Units and Measurements: NEET and JEE questions with solutions

Answer: (B) $[ML^2T^{-1}]$

Use $E = h\nu$, so $h = E/\nu$.
$$[h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$$
This is also the dimension of angular momentum.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $[\mathrm{ML^0T^{-1}}]$ and $[\mathrm{M^0LT^{-1}}]$

Concept. Only like quantities can be added. In the first bracket $\dfrac{A}{Bt^2}$ is added to $P$, so it is a pressure. In the second bracket $Bt$ is added to the height $h$, so it is a length. Find $B$ first.

Step 1. Dimensions of $B$.
$$[Bt]=[h]=L\ \Rightarrow\ [B]=\frac LT=M^0LT^{-1}$$

Step 2. Dimensions of $A$.
$$\left[\frac{A}{Bt^2}\right]=[P]=ML^{-1}T^{-2}\ \Rightarrow\ [A]=ML^{-1}T^{-2}\cdot LT^{-1}\cdot T^2=MT^{-1}$$

Answer. $[ML^0T^{-1}]$ and $[M^0LT^{-1}]$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $1.04$ cm

Least count $= \dfrac{1\ \text{mm}}{10} = 0.1$ mm.

The Vernier zero lies to the left of the main-scale zero, so the zero error is negative. Taking the zero error as $-(4 \times 0.1) = -0.4$ mm (the convention the options follow):

$$\text{Actual length} = \text{reading} - \text{zero error} = 10\ \text{mm} - (-0.4\ \text{mm}) = 10.4\ \text{mm} = 1.04\ \text{cm}$$

Note: in the stricter convention for a negative zero error, the error is $-(10 - 4) \times 0.1 = -0.6$ mm, which would give $1.06$ cm. That value is not among the options, so the intended answer is $1.04$ cm.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $[\mathrm{L^{-1}T^{-1}K^{-2}}]$

Stefan-Boltzmann constant ($E = \sigma T^4$, energy per unit area per unit time):
$$[\sigma_s] = \frac{[\mathrm{ML^2T^{-2}}]}{[\mathrm{L^2}][\mathrm{T}][\mathrm{K^4}]} = [\mathrm{MT^{-3}K^{-4}}]$$

Boltzmann constant (energy per kelvin): $[k_B] = [\mathrm{ML^2T^{-2}K^{-1}}]$

Wien's constant ($\lambda_m T = b$): $[b] = [\mathrm{LK}]$

$$[\sigma_s k_B^{-1} b] = \frac{[\mathrm{MT^{-3}K^{-4}}][\mathrm{LK}]}{[\mathrm{ML^2T^{-2}K^{-1}}]} = [\mathrm{L^{-1}T^{-1}K^{-2}}]$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $4.98$ cm

Least count $= 1\ \text{MSD} - 1\ \text{VSD} = 0.1 - \dfrac{9}{10}(0.1) = 0.01$ cm.

With the jaws closed, the vernier zero is at $+0.1$ cm, so the zero error is $+0.1$ cm.

Observed reading $= 5 + 8 \times 0.01 = 5.08$ cm.

Corrected diameter $= 5.08 - 0.1 = 4.98$ cm.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $13\%$

$$\frac{\Delta P}{P} = 3\frac{\Delta a}{a} + 2\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{1}{2}\frac{\Delta d}{d}$$

$$= 3(1\%) + 2(3\%) + 2\% + \frac{1}{2}(4\%) = 3 + 6 + 2 + 2 = 13\%$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{1}{100(N + 1)}$

$(N + 1)\ \text{VSD} = N\ \text{MSD}$, so $1\ \text{VSD} = \dfrac{N}{N + 1}\ \text{MSD}$.

$$\text{VC} = 1\ \text{MSD} - 1\ \text{VSD} = \frac{1}{N + 1}\ \text{MSD} = \frac{0.1\ \text{mm}}{N + 1} = \frac{0.01\ \text{cm}}{N + 1} = \frac{1}{100(N + 1)}\ \text{cm}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) strain and angle

Solid angle (area / radius$^2$) is dimensionless.

Strain (change in length / length) and plane angle (arc / radius) are also dimensionless.

Arc has dimension $[\mathrm{L}]$, angular speed $[\mathrm{T^{-1}}]$ and stress $[\mathrm{ML^{-1}T^{-2}}]$, so the other options are ruled out.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{\alpha t}{\beta}$

Terms that are added must have the same dimensions: $[\alpha t^2] = [\beta t]$.

Dividing both sides by $[\beta t]$:
$$\left[\frac{\alpha t}{\beta}\right] = 1$$

So $\dfrac{\alpha t}{\beta}$ is dimensionless.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $1.6\%$

$\rho = \dfrac{m}{\pi r^2 l}$, so
$$\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}$$

$$= \frac{0.002}{0.4} + 2 \times \frac{0.001}{0.3} + \frac{0.02}{5} = 0.005 + 0.00667 + 0.004 = 0.01567$$

Percentage error $\approx 1.6\%$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) Random errors

Errors that occur irregularly and unpredictably, for example from random fluctuations in temperature or supply voltage, are random errors. They can be reduced by taking the mean of many readings.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $[\mathrm{M^2}][\mathrm{L^{-1}}][\mathrm{T^0}]$

$[E] = [\mathrm{ML^2T^{-2}}]$.

From $F = \dfrac{Gm_1m_2}{r^2}$: $[G] = \dfrac{[\mathrm{MLT^{-2}}][\mathrm{L^2}]}{[\mathrm{M^2}]} = [\mathrm{M^{-1}L^3T^{-2}}]$.

$$\left[\frac{E}{G}\right] = \frac{[\mathrm{ML^2T^{-2}}]}{[\mathrm{M^{-1}L^3T^{-2}}]} = [\mathrm{M^2L^{-1}T^0}]$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $0.052$ cm

Least count $= \dfrac{1\ \text{mm}}{100} = 0.01$ mm.

$$d = 0 + 52 \times 0.01 = 0.52\ \text{mm} = 0.052\ \text{cm}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $[F][A][T^2]$

Energy $=$ force $\times$ distance, and distance has the dimensions of acceleration $\times$ time$^2$:
$$[E] = [F][L] = [F][A][T^2]$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $0.5$ mm

$\text{LC} = \dfrac{\text{pitch}}{\text{number of circular divisions}}$, so
$$\text{pitch} = 0.01 \times 50 = 0.5\ \text{mm}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $0.06$

$1\ \text{eV} = 1.6 \times 10^{-19}$ J, so
$$E = \frac{10^{-20}}{1.6 \times 10^{-19}} = 0.0625\ \text{eV} \approx 0.06\ \text{eV}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $[\mathrm{ML^{-1}T^{-2}}]$

$$[\text{stress}] = \frac{[\text{force}]}{[\text{area}]} = \frac{[\mathrm{MLT^{-2}}]}{[\mathrm{L^2}]} = [\mathrm{ML^{-1}T^{-2}}]$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $9.98$ m

$9.99 - 0.0099 = 9.9801$ m.

In subtraction, the result keeps as many decimal places as the least precise number. $9.99$ has two decimal places, so the answer is $9.98$ m.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $3$

Combine the right-hand side first: $\dfrac{\text{Pressure}}{\text{Density}} = \dfrac{[ML^{-1}T^{-2}]}{[ML^{-3}]} = [L^2T^{-2}]$

So the right-hand side is $\left(\dfrac{P}{\rho}\right)^{3/2} = [L^2T^{-2}]^{3/2} = [L^3T^{-3}] = [LT^{-1}]^3$

That is (velocity)$^3$, so $x = 3$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $a=1,\ b=2,\ c=1$

Write the dimensions of each side:
$$[MLT^{-2}] = [L^2]^a\,[LT^{-1}]^b\,[ML^{-3}]^c = M^{c}\,L^{2a+b-3c}\,T^{-b}$$

Compare powers:
M: $c = 1$
T: $-b = -2 \Rightarrow b = 2$
L: $2a + b - 3c = 1 \Rightarrow 2a + 2 - 3 = 1 \Rightarrow a = 1$

So $F = A v^2 d$, the familiar form of the force of a fluid jet.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $10$

From $T = 2\pi\sqrt{L/g}$, we get $g = \dfrac{4\pi^2 L}{T^2}$, so
$$\frac{\Delta g}{g}\times 100 = \frac{\Delta L}{L}\times 100 + 2\,\frac{\Delta t}{t}\times 100$$

The time for 30 oscillations is $t = 63\ \text{s}$ with error $3\ \text{s}$. (Using total time or time period gives the same fractional error.)
$$\frac{0.01}{1.01}\times 100 + 2\times\frac{3}{63}\times 100 \approx 0.99\% + 9.52\% \approx 10.5\%$$

The closest option is $10\%$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $[ML^{-1}T^{-2}]$

$\tfrac12\mu_0H^2$ is the energy stored per unit volume in a magnetic field (since $B = \mu_0 H$, it equals $\dfrac{B^2}{2\mu_0}$).

$$\left[\tfrac12\mu_0H^2\right] = \frac{[\text{Energy}]}{[\text{Volume}]} = \frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$$

Tip: energy density always has the same dimensions as pressure.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $[MLT^{-3}],\ [MLT^{-4}]$

Each term must have the dimensions of force, $[MLT^{-2}]$.

$[a][T] = [MLT^{-2}] \Rightarrow [a] = [MLT^{-3}]$

$[b][T^2] = [MLT^{-2}] \Rightarrow [b] = [MLT^{-4}]$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{b}{c}$

The argument of a sine must be dimensionless, so $bt$ and $cx$ are dimensionless. Also $y$ and $a$ are both lengths, so $y/a$ is dimensionless.

That leaves
$$\left[\frac{b}{c}\right] = \frac{[T^{-1}]}{[L^{-1}]} = [LT^{-1}]$$
which is a velocity. In fact $b/c = \omega/k$ is the speed of the wave.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $0$

Planck's constant has dimensions $[ML^2T^{-1}]$, which are exactly the dimensions of angular momentum (for example, $L = nh/2\pi$ in Bohr's model).

So $h = E^0\,L^1\,G^0$, and the power of $G$ is $0$.

You can confirm by writing $[ML^2T^{-1}] = [ML^2T^{-2}]^a[ML^2T^{-1}]^b[M^{-1}L^3T^{-2}]^c$ and solving: $a = 0$, $b = 1$, $c = 0$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $a + b$

For both a sum and a difference, the maximum possible absolute errors add:
$$Z = A \pm B \;\Rightarrow\; \Delta Z = \Delta A + \Delta B = a + b$$

Errors never cancel in the worst case, even when the quantities are subtracted. This is a very common trap.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $[LT^{-2}],\ [L],\ [T]$

Only like quantities can be added, so $c$ has the dimensions of time: $[c] = [T]$.

$at$ is a velocity: $[a] = \dfrac{[LT^{-1}]}{[T]} = [LT^{-2}]$

$\dfrac{b}{t+c}$ is a velocity: $[b] = [LT^{-1}][T] = [L]$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $2.2\%$

$\rho = \dfrac{M}{V}$, so percentage errors add:
$$\frac{\Delta\rho}{\rho}\times 100 = \frac{0.01}{22.42}\times 100 + \frac{0.1}{4.7}\times 100 \approx 0.045\% + 2.13\% \approx 2.2\%$$

Almost all of the error comes from the volume, which was measured much less precisely.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $[Ev^{-2}T^{-2}]$

Surface tension is force per unit length: $[S] = [MT^{-2}]$.

Let $S = E^a v^b T^c$:
$$[MT^{-2}] = [ML^2T^{-2}]^a\,[LT^{-1}]^b\,[T]^c$$

M: $a = 1$
L: $2a + b = 0 \Rightarrow b = -2$
T: $-2a - b + c = -2 \Rightarrow -2 + 2 + c = -2 \Rightarrow c = -2$

So $[S] = [Ev^{-2}T^{-2}]$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $P,\ T,\ h$

Write the power of time in each:

Power: $[ML^2T^{-3}]$, so $-3$
Surface tension: $[MT^{-2}]$, so $-2$
Planck's constant: $[ML^2T^{-1}]$, so $-1$

In ascending order: $-3 < -2 < -1$, which gives $P,\ T,\ h$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $13\%$

For a product of powers, multiply each percentage error by its power (ignore signs) and add:
$$\frac{\Delta Q}{Q}\times 100 = 2(1) + 3(2) + 1(3) + \tfrac12(4) = 2 + 6 + 3 + 2 = 13\%$$

Notice that $b$, with the largest power, contributes the most error. That's the quantity to measure most carefully.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $5\%$

$$\frac{\Delta K}{K}\times 100 = 2\,\frac{\Delta p}{p}\times 100 + \frac{\Delta m}{m}\times 100 = 2(2) + 1 = 5\%$$

The constant $\tfrac12$ has no error, so it contributes nothing.

Solution by Sreeraj P, M.Sc Physics

Answer: 2.1

$g = \dfrac{4\pi^2 L}{T^2}$, so
$$\frac{\Delta g}{g}\times 100 = \frac{\Delta L}{L}\times 100 + 2\,\frac{\Delta T}{T}\times 100$$
$$= \frac{0.001}{1.000}\times 100 + 2\times\frac{0.02}{2.00}\times 100 = 0.1\% + 2\% = 2.1\%$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $0.5\%$

Mean value: $\bar{l} = \dfrac{5.62 + 5.58 + 5.65 + 5.55 + 5.60}{5} = \dfrac{28.00}{5} = 5.60\ \text{cm}$

Absolute errors: $0.02,\ 0.02,\ 0.05,\ 0.05,\ 0.00$

Mean absolute error: $\Delta\bar{l} = \dfrac{0.14}{5} = 0.028\ \text{cm}$

Percentage error: $\dfrac{0.028}{5.60}\times 100 = 0.5\%$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $0.02\ \text{mm}$

$25\ \text{VSD} = 24\ \text{MSD} \Rightarrow 1\ \text{VSD} = \dfrac{24}{25}\ \text{MSD}$

$$\text{LC} = 1\ \text{MSD} - 1\ \text{VSD} = \frac{1}{25}\ \text{MSD} = \frac{0.5}{25}\ \text{mm} = 0.02\ \text{mm}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $2.28\ \text{cm}$

Vernier zero to the right of main scale zero means a positive zero error:
$$\text{Zero error} = +6 \times 0.01 = +0.06\ \text{cm}$$

Observed reading $= 2.3 + 4\times 0.01 = 2.34\ \text{cm}$

A positive zero error is subtracted:
$$\text{Corrected reading} = 2.34 - 0.06 = 2.28\ \text{cm}$$

Solution by Sreeraj P, M.Sc Physics

Answer: 1.83

Least count $= \dfrac{\text{pitch}}{\text{number of circular divisions}} = \dfrac{0.5}{50} = 0.01\ \text{mm}$

Observed reading $= 1.5 + 36\times 0.01 = 1.86\ \text{mm}$

Zero error $= +3\times 0.01 = +0.03\ \text{mm}$, which is subtracted:
$$d = 1.86 - 0.03 = 1.83\ \text{mm}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $0.19$

$\dfrac{4.36 \times 0.052}{1.2} = 0.18893\ldots$

In multiplication and division, the answer keeps as many significant figures as the least precise value. Here $0.052$ and $1.2$ each have 2 significant figures, so the answer is rounded to 2:
$$0.19$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $20.1\ \text{m}$

Actual sum $= 16.2 + 0.847 + 3.08 = 20.127\ \text{m}$

In addition and subtraction, the rule is about decimal places, not significant figures. The least precise value, $16.2$, has one decimal place, so the result is rounded to one decimal place:
$$20.1\ \text{m}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $65.4\ \text{cm}^3$

$$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (2.50)^3 = 65.449\ldots\ \text{cm}^3$$

The radius has 3 significant figures ($\tfrac43$ and $\pi$ are exact), so the volume is rounded to 3 significant figures: $65.4\ \text{cm}^3$.

Solution by Sreeraj P, M.Sc Physics

Answer: 200

Energy has dimensions $[ML^2T^{-2}]$, and $n_1 u_1 = n_2 u_2$:
$$n_2 = n_1\left(\frac{M_1}{M_2}\right)\left(\frac{L_1}{L_2}\right)^2\left(\frac{T_1}{T_2}\right)^{-2}$$
$$= 100\times\frac{1}{2}\times\left(\frac{1}{5}\right)^2\times\left(\frac{1}{10}\right)^{-2} = 100\times 0.5\times 0.04\times 100 = 200$$

The new unit of energy is smaller than a joule, so the number gets bigger.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $[L^{-1/2}T^{2}]$

Each term has the dimensions of force.

$[a][L^{1/2}] = [MLT^{-2}] \Rightarrow [a] = [ML^{1/2}T^{-2}]$

$[b][T^2] = [MLT^{-2}] \Rightarrow [b] = [MLT^{-4}]$

$$\left[\frac{a}{b}\right] = \frac{[ML^{1/2}T^{-2}]}{[MLT^{-4}]} = [L^{-1/2}T^{2}]$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $[M^2L^2T^{-4}]$

The exponent must be dimensionless. Since $kT$ is an energy:
$$[\beta][L^2] = [ML^2T^{-2}] \Rightarrow [\beta] = [MT^{-2}]$$

The exponential is dimensionless, so $A$ has the dimensions of energy: $[A] = [ML^2T^{-2}]$.
$$[A\beta] = [ML^2T^{-2}][MT^{-2}] = [M^2L^2T^{-4}]$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\sqrt{\dfrac{hG}{c^3}}$

Let length $= c^a G^b h^p$:
$$[L] = [LT^{-1}]^a\,[M^{-1}L^3T^{-2}]^b\,[ML^2T^{-1}]^p$$

M: $-b + p = 0$
L: $a + 3b + 2p = 1$
T: $-a - 2b - p = 0$

From M, $p = b$. Then T gives $a = -3b$, and L gives $-3b + 3b + 2b = 1 \Rightarrow b = \tfrac12$.

So $p = \tfrac12$ and $a = -\tfrac32$:
$$\text{Length} = \sqrt{\frac{hG}{c^3}}$$
This is the Planck length, about $10^{-35}\ \text{m}$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $[FL^{-1}T^{2}]$

From $F = ma$, we get $m = \dfrac{F}{a}$. Acceleration has dimensions $[LT^{-2}]$, so
$$[m] = \frac{[F]}{[LT^{-2}]} = [FL^{-1}T^{2}]$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $-1,\ \tfrac12,\ -\tfrac12$

$$[T^{-1}] = [L]^a\,[MLT^{-2}]^b\,[ML^{-1}]^c$$

M: $b + c = 0$
T: $-2b = -1 \Rightarrow b = \tfrac12$, so $c = -\tfrac12$
L: $a + b - c = 0 \Rightarrow a = -1$

So $f \propto \dfrac{1}{l}\sqrt{\dfrac{T}{\mu}}$, which matches $f = \dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}$ for the fundamental mode.

Solution by Sreeraj P, M.Sc Physics

Answer: 3

$[\eta] = [ML^{-1}T^{-1}]$
$$[MLT^{-2}] = [ML^{-1}T^{-1}]^a\,[L]^b\,[LT^{-1}]^c$$

M: $a = 1$
T: $-a - c = -2 \Rightarrow c = 1$
L: $-a + b + c = 1 \Rightarrow b = 1$

So $F = k\eta r v$ (Stokes' law, with $k = 6\pi$), and $a + b + c = 3$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $(100 \pm 6)\ \text{W}$

$P = \dfrac{50^2}{25} = 100\ \text{W}$
$$\frac{\Delta P}{P}\times 100 = 2\,\frac{\Delta V}{V}\times 100 + \frac{\Delta R}{R}\times 100 = 2(2\%) + 2\% = 6\%$$

So $\Delta P = 6\%$ of $100\ \text{W} = 6\ \text{W}$, and $P = (100 \pm 6)\ \text{W}$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $(60 \pm 1.7)\ \text{cm}^2$

$A = 12.0\times 5.0 = 60\ \text{cm}^2$

For a product, the fractional errors add:
$$\frac{\Delta A}{A} = \frac{0.1}{12.0} + \frac{0.1}{5.0} = 0.0083 + 0.020 = 0.0283$$
$$\Delta A = 0.0283\times 60 \approx 1.7\ \text{cm}^2$$

So $A = (60 \pm 1.7)\ \text{cm}^2$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $400\alpha$

Concept. A unit of length can be defined through a speed: here one "$\alpha$" is the distance light covers in one second. Then a distance in $\alpha$ units is simply the light travel time in seconds.

Step 1. Light travel time in seconds.
$$t=6\ \text{min}\ 40\ \text{s}=6\times60+40=400\ \text{s}$$

Step 2. Distance. In $1$ s light covers $1\alpha$, so in $400$ s it covers
$$d=c\,t=400\times(c\times1\ \text{s})=400\alpha$$

Answer. $400\alpha$ (Option B). In metres this is $400\times3\times10^8=1.2\times10^{11}$ m.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $1, 1, 1, 1$

Concept. Equate the dimensions of both sides. With four unknown powers we need four base dimensions: $M$, $L$, $T$ and $A$.

Step 1. Dimensions of each symbol.
$$[H]=\frac{[B]}{[\mu_0]}=L^{-1}A,\quad[\varepsilon]=M^{-1}L^{-3}T^4A^2,\quad[E]=MLT^{-3}A^{-1},\quad[x]=L,\quad[t]=T$$

Step 2. Equate powers in $L^{-1}A=L^p\,(M^{-1}L^{-3}T^4A^2)^q\,(MLT^{-3}A^{-1})^r\,T^{-s}$:

• $M$: $-q+r=0$
• $A$: $2q-r=1$
• $L$: $p-3q+r=-1$
• $T$: $4q-3r-s=0$

Step 3. Solve. From $M$: $r=q$. Then $A$: $q=1$, so $r=1$. $L$: $p-3+1=-1\Rightarrow p=1$. $T$: $4-3-s=0\Rightarrow s=1$.

Answer. $p=q=r=s=1$ (Option A). Indeed $\varepsilon E/t$ is a displacement current density, and multiplying by a length gives $H$ (Ampère's law).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $0.82\%$

Concept. Density of a cylinder $\rho=\dfrac{m}{V}=\dfrac{4m}{\pi d^2l}$. The relative errors add with their powers, so the diameter error counts twice.

Step 1. Error formula.
$$\frac{\Delta\rho}{\rho}=\frac{\Delta m}{m}+2\frac{\Delta d}{d}+\frac{\Delta l}{l}$$

Step 2. Each term.
$$\frac{\Delta m}{m}=\frac{0.02}{97.42}=0.00021$$
$$2\frac{\Delta d}{d}=2\times\frac{0.02}{20.20}=0.00198$$
$$\frac{\Delta l}{l}=\frac{0.05}{8.35}=0.00599$$

Step 3. Add.
$$\frac{\Delta\rho}{\rho}=0.00021+0.00198+0.00599=0.00818\approx0.82\%$$

Answer. $0.82\%$ (Option B). The short length ($8.35$ mm) gives the largest share of the error.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $[\text{M}^1\text{L}^{7/2}\text{T}^{-2}]$

Concept. In $x+B$, the constant $B$ is added to a distance, so by homogeneity $B$ is a length. With $B$ known, the dimensions of $A$ follow from requiring the whole expression to be an energy.

Step 1. Dimensions of $B$.
$$[B]=[x]=[L]$$

Step 2. Dimensions of $A$. The denominator $x+B$ has dimensions $L$, so
$$[V]=\frac{[A]\,L^{1/2}}{L}=[A]\,L^{-1/2}\ \Rightarrow\ [A]=[V]\,L^{1/2}$$
With $[V]=ML^2T^{-2}$ (energy):
$$[A]=ML^{2}T^{-2}\cdot L^{1/2}=[ML^{5/2}T^{-2}]$$

Step 3. Multiply.
$$[AB]=ML^{5/2}T^{-2}\cdot L=[ML^{7/2}T^{-2}]$$

Answer. $[M^1L^{7/2}T^{-2}]$ (Option D). Option A is the trap: it is $[A]$ alone.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $6$

$V = \dfrac{\pi d^3}{6}$, so $\dfrac{\Delta V}{V} = 3\dfrac{\Delta d}{d} = 3 \times 2\% = 6\%$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) A-III, B-IV, C-II, D-I

Concept. A constant gets its dimensions from the law in which it appears: rearrange the law so the constant stands alone, then put in the dimensions of everything else. Energy is $[ML^2T^{-2}]$ and force is $[MLT^{-2}]$.

Step 1. Boltzmann constant. The mean energy of a particle is of order $kT$, so $k=E/T$:
$$[k]=\frac{ML^2T^{-2}}{K}=[ML^2T^{-2}K^{-1}]\quad\to\ \text{III}$$

Step 2. Stefan's constant. Stefan's law gives power per unit area: $P/A=\sigma T^4$. Power per area is $\dfrac{ML^2T^{-3}}{L^2}=MT^{-3}$, so
$$[\sigma]=\frac{MT^{-3}}{K^4}=[ML^0T^{-3}K^{-4}]\quad\to\ \text{IV}$$

Step 3. Planck's constant. $E=h\nu$ with $[\nu]=T^{-1}$:
$$[h]=\frac{ML^2T^{-2}}{T^{-1}}=[ML^2T^{-1}]\quad\to\ \text{II}$$

Step 4. Gravitational constant. $F=\dfrac{Gm^2}{r^2}$ gives $G=\dfrac{Fr^2}{m^2}$:
$$[G]=\frac{MLT^{-2}\cdot L^2}{M^2}=[M^{-1}L^3T^{-2}]\quad\to\ \text{I}$$

Answer. A-III, B-IV, C-II, D-I (Option C). Note that $h$ has the same dimensions as angular momentum.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $0.07$ cm positive zero error

Concept. With the jaws closed, a vernier zero to the right of the main-scale zero means the instrument already reads more than zero: a positive zero error of (coinciding division) $\times$ LC.

Step 1. Least count. $1\,\text{MSD}=1$ mm, $10$ VSD: $\text{LC}=\dfrac{1}{10}=0.1$ mm $=0.01$ cm.

Step 2. Zero error. The $7$th vernier line coincides:
$$e=+7\times0.01\ \text{cm}=+0.07\ \text{cm}$$

Answer. $0.07$ cm positive zero error (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{R}{\sqrt{LC}}$

Concept. Rather than expanding every symbol, use known combinations: $\sqrt{LC}$ is a time (since $\omega=1/\sqrt{LC}$ in an LC circuit), $L/R$ and $RC$ are times, and $\sqrt{L/C}$ has the dimensions of resistance. Then only $[R]$ needs to be worked out.

Step 1. Dimensions of resistance. $R=\dfrac{V}{I}=\dfrac{P}{I^2}$ and power is $ML^2T^{-3}$:
$$[R]=[ML^2T^{-3}A^{-2}]$$

Step 2. Compare with the target. The target $ML^2T^{-4}A^{-2}$ is $[R]$ with one extra $T^{-1}$. So we need $R$ divided by a time.

Step 3. Check the options.

• (A) $\dfrac{R}{\sqrt{LC}}$: $\sqrt{LC}$ is a time, so this is $[R]\,T^{-1}=ML^2T^{-4}A^{-2}$. Matches.
• (B) $\dfrac{R}{LC}$: $LC$ is time$^2$, giving $[R]\,T^{-2}$. No.
• (D) $\dfrac1R\sqrt{\dfrac LC}$: $\sqrt{L/C}$ is a resistance, so this is dimensionless. No.
• (C) $\dfrac{C}{\sqrt{LR}}$: with $[C]=M^{-1}L^{-2}T^4A^2$ and $[\sqrt{LR}]=ML^2T^{-5/2}A^{-2}$ this is $M^{-2}L^{-4}T^{13/2}A^4$. No.

Answer. Option A, $\dfrac{R}{\sqrt{LC}}$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) A-IV, B-II, C-I, D-III

Concept. With $h$, $G$ and $c$ as base units, length, time and mass are the Planck units. Find each one by making combinations of $h$, $G$, $c$ in which the unwanted dimensions cancel.

Step 1. Mass. As in the previous question, $\left[\dfrac{hc}{G}\right]=M^2$, so
$$M=\sqrt{\frac{hc}{G}}\quad\to\ \text{I}$$

Step 2. Length. Try $\dfrac{Gh}{c^3}$:
$$\left[\frac{Gh}{c^3}\right]=\frac{M^{-1}L^3T^{-2}\cdot ML^2T^{-1}}{L^3T^{-3}}=\frac{L^5T^{-3}}{L^3T^{-3}}=L^2$$
So length $=\sqrt{\dfrac{Gh}{c^3}}\ \to$ IV.

Step 3. Time. Time $=$ length$/$speed $=\dfrac1c\sqrt{\dfrac{Gh}{c^3}}=\sqrt{\dfrac{Gh}{c^5}}\ \to$ II.

Step 4. Temperature. The only expression left, III, is the one that brings in the temperature; it is matched with kelvin by elimination.

Answer. A-IV, B-II, C-I, D-III (Option B). Remember: length $\sqrt{Gh/c^3}$, time $\sqrt{Gh/c^5}$, mass $\sqrt{hc/G}$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $0.18$

Concept. $R=\dfrac VI$, so $\dfrac{\Delta R}{R}=\dfrac{\Delta V}{V}+\dfrac{\Delta I}{I}$. For a single reading, the error is taken as the least count of the meter.

Step 1. Resistance.
$$R=\frac{10}{5}=2\ \Omega$$

Step 2. Errors of the readings. $\Delta V=500$ mV $=0.5$ V and $\Delta I=200$ mA $=0.2$ A.

Step 3. Relative error.
$$\frac{\Delta R}{R}=\frac{0.5}{10}+\frac{0.2}{5}=0.05+0.04=0.09$$

Step 4. Absolute error.
$$\Delta R=0.09\times2=0.18\ \Omega$$

Answer. $0.18\ \Omega$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) A-III, B-IV, C-I, D-II

Concept. All four quantities come from the photoelectric equation $h\nu=\phi+eV_0$: each term is an energy.

Step 1. Planck's constant. $[h]=\dfrac{[E]}{[\nu]}=\dfrac{ML^2T^{-2}}{T^{-1}}=ML^2T^{-1}$ (III).

Step 2. Stopping potential. A potential is energy per charge: $[V_0]=\dfrac{ML^2T^{-2}}{AT}=ML^2T^{-3}A^{-1}$ (IV).

Step 3. Work function. An energy: $ML^2T^{-2}$ (I).

Step 4. Threshold frequency. A frequency: $T^{-1}$ (II).

Answer. A-III, B-IV, C-I, D-II (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $5 \times 10^{-3}$

Concept. Pitch is the distance the spindle moves in one full rotation; the least count is the pitch divided by the number of circular divisions.

Step 1. Pitch.
$$p=\frac{2.5\ \text{mm}}{5}=0.5\ \text{mm}$$

Step 2. Least count.
$$\text{LC}=\frac{0.5}{100}=0.005\ \text{mm}=5\times10^{-3}\ \text{mm}$$

Answer. $5\times10^{-3}$ mm (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1$

$\frac{1}{2}\epsilon_0E^2$ is the energy stored per unit volume of an electric field, so it has the dimensions of energy/volume:

$$\left[\frac{\text{energy}}{\text{volume}}\right]=\frac{ML^2T^{-2}}{L^3}=ML^{-1}T^{-2}$$

So $a=1$, $b=-1$, $c=-2$ and $2a-b+c=2+1-2=1$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1.65$

Concept. Young's modulus from a stretched wire: $Y=\dfrac{\text{stress}}{\text{strain}}=\dfrac{F/A}{\Delta l/L}=\dfrac{4FL}{\pi d^2\,\Delta l}$. Each measured length carries an error equal to the least count of the instrument used, and relative errors add with their powers. The load is taken as exact.

Step 1. Value of $Y$ in SI units. $F=100$ N, $L=1.5$ m, $d=0.08$ cm $=8\times10^{-4}$ m, $\Delta l=0.5$ cm $=5\times10^{-3}$ m.
$$Y=\frac{4\times100\times1.5}{\pi\times(8\times10^{-4})^2\times5\times10^{-3}}=\frac{600}{\pi\times6.4\times10^{-7}\times5\times10^{-3}}$$
$$=\frac{600}{1.005\times10^{-8}}\approx5.97\times10^{10}\ \text{N m}^{-2}$$

Step 2. Relative error.
$$\frac{\Delta Y}{Y}=2\frac{\Delta d}{d}+\frac{\Delta L}{L}+\frac{\Delta(\Delta l)}{\Delta l}=2\times\frac{0.001}{0.08}+\frac{0.1}{150}+\frac{0.001}{0.5}$$
$$=0.025+0.00067+0.002=0.0277$$

Step 3. Absolute error.
$$\Delta Y=0.0277\times5.97\times10^{10}\approx1.65\times10^{9}\ \text{N m}^{-2}$$

Answer. $\alpha=1.65$ (Option B). The diameter term ($2.5\%$) is by far the largest, because $d$ is small and squared.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $[hT^{-1}LM^{-2}]$

Concept. Here $h$ is used together with $M$, $L$, $T$, so there are four symbols but only three independent dimensions; the answer is found by checking which option has the dimensions of $G$, $M^{-1}L^3T^{-2}$.

Step 1. Dimensions of $h$. $[h]=ML^2T^{-1}$.

Step 2. Test option B, $hT^{-1}LM^{-2}$.
$$ML^2T^{-1}\cdot T^{-1}\cdot L\cdot M^{-2}=M^{-1}L^3T^{-2}$$
This is $[G]$.

Step 3. Others. A: $hTLM^{-2}=M^{-1}L^3T^0$. C: $hTL^2M^{-2}=M^{-1}L^4T^0$. D: $h^{-1}T^{-1}LM^{-2}=M^{-3}L^{-1}T^0$. None match.

Answer. $[hT^{-1}LM^{-2}]$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $5.045$

Concept. Screw gauge: LC $=\dfrac{\text{pitch}}{N}$. With the studs in contact the reference line sits on the $5$th circular division, so the gauge reads $+5$ divisions when it should read zero: a positive zero error, subtracted from every reading.

Step 1. Least count.
$$\text{LC}=\frac{0.1}{100}=0.001\ \text{mm}$$

Step 2. Zero error.
$$+5\times0.001=+0.005\ \text{mm}$$

Step 3. Observed reading.
$$5+50\times0.001=5.050\ \text{mm}$$

Step 4. Corrected diameter.
$$5.050-0.005=5.045\ \text{mm}$$

Answer. $5.045$ mm (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) Voltage

Concept. A quantity is measurable only if an instrument can give its value without an arbitrary choice. Electric potential at a point depends on where we choose the zero of potential, so only potential *differences* have a definite value.

Step 1. Check each option.

• Resistance: measured with an ohmmeter or from $V/I$. Measurable.
• Voltage (potential) difference: read directly by a voltmeter across two points. Measurable.
• Displacement current $\varepsilon_0\,d\Phi_E/dt$: equal to the conduction current in the leads of a charging capacitor and produces a measurable magnetic field. Measurable.
• Voltage at a single point (absolute potential): needs a reference point; a voltmeter always has two terminals. Not measurable by itself.

Answer. Voltage (absolute potential) (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) 6.76

(For a spring–mass system the period is really $T = 2\pi\sqrt{m/k}$; either way $k \propto m\,T^{\pm 2}$, so the percentage error is the same.)

$$\frac{\Delta k}{k}\times100 = \frac{\Delta m}{m}\times100 + 2\,\frac{\Delta T}{T}\times100$$

Mass: $\dfrac{\Delta m}{m} = \dfrac{10\ \text{mg}}{10\ \text{g}} = 0.001 \Rightarrow 0.1\%$.

Time: for 50 oscillations, $\dfrac{\Delta t}{t} = \dfrac{2}{60}$, and the same fractional error applies to $T = t/50$, i.e. $3.33\%$.

$$\frac{\Delta k}{k}\times 100 = 0.1 + 2(3.33) \approx 6.77\% \approx 6.76\%$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) A-IV, B-III, C-I, D-II

• Viscosity: $\eta = \dfrac{F}{A\,(dv/dx)}$ → $[ML^{-1}T^{-1}]$ (IV)
• Surface tension: force/length → $[MT^{-2}] = [ML^0T^{-2}]$ (III)
• Pressure: force/area → $[ML^{-1}T^{-2}]$ (I)
• Surface energy (an energy) → $[ML^2T^{-2}]$ (II)

A-IV, B-III, C-I, D-II.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) 0.01

Concept. A reading is recorded to the smallest division the instrument can show, so the last decimal place of the readings shows the least count.

Step 1. Look at the readings. $1.24$, $1.25$, $1.23$, $1.21$ mm: all are given to the second decimal place in mm, and they differ in that place.

Step 2. Least count. The smallest step recorded is $0.01$ mm.

Answer. $0.01$ mm (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) 15.1

Zero error: the vernier zero is to the right of the main-scale zero, so the error is positive:
$$e = +4\times0.1 = +0.4\ \text{mm}$$
Observed reading (1 main-scale division = 1 mm):
$$15 + 5\times0.1 = 15.5\ \text{mm}$$
Corrected length $= 15.5 - 0.4 = 15.1$ mm.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) 0.002

$1\ \text{VSD} = \dfrac{48}{50}\ \text{MSD}$.
$$\text{LC} = 1\ \text{MSD} - 1\ \text{VSD} = \left(1 - \frac{48}{50}\right)\times0.05 = \frac{2}{50}\times0.05 = 0.002\ \text{mm}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) A-III, B-IV, C-I, D-II

Concept. All four magnetic quantities can be built step by step from the force on a current: $F=BIL$. Once $B$ is known, flux is $BA$, inductance is flux per current, and $\mu_0$ comes from the field of a long wire, $B=\dfrac{\mu_0I}{2\pi r}$.

Step 1. Magnetic induction $B$. From $F=BIL$,
$$[B]=\frac{[F]}{[I][L]}=\frac{MLT^{-2}}{A\cdot L}=[ML^0T^{-2}A^{-1}]\quad\to\ \text{III}$$

Step 2. Magnetic flux. $\Phi=BA$, so multiply by $L^2$:
$$[\Phi]=[ML^2T^{-2}A^{-1}]\quad\to\ \text{IV}$$

Step 3. Permeability. From $B=\dfrac{\mu_0I}{2\pi r}$, $\mu_0=\dfrac{2\pi rB}{I}$:
$$[\mu_0]=\frac{L\cdot MT^{-2}A^{-1}}{A}=[MLT^{-2}A^{-2}]\quad\to\ \text{I}$$

Step 4. Self inductance. $\Phi=LI$, so $L=\Phi/I$:
$$[L]=\frac{ML^2T^{-2}A^{-1}}{A}=[ML^2T^{-2}A^{-2}]\quad\to\ \text{II}$$

Answer. A-III, B-IV, C-I, D-II (Option D). Building each quantity from the previous one is faster and safer than recalling four separate formulas.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $[L^{-1}A]$

For a long straight wire $B = \dfrac{\mu_0 I}{2\pi r}$, so
$$\frac{B}{\mu_0} = \frac{I}{2\pi r}$$

This has the dimensions of current per length: $[L^{-1}A]$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) Both Statement I and Statement II are false

Concept. A vernier works because its divisions differ slightly from the main-scale divisions; the least count (vernier constant) is that difference: $\text{LC}=|1\,\text{MSD}-1\,\text{VSD}|$, which equals $\dfrac{1\,\text{MSD}}{N}$ for the usual $N$ VSD $=(N-1)$ MSD.

Step 1. Statement I. Usually $N$ VSD $=(N-1)$ MSD, so a VSD is shorter. But a vernier can also be made with $N$ VSD $=(N+1)$ MSD, where a VSD is *longer* than an MSD; the least count is still the difference. "Always smaller" is false.

Step 2. Statement II. The vernier constant is one MSD *divided by* the number of vernier divisions, not multiplied. False.

Answer. Both Statement I and Statement II are false (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $[\text{F}] = [C^2M^{-1}L^{-2}T^2]$

$C = \dfrac{Q}{V}$ and $V = \dfrac{W}{Q}$, so
$$C = \frac{Q^2}{W}$$

With $[W] = [ML^2T^{-2}]$:
$$[C] = \frac{[C^2]}{[ML^2T^{-2}]} = [C^2M^{-1}L^{-2}T^2]$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $5$

$\rho = \dfrac{m}{\pi r^2 l}$, so
$$\frac{\Delta\rho}{\rho}\times100 = \left(\frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta l}{l}\right)\times100$$
$$= \frac{0.003}{0.60}\times100 + 2\times\frac{0.01}{0.50}\times100 + \frac{0.05}{10.00}\times100 = 0.5 + 4 + 0.5 = 5\%$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $[L^2T^{-2}]$

Every term must have the dimension of length. $\sin t$ and $\cos^2 t$ are dimensionless, so
$$[A] = [B] = [D] = [L], \qquad [C] = \frac{[L]}{[T^2]} = [LT^{-2}]$$

$$\left[\frac{ABC}{D}\right] = \frac{[L][L][LT^{-2}]}{[L]} = [L^2T^{-2}]$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) displacement

Both terms have the dimension of flux, so $[\alpha\sigma] = [\beta\lambda]$:
$$\left[\frac{\alpha}{\beta}\right] = \left[\frac{\lambda}{\sigma}\right] = \frac{[Q/L]}{[Q/L^2]} = [L]$$

$\alpha/\beta$ has the dimension of length, i.e. displacement.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

• Magnetic field: $F = qvB \Rightarrow [B] = \dfrac{[MLT^{-2}]}{[AT][LT^{-1}]} = [MT^{-2}A^{-1}]$ (II).
• Permeability: $F/l = \dfrac{\mu_0I^2}{2\pi d} \Rightarrow [\mu_0] = [MLT^{-2}A^{-2}]$ (III).
• Magnetic moment: $m = IA \Rightarrow [L^2A]$ (IV).
• Torsional constant: $\tau = k\theta \Rightarrow [ML^2T^{-2}]$ (I).

(A)-(III), (B)-(II), (C)-(IV), (D)-(I).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $y = 1330$

Concept. In multiplication and division, the result is rounded to the same number of significant figures as the factor with the fewest significant figures.

Step 1. Count significant figures. $32.3$ has $3$, $1125$ has $4$, $27.4$ has $3$. So the result must have $3$.

Step 2. Calculate.
$$y=\frac{32.3\times1125}{27.4}=\frac{36337.5}{27.4}=1326.19\ldots$$

Step 3. Round to 3 significant figures. $1326.19\to1.33\times10^3=1330$.

Answer. $y=1330$ (Option B). The other options keep far more figures than the data justify.

Solution by Sreeraj P, M.Sc Physics

Answer: 35

Concept. For a screw gauge, $\text{LC}=\dfrac{\text{pitch}}{N}$, where $N$ is the number of circular-scale divisions. Changes in pitch and $N$ change the least count in proportion.

Step 1. New pitch and new $N$. Pitch increased by $75\%$: $p'=1.75\,p$. Divisions reduced by $50\%$: $N'=0.5\,N$.

Step 2. New least count.
$$\text{LC}'=\frac{1.75\,p}{0.5\,N}=3.5\times\frac{p}{N}=3.5\times0.01\ \text{mm}=0.035\ \text{mm}$$

Step 3. Express in the requested unit.
$$0.035\ \text{mm}=35\times10^{-3}\ \text{mm}$$

Answer. $35$.

Solution by Sreeraj P, M.Sc Physics

Answer: 0

Concept. Modulus of elasticity is stress/strain, with the dimensions of pressure. Torque has the dimensions of energy. Divide one by the other.

Step 1. Dimensions of each.
$$[Y]=ML^{-1}T^{-2},\qquad[\tau]=[rF]=ML^2T^{-2}$$

Step 2. Divide.
$$\left[\frac{Y}{\tau}\right]=\frac{ML^{-1}T^{-2}}{ML^2T^{-2}}=M^0L^{-3}T^0$$
So $a=0$, $b=-3$ (as given) and $c=0$.

Answer. $c=0$.

Solution by Sreeraj P, M.Sc Physics

Answer: 3

Concept. The least count of the screw gauge is the error in each single length reading. For an area $A=lb$, the relative errors add: $\dfrac{\Delta A}{A}=\dfrac{\Delta l}{l}+\dfrac{\Delta b}{b}$.

Step 1. Least count.
$$\text{LC}=\frac{0.75\ \text{mm}}{15}=0.05\ \text{mm}$$
So $\Delta l=\Delta b=0.05$ mm.

Step 2. Relative errors.
$$\frac{\Delta l}{l}=\frac{0.05}{5}=0.01,\qquad\frac{\Delta b}{b}=\frac{0.05}{2.5}=0.02$$

Step 3. Maximum fractional error in area.
$$\frac{\Delta A}{A}=0.01+0.02=0.03=\frac{3}{100}$$

Answer. $x=3$. The shorter side contributes twice as much error as the longer side.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

• Angular impulse = change in angular momentum: $[ML^2T^{-1}]$ (III).
• Latent heat = energy per mass: $[L^2T^{-2}]$ (I).
• Resistivity $\rho = RA/l$, with $[R] = [ML^2T^{-3}A^{-2}]$: $[ML^3T^{-3}A^{-2}]$ (IV).
• Emf = work per charge: $\dfrac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}]$ (II).

(A)-(III), (B)-(I), (C)-(IV), (D)-(II).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) Surface tension and impulse

• Pressure and Young's modulus: both $[ML^{-1}T^{-2}]$.
• Surface tension (force per length) $[MT^{-2}]$; impulse (force × time) $[MLT^{-1}]$: different.
• Torque and energy: both $[ML^2T^{-2}]$.
• Angular momentum and Planck's constant: both $[ML^2T^{-1}]$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

• Young's modulus (stress): $[ML^{-1}T^{-2}]$ (II).
• Torque: $[ML^2T^{-2}]$ (IV).
• Viscosity, $F = \eta A\dfrac{dv}{dx}$: $[ML^{-1}T^{-1}]$ (I).
• $G$, $F = \dfrac{Gm_1m_2}{r^2}$: $[M^{-1}L^3T^{-2}]$ (III).

(A)-(II), (B)-(IV), (C)-(I), (D)-(III).

Solution by Sreeraj P, M.Sc Physics

Answer: 7700

$$\frac{\Delta Q}{Q} = \frac{\Delta a}{a} + 4\frac{\Delta b}{b} + \frac{\Delta c}{c} + \frac{\Delta d}{d} = \frac{3}{60} + 4\times\frac{0.1}{20} + \frac{0.2}{40} + \frac{0.1}{50}$$
$$= 0.05 + 0.02 + 0.005 + 0.002 = 0.077 = 7.7\%$$

$\dfrac{x}{1000} = 7.7 \Rightarrow x = 7700$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $39.9\%$

Concept. Resistivity is $\rho=\dfrac{RA}{l}=\dfrac{R\pi r^2}{l}$. Form each relative error from the data, then add with the powers.

Step 1. Rule.
$$\frac{\Delta\rho}{\rho}=\frac{\Delta R}{R}+2\frac{\Delta r}{r}+\frac{\Delta l}{l}$$

Step 2. Each term.
$$\frac{\Delta R}{R}=\frac{10}{100}=0.100,\qquad 2\frac{\Delta r}{r}=2\times\frac{0.05}{0.35}=0.286,\qquad\frac{\Delta l}{l}=\frac{0.2}{15}=0.013$$

Step 3. Add.
$$\frac{\Delta\rho}{\rho}=0.100+0.286+0.013=0.399\approx39.9\%$$

Answer. $39.9\%$ (Option B). The radius, small and squared, gives most of the error.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $[\text{M}\text{L}^{2}\text{T}^{-1}]$

Angular impulse = torque × time = change in angular momentum.
$$[\tau t] = [\text{M}\text{L}^2\text{T}^{-2}][\text{T}] = [\text{M}\text{L}^2\text{T}^{-1}]$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{5}{11}$

$11\ \text{VSD} = 10\ \text{MSD} \Rightarrow 1\ \text{VSD} = \dfrac{10}{11}\ \text{MSD}$.
$$\text{LC} = 1\ \text{MSD} - 1\ \text{VSD} = \frac{1}{11}\ \text{MSD} = \frac{5}{11}\ \text{units}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Concept. Rules for counting significant figures:

• All non-zero digits count.
• Zeros between non-zero digits count.
• Leading zeros (before the first non-zero digit) never count.
• Trailing zeros after a decimal point count.

Step 1. $1001$. The two zeros lie between non-zero digits, so all four digits count: $4\ \to$ II.

Step 2. $010.1$. The leading zero does not count; $10.1$ has $3\ \to$ I.

Step 3. $100.100$. The zeros between $1$s count, and the trailing zeros after the decimal point count: $6\ \to$ IV.

Step 4. $0.0010010$. The three leading zeros do not count. The remaining $10010$ (including the final trailing zero after the decimal point) gives $5\ \to$ III.

Answer. (A)-(II), (B)-(I), (C)-(IV), (D)-(III), Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $f^2\left[\dfrac{\Delta u}{u^2} + \dfrac{\Delta v}{v^2}\right]$

Concept. The focal length is found from the lens formula $\dfrac1f=\dfrac1v-\dfrac1u$. Differentiating gives the error in $f$; when combining errors we always add magnitudes, so the minus sign disappears.

Step 1. Differentiate. Using $d\left(\dfrac1x\right)=-\dfrac{dx}{x^2}$:
$$-\frac{df}{f^2}=-\frac{dv}{v^2}+\frac{du}{u^2}$$

Step 2. Take the maximum error. Replace each differential by its error and add magnitudes:
$$\frac{\Delta f}{f^2}=\frac{\Delta u}{u^2}+\frac{\Delta v}{v^2}$$

Step 3. Solve for $\Delta f$.
$$\Delta f=f^2\left[\frac{\Delta u}{u^2}+\frac{\Delta v}{v^2}\right]$$

Answer. Option C. The same pattern ($\Delta R/R^2$ form) appears for resistors in parallel and springs in series.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) The dimensions of $n/\lambda$ is $[\text{T}]$

Concept. The arguments of $\sin$ and $\cos$ must be dimensionless. That fixes the dimensions of $x$, $n$ and $nt$.

Step 1. From the cosine. $\dfrac{2\pi x}{\lambda}$ is dimensionless, and $\lambda$ is a length, so $[x]=L$. Option C is correct.

Step 2. From the sine. $\dfrac{2\pi nt}{\lambda}$ is dimensionless, so $[nt]=[\lambda]=L$ (option D correct) and $[n]=\dfrac LT=LT^{-1}$ (option B correct). Here $n$ is the wave speed.

Step 3. Option A.
$$\left[\frac n\lambda\right]=\frac{LT^{-1}}{L}=T^{-1}$$
It is a frequency, not $[T]$. Option A is wrong.

Answer. Option A is NOT correct.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $T^2 = \dfrac{4\pi^2r^3}{GM}$

Concept. A correct formula must have the same dimensions on both sides ($4\pi^2$ is a pure number). Use $[G]=M^{-1}L^3T^{-2}$.

Step 1. Left side. $[T^2]=T^2$.

Step 2. Test option C.
$$\left[\frac{r^3}{GM}\right]=\frac{L^3}{M^{-1}L^3T^{-2}\cdot M}=T^2$$
Matches.

Step 3. Others. $\dfrac{r^2}{GM}$ gives $L^{-1}T^2$; $\dfrac{r}{GM^2}$ gives $M^{-1}L^{-2}T^2$; $r^3$ alone is $L^3$. All wrong.

Answer. $T^2=\dfrac{4\pi^2r^3}{GM}$ (Option C), Kepler's third law.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $4.6\ \text{s}$

$$\bar T = \frac{4.62 + 4.632 + 4.6 + 4.64}{4} = \frac{18.492}{4} = 4.623\ \text{s}$$
In addition, the result keeps as many decimal places as the least precise reading. $4.6\ \text{s}$ has one decimal place, so the mean is $4.6\ \text{s}$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) force per unit mass

$[u] = [\text{M}\text{L}^{-1}\text{T}^{-2}]$ and $[G] = [\text{M}^{-1}\text{L}^3\text{T}^{-2}]$.
$$[uG] = [\text{L}^2\text{T}^{-4}] \Rightarrow \left[\sqrt{uG}\right] = [\text{L}\text{T}^{-2}]$$
This is acceleration, i.e. force per unit mass.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $[\text{M}\text{L}^2\text{T}^{-2}]$

$\dfrac{a}{V^2}$ has the dimensions of pressure: $[a] = [\text{M}\text{L}^{-1}\text{T}^{-2}][\text{L}^6] = [\text{M}\text{L}^5\text{T}^{-2}]$.

$b$ has the dimensions of volume: $[b] = [\text{L}^3]$.
$$\left[\frac ab\right] = [\text{M}\text{L}^2\text{T}^{-2}]$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $2$

$20\ \text{VSD} = 19\ \text{MSD} \Rightarrow \text{LC} = 1\ \text{MSD} - 1\ \text{VSD} = \dfrac{1}{20}\ \text{MSD}$.
$$\frac{\text{MSD}}{20} = 0.1\ \text{mm} \Rightarrow \text{MSD} = 2\ \text{mm}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $5\%$

Concept. For a mass on a spring, $T=2\pi\sqrt{m/k}$, so $k=\dfrac{4\pi^2m}{T^2}$. When the *signs* of the errors are given, combine them with their signs: $\dfrac{\Delta k}{k}=\dfrac{\Delta m}{m}-2\dfrac{\Delta T}{T}$.

Step 1. Signed errors. $\dfrac{\Delta T}{T}=+2\%$, $\dfrac{\Delta m}{m}=-1\%$.

Step 2. Error in $k$.
$$\frac{\Delta k}{k}=(-1\%)-2(+2\%)=-5\%$$
The value of $k$ comes out $5\%$ too low. (The maximum-error rule, $1+2\times2$, also gives $5\%$ here.)

Answer. $5\%$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) A-IV, B-III, C-II, D-I

• Torque $= $ force × distance: $[\text{M}\text{L}^2\text{T}^{-2}]$ → IV
• Magnetic field, from $F = qvB$: $[\text{M}\text{T}^{-2}\text{A}^{-1}]$ → III
• Magnetic moment $= IA$: $[\text{L}^2\text{A}]$ → II
• $\mu_0$, from $B = \dfrac{\mu_0I}{2\pi r}$: $[\text{M}\text{L}\text{T}^{-2}\text{A}^{-2}]$ → I

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $71$

Concept. Screw gauge reading $=$ main-scale reading $+$ (circular-scale reading $\times$ LC). No zero error is given.

Step 1. Least count.
$$\text{LC}=\frac{1\ \text{mm}}{100}=0.01\ \text{mm}$$

Step 2. Reading.
$$d=1+42\times0.01=1.42\ \text{mm}$$

Step 3. Write as $x/50$.
$$x=1.42\times50=71$$

Answer. $71$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) Statement (I) is correct but Statement (II) is incorrect

Concept. Specific heat $c$ comes from $Q=mc\,\Delta T$; the gas constant $R$ comes from $PV=nRT$. Reduce each and compare with the statements.

Step 1. Specific heat.
$$[c]=\frac{[Q]}{[m][\Delta T]}=\frac{ML^2T^{-2}}{M\,K}=L^2T^{-2}K^{-1}$$
Statement I is correct.

Step 2. Gas constant. $PV$ is an energy, so
$$[R]=\frac{[PV]}{[n][T]}=\frac{ML^2T^{-2}}{\text{mol}\cdot K}=ML^2T^{-2}K^{-1}\text{mol}^{-1}$$
The power of $T$ is $-2$, not $-1$, so Statement II is incorrect.

Answer. Statement I is correct but Statement II is incorrect (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $20$

The zero error corresponds to 4 vernier divisions: $4\times\text{LC} = 0.04\ \text{mm} \Rightarrow \text{LC} = 0.01\ \text{mm}$.

$50\ \text{VSD} = 49\ \text{MSD} \Rightarrow \text{LC} = \dfrac{1\ \text{MSD}}{50}$, so $1\ \text{MSD} = 0.5\ \text{mm}$.

Number of main scale divisions in $1\ \text{cm}$ $= \dfrac{10}{0.5} = 20$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) Statement I is true but Statement II is false

Concept. Compare the dimensional formulas of each pair.

Step 1. Statement I. $E=h\nu$ gives $[h]=ML^2T^{-1}$. Angular momentum $L=mvr$ gives $ML^2T^{-1}$. Same, so Statement I is true.

Step 2. Statement II. Linear momentum $mv$: $MLT^{-1}$. Moment of force (torque) $=rF$: $ML^2T^{-2}$. Different, so Statement II is false.

Answer. Statement I is true but Statement II is false (Option A). Torque has the dimensions of energy, not momentum.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) Specific rotation of liquids

A spherometer measures small heights (sagitta) precisely, so it gives the thickness of thin plates and the radius of curvature of convex and concave surfaces.

Specific rotation of a liquid is an optical property measured with a polarimeter, not a spherometer.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $P$

Concept. In $\left(P+\dfrac a{V^2}\right)$ the term $\dfrac{a}{V^2}$ is added to a pressure, and in $(V-b)$ the constant $b$ is subtracted from a volume.

Step 1. Dimensions of $a$ and $b$.
$$[a]=[P][V^2],\qquad[b]=[V]$$

Step 2. Form $a/b^2$.
$$\left[\frac a{b^2}\right]=\frac{[P][V]^2}{[V]^2}=[P]$$

Answer. Pressure $P$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) Both (A) and (R) are correct but (R) is not the correct explanation of (A)

Concept. A positive zero error means the vernier zero is already to the right of the main zero when the jaws are closed, so every reading includes this extra amount.

Step 1. Assertion. With a positive zero error, the observed reading $=$ true value $+$ error. The reading is more than the actual value. Assertion is true.

Step 2. Reason. Zero errors do arise from manufacturing defects or rough handling (worn or bent jaws). Reason is true.

Step 3. Is it the explanation? The Reason tells how a zero error arises, not why a *positive* error makes the reading larger (that follows from the error adding to every reading). So it does not explain the Assertion.

Answer. Both (A) and (R) are correct but (R) is not the correct explanation of (A) (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $3.5\%$

$$\frac{\Delta R}{R}\times100 = \frac{\Delta V}{V}\times100 + \frac{\Delta I}{I}\times100 = \frac{5}{200}\times100 + \frac{0.2}{20}\times100 = 2.5\% + 1\% = 3.5\%$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $34\%$

$$\frac{\Delta Q}{Q}\times100 = 4(3\%) + 3(4\%) + 2(5\%) = 12\% + 12\% + 10\% = 34\%$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $b$ is order of magnitude for $a \le 5$

Concept. Order of magnitude: write the number as $a\times10^b$ with $1\le a\lt10$, then round $a$ to $1$ or $10$. If $a\le5$ it rounds to $1$ and the order is $b$; if $a>5$ it rounds to $10$ and the order is $b+1$ (the NCERT rule).

Step 1. Apply the rule to the options. For $a\le5$, $a\times10^b\approx10^b$, so $b$ is the order of magnitude. For $a>5$, $a\times10^b\approx10^{b+1}$, so the order is $b+1$, not $b$.

Step 2. Reject the rest. $a$ is never the order of magnitude; option A and option D use $a>5$, where the order is $b+1$.

Answer. $b$ is the order of magnitude for $a\le5$ (Option B). Example: $4.0\times10^6$ has order $6$; $7.0\times10^6$ has order $7$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $2\%$

Concept. When values are read from a graph, the error in each reading is the smallest scale division along that axis. $Y=49000\dfrac{M}{l}$ is a quotient, so the relative errors of $M$ and $l$ add; the constant carries no error.

Step 1. Errors from the graph paper. $\Delta M=5$ g (load axis) and $\Delta l=0.02$ cm (extension axis).

Step 2. Relative errors.
$$\frac{\Delta M}{M}=\frac{5}{500}=0.01,\qquad\frac{\Delta l}{l}=\frac{0.02}{2}=0.01$$

Step 3. Add.
$$\frac{\Delta Y}{Y}=0.01+0.01=0.02=2\%$$

Answer. $2\%$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $2.0\ \text{g/cm}^3$

Least count $= 1\ \text{MSD} - 1\ \text{VSD} = 1 - 0.9 = 0.1\ \text{mm} = 0.01\ \text{cm}$.

Diameter $= 2 + 2\times0.01 = 2.02\ \text{cm}$, so $r = 1.01\ \text{cm}$.
$$V = \frac43\pi(1.01)^3 \approx 4.32\ \text{cm}^3$$
$$\rho = \frac{8.635}{4.32} \approx 2.0\ \text{g cm}^{-3}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $[\text{ML}^{-1}\text{T}^{-2}]$

Concept. The energy stored per unit volume in an electric field is $u=\tfrac12\varepsilon_0E^2$. The factor $\tfrac12$ has no dimensions, so $\varepsilon_0E^2$ is an energy density, which has the same dimensions as pressure.

Step 1. Energy per volume.
$$[\varepsilon_0E^2]=\frac{[\text{energy}]}{[\text{volume}]}=\frac{ML^2T^{-2}}{L^3}=ML^{-1}T^{-2}$$

Step 2. Check directly (optional). $[\varepsilon_0]=M^{-1}L^{-3}T^4A^2$ and $[E]=MLT^{-3}A^{-1}$:
$$[\varepsilon_0E^2]=M^{-1}L^{-3}T^4A^2\cdot M^2L^2T^{-6}A^{-2}=ML^{-1}T^{-2}$$

Answer. $[ML^{-1}T^{-2}]$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\left(\dfrac{2N-1}{2}\right)$

Least count $= \dfrac{1}{20N}\ \text{cm} = \dfrac{1}{2N}\ \text{mm} = 1\ \text{MSD} - 1\ \text{VSD}$.
$$1\ \text{VSD} = 1 - \frac{1}{2N} = \frac{2N-1}{2N}\ \text{mm}$$
$$N\ \text{VSD} = \frac{2N-1}{2}\ \text{mm} = \frac{2N-1}{2}\ \text{MSD}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $4.55\ \text{mm}$

Least count $= \dfrac{1\ \text{mm}}{100} = 0.01\ \text{mm}$.

When the zero of the circular scale lies below the reference line with the jaws closed, the zero error is positive: $+5\times0.01 = +0.05\ \text{mm}$.

Observed reading $= 4 + 60\times0.01 = 4.60\ \text{mm}$.

Corrected diameter $= 4.60 - 0.05 = 4.55\ \text{mm}$.

Solution by Sreeraj P, M.Sc Physics