \text{s}$ is ______ J.
Numerical value type. Enter your answer.
Answer: 450
$300 = (2)^2R(15)\Rightarrow R = 5\ \Omega$. Then $H = (3)^2\times5\times10 = 450\ \text{J}$.
Solution by Sreeraj P, M.Sc Physics
The total current supplied to the circuit as shown in figure by the $5\ \text{V}$ battery is ______ A.
Numerical value type. Enter your answer.
Answer: 2
Concept. The rim is plain wire except for the cell and one $5\ \Omega$ resistor. Mark the two nodes the rim wire creates, then see which spokes run between them.
Step 1. Nodes. Call $P$ the rim from the cell's terminal round to the $5\ \Omega$ rim resistor, and $Q$ the rim beyond it, back to the cell. The two $5\ \Omega$ spokes join $P$ to the centre $O$; the $2.5\ \Omega$ spoke joins $O$ to $Q$.
Step 2. Spoke path. $(5\parallel5)+2.5=2.5+2.5=5\ \Omega$ from $P$ to $Q$.
Step 3. With the rim resistor. $5\ \Omega$ (rim) in parallel with $5\ \Omega$ (spokes): $2.5\ \Omega$.
Step 4. Current. $I=\dfrac{5}{2.5}=2$ A.
Answer. $2$.
Solution by Sreeraj P, M.Sc Physics
Two cells of the same emf $E$ but different internal resistances $r_1$ and $r_2$ are connected in series with an external resistance $R$ as shown in the figure. The terminal potential difference across the second cell is found to be zero. The external resistance $R$ must then be:
Answer: (C) $r_2 - r_1$
Current: $i = \dfrac{2E}{R + r_1 + r_2}$. Terminal voltage of the second cell: $E - ir_2 = 0\Rightarrow E = ir_2$.
$$E = \frac{2Er_2}{R + r_1 + r_2}\ \Rightarrow\ R + r_1 + r_2 = 2r_2\ \Rightarrow\ R = r_2 - r_1$$
Solution by Sreeraj P, M.Sc Physics
The length of a given cylindrical wire is increased to double of its original length. The percentage increase in the resistance of the wire will be ______ %.
Numerical value type. Enter your answer.
Answer: 300
The volume stays constant, so $A\propto\dfrac1l$ and $R = \dfrac{\rho l}{A}\propto l^2$. Doubling $l$ makes $R' = 4R$, an increase of $300\%$.
Solution by Sreeraj P, M.Sc Physics
An aluminium wire is stretched to make its length $0.4\%$ larger. The percentage change in resistance is
Answer: (C) $0.8\%$
With volume constant, $R\propto l^2$, so $\dfrac{\Delta R}{R} = 2\dfrac{\Delta l}{l} = 2\times0.4\% = 0.8\%$.
Solution by Sreeraj P, M.Sc Physics
Find the equivalent resistance between points $A$ and $B$.
Answer: (C) $5\ \Omega$
Reduce from the far end, always measuring resistance to $A$:
$P$–$A$: $5\ \Omega$. $Q$–$A$: $(5 + 5)\parallel10 = 5\ \Omega$.
$R$–$A$: $(5 + 5)\parallel10 = 5\ \Omega$. $S$–$A$: $(5 + 5)\parallel10 = 5\ \Omega$.
$B$–$A$: $(5 + 5)\parallel10 = 5\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
In the given figure, the value of $V_0$ will be ______ V.
Numerical value type. Enter your answer.
Answer: 4
Concept. Several branches, each a cell and a resistor, joined between the same two nodes: the common voltage is $V=\dfrac{\sum E_i/R_i}{\sum 1/R_i}$ (from the junction law at the output node, with no current drawn out).
Step 1. Junction law at $V_0$. $\dfrac{2-V_0}{1}+\dfrac{4-V_0}{1}+\dfrac{6-V_0}{1}=0$ (resistances in k$\Omega$).
Step 2. Solve. $12=3V_0\Rightarrow V_0=4$ V.
Answer. $4$.
Solution by Sreeraj P, M.Sc Physics
Eight copper wires of length $l$ and diameter $d$ are joined in parallel to form a single composite conductor of resistance $R$. If a single copper wire of length $2l$ has the same resistance $R$, then its diameter will be ______ $d$.
Numerical value type. Enter your answer.
Answer: 4
Parallel: $R = \dfrac18\cdot\dfrac{\rho l}{\pi d^2/4}$. Single wire: $R = \dfrac{\rho(2l)}{\pi D^2/4}$.
$$\frac{2}{D^2} = \frac{1}{8d^2}\ \Rightarrow\ D^2 = 16d^2\ \Rightarrow\ D = 4d$$
Solution by Sreeraj P, M.Sc Physics
A current of $15\ \text{mA}$ flows in the circuit as shown in figure. The value of potential difference between the points $A$ and $B$ will be
Answer: (D) $275\ \text{V}$
Between $A$ and $B$ the current passes through $5\ \text{k}\Omega$, then $10\parallel5 = \dfrac{10}{3}\ \text{k}\Omega$, then $10\ \text{k}\Omega$:
$$R = 5 + \frac{10}{3} + 10 = \frac{55}{3}\ \text{k}\Omega,\qquad V = 15\ \text{mA}\times\frac{55}{3}\ \text{k}\Omega = 275\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics
In a potentiometer arrangement, a cell of emf $1.20\ \text{V}$ gives a balance point at $36\ \text{cm}$ length of wire. This cell is now replaced by another cell of emf $1.80\ \text{V}$. The difference in balancing length of potentiometer wire in the above conditions will be ______ cm.
Numerical value type. Enter your answer.
Answer: 18
$l\propto E$: $l_2 = 36\times\dfrac{1.80}{1.20} = 54\ \text{cm}$. Difference $= 18\ \text{cm}$.
Solution by Sreeraj P, M.Sc Physics
The current $I$ in the given circuit will be
Answer: (A) $10\ \text{A}$
Label the nodes: left $L$, right $R$, and the two junctions $M$, $N$ on either side of the $2\ \Omega$ resistor. Then $L$–$M = 4\ \Omega$, $L$–$N = 4\ \Omega$ (top bypass), $M$–$N = 2\ \Omega$, $N$–$R = 4\ \Omega$ and $M$–$R = 4\ \Omega$ (bottom bypass).
This is a balanced Wheatstone bridge ($\frac44 = \frac44$), so no current flows in the $2\ \Omega$ resistor:
$$R_{eq} = (4+4)\parallel(4+4) = 4\ \Omega,\qquad I = \frac{40}{4} = 10\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics
Resistances are connected in a meter bridge circuit as shown in the figure. The balancing length $l_1$ is $40\ \text{cm}$. Now an unknown resistance $x$ is connected in series with $P$ and the new balancing length is found to be $80\ \text{cm}$ measured from the same end. Then the value of $x$ will be ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 20
Balance: $\dfrac PQ = \dfrac{40}{60}\Rightarrow Q = \dfrac{4\times60}{40} = 6\ \Omega$.
With $x$ added: $\dfrac{4 + x}{6} = \dfrac{80}{20} = 4\Rightarrow x = 20\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
A battery of $6\ \text{V}$ is connected to the circuit as shown below. The current $I$ drawn from the battery is
Answer: (A) $1\ \text{A}$
The bridge is balanced ($\frac33 = \frac66$), so the $5\ \Omega$ carries no current.
$$R_{\text{bridge}} = (3+3)\parallel(6+6) = \frac{6\times12}{18} = 4\ \Omega,\qquad I = \frac{6}{4 + 2} = 1\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics
A potentiometer wire of length $300\ \text{cm}$ is connected in series with a resistance $780\ \Omega$ and a standard cell of emf $4\ \text{V}$. A constant current flows through the potentiometer wire. The length of the null point for a cell of emf $20\ \text{mV}$ is found to be $60\ \text{cm}$. The resistance of the potentiometer wire is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 20
Concept. Potential gradient $=\dfrac{V_{\text{wire}}}{L}$. Find the wire's voltage from the null point, then the current.
Step 1. Wire voltage. $20$ mV over $60$ cm, so $100$ mV over $300$ cm.
Step 2. Divider. $\dfrac{4R_w}{780+R_w}=0.1\Rightarrow3.9R_w=78$.
Answer. $R_w=20\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
A cell, shunted by a $8\ \Omega$ resistance, is balanced across a potentiometer wire of length $3$ m. The balancing length is $2$ m when the cell is shunted by $4\ \Omega$ resistance. The value of internal resistance of the cell will be ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 8
Concept. Same method as above.
Step 1. Ratio. $\dfrac32=\dfrac{8(4+r)}{4(8+r)}$.
Step 2. Solve. $12(8+r)=16(4+r)\Rightarrow r=8\ \Omega$.
Answer. $8$.
Solution by Sreeraj P, M.Sc Physics
The current density in a cylindrical wire of radius $4$ mm is $4\times10^6\ \text{A m}^{-2}$. The current through the outer portion of the wire between radial distances $\dfrac R2$ and $R$ is ______ $\pi$ A.
Numerical value type. Enter your answer.
Answer: 48
Concept. For uniform current density, current $=J\times$ area. The outer ring has area $\pi(R^2-R^2/4)$.
Step 1. Area. $\pi\times\tfrac34\times(4\times10^{-3})^2=12\pi\times10^{-6}$ m$^2$.
Step 2. Current. $4\times10^6\times12\pi\times10^{-6}=48\pi$ A.
Answer. $48$.
Solution by Sreeraj P, M.Sc Physics
The current density in a cylindrical wire of radius $r = 4.0$ mm is $1.0\times10^6\ \text{A m}^{-2}$. The current through the outer portion of the wire between radial distances $\dfrac r2$ and $r$ is $x\pi$ A; where $x$ is ______ .
Numerical value type. Enter your answer.
Answer: 12
$$I = J\pi\left(r^2 - \frac{r^2}{4}\right) = \frac34 J\pi r^2 = \frac34\times10^6\times\pi\times16\times10^{-6} = 12\pi\ \text{A}$$
$x = 12$.
Solution by Sreeraj P, M.Sc Physics
In the given circuit $'a'$ is an arbitrary constant. The value of $m$ for which the equivalent circuit resistance is minimum, will be $\sqrt{\dfrac x2}$. The value of $x$ is ______ .
Numerical value type. Enter your answer.
Answer: 3
Concept. Write the equivalent resistance as a function of $m$ and minimise it. A sum $pm+\dfrac qm$ is least when the two terms are equal.
Step 1. Equivalent resistance. Three $ma$ in parallel: $\dfrac{ma}{3}$. Two $\dfrac am$ in parallel: $\dfrac{a}{2m}$. So $R=\dfrac{ma}3+\dfrac{a}{2m}$.
Step 2. Minimise. $\dfrac{dR}{dm}=\dfrac a3-\dfrac{a}{2m^2}=0\Rightarrow m^2=\dfrac32$.
Step 3. Compare. $m=\sqrt{\dfrac32}=\sqrt{\dfrac x2}$ gives $x=3$.
Answer. $3$.
Solution by Sreeraj P, M.Sc Physics
Two sources of equal emfs are connected in series. This combination is connected to an external resistance $R$. The internal resistances of the two sources are $r_1$ and $r_2$ $(r_1 > r_2)$. If the potential difference across the source of internal resistance $r_1$ is zero then the value of $R$ will be
Answer: (A) $r_1 - r_2$
Concept. Terminal voltage of a cell $=E-Ir$. Zero means $Ir_1=E$.
Step 1. Current. $I=\dfrac{2E}{R+r_1+r_2}$.
Step 2. Condition. $\dfrac{2Er_1}{R+r_1+r_2}=E\Rightarrow R=r_1-r_2$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
A $1$ m long copper wire carries a current of $1$ A. If the cross section of the wire is $2.0\ \text{mm}^2$ and the resistivity of copper is $1.7\times10^{-8}\ \Omega$ m. The force experienced by moving electron in the wire is ______ $\times10^{-23}$ N
(Charge of electron $= 1.6\times10^{-19}$ C)
Numerical value type. Enter your answer.
Answer: 136
Field inside the wire:
$$E = \rho J = 1.7\times10^{-8}\times\frac{1}{2\times10^{-6}} = 8.5\times10^{-3}\ \text{V m}^{-1}$$
$$F = eE = 1.6\times10^{-19}\times8.5\times10^{-3} = 1.36\times10^{-21} = 136\times10^{-23}\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics
In meter bridge experiment for measuring unknown resistance $S$, the unknown resistance $S$ is in the left gap and $R$ in the right gap. The null point is obtained at a distance $30$ cm from the left side, at point $D$. If $R$ is $5.6\ \text{k}\Omega$, then the value of unknown resistance $S$ will be ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 2400
At balance,
$$\frac{S}{R} = \frac{l}{100 - l} = \frac{30}{70}$$
$$S = 5600\times\frac37 = 2400\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
A. The drift velocity of electrons decreases with the increase in the temperature of conductor.
B. The drift velocity is inversely proportional to the area of cross-section of given conductor.
C. The drift velocity does not depend on the applied potential difference to the conductor.
D. The drift velocity of electron is inversely proportional to the length of the conductor.
E. The drift velocity increases with the increase in the temperature of conductor.
Choose the correct answer from the options given below:
Answer: (B) A and D only
For a conductor across a potential difference $V$:
$$v_d = \frac{eE\tau}{m} = \frac{eV\tau}{mL}$$
A: True. A higher temperature shortens the relaxation time $\tau$, so $v_d$ decreases (E is false).
B: False. For a given $V$, $v_d$ does not depend on the area.
C: False. $v_d \propto V$.
D: True. $v_d \propto \dfrac1L$.
A and D only.
Solution by Sreeraj P, M.Sc Physics
As shown in the figure, in steady state, the charge stored in the capacitor is ______ $\times10^{-6}$ C.
Numerical value type. Enter your answer.
Answer: 10
In steady state no current flows through the capacitor branch, so the cell drives current only through $R$:
$$I = \frac{E}{R + r} = \frac{10}{110} = \frac{1}{11}\ \text{A}$$
The capacitor branch is in parallel with $R$, and with no current $R'$ has no drop, so
$$V_C = IR = \frac{100}{11}\ \text{V}, \qquad Q = CV_C = 1.1\times10^{-6}\times\frac{100}{11} = 10\times10^{-6}\ \text{C}$$
Solution by Sreeraj P, M.Sc Physics
In a meter bridge experiment with resistances $R_1$ and $R_2$ in the two gaps, the balancing length $AC$ corresponding to null deflection of the galvanometer is $40$ cm. The balancing length, if the radius of the wire $AB$ is doubled, will be ______ cm.
Numerical value type. Enter your answer.
Answer: 40
At balance $\dfrac{R_1}{R_2} = \dfrac{l}{100 - l}$. The wire is uniform, so this ratio of lengths does not depend on the wire's radius.
Doubling the radius changes the resistance per cm equally everywhere, so the balancing length stays $40$ cm.
Solution by Sreeraj P, M.Sc Physics
A wire of resistance $R_1$ is drawn out so that its length is increased by twice of its original length. The ratio of new resistance to original resistance is:
Answer: (A) $9 : 1$
Concept. When a wire is drawn out its volume stays the same, so $A\propto\dfrac1l$ and $R=\dfrac{\rho l}{A}\propto l^2$.
Step 1. New length. "Increased by twice of its original length" means the increase is $2l$, so the new length is $l+2l=3l$.
Step 2. Ratio. $\dfrac{R_2}{R_1}=\left(\dfrac{3l}{l}\right)^2=9$.
Answer. $9:1$ (Option A).
Solution by Sreeraj P, M.Sc Physics
A potentiometer wire $AB$ of resistance $20\ \Omega$ and length $300$ cm is connected in series with a resistance box (R.B.) and a standard cell of emf $4$ V. For a resistance $R$ of resistance box introduced into the circuit, the null point for a cell of $20$ mV is found to be $60$ cm. The value of $R$ is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 780
Potential gradient $= \dfrac{20\ \text{mV}}{60\ \text{cm}}$, so the full $300$ cm wire carries $100$ mV $= 0.1$ V.
Current in the wire: $I = \dfrac{0.1}{20} = 5\times10^{-3}$ A.
$$4 = 5\times10^{-3}(20 + R) \Rightarrow 20 + R = 800 \Rightarrow R = 780\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
In the given figure, a battery of emf $E$ is connected across a conductor $PQ$ of length $l$ and different area of cross-sections having radii $r_1$ and $r_2$ $(r_2 < r_1)$.
Choose the correct option as one moves from $P$ to $Q$.
Answer: (A) Drift velocity of electron increases.
Concept. In a series circuit the same current flows through every cross-section of the conductor. What changes along a tapered wire is how that current is shared over the area.
Step 1. Current. $I$ (and the electron current) is the same at every section from $P$ to $Q$.
Step 2. Drift velocity. $v_d=\dfrac{I}{neA}$. From $P$ to $Q$ the radius falls from $r_1$ to $r_2$, so $A$ falls and $v_d$ rises.
Step 3. Electric field. $E=\rho J=\dfrac{\rho I}{A}$, so $E$ also rises towards $Q$. Options B and C are false, so "all of these" is false too.
Answer. Drift velocity of electrons increases (Option A).
Solution by Sreeraj P, M.Sc Physics
The resistance of a conductor at $15^\circ\text{C}$ is $16\ \Omega$ and at $100^\circ\text{C}$ is $20\ \Omega$. What will be the temperature coefficient of resistance of the conductor?
Answer: (C) $0.003\ ^\circ\text{C}^{-1}$
$R_2 = R_1\left[1 + \alpha(T_2 - T_1)\right]$:
$$\alpha = \frac{R_2 - R_1}{R_1(T_2 - T_1)} = \frac{20 - 16}{16\times85} = \frac{4}{1360} \approx 0.003\ ^\circ\text{C}^{-1}$$
Solution by Sreeraj P, M.Sc Physics
For the circuit shown, the value of current at time $t = 3.2$ s will be ______ A.
[Voltage distribution $V(t)$ is shown by Fig. (1) and the circuit is shown in Fig. (2)]
Numerical value type. Enter your answer.
Answer: 1
Concept. The source $V(t)$ and the 5 V battery act in the same loop with $R$, so at each instant $I=\dfrac{V(t)-5}{R}$.
Step 1. Read the graph. From $t=3$ s to $4$ s, $V$ rises from 5 V to 10 V, i.e. 5 V per second. At $t=3.2$ s: $V=5+0.2\times5=6$ V.
Step 2. Current. $I=\dfrac{6-5}{1}=1$ A.
Answer. $1$.
Solution by Sreeraj P, M.Sc Physics
The voltage drop across $15\ \Omega$ resistance in the given figure will be ______ V.
Numerical value type. Enter your answer.
Answer: 6
Upper branch: $(4\parallel4) + 2 + (15\parallel10) = 2 + 2 + 6 = 10\ \Omega$.
Lower branch: $(8\parallel8) + (12\parallel12) = 4 + 6 = 10\ \Omega$.
The two branches in parallel: $5\ \Omega$. Total with $1\ \Omega$: $6\ \Omega$, so $I = \dfrac{12}{6} = 2$ A, and each branch carries $1$ A.
Voltage across the $15\parallel10$ combination $= 1\times6 = 6$ V.
Solution by Sreeraj P, M.Sc Physics
A square-shaped wire with a resistance of each side $3\ \Omega$ is bent to form a complete circle. The resistance between two diametrically opposite points of the circle in a unit of $\Omega$ will be ______.
Numerical value type. Enter your answer.
Answer: 3
Total resistance of the wire $= 4\times3 = 12\ \Omega$. Diametrically opposite points split it into two halves of $6\ \Omega$ each, in parallel:
$$R = \frac{6\times6}{6 + 6} = 3\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
The equivalent resistance of the given circuit between the terminals $A$ and $B$ is:
Answer: (D) $1\ \Omega$
Work from the left end. The plain wire at the far left shorts the $5\ \Omega$ resistor, so the left end of the first $2\ \Omega$ is connected to the bottom line.
That $2\ \Omega$ is then in parallel with the vertical $2\ \Omega$: $1\ \Omega$.
Adding the next series $2\ \Omega$: $3\ \Omega$.
The two $3\ \Omega$ resistors across $AB$ give $1.5\ \Omega$. Finally:
$$R_{AB} = \frac{3\times1.5}{3 + 1.5} = 1\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
A resistor dissipates $192$ J of energy in $1$ s when a current of $4$ A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in $5$ s is ______ J.
Numerical value type. Enter your answer.
Answer: 3840
$I^2R = 192 \Rightarrow R = \dfrac{192}{16} = 12\ \Omega$.
$$H = (8)^2\times12\times5 = 3840\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics
A current through a wire depends on time as $i = \alpha_0 t + \beta t^2$, where $\alpha_0 = 20\ \text{A s}^{-1}$ and $\beta = 8\ \text{A s}^{-2}$. Find the charge crossed through a section of the wire in 15 s.
Answer: (B) 11250 C
$$q = \int_0^{15}(20t + 8t^2)\,dt = 10t^2 + \frac{8}{3}t^3\Big|_0^{15} = 2250 + 9000 = 11250\ \text{C}$$
Solution by Sreeraj P, M.Sc Physics
A cell $E_1$ of emf 6 V and internal resistance $2\ \Omega$ is connected with another cell $E_2$ of emf 4 V and internal resistance $8\ \Omega$ (as shown in the figure). The potential difference across points $X$ and $Y$ is:
Answer: (B) 5.6 V
The two cells are connected in opposition in a single loop:
$$I = \frac{6 - 4}{2 + 8} = 0.2\ \text{A}$$
The larger emf $E_1$ drives the current, so the current passes through $E_2$ against its emf (charging it). The p.d. across $E_2$, i.e. between $X$ and $Y$:
$$V_{XY} = E_2 + Ir_2 = 4 + 0.2\times 8 = 5.6\ \text{V}$$
(Check through $E_1$: $6 - 0.2\times2 = 5.6$ V.)
Solution by Sreeraj P, M.Sc Physics
A cylindrical wire of radius 0.5 mm and conductivity $5\times10^7\ \text{S m}^{-1}$ is subjected to an electric field of 10 mV m$^{-1}$. The expected value of current in the wire will be $x^3\pi$ mA. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 5
$J = \sigma E = 5\times10^7\times10^{-2} = 5\times10^5\ \text{A m}^{-2}$
$$I = J\pi r^2 = 5\times10^5\times\pi\times(0.5\times10^{-3})^2 = 0.125\pi\ \text{A} = 125\pi\ \text{mA} = 5^3\pi\ \text{mA}$$
So $x = 5$.
Solution by Sreeraj P, M.Sc Physics
In the given circuit of potentiometer, the potential difference $E$ across $AB$ (10 m length) is larger than $E_1$ and $E_2$ as well. For key $K_1$ (closed), the jockey is adjusted to touch the wire at point $J_1$ so that there is no deflection in the galvanometer. Now the first battery $(E_1)$ is replaced by second battery $(E_2)$ for working by making $K_1$ open and $K_2$ closed. The galvanometer gives then null deflection at $J_2$. The value of $\dfrac{E_1}{E_2}$ is $\dfrac{a}{2}$, where $a$ = ______.
Numerical value type. Enter your answer.
Answer: 1
Concept. A folded potentiometer wire is one continuous wire: measure each balance point as the length of wire from $A$. The emfs are then in the ratio of these lengths.
Step 1. Direction of each 1 m length. Starting from $A$, the lengths run alternately left-to-right and right-to-left.
Step 2. $J_1$. It is on the 4th length, which runs right-to-left, 20 cm from its left end, i.e. 80 cm along it: $3+0.8=3.8$ m from $A$.
Step 3. $J_2$. It is on the 8th length (right-to-left), 60 cm from its right end, i.e. 60 cm along it: $7+0.6=7.6$ m from $A$.
Step 4. Ratio. $\dfrac{E_1}{E_2}=\dfrac{3.8}{7.6}=\dfrac12$, so $a=1$.
Answer. $1$.
Solution by Sreeraj P, M.Sc Physics
Consider a 72 cm long wire $AB$ as shown in the figure. The galvanometer jockey is placed at $P$ on $AB$ at a distance $x$ cm from $A$. The galvanometer shows zero deflection.
The value of $x$, to the nearest integer, is ______.
Numerical value type. Enter your answer.
Answer: 48
Balanced Wheatstone bridge: $\dfrac{12}{6} = \dfrac{x}{72 - x}$
$$x = 2(72 - x) \Rightarrow 3x = 144 \Rightarrow x = 48\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics
Two wires of same length and thickness having specific resistances $6\ \Omega$ cm and $3\ \Omega$ cm respectively are connected in parallel. The effective resistivity is $\rho\ \Omega$ cm. The value of $\rho$, to the nearest integer, is ______.
Numerical value type. Enter your answer.
Answer: 4
Concept. Two equal wires in parallel are equivalent to one wire of the same length and double the area. Find the resistivity that gives the same resistance.
Step 1. Parallel value. $\dfrac{6\cdot3}{9}\dfrac LA=2\dfrac LA$.
Step 2. Equivalent wire. $\rho\dfrac{L}{2A}=2\dfrac LA\Rightarrow\rho=4\ \Omega$ cm.
Answer. $4$.
Solution by Sreeraj P, M.Sc Physics
The value of current in the $6\ \Omega$ resistance is:
Answer: (C) 10 A
Take the bottom wire as 0 V and let the top junction be at $V$. Kirchhoff's current law at the junction:
$$\frac{140 - V}{20} + \frac{90 - V}{5} = \frac{V}{6}$$
Multiply by 60: $420 - 3V + 1080 - 12V = 10V \Rightarrow V = 60$ V.
Current in $6\ \Omega$: $\dfrac{60}{6} = 10$ A.
Solution by Sreeraj P, M.Sc Physics
A current of 5 A is passing through a non-linear magnesium wire of cross-section $0.04\ \text{m}^2$. At every point the direction of current density is at an angle of $60^\circ$ with the unit vector of area of cross-section.
The magnitude of electric field at every point of the conductor is: (resistivity of magnesium $\rho = 44\times10^{-8}\ \Omega$ m)
Answer: (C) $11\times10^{-5}\ \text{V m}^{-1}$
Concept. Current is the flux of current density through the area: $I=\vec J\cdot\vec A=JA\cos\theta$. Then Ohm's law in field form gives $E=\rho J$.
Step 1. Current density. $J=\dfrac{I}{A\cos60^\circ}=\dfrac{5}{0.04\times0.5}=250\ \text{A m}^{-2}$.
Step 2. Field. $E=\rho J=44\times10^{-8}\times250=1.1\times10^{-4}\ \text{V m}^{-1}$.
Answer. $11\times10^{-5}\ \text{V m}^{-1}$ (Option C).
Solution by Sreeraj P, M.Sc Physics
In the given figure switches $S_1$ and $S_2$ are in open condition. The resistance across $ab$ when the switches $S_1$ and $S_2$ are closed is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 10
Concept. A closed switch is a plain wire: it joins the top and bottom rows into one node at each switch. The network becomes three parallel pairs in series.
Step 1. First section. $12\parallel6=\dfrac{72}{18}=4\ \Omega$.
Step 2. Second section. $4\parallel4=2\ \Omega$.
Step 3. Third section. $6\parallel12=4\ \Omega$.
Step 4. Series. $4+2+4=10\ \Omega$.
Answer. $10$.
Solution by Sreeraj P, M.Sc Physics
A Copper (Cu) rod of length 25 cm and cross-sectional area $3\ \text{mm}^2$ is joined with a similar Aluminium (Al) rod as shown in figure. Find the resistance of the combination between the ends $A$ and $B$.
(Take resistivity of Copper $= 1.7\times10^{-8}\ \Omega$ m, Resistivity of aluminium $= 2.6\times10^{-8}\ \Omega$ m)
Answer: (D) 0.858 m$\Omega$
Concept. The two rods share the same two ends $A$ and $B$, so they are in parallel. Each rod's resistance is $R=\dfrac{\rho l}{A}$.
Step 1. Copper. $R_{Cu}=\dfrac{1.7\times10^{-8}\times0.25}{3\times10^{-6}}=1.417\times10^{-3}\ \Omega$.
Step 2. Aluminium. $R_{Al}=\dfrac{2.6\times10^{-8}\times0.25}{3\times10^{-6}}=2.167\times10^{-3}\ \Omega$.
Step 3. Parallel. $R=\dfrac{R_{Cu}R_{Al}}{R_{Cu}+R_{Al}}=\dfrac{1.417\times2.167}{3.583}\times10^{-3}\approx0.858\times10^{-3}\ \Omega$.
Answer. $0.858$ m$\Omega$ (Option D).
Solution by Sreeraj P, M.Sc Physics
In an electric circuit, a cell of certain emf provides a potential difference of 1.25 V across a load resistance of $5\ \Omega$. However, it provides a potential difference of 1 V across a load resistance of $2\ \Omega$. The emf of the cell is given by $\dfrac{x}{10}$ V. Then the value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 15
$V = \dfrac{ER}{R + r}$: $1.25 = \dfrac{5E}{5 + r}$ and $1 = \dfrac{2E}{2 + r}$.
So $E = 0.25(5 + r) = 0.5(2 + r) \Rightarrow r = 1\ \Omega$ and $E = 1.5$ V $= \dfrac{15}{10}$ V. So $x = 15$.
Solution by Sreeraj P, M.Sc Physics
Five equal resistances are connected in a network as shown in figure. The net resistance between the points $A$ and $B$ is:
Answer: (A) $R$
Concept. Look for symmetry. Points placed symmetrically with respect to the two terminals are at equal potential, and a resistor between equal potentials carries no current.
Step 1. Nodes. $B$ is joined to $C$ by a wire, so the terminals are $A$ and $C$. $A$ connects to $D$ and $E$; $D$ and $E$ each connect to $C$; $R$ also joins $D$ and $E$.
Step 2. Symmetry. The paths $A\to D\to C$ and $A\to E\to C$ are identical, so $V_D=V_E$ and the resistor $DE$ carries no current.
Step 3. Two equal paths. Each path is $R+R=2R$; two in parallel give $R$.
Answer. $R$ (Option A).
Solution by Sreeraj P, M.Sc Physics
In an electrical circuit, a battery is connected to pass $20$ C of charge through it in a certain given time. The potential difference between two plates of the battery is maintained at $15$ V. The work done by the battery is ______ J
Numerical value type. Enter your answer.
Answer: 300
$W = qV = 20\times15 = 300$ J.
Solution by Sreeraj P, M.Sc Physics
A wire of $1\ \Omega$ has a length of $1$ m. It is stretched till its length increases by $25\%$. The percentage change in resistance to the nearest integer is:
Answer: (D) $56\%$
When a wire is stretched its volume stays constant, so $A \propto 1/l$ and $R = \rho l/A \propto l^2$.
$$\frac{R'}{R} = (1.25)^2 = 1.5625$$
Percentage change $= 56.25\% \approx 56\%$.
Solution by Sreeraj P, M.Sc Physics
In the given potentiometer circuit arrangement, the balancing length $AC$ is measured to be $250$ cm. When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes $400$ cm. The ratio of the emf of two cells $\frac{\varepsilon_1}{\varepsilon_2}$ is:
Answer: (A) $\frac{5}{3}$
Concept. At point (1) the galvanometer balances $\varepsilon_1$ alone; at point (2) it balances the two cells together. Both cells point the same way, so together they give $\varepsilon_1+\varepsilon_2$.
Step 1. Two balances. $\varepsilon_1=k\times250$, $\varepsilon_1+\varepsilon_2=k\times400$.
Step 2. Subtract. $\varepsilon_2=k\times150$.
Step 3. Ratio. $\dfrac{\varepsilon_1}{\varepsilon_2}=\dfrac{250}{150}=\dfrac53$.
Answer. $\dfrac53$ (Option A).
Solution by Sreeraj P, M.Sc Physics
The given potentiometer has its wire of resistance $10\ \Omega$. When the sliding contact is in the middle of the potentiometer wire, the potential drop across $2\ \Omega$ resistor is:
Answer: (C) $\frac{40}{9}$ V
Concept. A load connected across part of a potential divider is in parallel with that part, so it lowers that part's resistance and its share of the voltage.
Step 1. Halves of the wire. With the contact in the middle, each half is $5\ \Omega$.
Step 2. Loaded half. The $2\ \Omega$ resistor is in parallel with the half from $A$: $\dfrac{5\times2}{7}=\dfrac{10}{7}\ \Omega$.
Step 3. Current. Total $\dfrac{10}7+5=\dfrac{45}{7}\ \Omega$, so $I=\dfrac{20\times7}{45}=\dfrac{28}{9}$ A.
Step 4. Voltage across $2\ \Omega$. $V=\dfrac{28}{9}\times\dfrac{10}{7}=\dfrac{40}{9}$ V.
Answer. $\dfrac{40}{9}$ V (Option C).
Solution by Sreeraj P, M.Sc Physics
A $16\ \Omega$ wire is bent to form a square loop. A $9$ V supply having an internal resistance of $1\ \Omega$ is connected across one of its sides. The potential drop across the diagonals of the square loop is ______ $\times10^{-1}$ V.
Numerical value type. Enter your answer.
Answer: 45
Each side is $4\ \Omega$. Across the chosen side, $4\ \Omega$ is in parallel with the other three sides ($12\ \Omega$): $R = 3\ \Omega$.
$I = \dfrac{9}{3 + 1} = 2.25$ A, so the voltage across the side is $6.75$ V.
The current in the $12\ \Omega$ path is $\dfrac{6.75}{12} = 0.5625$ A, giving $2.25$ V per side. Each diagonal spans two sides of this path:
$$V_{diagonal} = 2\times2.25 = 4.5\ \text{V} = 45\times10^{-1}\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics
The material filled between the plates of a parallel plate capacitor has resistivity $200\ \Omega$ m. The value of capacitance of the capacitor is $2$ pF. If a potential difference of $40$ V is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is: (given the value of relative permittivity of material is $50$)
Answer: (A) $0.9$ mA
Concept. The same slab of material sets both the capacitance and the leakage resistance: $C=\dfrac{K\varepsilon_0A}{d}$ and $R=\dfrac{\rho d}{A}$, so $RC=\rho K\varepsilon_0$, independent of the plate size.
Step 1. Leakage resistance. $R=\dfrac{\rho K\varepsilon_0}{C}=\dfrac{200\times50\times8.85\times10^{-12}}{2\times10^{-12}}\approx4.4\times10^4\ \Omega$.
Step 2. Current. $I=\dfrac VR=\dfrac{40}{4.4\times10^4}\approx9\times10^{-4}$ A.
Answer. $0.9$ mA (Option A).
Solution by Sreeraj P, M.Sc Physics
In the given figure, the emf of the cell is $2.2$ V and if internal resistance is $0.6\ \Omega$. Calculate the power dissipated in the whole circuit:
Answer: (D) $2.2$ W
The left side of the rectangle is a plain wire joined to $A$ and the right side a plain wire joined to $B$, so every resistor path runs from $A$ to $B$. The two diagonals cross without touching.
Four parallel paths between $A$ and $B$: $4\ \Omega$ (top), $8\ \Omega$ (bottom), $4 + 8 = 12\ \Omega$ and $2 + 4 = 6\ \Omega$ (diagonals).
$$\frac{1}{R} = \frac{1}{4} + \frac{1}{8} + \frac{1}{12} + \frac{1}{6} = \frac{15}{24} \Rightarrow R = 1.6\ \Omega$$
$I = \dfrac{2.2}{1.6 + 0.6} = 1$ A. Power dissipated in the whole circuit (including the cell) $= \varepsilon I = 2.2$ W.
Solution by Sreeraj P, M.Sc Physics
What equal length of an iron wire and a copper-nickel alloy wire, each of $2$ mm diameter connected parallel to give an equivalent resistance of $3\ \Omega$? (Given resistivities of iron and copper-nickel alloy wire are $12\ \mu\Omega$ cm and $51\ \mu\Omega$ cm respectively)
Answer: (A) $97$ m
For equal $l$ and $A$, $R_{eq} = \dfrac{l}{A}\cdot\dfrac{\rho_1\rho_2}{\rho_1 + \rho_2}$.
$\dfrac{\rho_1\rho_2}{\rho_1+\rho_2} = \dfrac{12\times51}{63} = 9.71\ \mu\Omega$ cm $= 9.71\times10^{-8}\ \Omega$ m; $A = \pi(10^{-3})^2 = 3.14\times10^{-6}$ m$^2$.
$$l = \frac{3\times3.14\times10^{-6}}{9.71\times10^{-8}} \approx 97\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics
An electric bulb of $500$ W at $100$ V is used in a circuit having a $200$ V supply. Calculate the resistance $R$ to be connected in series with the bulb so that the power delivered by the bulb is $500$ W.
Answer: (C) $20\ \Omega$
Concept. The bulb must still get its rated $100$ V and current; the series resistor drops the rest.
Step 1. Bulb. $I=\dfrac{500}{100}=5$ A.
Step 2. Resistor. $R=\dfrac{200-100}{5}=20\ \Omega$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
If you are provided a set of resistances, $2\ \Omega$, $4\ \Omega$, $6\ \Omega$ and $8\ \Omega$. Connect these resistances to obtain an equivalent resistance of $\frac{46}{3}\ \Omega$.
Answer: (C) $2\ \Omega$ and $4\ \Omega$ are in parallel with $6\ \Omega$ and $8\ \Omega$ in series
Each option means: the first two resistors in parallel, and this combination in series with the other two.
$2\ \Omega \parallel 4\ \Omega = \dfrac{8}{6} = \dfrac{4}{3}\ \Omega$; adding $6 + 8 = 14\ \Omega$ in series:
$$R = \frac{4}{3} + 14 = \frac{46}{3}\ \Omega$$
(The others give $13.5\ \Omega$, $12.4\ \Omega$ and $\frac{66}{7}\ \Omega$.)
Solution by Sreeraj P, M.Sc Physics
Five identical cells each of internal resistance $1\ \Omega$ and emf $5$ V are connected in series and in parallel with an external resistance $R$. For what value of $R$, current in series and parallel combination will remain the same?
Answer: (A) $1\ \Omega$
Series: $I_s = \dfrac{25}{R + 5}$. Parallel: $I_p = \dfrac{5}{R + \frac{1}{5}}$.
$$25\left(R + \frac{1}{5}\right) = 5(R + 5) \Rightarrow 20R = 20 \Rightarrow R = 1\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
First, a set of $n$ equal resistors of $10\ \Omega$ each are connected in series to a battery of E.M.F. $20$ V and internal resistance $10\ \Omega$. A current $I$ is observed to flow. Then, the $n$ resistors are connected in parallel to the same battery. It is observed that the current is increased $20$ times, then the value of $n$ is ______.
Numerical value type. Enter your answer.
Answer: 20
Concept. Write the current for each arrangement with $I=\dfrac{E}{R_{ext}+r}$ and use the given ratio.
Step 1. Series. $R_{ext}=10n$: $I=\dfrac{20}{10n+10}$.
Step 2. Parallel. $R_{ext}=\dfrac{10}{n}$: $I'=\dfrac{20}{10/n+10}$.
Step 3. Ratio. $\dfrac{I'}{I}=\dfrac{10n+10}{10/n+10}=\dfrac{n+1}{(1+n)/n}=n$. So $n=20$.
Answer. $20$.
Solution by Sreeraj P, M.Sc Physics
The colour coding on a carbon resistor is shown in the given figure. The resistance value of the given resistor is:
Answer: (A) $(7500 \pm 375)\ \Omega$
Concept. Colour code: first band = first digit, second band = second digit, third band = power of ten, fourth band = tolerance (gold 5%). Digits: black 0, brown 1, red 2, orange 3, yellow 4, green 5, blue 6, violet 7, grey 8, white 9.
Step 1. Digits. Violet $=7$, green $=5$: $75$.
Step 2. Multiplier. Red $=10^2$: $75\times100=7500\ \Omega$.
Step 3. Tolerance. Gold $=5\%$ of $7500=375\ \Omega$.
Answer. $(7500\pm375)\ \Omega$ (Option A).
Solution by Sreeraj P, M.Sc Physics
The ratio of the equivalent resistance of the network (shown in figure) between the points $a$ and $b$ when switch is open and switch is closed is $x : 8$. The value of $x$ is
Numerical value type. Enter your answer.
Answer: 9
Switch open: two branches of $R + 2R = 3R$ in parallel: $R_{open} = \dfrac{3R}{2}$.
Switch closed: the midpoints are joined, giving $(R \parallel 2R)$ in series with $(2R \parallel R)$:
$$R_{closed} = \frac{2R}{3} + \frac{2R}{3} = \frac{4R}{3}$$
$$\frac{R_{open}}{R_{closed}} = \frac{3/2}{4/3} = \frac{9}{8} \Rightarrow x = 9$$
Solution by Sreeraj P, M.Sc Physics
The series combination of two batteries, both of the same emf $10$ V, but different internal resistance of $20\ \Omega$ and $5\ \Omega$, is connected to the parallel combination of two resistors $30\ \Omega$ and $x\ \Omega$. The voltage difference across the battery of internal resistance $20\ \Omega$ is zero, the value of $x$ (in $\Omega$) is ______.
Numerical value type. Enter your answer.
Answer: 30
Terminal voltage of the $20\ \Omega$ battery: $10 - 20I = 0$, so $I = 0.5$ A.
The total emf is $20$ V and the total internal resistance $25\ \Omega$:
$$I = \frac{20}{25 + R_{\text{ext}}} = 0.5 \Rightarrow R_{\text{ext}} = 15\ \Omega$$
$$\frac{30x}{30 + x} = 15 \Rightarrow x = 30\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
In a metre bridge experiment $S$ is a standard resistance. $R$ is a resistance wire. It is found that the balancing length, measured from the end of the bridge wire on the side of $R$, is $l = 25$ cm. If $R$ is replaced by a wire of half length and half diameter that of $R$ of same material, then the balancing distance $l'$ (in cm) will now be ______.
Numerical value type. Enter your answer.
Answer: 40
Concept. Balance in a metre bridge: $\dfrac RS=\dfrac{l}{100-l}$, with $l$ measured from the end on the side of $R$. For a wire, $R=\dfrac{\rho l}{\pi d^2/4}\propto\dfrac{l}{d^2}$.
Step 1. First balance. $\dfrac RS=\dfrac{25}{75}=\dfrac13$.
Step 2. New wire. Half the length and half the diameter: $R'=R\times\dfrac{1/2}{(1/2)^2}=2R$, so $\dfrac{R'}{S}=\dfrac23$.
Step 3. New balance. $\dfrac{l'}{100-l'}=\dfrac23\Rightarrow3l'=200-2l'\Rightarrow l'=40$ cm.
Answer. $40$.
Solution by Sreeraj P, M.Sc Physics
In the circuit given in the figure, the currents in different branches and the value of one resistor are shown. Then the potential at point $B$ with respect to the point $A$ is:
Answer: (D) $+1$ V
Concept. Find the current in the middle branch with the junction law, then walk from $A$ to $B$ adding each rise and drop in potential.
Step 1. Current in $DC$. At $C$: $1$ A arrives from $A$ (the $1$ A in branch $EA$ continues through $AC$) and $2$ A leaves towards $F$. So $1$ A comes from $D$ to $C$ through the $2\ \Omega$ resistor.
Step 2. Walk $A\to C$. The 1 V cell has its positive terminal towards $C$: rise of $1$ V.
Step 3. Walk $C\to D$. Against the $1$ A current through $2\ \Omega$: rise of $2$ V.
Step 4. Walk $D\to B$. The 2 V cell has its positive terminal towards $D$: drop of $2$ V.
Step 5. Total. $V_B-V_A=+1+2-2=+1$ V.
Answer. $+1$ V (Option D).
Solution by Sreeraj P, M.Sc Physics
In the figure shown, the current in the $10$ V battery is close to:
Answer: (C) $0.21$ A from positive to negative terminal
Concept. Three branches join the same two nodes (top junction $T$ and bottom junction $M$). Find $V_{TM}$ with the junction law, then the current in the 10 V branch.
Step 1. Branches from $M$ to $T$. Left: 20 V cell (raising the potential towards $T$) with $2+5=7\ \Omega$. Middle: $10\ \Omega$. Right: 10 V cell (positive towards $T$) with $4\ \Omega$.
Step 2. Junction law at $T$. $\dfrac{20-V}{7}+\dfrac{0-V}{10}+\dfrac{10-V}{4}=0$, with $V=V_T-V_M$.
Step 3. Solve (×140). $400-20V-14V+350-35V=0\Rightarrow V=\dfrac{750}{69}\approx10.87$ V.
Step 4. 10 V branch. $I=\dfrac{10-10.87}{4}\approx-0.217$ A. The minus sign means the current flows against the cell's own push: it enters at the positive terminal, so inside the cell it flows from positive to negative (the cell is being charged). Its size, about $0.22$ A, matches $0.21$ A.
Answer. $0.21$ A from positive to negative terminal (Option C).
Solution by Sreeraj P, M.Sc Physics
A circuit to verify Ohm's law uses an ammeter and a voltmeter, connected correctly to the resistor in series or parallel. In the circuit:
Answer: (C) The ammeter is always connected in series and the voltmeter in parallel
The ammeter must carry the same current as the resistor, so it is placed in series with it. The voltmeter must measure the potential difference across the resistor, so it is placed in parallel with it.
Solution by Sreeraj P, M.Sc Physics
The current $I_1$ (in A) flowing through the $1\ \Omega$ resistor in the following circuit is:
Answer: (C) $0.2$
The ideal $1$ V cell is connected directly across both the $2\ \Omega$ resistor and the upper branch, so each branch has $1$ V across it.
Upper branch: the two $1\ \Omega$ resistors in parallel ($0.5\ \Omega$) in series with $2\ \Omega$, total $2.5\ \Omega$. Its current is $\dfrac{1}{2.5} = 0.4$ A.
This divides equally between the two identical $1\ \Omega$ resistors:
$$I_1 = 0.2\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics
In a building there are $15$ bulbs of $45$ W, $15$ bulbs of $100$ W, $15$ small fans of $10$ W and $2$ heaters of $1$ kW. The voltage of the electric main supply is $220$ V. The minimum fuse capacity (rated value) of the building will be:
Answer: (D) $20$ A
Concept. The fuse must carry the total current when everything is on: $I=\dfrac{P_{total}}{V}$. Choose the next rating above it.
Step 1. Total power. $15\times45+15\times100+15\times10+2\times1000=675+1500+150+2000=4325$ W.
Step 2. Current. $I=\dfrac{4325}{220}\approx19.7$ A.
Step 3. Fuse. The smallest rating above 19.7 A is 20 A.
Answer. $20$ A (Option D).
Solution by Sreeraj P, M.Sc Physics
The balancing length for a cell is $560$ cm in a potentiometer experiment. When an external resistance of $10\ \Omega$ is connected in parallel to the cell, the balancing length changes by $60$ cm. If the internal resistance of the cell is $\dfrac{n}{10}\ \Omega$, where $n$ is an integer, then the value of $n$ is ______.
Numerical value type. Enter your answer.
Answer: 12
With the shunt, the terminal voltage drops, so the new balancing length is $560 - 60 = 500$ cm.
$$r = R\left(\frac{l_1}{l_2} - 1\right) = 10\left(\frac{560}{500} - 1\right) = 1.2\ \Omega = \frac{12}{10}\ \Omega$$
So $n = 12$.
Solution by Sreeraj P, M.Sc Physics
The length of a potentiometer wire is $1200$ cm and it carries a current of $60$ mA. For a cell of emf $5$ V and internal resistance of $20\ \Omega$, the null point on it is found to be at $1000$ cm. The resistance of the whole wire is:
Answer: (D) $100\ \Omega$
Concept. At balance no current flows in the cell, so its full emf $5$ V matches the wire voltage over $1000$ cm.
Step 1. Gradient. $\dfrac{5}{1000}$ V/cm, so the whole wire has $6$ V.
Step 2. Resistance. $\dfrac{6}{0.06}=100\ \Omega$.
Answer. Option D. (Internal resistance plays no part at balance.)
Solution by Sreeraj P, M.Sc Physics
Four resistances of $15\ \Omega$, $12\ \Omega$, $4\ \Omega$ and $10\ \Omega$ respectively, in cyclic order, form a Wheatstone's network. The resistance that is to be connected in parallel with the resistance of $10\ \Omega$ to balance the network is ______ $\Omega$.
Numerical value type. Enter your answer.
Answer: 10
Concept. A Wheatstone network $P,Q,R,S$ in cyclic order balances when the products of opposite arms are equal: $PR=QS$.
Step 1. Condition. $15\times4=12\times S\Rightarrow S=5\ \Omega$.
Step 2. Shunt on $10\ \Omega$. $\dfrac{10x}{10+x}=5\Rightarrow x=10\ \Omega$.
Answer. $10$.
Solution by Sreeraj P, M.Sc Physics
Consider four conducting materials copper, tungsten, mercury and aluminium with resistivity $\rho_C$, $\rho_T$, $\rho_M$ and $\rho_A$ respectively. Then:
Answer: (B) $\rho_M > \rho_A > \rho_C$
Concept. Typical resistivities at room temperature: copper $1.7\times10^{-8}$, aluminium $2.7\times10^{-8}$, tungsten $5.6\times10^{-8}$ and mercury about $98\times10^{-8}\ \Omega$ m.
Step 1. Order. $\rho_M>\rho_T>\rho_A>\rho_C$.
Step 2. Check the options. Only $\rho_M>\rho_A>\rho_C$ agrees with this order.
Answer. $\rho_M>\rho_A>\rho_C$ (Option B).
Solution by Sreeraj P, M.Sc Physics
A potentiometer wire $PQ$ of $1\ \text{m}$ length is connected to a standard cell $E_1$. Another cell $E_2$ of emf $1.02\ \text{V}$ is connected with a resistance $r$ and switch $S$ (as shown in figure). With switch $S$ open, the null position is obtained at a distance of $49\ \text{cm}$ from $Q$. The potential gradient in the potentiometer wire is:
Answer: (A) $0.02\ \text{V/cm}$
At the null point no current flows through $E_2$, so the resistance $r$ has no effect and the balancing length corresponds to the emf $1.02$ V:
$$k = \frac{1.02\ \text{V}}{49\ \text{cm}} \approx 0.02\ \text{V/cm}$$
Solution by Sreeraj P, M.Sc Physics
An ideal cell of emf $10\ \text{V}$ is connected in the circuit shown in figure. Each resistance is $2\ \Omega$. The potential difference (in V) across the capacitor when it is fully charged is ______.
Numerical value type. Enter your answer.
Answer: 8
Concept. In steady state the capacitor branch (with $R_5$) carries no current. Find the potentials at the two ends of the capacitor from the conducting part of the circuit.
Step 1. Conducting network. Take the left node at 10 V and the right node at 0. $R_1+R_2=4\ \Omega$ is in parallel with $R_3=2\ \Omega$ (both from the left node to the middle node), then $R_4$ to the right node.
Step 2. Current. $(4\parallel2)+2=\dfrac43+2=\dfrac{10}3\ \Omega$, so $I=3$ A and the middle node is at $10-3\times\dfrac43=6$ V.
Step 3. Top node. The $R_1R_2$ path carries $\dfrac{4\ \text{V}}{4\ \Omega}=1$ A, so the node between $R_1$ and $R_2$ is at $10-1\times2=8$ V.
Step 4. Capacitor. One plate is at 8 V; the other connects through $R_5$ (no current) to the right node at 0 V. $V_C=8$ V.
Answer. $8$.
Solution by Sreeraj P, M.Sc Physics
Model a torch battery of length $l$ to be made up of a thin cylindrical bar of radius $a$ and a concentric thin cylindrical shell of radius $b$, with the space in between filled with an electrolyte of resistivity $\rho$. If the battery is connected to a resistance of value $R$, the maximum Joule heating in $R$ will take place for:
Answer: (B) $R = \dfrac{\rho}{2\pi l}\ln\left(\dfrac ba\right)$
Concept. The electrolyte between the bar and the shell is the battery's internal resistance, with current flowing radially. Maximum power goes to the load when $R$ equals the internal resistance.
Step 1. Internal resistance. A thin shell of radius $x$ and thickness $dx$ has $dr=\dfrac{\rho\,dx}{2\pi xl}$, so $r=\displaystyle\int_a^b\frac{\rho\,dx}{2\pi lx}=\frac{\rho}{2\pi l}\ln\frac ba$.
Step 2. Maximum power transfer. Joule heating in $R$ is greatest when $R=r$.
Answer. $R=\dfrac{\rho}{2\pi l}\ln\left(\dfrac ba\right)$ (Option B).
Solution by Sreeraj P, M.Sc Physics
Two resistors $400\ \Omega$ and $800\ \Omega$ are connected in series across a $6\ \text{V}$ battery. The potential difference measured by a voltmeter of $10\ \text{k}\Omega$ across $400\ \Omega$ resistor is close to:
Answer: (D) $1.95\ \text{V}$
$400\ \Omega$ in parallel with $10\ \text{k}\Omega$: $R_p = \dfrac{400\times10000}{10400} \approx 384.6\ \Omega$.
$$V = 6\times\frac{384.6}{384.6 + 800} \approx 1.95\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics
A battery of $3.0\ \text{V}$ is connected to a resistor dissipating $0.5\ \text{W}$ of power. If the terminal voltage of the battery is $2.5\ \text{V}$, the power dissipated within the internal resistance is:
Answer: (C) $0.10\ \text{W}$
Concept. The lost voltage $E-V$ is across the internal resistance; the same current flows through it.
Step 1. Current. $\dfrac{0.5}{2.5}=0.2$ A.
Step 2. Internal power. $(3-2.5)\times0.2=0.1$ W.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
In the circuit diagram shown, the value of current $i_1$ flowing from $A$ to $C$ through the two $4\ \Omega$ resistors along $AC$ is:
Answer: (D) $1\ \text{A}$
The network is symmetric about the line $AC$ and also about the line $BD$, so $B$, $D$ and the centre $O$ are all at the same potential (midway between $A$ and $C$). No current flows in the two $5\ \Omega$ resistors.
The path $A \to O \to C$ is then just $4 + 4 = 8\ \Omega$ directly across the $8$ V cell:
$$i_1 = \frac{8}{8} = 1\ \text{A}$$
(The other two paths, $ABC$ and $ADC$, carry $2$ A each, so the cell supplies $5$ A in all.)
Solution by Sreeraj P, M.Sc Physics
Four resistances $40\ \Omega$, $60\ \Omega$, $90\ \Omega$ and $110\ \Omega$ make the arms of a quadrilateral $ABCD$ ($AB = 40\ \Omega$, $BC = 60\ \Omega$, $AD = 90\ \Omega$, $DC = 110\ \Omega$). Across $AC$ is a battery of emf $40\ \text{V}$ and internal resistance negligible. The potential difference across $BD$ in V is ______.
Numerical value type. Enter your answer.
Answer: 2
Concept. With the battery across $AC$, the paths $ABC$ and $ADC$ are two potential dividers. The potential difference across $BD$ is the difference of the two midpoint potentials.
Step 1. Path $ABC$. Take $V_A=40$ V, $V_C=0$: $V_B=40\times\dfrac{60}{40+60}=24$ V.
Step 2. Path $ADC$. $V_D=40\times\dfrac{110}{90+110}=22$ V.
Step 3. Difference. $V_B-V_D=2$ V.
Answer. $2$.
Solution by Sreeraj P, M.Sc Physics
An electrical power line, having a total resistance of $2\ \Omega$, delivers $1\ \text{kW}$ at $220\ \text{V}$. The efficiency of the transmission line is approximately:
Answer: (D) $96\%$
Concept. Line loss $I^2R$; efficiency $=\dfrac{P_{\text{delivered}}}{P_{\text{delivered}}+P_{\text{loss}}}$.
Step 1. Current. $\dfrac{1000}{220}=4.55$ A.
Step 2. Loss. $4.55^2\times2=41.3$ W.
Step 3. Efficiency. $\dfrac{1000}{1041.3}=96\%$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
A carbon resistor has colour bands green, orange, yellow and gold, in that order. What is the value of the resistance?
Answer: (C) $530\ \text{k}\Omega \pm 5\%$
Colour code: green $= 5$, orange $= 3$, yellow $=$ multiplier $10^4$, gold $= \pm5\%$.
$$R = 53\times10^4\ \Omega = 530\ \text{k}\Omega \pm 5\%$$
Solution by Sreeraj P, M.Sc Physics
In the given circuit the internal resistance of the $18\ \text{V}$ cell is negligible. If $R_1 = 400\ \Omega$, $R_3 = 100\ \Omega$ and $R_4 = 500\ \Omega$ and the reading of an ideal voltmeter across $R_4$ is $5\ \text{V}$, then the value of $R_2$ will be:
Answer: (B) $300\ \Omega$
Concept. An ideal voltmeter across $R_4$ gives the current in the $R_3$–$R_4$ branch. That branch is in parallel with $R_2$, which gives the voltage across $R_2$, and the rest of the 18 V is across $R_1$.
Step 1. Branch current. $I_{34}=\dfrac{5}{500}=0.01$ A.
Step 2. Voltage across the parallel part. $V=0.01\times(100+500)=6$ V.
Step 3. Main current. $R_1$ takes $18-6=12$ V, so $I=\dfrac{12}{400}=0.03$ A.
Step 4. $R_2$. It carries $0.03-0.01=0.02$ A at 6 V: $R_2=\dfrac{6}{0.02}=300\ \Omega$.
Answer. $300\ \Omega$ (Option B).
Solution by Sreeraj P, M.Sc Physics
In the given circuit the cells have zero internal resistance. The currents (in amperes) passing through resistance $R_1$ and $R_2$ respectively, are:
Answer: (C) $0.5,\ 0$
Take the top wire as $0\ \text{V}$. The left vertical wire connects it to the negative terminal of the left cell, so the junction between the two cells (bottom end of $R_1$) is at $+10\ \text{V}$.
$R_1$ has $10\ \text{V}$ across it: $I_1 = \dfrac{10}{20} = 0.5\ \text{A}$.
Going from that junction through the right cell (positive to negative) the potential falls by $10\ \text{V}$, so the bottom of $R_2$ is at $0\ \text{V}$, the same as its top. Hence $I_2 = 0$.
Solution by Sreeraj P, M.Sc Physics
A uniform metallic wire has a resistance of $18\ \Omega$ and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is:
Answer: (D) $4\ \Omega$
Each side is $6\ \Omega$. Between two vertices one side ($6\ \Omega$) is in parallel with the other two in series ($12\ \Omega$):
$$R = \frac{6\times12}{6+12} = 4\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
A $2\ \text{W}$ carbon resistor is colour coded with green, black, red and silver respectively. The maximum current which can be passed through this resistor is:
Answer: (C) $20\ \text{mA}$
Concept. Read the resistance from the colour code, then the power rating limits the current: $P=I^2R\Rightarrow I_{\max}=\sqrt{P/R}$.
Step 1. Resistance. Green $5$, black $0$, red $10^2$: $R=50\times100=5000\ \Omega$ (silver is the tolerance).
Step 2. Maximum current. $I_{\max}=\sqrt{\dfrac{2}{5000}}=\sqrt{4\times10^{-4}}=0.02$ A.
Answer. $20$ mA (Option C).
Solution by Sreeraj P, M.Sc Physics
A potentiometer wire $AB$ having length $L$ and resistance $12r$ is joined to a cell $D$ of emf $\varepsilon$ and internal resistance $r$. A cell $C$ having emf $\varepsilon/2$ and internal resistance $3r$ is connected. The length $AJ$, at which the galvanometer, as shown in the figure, shows no deflection is
Answer: (D) $\dfrac{13}{24}L$
Concept. Find the potential gradient from the primary circuit. At the null point no current flows through cell $C$, so its internal resistance does not matter and the wire balances its full emf.
Step 1. Primary current. $I=\dfrac{\varepsilon}{12r+r}=\dfrac{\varepsilon}{13r}$.
Step 2. Gradient. Voltage across the whole wire $=\dfrac{12\varepsilon}{13}$, so $k=\dfrac{12\varepsilon}{13L}$.
Step 3. Balance. $\dfrac{\varepsilon}{2}=k\,(AJ)\Rightarrow AJ=\dfrac{\varepsilon}{2}\times\dfrac{13L}{12\varepsilon}=\dfrac{13}{24}L$.
Answer. $\dfrac{13}{24}L$ (Option D).
Solution by Sreeraj P, M.Sc Physics
The actual value of a resistance $R$ is $30\ \Omega$. It is measured using the standard formula $R = \dfrac VI$, where $V$ and $I$ are the readings of a voltmeter connected directly across $R$ and an ammeter connected in series with this combination and a cell. If the measured value of $R$ is $5\%$ less, then the internal resistance of the voltmeter is:
Answer: (C) $570\ \Omega$
Concept. The voltmeter across $R$ takes some current, so the ammeter reads the current through $R$ and $R_V$ together. The measured value $V/I$ is $R$ in parallel with $R_V$.
Step 1. Measured value. $5\%$ less than $30\ \Omega$: $28.5\ \Omega$.
Step 2. Parallel equation. $\dfrac{30R_V}{30+R_V}=28.5\Rightarrow30R_V=855+28.5R_V$.
Step 3. Solve. $1.5R_V=855\Rightarrow R_V=570\ \Omega$.
Answer. $570\ \Omega$ (Option C).
Solution by Sreeraj P, M.Sc Physics
The Wheatstone bridge shown in the figure below, gets balanced when the carbon resistor used as $R_1$ has the colour code (orange, red, brown). The resistors $R_2$ and $R_4$ are $80\ \Omega$ and $40\ \Omega$, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as $R_3$, would be
Answer: (B) brown, blue, brown
Orange $= 3$, red $= 2$, brown $= \times10$: $R_1 = 320\ \Omega$.
At balance $\dfrac{R_1}{R_2} = \dfrac{R_3}{R_4}$:
$$R_3 = \frac{320\times40}{80} = 160\ \Omega = 16\times10^1\ \Omega$$
Colour code: brown $(1)$, blue $(6)$, brown $(\times10)$.
Solution by Sreeraj P, M.Sc Physics
A current of $2\ \text{mA}$ was passed through an unknown resistor which dissipated a power of $4.4\ \text{W}$. Dissipated power when an ideal power supply of $11\ \text{V}$ is connected across it is:
Answer: (C) $11\times10^{-5}\ \text{W}$
$R = \dfrac{P}{I^2} = \dfrac{4.4}{(2\times10^{-3})^2} = 1.1\times10^6\ \Omega$.
$$P' = \frac{V^2}{R} = \frac{121}{1.1\times10^6} = 11\times10^{-5}\ \text{W}$$
Solution by Sreeraj P, M.Sc Physics
A current of $5\ \text{A}$ passes through a copper conductor (resistivity $= 1.7\times10^{-8}\ \Omega\,\text{m}$) of radius of cross-section $5\ \text{mm}$. Find the mobility of the charges if their drift velocity is $1.1\times10^{-3}\ \text{m s}^{-1}$.
Answer: (C) $1.0\ \text{m}^2\text{V}^{-1}\text{s}^{-1}$
Concept. Mobility $\mu=\dfrac{v_d}{E}$; the field inside is $E=\rho J=\rho\dfrac{I}{A}$.
Step 1. Field. $E=\dfrac{1.7\times10^{-8}\times5}{\pi\times25\times10^{-6}}=1.08\times10^{-3}$ V/m.
Step 2. Mobility. $\mu=\dfrac{1.1\times10^{-3}}{1.08\times10^{-3}}\approx1.0$ m$^2$V$^{-1}$s$^{-1}$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Two cells, each of emf $1.5\ \text{V}$ and internal resistance $r$, are connected in series with a $2\ \Omega$ resistor and a parallel combination of $15\ \Omega$ and $10\ \Omega$ resistors. An ideal voltmeter connected across the $10\ \Omega$ resistance reads $2\ \text{V}$. The internal resistance $r$ of each cell is:
Answer: (C) $0.5\ \Omega$
Concept. The voltmeter reading gives the current in the parallel pair; then $2E=I(R_{\text{total}}+2r)$.
Step 1. Parallel. $15\parallel10=6\ \Omega$; current $\dfrac26=\dfrac13$ A.
Step 2. Loop. $3=\dfrac13(2+6+2r)\Rightarrow r=0.5\ \Omega$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
In an experiment, the resistance of a material is measured as a function of temperature (in some range). A graph of $\ln R(T)$ against $\dfrac{1}{T^2}$ is found to be a straight line with a negative slope. One may conclude that
Answer: (D) $R(T) = R_0e^{-T_0^2/T^2}$
Concept. A straight line of $y$ against $x$ means $y=c+mx$. Here $y=\ln R$ and $x=\dfrac{1}{T^2}$, with a negative slope.
Step 1. Equation of the line. $\ln R=\ln R_0-\dfrac{T_0^2}{T^2}$, where $-T_0^2$ is the (negative) slope.
Step 2. Take the exponential. $R=R_0\,e^{-T_0^2/T^2}$.
Answer. $R(T)=R_0e^{-T_0^2/T^2}$ (Option D).
Solution by Sreeraj P, M.Sc Physics
In a meter bridge experiment, a resistance box $R$ is in the left gap and an unknown resistance $X$ in the right gap; the balance length $l$ is measured from the left end of the wire. The observations are:
$$\begin{array}{ccc} \text{S. No.} & R\,(\Omega) & l\,(\text{cm}) \\ \hline 1 & 1000 & 60 \\ 2 & 100 & 13 \\ 3 & 10 & 1.5 \\ 4 & 1 & 1.0 \end{array}$$
Which of the readings is inconsistent?
Answer: (D) 4
Concept. With $R$ in the left gap, $X=R\,\dfrac{100-l}{l}$. A consistent set of readings gives roughly the same $X$ each time.
Step 1. Reading 1. $1000\times\dfrac{40}{60}\approx667\ \Omega$.
Step 2. Reading 2. $100\times\dfrac{87}{13}\approx669\ \Omega$.
Step 3. Reading 3. $10\times\dfrac{98.5}{1.5}\approx657\ \Omega$.
Step 4. Reading 4. $1\times\dfrac{99}{1}=99\ \Omega$: far from the others.
Answer. Reading 4 (Option D).
Solution by Sreeraj P, M.Sc Physics
Space between two concentric conducting spheres of radii $a$ and $b$ $(b > a)$ is filled with a medium of resistivity $\rho$. The resistance between the two spheres will be:
Answer: (C) $\dfrac{\rho}{4\pi}\left(\dfrac1a - \dfrac1b\right)$
Concept. Current flows radially, so thin shells of area $4\pi r^2$ and thickness $dr$ are in series: $R=\int\dfrac{\rho\,dr}{4\pi r^2}$.
Step 1. Integrate. $\dfrac{\rho}{4\pi}\left(\dfrac1a-\dfrac1b\right)$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
A copper wire is stretched to make it $0.5\%$ longer. The percentage change in its electrical resistance if its volume remains unchanged is
Answer: (B) $1.0\%$
With volume $V = Al$ fixed, $A = V/l$, so
$$R = \frac{\rho l}{A} = \frac{\rho l^2}{V} \propto l^2$$
For a small change, $\dfrac{\Delta R}{R} = 2\dfrac{\Delta l}{l} = 2 \times 0.5\% = 1.0\%$.
Solution by Sreeraj P, M.Sc Physics
When the switch $S$, in the circuit shown, is closed, then the value of current $i$ will be
Answer: (B) $5\ \text{A}$
Concept. With the switch closed, $C$ is joined to $V=0$ through $2\ \Omega$. Apply the junction law at $C$, the only unknown node.
Step 1. Junction law at $C$. $\dfrac{20-V_C}{2}+\dfrac{10-V_C}{4}=\dfrac{V_C-0}{2}$.
Step 2. Solve (×4). $40-2V_C+10-V_C=2V_C\Rightarrow V_C=10$ V.
Step 3. Current. $i=\dfrac{10}{2}=5$ A.
Answer. $5$ A (Option B).
Solution by Sreeraj P, M.Sc Physics
Drift speed of electrons, when $1.5\ \text{A}$ current flows in a copper wire of cross section $5\ \text{mm}^2$, is $v_d$. If the electron density in copper is $9\times10^{28}\ \text{m}^{-3}$ the value of $v_d$ in $\text{mm s}^{-1}$ is close to (Take charge of an electron to be $1.6\times10^{-19}\ \text{C}$)
Answer: (D) $0.02$
$$v_d = \frac{I}{neA} = \frac{1.5}{9\times10^{28}\times1.6\times10^{-19}\times5\times10^{-6}} = \frac{1.5}{7.2\times10^{4}}$$
$$v_d \approx 2.1\times10^{-5}\ \text{m s}^{-1} = 0.02\ \text{mm s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics
A carbon resistor has colour bands red, violet, orange and silver, in that order. Its value and tolerance are given respectively by
Answer: (A) $27\ \text{k}\Omega,\ 10\%$
Colour code: red $= 2$, violet $= 7$, orange $= $ multiplier $10^3$, silver $=$ tolerance $\pm10\%$.
$$R = 27\times10^3\ \Omega = 27\ \text{k}\Omega,\quad \pm10\%$$
Solution by Sreeraj P, M.Sc Physics
For the circuit shown, with $R_1 = 1.0\ \Omega$, $R_2 = 2.0\ \Omega$, $E_1 = 2\ \text{V}$ and $E_2 = E_3 = 4\ \text{V}$, the potential difference between the points $a$ and $b$ is approximately (in V)
Answer: (A) $3.3$
Concept. All three branches run between $a$ and $b$, each with a cell and a resistance. The junction law at $a$ gives $V_{ab}=\dfrac{\sum E_i/R_i}{\sum1/R_i}$.
Step 1. Branches (from $b$ to $a$). $E_1=2$ V with $2R_1=2\ \Omega$; $E_2=4$ V with $R_2=2\ \Omega$; $E_3=4$ V with $2R_1=2\ \Omega$. All three cells have their positive terminals towards $a$.
Step 2. Junction law. $V_{ab}=\dfrac{\frac22+\frac42+\frac42}{\frac12+\frac12+\frac12}=\dfrac{5}{1.5}$.
Step 3. Value. $V_{ab}\approx3.3$ V.
Answer. $3.3$ (Option A).
Solution by Sreeraj P, M.Sc Physics
A $200\ \Omega$ resistor has certain colour code. If one replaces the red colour by green in the code, the new resistance will be
Answer: (D) $500\ \Omega$
$200\ \Omega = 20\times10^1$: bands red (2), black (0), brown ($\times10$). Replacing red by green (5) gives $50\times10 = 500\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
A four-wire potentiometer is made of a $400\ \text{cm}$ long wire, which extends between A and B. The resistance per unit length of the potentiometer wire is $r = 0.01\ \Omega/\text{cm}$. The wire is connected in series with a $1\ \Omega$ resistor and two cells, each of emf $1.5\ \text{V}$ and internal resistance $0.5\ \Omega$. If an ideal voltmeter is connected between end A and a jockey J placed at $50\ \text{cm}$ from end A (along the wire), the expected reading of the voltmeter will be
Answer: (B) $0.25\ \text{V}$
Concept. The potentiometer wire carries the current of the primary circuit. An ideal voltmeter across part of the wire reads the current times the resistance of that part.
Step 1. Wire resistance. $400\times0.01=4\ \Omega$.
Step 2. Current. Two cells in series: $E=3$ V, internal $1\ \Omega$. $I=\dfrac{3}{4+1+1}=0.5$ A.
Step 3. Reading. 50 cm of wire is $0.5\ \Omega$: $V=0.5\times0.5=0.25$ V.
Answer. $0.25$ V (Option B).
Solution by Sreeraj P, M.Sc Physics
A cell of internal resistance $r$ drives current through an external resistance $R$. The power delivered by the cell to the external resistance will be maximum when
Answer: (B) $R = r$
$$P = \frac{E^2R}{(R + r)^2},\quad \frac{dP}{dR} = E^2\frac{(R + r) - 2R}{(R + r)^3} = 0 \Rightarrow R = r$$
This is the maximum power transfer condition.
Solution by Sreeraj P, M.Sc Physics
In the figure shown, what is the current (in ampere) drawn from the battery? You are given: $R_1 = 15\ \Omega$, $R_2 = 10\ \Omega$, $R_3 = 20\ \Omega$, $R_4 = 5\ \Omega$, $R_5 = 25\ \Omega$, $R_6 = 30\ \Omega$, $E = 15\ \text{V}$
Answer: (A) $9/32$
$R_3$, $R_4$ and $R_5$ are in series: $20 + 5 + 25 = 50\ \Omega$. This is in parallel with $R_2$:
$$\frac{50\times10}{60} = \frac{25}{3}\ \Omega$$
Adding $R_1$ and $R_6$ in series: $R_{eq} = 15 + \frac{25}3 + 30 = \frac{160}{3}\ \Omega$.
$$I = \frac{15}{160/3} = \frac{45}{160} = \frac{9}{32}\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics
A wire of resistance $R$ is bent to form a square $ABCD$ as shown in the figure. The effective resistance between $E$ and $C$ is ($E$ is mid-point of arm $CD$)
Answer: (B) $\dfrac{7}{64}R$
Concept. Between two points on a closed loop of uniform wire, the two arcs are in parallel. If one arc is a fraction $f$ of the whole resistance $R$, the equivalent is $f(1-f)R$.
Step 1. Arc $EC$. Each side is $\dfrac R4$, and $E$ is the midpoint of $CD$, so arc $EC=\dfrac R8$.
Step 2. Other arc. $E\to D\to A\to B\to C$: $R-\dfrac R8=\dfrac{7R}8$.
Step 3. Parallel. $\dfrac{(R/8)(7R/8)}{R}=\dfrac{7R}{64}$.
Answer. $\dfrac{7}{64}R$ (Option B).
Solution by Sreeraj P, M.Sc Physics
Determine the charge on the capacitor in the following circuit:
Answer: (D) $200\ \mu\text{C}$
In steady state no current flows through the capacitor branch.
$2\ \Omega$ and $10\ \Omega$ in series ($12\ \Omega$) are in parallel with $4\ \Omega$: $\dfrac{12\times4}{16} = 3\ \Omega$. Total resistance $6 + 3 = 9\ \Omega$, so $I = 72/9 = 8\ \text{A}$.
Voltage across the $4\ \Omega$ resistor: $72 - 8\times6 = 24\ \text{V}$. Current through the $2\ \Omega$–$10\ \Omega$ branch: $24/12 = 2\ \text{A}$, so the capacitor (in parallel with $10\ \Omega$) has $V = 20\ \text{V}$:
$$Q = CV = 10\ \mu\text{F}\times20\ \text{V} = 200\ \mu\text{C}$$
Solution by Sreeraj P, M.Sc Physics
A metal wire of resistance $3\ \Omega$ is elongated to make a uniform wire of double its previous length. This new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle $60^\circ$ at the center, the equivalent resistance between these two points will be
Answer: (A) $\dfrac53\ \Omega$
Concept. Doubling length (constant volume) gives $R\times4=12\ \Omega$. A $60^\circ$ arc is $\tfrac16$ of the circle; the two arcs are in parallel.
Step 1. Arcs. $2\ \Omega$ and $10\ \Omega$.
Step 2. Parallel. $\dfrac{20}{12}=\dfrac53\ \Omega$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
In a conductor, if the number of conduction electrons per unit volume is $8.5\times10^{28}\ \text{m}^{-3}$ and mean free time is $25\ \text{fs}$ (femto second), its approximate resistivity is ($m_e = 9.1\times10^{-31}\ \text{kg}$)
Answer: (D) $10^{-8}\ \Omega\,\text{m}$
Concept. In the free-electron model the resistivity is $\rho=\dfrac{m}{ne^2\tau}$, where $\tau$ is the mean free time.
Step 1. Substitute. $\rho=\dfrac{9.1\times10^{-31}}{8.5\times10^{28}\times(1.6\times10^{-19})^2\times25\times10^{-15}}$.
Step 2. Denominator. $8.5\times10^{28}\times2.56\times10^{-38}\times2.5\times10^{-14}=5.44\times10^{-23}$.
Step 3. Result. $\rho=\dfrac{9.1\times10^{-31}}{5.44\times10^{-23}}\approx1.7\times10^{-8}\ \Omega$ m, of order $10^{-8}\ \Omega$ m.
Answer. $10^{-8}\ \Omega$ m (Option D).
Solution by Sreeraj P, M.Sc Physics
Two equal resistances when connected in series to a battery, consume electric power of $60$ W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be:
Answer: (B) $240$ W
Concept. Same battery: $P=\dfrac{V^2}{R_{\text{eq}}}$. Series $2R$, parallel $\dfrac R2$: four times smaller resistance.
Step 1. Power. $4\times60=240$ W.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
In a Wheatstone bridge (see figure), resistances $P$ and $Q$ are approximately equal. When $R = 400\ \Omega$, the bridge is balanced. On interchanging $P$ and $Q$, the value of $R$, for balance, is $405\ \Omega$. The value of $X$ is close to
Answer: (D) $402.5\ \Omega$
Concept. At balance $\dfrac PQ=\dfrac RX$. Interchanging $P$ and $Q$ gives a second balance with the ratio inverted; multiplying the two equations removes the unknown ratio $\dfrac PQ$.
Step 1. First balance. $\dfrac PQ=\dfrac{400}{X}$.
Step 2. After interchanging. $\dfrac QP=\dfrac{405}{X}$.
Step 3. Multiply. $1=\dfrac{400\times405}{X^2}\Rightarrow X=\sqrt{400\times405}\approx402.5\ \Omega$.
Answer. $402.5\ \Omega$ (Option D).
Solution by Sreeraj P, M.Sc Physics
The resistance of the meter bridge AB in given figure is $4\ \Omega$. With a cell of emf $\varepsilon = 0.5$ V and rheostat resistance $R_h = 2\ \Omega$ the null point is obtained at some point J. When the cell is replaced by another one of emf $\varepsilon = \varepsilon_2$ the same null point J is found for $R_h = 6\ \Omega$. The emf $\varepsilon_2$ is:
Answer: (B) $0.3$ V
Current in the wire $= \dfrac{6}{4 + R_h}$ (the 6 V cell taken as ideal). Let the resistance of AJ be $r$.
With $R_h = 2\ \Omega$: $I = 1$ A, so $\varepsilon = 1\times r = 0.5$ V gives $r = 0.5\ \Omega$.
With $R_h = 6\ \Omega$: $I = 0.6$ A, so
$$\varepsilon_2 = 0.6\times0.5 = 0.3\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics
In the circuit shown, the potential difference between A and B is
Answer: (B) $2$ V
A and B are open ends, so no current flows in the $5\ \Omega$ and $10\ \Omega$ resistors; $V_{AB} = V_{DC}$.
The three cells with equal internal-type resistances ($1\ \Omega$ each) are in parallel between D and C. With all cells facing the same way, the common voltage is
$$V_{DC} = \frac{\varepsilon_1/r + \varepsilon_2/r + \varepsilon_3/r}{3/r} = \frac{1 + 2 + 3}{3} = 2\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics
In the experimental set up of metre bridge shown in the figure, the null point is obtained at a distance of $40$ cm from A. If a $10\ \Omega$ resistor is connected in series with $R_1$, the null point shifts by $10$ cm. The resistance that should be connected in parallel with $(R_1 + 10)\ \Omega$ such that the null point shifts back to its initial position is
Answer: (C) $60\ \Omega$
Concept. Write the balance condition before and after each change; the unknowns are $R_1$ and $R_2$.
Step 1. First balance. $\dfrac{R_1}{R_2}=\dfrac{40}{60}=\dfrac23$.
Step 2. $10\ \Omega$ in series with $R_1$. The null point moves to 50 cm (a larger left arm moves it away from $A$): $\dfrac{R_1+10}{R_2}=1$.
Step 3. Solve. $R_2=R_1+10$ and $R_1=\dfrac23R_2$: $\dfrac{R_2}{3}=10\Rightarrow R_2=30\ \Omega$, $R_1=20\ \Omega$, so the left arm is now $30\ \Omega$.
Step 4. Shunt to return to 40 cm. The left arm must again be $20\ \Omega$: $\dfrac{30X}{30+X}=20\Rightarrow X=60\ \Omega$.
Answer. $60\ \Omega$ (Option C).
Solution by Sreeraj P, M.Sc Physics
An ideal battery of emf $4$ V and resistance $R$ are connected in series in the primary circuit of a potentiometer of length $1$ m and resistance $5\ \Omega$. The value of $R$, to give a potential difference of $5$ mV across $10$ cm of potentiometer wire, is:
Answer: (C) $395\ \Omega$
$5$ mV across $10$ cm means $50$ mV across the whole $1$ m wire, so the current is $\dfrac{0.05}{5} = 0.01$ A.
$$\frac{4}{5 + R} = 0.01 \Rightarrow R = 395\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
Two electric bulbs, rated at $(25\ \text{W}, 220\ \text{V})$ and $(100\ \text{W}, 220\ \text{V})$, are connected in series across a $220$ V voltage source. If the $25$ W and $100$ W bulbs draw powers $P_1$ and $P_2$ respectively, then:
Answer: (D) $P_1 = 16$ W, $P_2 = 4$ W
Concept. Ratings give resistances, $R=\dfrac{V^2}{P}$. In series the current is the same, so $P\propto R$.
Step 1. Resistances. $R_{25}=4R_{100}$.
Step 2. Current. $I=\dfrac{220}{5R_{100}}$.
Step 3. Powers. $P_{100}=I^2R_{100}=\dfrac{100}{25}=4$ W; $P_{25}=4\times4=16$ W.
Answer. Option D. (The lower-rated bulb glows brighter in series.)
Solution by Sreeraj P, M.Sc Physics
In a meter bridge, the wire of length $1$ m has a non-uniform cross-section such that, the variation $\dfrac{dR}{dl}$ of its resistance $R$ with length $l$ is $\dfrac{dR}{dl} \propto \dfrac{1}{\sqrt l}$. Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP?
Answer: (D) $0.25$ m
Concept. With equal resistances in the two gaps, balance needs equal wire resistances on the two sides: $R_{AP}=R_{PB}$. For a non-uniform wire, find $R$ along the length by integrating $\dfrac{dR}{dl}$.
Step 1. Resistance up to $l$. $\dfrac{dR}{dl}=\dfrac{k}{\sqrt l}\Rightarrow R(l)=2k\sqrt l$.
Step 2. Balance. $R_{AP}=\dfrac12R_{AB}$: $2k\sqrt{l}=\dfrac12\times2k\sqrt1\Rightarrow\sqrt l=\dfrac12$.
Step 3. Length. $l=0.25$ m.
Answer. $0.25$ m (Option D).
Solution by Sreeraj P, M.Sc Physics
The charge on a capacitor plate in a circuit, as a function of time, is shown in the figure. What is the value of current at $t = 4$ s?
Answer: (C) Zero
Concept. Current is the rate of flow of charge, $I=\dfrac{dq}{dt}$: the slope of the $q$–$t$ graph.
Step 1. Read the graph at $t=4$ s. Between $t=2$ s and $t=6$ s the charge stays at $3\ \mu$C, so the graph is flat there.
Step 2. Slope. A flat line has zero slope, so $I=0$ at $t=4$ s.
Answer. Zero (Option C).
Solution by Sreeraj P, M.Sc Physics
In the given circuit diagram, the currents, $I_1 = -0.3$ A, $I_4 = 0.8$ A and $I_5 = 0.4$ A, are flowing as shown. The currents $I_2$, $I_3$ and $I_6$, respectively, are:
Answer: (C) $1.1$ A, $0.4$ A, $0.4$ A
Concept. Kirchhoff's junction law: at every node, current in equals current out. A negative value means the current flows opposite to the arrow.
Step 1. Node $S$. In: $I_4=0.8$ A. Out: $I_5=0.4$ A and $I_3$. So $I_3=0.4$ A.
Step 2. Node $P$. In: $I_5=0.4$ A. Out: $I_6$. So $I_6=0.4$ A.
Step 3. Node $Q$. In: $I_6+I_3=0.8$ A. Out: $I_2+I_1$, with $I_1=-0.3$ A. So $I_2=0.8+0.3=1.1$ A.
Step 4. Check at $R$. In: $I_2+I_1=1.1-0.3=0.8$ A $=I_4$. ✓
Answer. $1.1$ A, $0.4$ A, $0.4$ A (Option C).
Solution by Sreeraj P, M.Sc Physics
To verify Ohm's law, a student connects the voltmeter across the battery as shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained. If $V_0$ is almost zero, identify the correct statement:
Answer: (A) The emf of the battery is $1.5$ V and its internal resistance is $1.5\ \Omega$
Concept. A voltmeter across a battery reads its terminal voltage, $V=E-Ir$. A graph of $V$ against $I$ is a straight line with intercept $E$ (at $I=0$) and slope $-r$.
Step 1. emf. At $I=0$, $V=1.5$ V, so $E=1.5$ V.
Step 2. Internal resistance. At $I=1000$ mA $=1$ A, $V\approx0$: $0=1.5-1\times r\Rightarrow r=1.5\ \Omega$.
Step 3. Check the other options. The graph says nothing about $R$ (the rheostat changes the current), and at 1 A the terminal voltage is about zero, not 1.5 V.
Answer. emf $1.5$ V and internal resistance $1.5\ \Omega$ (Option A).
Solution by Sreeraj P, M.Sc Physics
The resistive network shown below is connected to a D.C. source of $16$ V. The power consumed by the network is $4$ Watt. The value of $R$ is:
Answer: (B) $8\ \Omega$
$4R \parallel 4R = 2R$ and $6R \parallel 12R = 4R$, so the total is $2R + R + 4R + R = 8R$.
$$P = \frac{V^2}{8R} = \frac{256}{8R} = 4 \Rightarrow R = 8\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
In a meter bridge, resistance $X$ is in the left gap and resistance $Y$ in the right gap. It is given that resistance $Y = 12.5\ \Omega$ and that the balance is obtained at a distance $39.5\ \text{cm}$ from end $A$ (the left end of the wire) by jockey $J$. After interchanging the resistances $X$ and $Y$ a new balance point is found at a distance $l_2$ from end $A$. What are the values of $X$ (in $\Omega$) and $l_2$?
Answer: (B) $8.16\ \Omega$ and $60.5\ \text{cm}$
At balance,
$$\frac{X}{Y} = \frac{l_1}{100 - l_1} = \frac{39.5}{60.5}$$
$$X = 12.5 \times \frac{39.5}{60.5} = 8.16\ \Omega$$
After interchanging, $\dfrac{Y}{X} = \dfrac{l_2}{100 - l_2}$, which is satisfied by $l_2 = 100 - 39.5 = 60.5\ \text{cm}$.
Solution by Sreeraj P, M.Sc Physics
In the given circuit all resistance is of the value of $R$ ohm each. The equivalent resistance between $A$ and $B$ is:
Answer: (D) $2R$
Look for plain wires (no resistor). The left-middle junction, the centre junction, the bottom-middle junction and the bottom-left junction are all connected to one another by plain wires, so they are a single node.
The junction after the resistor from $A$ and the junction after the resistor from $B$ are therefore at the same potential. Every other resistor joins this node to the upper and right-hand junctions, which connect only back to the same node, so no current flows through any of them.
Only the two resistors attached to the terminals remain, in series:
$$R_{AB} = R + R = 2R$$
Solution by Sreeraj P, M.Sc Physics
A copper rod of cross-sectional area $A$ carries a uniform current $I$ through it. At temperature $T$, if the volume charge density of the rod is $\rho$, how long will the charges take to travel a distance $d$?
Answer: (C) $\dfrac{\rho dA}{I}$
Current $I = \rho A v_d$, where $\rho$ is the charge per unit volume of the moving carriers. So
$$v_d = \frac{I}{\rho A}, \qquad t = \frac{d}{v_d} = \frac{\rho dA}{I}$$
Solution by Sreeraj P, M.Sc Physics
A constant voltage is applied between two ends of a metallic wire. If the length is halved and the radius of the wire is doubled, the rate of heat developed in the wire will be:
Answer: (A) Increased 8 times
$R = \dfrac{\rho l}{\pi r^2}$. Halving $l$ and doubling $r$:
$$R' = \frac{\rho(l/2)}{\pi(2r)^2} = \frac{R}{8}$$
At constant voltage, $P = \dfrac{V^2}{R}$, so the rate of heating increases 8 times.
Solution by Sreeraj P, M.Sc Physics
A heating element has a resistance of $100\ \Omega$ at room temperature. When it is connected to a supply of $220\ \text{V}$, a steady current of $2\ \text{A}$ passes in it and temperature is $500^\circ\text{C}$ more than the room temperature. The temperature coefficient of resistance of the heating element is
Answer: (B) $2\times10^{-4}\ ^\circ\text{C}^{-1}$
Hot resistance: $R = \dfrac{220}{2} = 110\ \Omega$.
$$R = R_0(1 + \alpha\Delta T) \;\Rightarrow\; 110 = 100(1 + 500\alpha) \;\Rightarrow\; \alpha = \frac{0.1}{500} = 2\times10^{-4}\ ^\circ\text{C}^{-1}$$
Solution by Sreeraj P, M.Sc Physics
In the given circuit diagram, when the current reaches a steady-state in the circuit, the charge on the capacitor of capacitance $C$ will be:
Answer: (D) $CE\dfrac{r_2}{r + r_2}$
Concept. In steady state no current flows through a capacitor branch, so the resistor in series with it ($r_1$) has no drop, and the capacitor takes the voltage across the branch it is parallel to.
Step 1. Current. Only the loop through $E$, $r$ and $r_2$ carries current: $I=\dfrac{E}{r+r_2}$.
Step 2. Voltage across the capacitor branch. It is parallel to $r_2$: $V=Ir_2=\dfrac{Er_2}{r+r_2}$.
Step 3. Charge. $q=CV=CE\dfrac{r_2}{r+r_2}$.
Answer. $CE\dfrac{r_2}{r+r_2}$ (Option D).
Solution by Sreeraj P, M.Sc Physics
In the circuit shown, the current in each resistance is:
Answer: (A) $0$ A
Apply Kirchhoff's loop law to the first loop: left wire, top $2$ V cell, first $1\ \Omega$ resistor and bottom $2$ V cell. The top and bottom cells face the same way, so going round the loop one is crossed from $-$ to $+$ and the other from $+$ to $-$: their emfs cancel. Hence
$$I_1(1\ \Omega) = 2 - 2 = 0 \;\Rightarrow\; I_1 = 0$$
The same argument applies to the loop through the first and second resistors, and then to the second and third: the cell emfs cancel in every loop and the resistors start with equal potential differences of zero. So the current in every resistor is $0$ A.
Solution by Sreeraj P, M.Sc Physics
Which of the following statements is false?
Answer: (C) In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null point is disturbed
Concept. Check each statement against the physics of Kirchhoff's laws and the Wheatstone bridge.
Step 1. A. The loop law says the sum of emfs equals the sum of drops round a loop: energy conservation. True.
Step 2. B. A Wheatstone bridge is most sensitive when all four arms are of the same order. True.
Step 3. C. The balance condition $\frac PQ=\frac RS$ is the same when the cell and galvanometer are interchanged, so the null point is not disturbed. False.
Step 4. D. A rheostat with a sliding contact can be used as a potential divider. True.
Answer. Option C is the false statement.
Solution by Sreeraj P, M.Sc Physics
A $9$ V battery with an internal resistance of $0.5\ \Omega$ is connected across an infinite network, as shown in the figure. All ammeters $A_1$, $A_2$, $A_3$ and voltmeter $V$ are ideal. Choose the correct statement.
Answer: (D) Reading of $A_1$ is $2$ A.
Concept. An infinite ladder looks the same after one section is removed, so its resistance $R_\infty$ satisfies: one section followed by $R_\infty$ equals $R_\infty$.
Step 1. Equation. Each section has $1\ \Omega$ in the top and bottom wires and a $4\ \Omega$ shunt: $R_\infty=2+\dfrac{4R_\infty}{4+R_\infty}$.
Step 2. Solve. $R_\infty^2+4R_\infty=8+2R_\infty+4R_\infty\Rightarrow R_\infty^2-2R_\infty-8=0\Rightarrow R_\infty=4\ \Omega$.
Step 3. Battery current. $I=\dfrac{9}{4+0.5}=2$ A; this is the reading of $A_1$.
Step 4. Check $V$. $V=9-2\times0.5=8$ V, so the options for $V$ are wrong.
Answer. Reading of $A_1$ is $2$ A (Option D).
Solution by Sreeraj P, M.Sc Physics
A potentiometer $PQ$ is set up to compare two resistances, as shown in the figure. The ammeter $A$ in the circuit reads $1.0$ A when the two-way key $K_3$ is open. The balance point is at a length $l_1$ cm from $P$ when the two-way key $K_3$ is plugged in between $2$ and $1$, while the balance point is at a length $l_2$ cm from $P$ when the key $K_3$ is plugged in between $3$ and $1$. The ratio of two resistances $\dfrac{R_1}{R_2}$ is found to be:
Answer: (D) $\dfrac{l_1}{l_2 - l_1}$
Concept. The current through $R_1$ and $R_2$ stays the same, so the potentiometer balances voltages proportional to resistances. With $K_3$ at 2 it measures the voltage across $R_1$; at 3, across $R_1+R_2$.
Step 1. Two balances. $IR_1=kl_1$ and $I(R_1+R_2)=kl_2$.
Step 2. Subtract. $IR_2=k(l_2-l_1)$.
Step 3. Ratio. $\dfrac{R_1}{R_2}=\dfrac{l_1}{l_2-l_1}$.
Answer. $\dfrac{l_1}{l_2-l_1}$ (Option D).
Solution by Sreeraj P, M.Sc Physics
In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance $P = 4\ \Omega$ and the neutral point $N$ is at $60$ cm from $A$. Now an unknown resistance $R$ is connected in series to $P$ and the new position of the neutral point is at $80$ cm from $A$. The value of unknown resistance $R$ is:
Answer: (C) $\dfrac{20}{3}\ \Omega$
Balance condition: $\dfrac{P}{Q} = \dfrac{l}{100 - l}$.
Initially: $\dfrac{4}{Q} = \dfrac{60}{40} \Rightarrow Q = \dfrac83\ \Omega$.
With $R$ in series with $P$:
$$\frac{4 + R}{8/3} = \frac{80}{20} = 4 \;\Rightarrow\; 4 + R = \frac{32}{3} \;\Rightarrow\; R = \frac{20}{3}\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
A uniform wire of length $l$ and radius $r$ has a resistance of $100\ \Omega$. It is recast into a wire of radius $\dfrac r2$. The resistance of new wire will be:
Answer: (A) $1600\ \Omega$
The volume is unchanged, so halving the radius makes the area $\frac14$ and the length $4$ times:
$$R' = \rho\frac{4l}{A/4} = 16R = 1600\ \Omega$$
Solution by Sreeraj P, M.Sc Physics
The figure shows three circuits I, II and III which are connected to a $3$ V battery. If the powers dissipated by the configurations I, II and III are $P_1$, $P_2$ and $P_3$ respectively, then:
Answer: (A) $P_2 > P_1 > P_3$
Concept. With the same 3 V battery, $P=\dfrac{V^2}{R_{eq}}$: the smaller the equivalent resistance, the larger the power. Find $R_{eq}$ of each circuit.
Step 1. Circuit I. Name the nodes top $T$, bottom $B$ and the two middle junctions $N_1$, $N_2$. The resistors are $TN_1$, $N_1B$, $TN_2$, $N_2B$ and $N_1N_2$: a balanced bridge of equal resistors, so $N_1N_2$ carries no current and $R_I=2\parallel2=1\ \Omega$.
Step 2. Circuit II. The battery is across the top and bottom vertices. The left path ($1+1$), the right path ($1+1$) and the middle $1\ \Omega$ are in parallel: $R_{II}=2\parallel2\parallel1=0.5\ \Omega$.
Step 3. Circuit III. The diamond between its top and bottom vertices is $2\parallel2=1\ \Omega$, in series with $1\ \Omega$: $R_{III}=2\ \Omega$.
Step 4. Powers. $P_1=9$ W, $P_2=18$ W, $P_3=4.5$ W, so $P_2>P_1>P_3$.
Answer. $P_2>P_1>P_3$ (Option A).
Solution by Sreeraj P, M.Sc Physics
Two batteries with e.m.f. $12$ V and $13$ V are connected in parallel across a load resistor of $10\ \Omega$. The internal resistances of the two batteries are $1\ \Omega$ and $2\ \Omega$ respectively. The voltage across the load lies between:
Answer: (C) $11.5$ V and $11.6$ V
Concept. Replace the two parallel cells by one equivalent cell, then use a divider.
Step 1. Equivalent cell. $E=\dfrac{12/1+13/2}{1+1/2}=12.33$ V, $r=\dfrac23\ \Omega$.
Step 2. Load voltage. $12.33\times\dfrac{10}{10.67}=11.56$ V.
Answer. Between $11.5$ and $11.6$ V (Option C).
Solution by Sreeraj P, M.Sc Physics
On interchanging the resistances, the balance point of a meter bridge shifts to the left by $10$ cm. The resistance of their series combination is $1\ \text{k}\Omega$. How much was the resistance on the left slot before interchanging the resistances?
Answer: (D) $550\ \Omega$
Concept. Before: $\dfrac XY=\dfrac{l}{100-l}$. After swapping: the balance moves to the mirror position $100-l$. A shift of $10$ cm to the left means $100-l=l-10$.
Step 1. Balance length. $l=55$ cm.
Step 2. Ratio. $\dfrac XY=\dfrac{55}{45}$; with $X+Y=1000\ \Omega$: $X=550\ \Omega$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across $52$ cm of the potentiometer wire. If the cell is shunted by a resistance of $5\ \Omega$, a balance is found when the cell is connected across $40$ cm of the wire. Find the internal resistance of the cell.
Answer: (C) $1.5\ \Omega$
Concept. Open cell balances at $l_1$ (its emf); shunted by $R$ it balances at $l_2$. Then $r=R\left(\dfrac{l_1}{l_2}-1\right)$.
Step 1. Apply. $5\left(\dfrac{52}{40}-1\right)=1.5\ \Omega$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Two resistors of resistance, $100\ \Omega$ and $200\ \Omega$ are connected in parallel in an electrical circuit. The ratio of the thermal energy developed in $100\ \Omega$ to that in $200\ \Omega$ in a given time is
Answer: (C) $2 : 1$
In parallel both resistors have the same voltage $V$, so the heat produced in time $t$ is $H = \dfrac{V^2t}{R}$.
$$\frac{H_{100}}{H_{200}} = \frac{200}{100} = 2 : 1$$
Solution by Sreeraj P, M.Sc Physics
As the temperature increases, the electrical resistance
Answer: (D) Increases for conductors but decreases for semiconductors
In a metal the number of free electrons is fixed; heating makes the ions vibrate more, so collisions are more frequent, the relaxation time falls and the resistance rises.
In a semiconductor heating breaks more covalent bonds, so the number of charge carriers rises sharply. This outweighs the extra collisions and the resistance falls.
Solution by Sreeraj P, M.Sc Physics
A copper wire of length $10$ m and radius $\left(\dfrac{10^{-2}}{\sqrt{\pi}}\right)$ m has electrical resistance of $10\ \Omega$. The current density in the wire for an electric field strength of $10$ (V/m) is
Answer: (A) $10^5\ \text{A/m}^2$
Area: $A = \pi r^2 = \pi \times \dfrac{10^{-4}}{\pi} = 10^{-4}\ \text{m}^2$.
Resistivity: $\rho = \dfrac{RA}{l} = \dfrac{10 \times 10^{-4}}{10} = 10^{-4}\ \Omega\,\text{m}$.
$$J = \frac{E}{\rho} = \frac{10}{10^{-4}} = 10^5\ \text{A/m}^2$$
Solution by Sreeraj P, M.Sc Physics
A wheatstone bridge is used to determine the value of unknown resistance $X$ by adjusting the variable resistance $Y$ as shown in the figure. For the most precise measurement of $X$, the resistances $P$ and $Q$
Answer: (C) Should be approximately equal and are small
A Wheatstone bridge is most sensitive when all four arms have resistances of the same order. Then a small change in $Y$ near balance gives the largest galvanometer deflection, so the balance point is found most precisely.
So $P$ and $Q$ should be about equal to each other (and comparable to $X$, i.e. not very large). Of the given choices, "approximately equal and small" is the right one.
Solution by Sreeraj P, M.Sc Physics
The battery of a mobile phone is rated as $4.2\ \text{V}$, $5800\ \text{mAh}$. How much energy is stored in it when fully charged?
Answer: (C) $87.7\ \text{kJ}$
Concept. A battery's charge capacity in ampere-hours times its voltage gives the stored energy: $W=qV$.
Step 1. Charge. $5800$ mAh $=5.8\ \text{A}\times3600\ \text{s}=20880$ C.
Step 2. Energy. $W=20880\times4.2\approx8.77\times10^4$ J.
Answer. $87.7$ kJ (Option C).
Solution by Sreeraj P, M.Sc Physics
A motor operating on $100\ \text{V}$ draws a current of $1\ \text{A}$. If the efficiency of the motor is $91.6\%$, then the loss of power in units of cal/s is:
Answer: (C) $2$
Input power $= VI = 100\ \text{W}$; useful output $= 91.6\ \text{W}$.
Power lost $= 8.4\ \text{W} = 8.4\ \text{J/s} = \dfrac{8.4}{4.2}\ \text{cal/s} = 2\ \text{cal/s}$.
Solution by Sreeraj P, M.Sc Physics
Two cells of emf $1\ \text{V}$ and $2\ \text{V}$ and internal resistance $2\ \Omega$ and $1\ \Omega$, respectively, are connected in series with an external resistance of $6\ \Omega$. The total current in the circuit is $I_1$. Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is $I_2$. The value of $\left(\dfrac{I_1}{I_2}\right)$ is $\dfrac x3$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 4
Series (emfs aiding): $I_1 = \dfrac{1 + 2}{2 + 1 + 6} = \dfrac13\ \text{A}$.
Parallel (like terminals together): the equivalent cell has
$$\varepsilon_{eq} = \frac{\frac{1}{2} + \frac{2}{1}}{\frac12 + \frac11} = \frac{5}{3}\ \text{V},\qquad r_{eq} = \frac{2\times1}{2 + 1} = \frac23\ \Omega$$
$$I_2 = \frac{5/3}{2/3 + 6} = \frac{5/3}{20/3} = \frac14\ \text{A}$$
$$\frac{I_1}{I_2} = \frac{1/3}{1/4} = \frac43 \Rightarrow x = 4$$
Solution by Sreeraj P, M.Sc Physics
The current passing through a wire as a function of time is given as $I(t) = 0.02t + 0.01\ \text{A}$. The charge that will flow through the wire from $t = 1\ \text{s}$ to $t = 2\ \text{s}$ is:
Answer: (D) $0.04\ \text{C}$
$$q = \int_1^2(0.02t + 0.01)\,dt = \left[0.01t^2 + 0.01t\right]_1^2 = 0.01(4 - 1) + 0.01(2 - 1) = 0.04\ \text{C}$$
Solution by Sreeraj P, M.Sc Physics
There are $n$ identical electric bulbs, each designed to draw a power $p$ independently from the mains supply. They are now joined in series across the mains supply. The total power drawn by the combination is:
Answer: (C) $\dfrac pn$
Each bulb has resistance $R = \dfrac{V^2}{p}$. In series the total resistance is $nR$, so across the same supply
$$P = \frac{V^2}{nR} = \frac pn$$
(assuming the resistance of each bulb does not change).
Solution by Sreeraj P, M.Sc Physics
From the combination of resistors with resistance values $R_1 = R_2 = R_3 = 5\ \Omega$ and $R_4 = 10\ \Omega$, which of the combinations shown in the figure is the best circuit to get an equivalent resistance of $6\ \Omega$?
Answer: (A) Circuit (1)
Work out each circuit:
Circuit (1) gives $6\ \Omega$.
Solution by Sreeraj P, M.Sc Physics
A wire of resistance $R$ is bent into a triangular pyramid as shown in the figure, with each segment having the same length. The resistance between points A and B is $R/n$. The value of $n$ is:
Answer: (D) $12$
Concept. A triangular pyramid has 6 equal edges. Between two corners $A$ and $B$, the other two corners are symmetric, so they are at the same potential and the edge joining them carries no current.
Step 1. Edge resistance. The wire is cut into 6 equal edges: $r=\dfrac R6$.
Step 2. Paths from $A$ to $B$. The direct edge $r$, and two paths through the other corners, each $2r$.
Step 3. Parallel. $\dfrac1{R_{AB}}=\dfrac1r+\dfrac1{2r}+\dfrac1{2r}=\dfrac2r$, so $R_{AB}=\dfrac r2=\dfrac{R}{12}$.
Answer. $n=12$ (Option D).
Solution by Sreeraj P, M.Sc Physics
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