Mechanical Properties of Solids: NEET and JEE questions with solutions

Answer: (C) $4$ mm

Concept. The largest stress the wire can take without permanent deformation is the elastic limit; Hooke's law then gives the maximum strain, $\text{strain} = \dfrac{\text{stress}}{Y}$.

Step 1. Maximum strain.
$$\frac{\Delta l}{l} = \frac{8 \times 10^8}{2 \times 10^{11}} = 4 \times 10^{-3}$$

Step 2. Maximum elongation.
$$\Delta l = 4 \times 10^{-3} \times 1\ \text{m} = 4 \times 10^{-3}\ \text{m} = 4\ \text{mm}$$

Answer. $4$ mm (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $W/A$

Concept. Longitudinal stress is the internal force per unit cross-sectional area.

Step 1. Force in the wire. The light wire carries the tension $W$ at every point along its length.

Step 2. Stress.
$$\text{stress} = \frac{W}{A}$$

Answer. $W/A$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{MgL}{A(L_1 - L)}$

Concept. Young's modulus is $Y = \dfrac{\text{stress}}{\text{strain}}$.

Step 1. Stress and strain. Stress $= \dfrac{Mg}{A}$; strain $= \dfrac{L_1 - L}{L}$.

Step 2. Young's modulus.
$$Y = \frac{Mg/A}{(L_1 - L)/L} = \frac{MgL}{A(L_1 - L)}$$

Answer. $\dfrac{MgL}{A(L_1 - L)}$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $1$ mm

Concept. From $Y = \dfrac{F/A}{\Delta L/L}$, the extension is $\Delta L = \dfrac{FL}{AY}$.

Step 1. Substitute. $A = 1\ \text{mm}^2 = 10^{-6}\ \text{m}^2$:
$$\Delta L = \frac{100 \times 2}{10^{-6} \times 2 \times 10^{11}} = 10^{-3}\ \text{m} = 1\ \text{mm}$$

Answer. $1$ mm (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1 : 8$

Concept. $\Delta L = \dfrac{FL}{AY}$ with $A = \dfrac{\pi d^2}{4}$. Same material and load, so $\Delta L \propto \dfrac{L}{d^2}$.

Step 1. Set up. P: length $L$, diameter $2d$. Q: length $2L$, diameter $d$.

Step 2. Ratio.
$$\frac{\Delta L_P}{\Delta L_Q} = \frac{L/(2d)^2}{2L/d^2} = \frac{1/4}{2} = \frac18$$

Answer. $1 : 8$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{2Y_1Y_2}{Y_1 + Y_2}$

Concept. Rods joined end to end carry the same force, and their extensions add.

Step 1. Extensions. Each rod has length $L$: $\Delta L_1 = \dfrac{FL}{AY_1}$, $\Delta L_2 = \dfrac{FL}{AY_2}$.

Step 2. Equivalent rod. Length $2L$, same area, same force:
$$\frac{F(2L)}{AY} = \frac{FL}{AY_1} + \frac{FL}{AY_2} \;\Rightarrow\; \frac{2}{Y} = \frac{1}{Y_1} + \frac{1}{Y_2}$$

Step 3. Solve.
$$Y = \frac{2Y_1Y_2}{Y_1 + Y_2}$$

Answer. $\dfrac{2Y_1Y_2}{Y_1 + Y_2}$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: 100

Concept. Elastic energy per unit volume is $u = \dfrac12 \times \text{stress} \times \text{strain} = \dfrac{\text{stress}^2}{2Y}$.

Step 1. Substitute.
$$u = \frac{(2 \times 10^8)^2}{2 \times 2 \times 10^{11}} = \frac{4 \times 10^{16}}{4 \times 10^{11}} = 10^5\ \text{J m}^{-3}$$

Step 2. Convert. $10^5\ \text{J m}^{-3} = 100\ \text{kJ m}^{-3}$.

Answer. $100$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $0.5\%$

Concept. Bulk modulus: $B = \dfrac{\Delta p}{\Delta V/V}$, so $\dfrac{\Delta V}{V} = \dfrac{\Delta p}{B}$.

Step 1. Substitute.
$$\frac{\Delta V}{V} = \frac{1.1 \times 10^7}{2.2 \times 10^9} = 5 \times 10^{-3} = 0.5\%$$

Answer. $0.5\%$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $4\ \mu$m

Concept. Shear modulus $\eta = \dfrac{F/A}{\Delta x/L}$, so the sideways shift of the top face is $\Delta x = \dfrac{F}{A\eta}L$.

Step 1. Shear stress. $A = (0.1)^2 = 0.01\ \text{m}^2$:
$$\frac{F}{A} = \frac{10^4}{0.01} = 10^6\ \text{N m}^{-2}$$

Step 2. Shear strain.
$$\frac{10^6}{2.5 \times 10^{10}} = 4 \times 10^{-5}$$

Step 3. Displacement.
$$\Delta x = 4 \times 10^{-5} \times 0.1 = 4 \times 10^{-6}\ \text{m} = 4\ \mu\text{m}$$

Answer. $4\ \mu$m (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $0.5$

Concept. Poisson's ratio is $\sigma = -\dfrac{\Delta r/r}{\Delta L/L}$. Use the constant-volume condition to relate the two strains.

Step 1. Constant volume. $V = \pi r^2L$, so
$$\frac{\Delta V}{V} = 2\frac{\Delta r}{r} + \frac{\Delta L}{L} = 0 \;\Rightarrow\; \frac{\Delta r}{r} = -\frac12\frac{\Delta L}{L}$$

Step 2. Poisson's ratio. $\sigma = \dfrac12 = 0.5$.

Answer. $0.5$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $1.2 \times 10^8$ N m$^{-2}$

Concept. The walls stop the rod expanding, so they compress it by the strain it would have expanded, $\alpha\Delta T$. Stress $= Y \times$ strain.

Step 1. Prevented strain. $\alpha\Delta T = 1.2 \times 10^{-5} \times 50 = 6 \times 10^{-4}$.

Step 2. Stress.
$$\text{stress} = Y\alpha\Delta T = 2 \times 10^{11} \times 6 \times 10^{-4} = 1.2 \times 10^8\ \text{N m}^{-2}$$
It does not depend on the rod's length or area.

Answer. $1.2 \times 10^8$ N m$^{-2}$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{\rho g L^2}{2Y}$

Concept. The tension in a hanging wire varies along it: a point at height $x$ above the free end supports only the wire below it. Add the stretches of thin slices.

Step 1. Tension at height $x$. The part below weighs $\rho Axg$, so the tension there is $T(x) = \rho Agx$.

Step 2. Stretch of a slice $dx$.
$$d(\Delta L) = \frac{T(x)\,dx}{AY} = \frac{\rho gx}{Y}dx$$

Step 3. Integrate.
$$\Delta L = \int_0^L \frac{\rho gx}{Y}dx = \frac{\rho gL^2}{2Y}$$
(The same as the whole weight acting through half the length.)

Answer. $\dfrac{\rho gL^2}{2Y}$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) ductile

Concept. The gap between the elastic limit and the fracture point on a stress–strain curve measures how much permanent (plastic) deformation a material can take.

Step 1. Classify. A large plastic region means the material can be drawn into wires or beaten into sheets: it is ductile (copper, aluminium). Brittle materials break soon after the elastic limit.

Answer. Ductile (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) steel needs a much larger stress for the same strain

Concept. In physics, "more elastic" means a larger modulus of elasticity: more stress is needed for the same strain.

Step 1. Compare. Steel's Young's modulus ($\sim 2 \times 10^{11}$ Pa) is far larger than rubber's ($\sim 10^6$ Pa), so steel needs a much larger stress for the same strain.

Answer. Steel needs a much larger stress for the same strain (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $400$ kg

Concept. Breaking load $=$ breaking stress $\times$ area. The breaking stress depends only on the material; the length plays no part.

Step 1. Area. Twice the radius gives $4$ times the area.

Step 2. Load. $4 \times 100 = 400$ kg.

Answer. $400$ kg (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) infinite

Concept. $Y = \dfrac{\text{stress}}{\text{strain}}$.

Step 1. Rigid body. A perfectly rigid body shows zero strain for any stress, so $Y = \dfrac{\text{stress}}{0}$ is infinite.

Answer. Infinite (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $0.2$ J

Concept. The stretching force grows linearly from $0$ to $F$, so the work done (energy stored) is $U = \dfrac12F\Delta L$.

Step 1. Substitute.
$$U = \frac12 \times 200 \times 2 \times 10^{-3} = 0.2\ \text{J}$$

Answer. $0.2$ J (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $2.56$

Concept. The wire breaks if its stress exceeds the maximum stress, so the largest tension it can take is $T_{\max}=\sigma_{\max}A$. When the lift accelerates upward, Newton's law gives $T-mg=ma$, so the tension is largest then.

Step 1. Largest tension.
$$T_{\max}=4\times10^8\times\pi(4\times10^{-3})^2=4\times10^8\times3.14\times16\times10^{-6}=20\,096\ \text{N}$$

Step 2. Newton's law for the lift (upward acceleration $a$):
$$T_{\max}=m(g+a)\ \Rightarrow\ g+a=\frac{20\,096}{1600}=12.56$$

Step 3. Acceleration. $a=12.56-10=2.56$ m s$^{-2}$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $1.21$

Concept. Joined wires carry the same tension. With equal areas, $\Delta L = \dfrac{FL}{AY}$ is equal when $\dfrac{L}{Y}$ is equal.

Step 1. Condition. $\dfrac{L_A}{Y_A} = \dfrac{L_B}{Y_B}$.

Step 2. Solve.
$$L_B = L_A\frac{Y_B}{Y_A} = 2.2 \times \frac{11}{20} = 1.21\ \text{m}$$

Answer. $1.21$ m (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $1.0 \times 10^{10}$

Concept. $\Delta l=\dfrac{L}{AY}W$: the slope of the extension–load line is $\dfrac{L}{AY}$.

Step 1. Slope. From the graph, $7\times10^{-4}$ m at $70$ N: slope $=10^{-5}$ m/N.

Step 2. Young's modulus. $Y=\dfrac{L}{A\times\text{slope}}=\dfrac{1}{10^{-5}\times10^{-5}}=10^{10}$ N/m$^2$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) remains unchanged

Concept. Young's modulus $Y=\dfrac{\text{stress}}{\text{strain}}$ is a property of the material, like density. Changing the size or shape of a wire changes how much it stretches, but not $Y$.

Answer. $Y$ remains unchanged (Option C). Only temperature and the material change $Y$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $38$

Concept. Each wire carries everything hanging below it. Find the largest pan load allowed by each wire; the smaller one decides. From the figure: upper wire, $30$ kg block, lower wire, $10$ kg block, then the pan (mass neglected).

Step 1. Breaking loads. Upper: $12\times10^8\times0.008\times10^{-4}=960$ N ($96$ kg). Lower: $12\times10^8\times0.004\times10^{-4}=480$ N ($48$ kg).

Step 2. Upper wire. Carries $30+10+m\le96\Rightarrow m\le56$ kg.

Step 3. Lower wire. Carries $10+m\le48\Rightarrow m\le38$ kg.

Answer. $38$ kg (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. Shear modulus $\eta=\dfrac{F/A}{\theta}$ with $\theta=\dfrac{\Delta x}{L}$. For a cube, $\Delta x=\dfrac{FL}{\eta A}=\dfrac{F}{\eta L}$.

Step 1. Substitute. $\Delta x=\dfrac{10}{10^5\times0.05}=2\times10^{-3}$ m.

Answer. $2$ mm.

Solution by Sreeraj P, M.Sc Physics

Answer: 33

Concept. Energy stored in a stretched wire is $\tfrac12\times\text{stress}\times\text{strain}\times\text{volume}=\tfrac12Y(\text{strain})^2V$.

Step 1. Strain. $\dfrac{3\times10^{-3}}{3}=10^{-3}$.

Step 2. Energy.
$$U=\frac12\times1.1\times10^{11}\times(10^{-3})^2\times600\times10^{-6}=\frac12\times1.1\times10^{11}\times6\times10^{-10}=33\ \text{J}$$

Answer. $33$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $6$

Concept. Each string carries the weight of everything below it, then $\Delta l=\dfrac{TL}{AY}$. Take the ratio so that the common quantities cancel.

Step 1. Tensions. String $A$ holds both blocks: $T_A=3Mg$. String $B$ holds only the lower block: $T_B=2Mg$.

Step 2. Ratio (same area):
$$\frac{\Delta l_A}{\Delta l_B}=\frac{T_A}{T_B}\cdot\frac{L_A}{L_B}\cdot\frac{Y_B}{Y_A}=\frac32\times2\times\frac1{0.5}=6$$

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $3\%$

Concept. $\dfrac{\Delta V}{V}=\dfrac{\Delta P}{K}$.

Step 1. Substitute. $\dfrac{6.3\times10^7}{2.1\times10^9}=0.03=3\%$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $2$

Concept. In series both strings carry the full load; add the two extensions $\dfrac{FL}{AY}$.

Step 1. Area and load. $A=\pi(2\times10^{-4})^2=1.257\times10^{-7}$ m$^2$; $F=8$ N.

Step 2. String A. $\dfrac{8\times0.314}{1.257\times10^{-7}\times2\times10^{10}}=1.0\times10^{-3}$ m.

Step 3. String B. Length and $Y$ both doubled: the same $1.0$ mm.

Step 4. Total. $2$ mm.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 3

Concept. The tension at any point of a hanging wire equals the weight of everything below that point. Divide by the area to get the stress there.

Step 1. What hangs below the point. At $\dfrac l3$ from the top, the lower $\dfrac23$ of the wire (weight $\dfrac23w$) and the load $W$ hang below.

Step 2. Stress there.
$$\sigma=\frac{W+\frac23w}{A}=\frac WA+\frac23\cdot\frac wA$$

Step 3. Compare with the given form. $\dfrac{2}{\gamma}=\dfrac23\Rightarrow\gamma=3$.

Answer. $3$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) 5

Concept. $Y = \dfrac{FL}{A\,\Delta L}$. With the same load and the same extension, $Y \propto \dfrac{L}{A}$.

Step 1. Ratio.
$$\frac{Y_A}{Y_B} = \frac{L_A}{L_B} \cdot \frac{A_B}{A_A} = \frac{6.0}{5.4} \times \frac{4.5}{3.0} = \frac{27}{16.2} = \frac53$$

Step 2. Compare with $x : 3$. $x = 5$.

Answer. $5$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: 43

Concept. $\Delta P=K\dfrac{\Delta V}{V}$.

Step 1. Substitute. $2.15\times10^9\times0.002=4.3\times10^6=43\times10^5$ N m$^{-2}$.

Answer. $43$.

Solution by Sreeraj P, M.Sc Physics

Answer: 50

Concept. $\Delta V = \dfrac{pV}{B}$.

Step 1. Volume. $V = (0.1)^3 = 10^{-3}\ \text{m}^3$.

Step 2. Contraction.
$$\Delta V = \frac{7 \times 10^6 \times 10^{-3}}{1.4 \times 10^{11}} = 5 \times 10^{-8}\ \text{m}^3 = 50\ \text{mm}^3$$

Answer. $50$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $1.25$

Concept. The extra pressure at depth $h$ is $p = \rho gh$, and $\dfrac{\Delta V}{V} = \dfrac{p}{B}$.

Step 1. Pressure.
$$p = 10^3 \times 10 \times 2500 = 2.5 \times 10^7\ \text{Pa}$$

Step 2. Fractional compression.
$$\frac{\Delta V}{V} = \frac{2.5 \times 10^7}{2 \times 10^9} = 0.0125 = 1.25\%$$

Answer. $1.25\%$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) decreases

Concept. Interatomic bonds weaken as temperature rises, so the restoring force per unit strain, and $Y$, falls.

Answer. Decreases (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: 3

Concept. Identical wires: strain $\propto$ tension, and each wire's tension is the weight below it.

Step 1. Tensions. Upper: $(2+1)g$; lower: $1g$.

Step 2. Ratio. $3$.

Answer. $3$.

Solution by Sreeraj P, M.Sc Physics

Answer: 5

Concept. A spring (or wire) obeys $l=L_0+\dfrac Tk$: length is linear in tension, so any linear combination of lengths corresponds to the same combination of tensions.

Step 1. Lengths. $a=L_0+\dfrac3k$, $b=L_0+\dfrac2k$.

Step 2. Combine. $3a-2b=L_0+\dfrac{9-4}{k}=L_0+\dfrac5k$: the length under $5$ N.

Answer. $5$.

Solution by Sreeraj P, M.Sc Physics

Answer: 600

Concept. A hanging wire holds its own weight. The stress is largest at the top, where the whole wire hangs below: $\sigma=\dfrac{\rho ALg}{A}=\rho gL$. It breaks when this reaches the breaking stress. The area cancels.

Step 1. Gravity on the planet. $g'=\dfrac{10}{3}$ m s$^{-2}$.

Step 2. Longest wire.
$$L_{\max}=\frac{\sigma_b}{\rho g'}=\frac{1.2\times10^8}{6\times10^4\times\frac{10}{3}}=\frac{1.2\times10^8}{2\times10^5}=600\ \text{m}$$

Answer. $600$ m. The maximum length does not depend on the thickness of the wire.

Solution by Sreeraj P, M.Sc Physics

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

Concept. Stress is the restoring force per unit area. Each modulus belongs to one kind of stress: normal forces on opposite faces (Young's), tangential forces on opposite faces (shear), equal normal pressure everywhere (bulk).

Step 1. Match. A–III (stress), B–IV (shear), C–I (bulk), D–II (Young's).

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 1

Concept. The sagging wire stretches a little, and its tension supports the mass through the vertical components. Find $T$ from equilibrium, the strain from the geometry, then $A$ from $T=YA\times\text{strain}$.

Step 1. Tension. At the middle, two tensions at angle $\theta$ below the horizontal hold the weight:
$$2T\sin\theta=mg\ \Rightarrow\ T=\frac{20}{2\times0.01}=1000\ \text{N}$$

Step 2. Strain. Each half of length $L$ becomes $\dfrac{L}{\cos\theta}\approx L\left(1+\dfrac{\theta^2}{2}\right)$, so the strain is
$$\frac{\theta^2}{2}=\frac{10^{-4}}{2}=5\times10^{-5}$$

Step 3. Area.
$$A=\frac{T}{Y\times\text{strain}}=\frac{1000}{2\times10^{11}\times5\times10^{-5}}=10^{-4}\ \text{m}^2$$

Answer. $1$.

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. Bulk modulus $K=\dfrac{\Delta P}{\Delta V/V}$, so the fractional compression is $\dfrac{\Delta V}{V}=\dfrac{\Delta P}{K}$. At depth $h$ the extra pressure is $\rho gh$.

Step 1. Pressure at the bottom. $\Delta P=1000\times10\times4000=4\times10^7$ Pa.

Step 2. Fractional compression. $\dfrac{4\times10^7}{2\times10^9}=2\times10^{-2}$.

Answer. $\alpha=2$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) Both (A) and (R) are true and (R) is the correct explanation of (A)

Concept. Elasticity is the tendency to regain shape; the restoring force comes from the interatomic (and intermolecular) forces in the solid, which is why a body is elastic.

Answer. Both true and R is the correct explanation of A (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) 2 times

Concept. $\Delta L=\dfrac{FL}{\pi r^2Y}\propto\dfrac{F}{r^2}$ at fixed $L$ and material.

Step 1. Scale. $\dfrac{1/2}{1/4}=2$.

Answer. 2 times (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: 16

Concept. Same volume means $A_1L_1=A_2L_2$, so a wire with four times the area is a quarter as long. The force for a given extension is $F=\dfrac{YA\,\Delta l}{L}$, so $F\propto\dfrac AL$.

Step 1. Lengths. $\dfrac{A_P}{A_Q}=4\Rightarrow\dfrac{L_P}{L_Q}=\dfrac14$.

Step 2. Force ratio (same $Y$ and $\Delta l$):
$$\frac{F_1}{F_2}=\frac{A_P/L_P}{A_Q/L_Q}=4\times4=16$$

Answer. $16$. Equivalently, at fixed volume $F\propto A^2$.

Solution by Sreeraj P, M.Sc Physics

Answer: 20

Concept. A wire pulled by $F$ at each end has tension $F$ (not $2F$); the ends are just in equilibrium.

Step 1. Extension. $\Delta L=\dfrac{200\times2}{2\times10^{-4}\times10^{11}}=2\times10^{-5}$ m $=20\ \mu$m.

Answer. $20$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $Y$

Concept. Young's modulus is a property of the material, not of the specimen's size or shape.

Step 1. Apply. Changing the length and area of the wire changes its extension under a load, but not $Y$.

Answer. $Y$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. Find the tension in wire $B$ from Newton's laws, then strain $=\dfrac{T}{AY}$. From the figure, $P$ and $Q$ are on the smooth table (joined by $A$) and $B$ runs from $Q$ over the pulley to the hanging block $R$.

Step 1. Acceleration. Only $R$'s weight drives all three: $a=\dfrac{30}{9}=\dfrac{10}{3}$ m s$^{-2}$.

Step 2. Tension in $B$. It pulls $P$ and $Q$: $T_B=6\times\dfrac{10}{3}=20$ N.

Step 3. Strain. $A=0.005$ cm$^2=5\times10^{-7}$ m$^2$:
$$\varepsilon=\frac{20}{5\times10^{-7}\times2\times10^{11}}=2\times10^{-4}$$

Answer. $2$.

Solution by Sreeraj P, M.Sc Physics

Answer: 18

Concept. The extra pressure at depth $h$ is $\rho gh$; it compresses the ball by $\dfrac{\Delta V}{V}=\dfrac{\rho gh}{K}$.

Step 1. Pressure needed. $9\times10^8\times2\times10^{-4}=1.8\times10^5$ Pa.

Step 2. Depth. $h=\dfrac{1.8\times10^5}{10^3\times10}=18$ m.

Answer. $18$.

Solution by Sreeraj P, M.Sc Physics

Answer: 12

Concept. First find the tension from the dynamics (Atwood machine), then strain $=\dfrac{T}{AY}$.

Step 1. Tension. $T=\dfrac{2m_1m_2g}{m_1+m_2}=\dfrac{2(2)(4)(10)}{6}=\dfrac{80}{3}$ N.

Step 2. Area. $A=\pi(4\times10^{-5})^2=16\pi\times10^{-10}$ m$^2$.

Step 3. Strain. $\dfrac{80/3}{16\pi\times10^{-10}\times2\times10^{11}}=\dfrac{80/3}{320\pi}=\dfrac{1}{12\pi}$.

Answer. $\alpha=12$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\sigma=\dfrac{3K-2\eta}{6K+2\eta}$

Concept. From $Y=3K(1-2\sigma)$ and $Y=2\eta(1+\sigma)$, eliminate $Y$.

Step 1. Equate. $3K-6K\sigma=2\eta+2\eta\sigma$.

Step 2. Solve. $\sigma(6K+2\eta)=3K-2\eta\Rightarrow\sigma=\dfrac{3K-2\eta}{6K+2\eta}$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $6.9\times10^{-2}\ \text{mm}$

Concept. From $Y=\dfrac{FL}{A\,\Delta l}$, the extension is $\Delta l=\dfrac{FL}{AY}$. Same material and same force, so $\Delta l\propto\dfrac{L}{d^2}$.

Step 1. Ratio.
$$\frac{\Delta l_B}{\Delta l_A}=\frac{L_B}{L_A}\left(\frac{d_A}{d_B}\right)^2=2\times\frac{1}{(2.4)^2}=\frac{2}{5.76}=0.347$$

Step 2. Extension of $B$. $0.347\times0.2=0.0694$ mm $=6.9\times10^{-2}$ mm.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: 60

Concept. Cooling shortens the rod by $L\alpha\Delta T$. The hanging mass must stretch it back by the same amount: $\dfrac{MgL}{AY}=L\alpha\Delta T$, so $Mg=YA\alpha\Delta T$.

Step 1. Shortening. $\Delta L=1\times2\times10^{-5}\times50=10^{-3}$ m.

Step 2. Force to stretch it back.
$$Mg=\frac{YA\,\Delta L}{L}=\frac{2\times10^{11}\times3\times10^{-6}\times10^{-3}}{1}=600\ \text{N}$$

Step 3. Mass. $M=60$ kg.

Answer. $60$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1:1$

Concept. $Y = \dfrac{FL}{A\,\Delta L}$; with the same load and extension, $Y \propto \dfrac{L}{A}$.

Step 1. Ratio.
$$\frac{Y_A}{Y_B} = \frac{5.0/2.5}{6.0/3.0} = \frac{2}{2} = 1$$

Answer. $1 : 1$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $4\times10^{-3}\ \text{m}$

Concept. $\Delta L=\dfrac{FL}{AY}$.

Step 1. Substitute. $\Delta L=\dfrac{250\times100}{6.25\times10^{-4}\times10^{10}}=\dfrac{25\,000}{6.25\times10^6}=4\times10^{-3}$ m.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) Both A and R are correct and R is the correct explanation of A

Concept. Steel is chosen for structures because it is highly elastic (large $Y$, so small strain under load) and has a high elastic limit, so it returns to shape under large loads.

Answer. Both correct and R is the correct explanation of A (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: 5

Concept. $Y=\dfrac{FL}{A\,\Delta l}=\dfrac LA\cdot\dfrac{F}{\Delta l}$. On an extension–load graph the slope is $\dfrac{\Delta l}{F}$, so $Y=\dfrac{L}{A\times\text{slope}}$.

Step 1. Slope. The line makes $45^\circ$ with the load axis: $\dfrac{\Delta l}{F}=\tan45^\circ=1$ m N$^{-1}$.

Step 2. Area. Diameter $4$ mm: $A=\pi(2\times10^{-3})^2=1.256\times10^{-5}$ m$^2$.

Step 3. Young's modulus.
$$Y=\frac{0.628}{1.256\times10^{-5}\times1}=5\times10^4\ \text{N m}^{-2}$$

Answer. $x=5$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $l$

Concept. $\Delta L\propto\dfrac{FL}{r^2}$ for one material.

Step 1. Scale. $\dfrac{2\times2}{4}=1$: the same $l$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $12:1$

Concept. Same load and length: $\Delta L\propto\dfrac{1}{AY}$.

Step 1. Ratio. $\dfrac{\Delta L_A}{\Delta L_B}=\dfrac{A_BY_B}{A_AY_A}=3\times4=12$.

Answer. $12:1$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. $l=L_0+\dfrac Tk$. Two readings give $L_0$.

Step 1. Difference. $l_2-l_1=\dfrac{20}{k}$, and $l_2=1.1l_1$, so $\dfrac{20}{k}=0.1l_1\Rightarrow\dfrac1k=\dfrac{l_1}{200}$.

Step 2. Natural length. $L_0=l_1-\dfrac{100}{k}=l_1-\dfrac{l_1}{2}=\dfrac{l_1}{2}$.

Answer. $x=2$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $3P$

Concept. Bulk modulus in calculus form: $K=-V\dfrac{dP}{dV}$. Differentiate the given $P(V)$.

Step 1. Differentiate. $P=aV^{-3}\Rightarrow\dfrac{dP}{dV}=-3aV^{-4}$.

Step 2. Bulk modulus.
$$K=-V(-3aV^{-4})=3aV^{-3}=3P$$

Answer. Option A. (For an ideal gas at constant temperature, $PV=$ constant gives $K=P$; at constant entropy, $K=\gamma P$.)

Solution by Sreeraj P, M.Sc Physics

Answer: 40

Concept. Elastic energy $U=\dfrac{YA(\Delta L)^2}{2L}$.

Step 1. Solve for $A$. $A=\dfrac{2UL}{Y(\Delta L)^2}=\dfrac{2\times80\times20}{2\times10^{11}\times4\times10^{-4}}=4\times10^{-5}$ m$^2$.

Answer. $40$ mm$^2$.

Solution by Sreeraj P, M.Sc Physics

Answer: 25

Concept. Strain $=\dfrac{\text{stress}}{Y}=\dfrac{F}{AY}$.

Step 1. Stress. $A=\pi(0.02)^2=1.256\times10^{-3}$ m$^2$; $\sigma=\dfrac{62\,800}{1.256\times10^{-3}}=5.0\times10^7$ N m$^{-2}$.

Step 2. Strain. $\dfrac{5.0\times10^7}{2.0\times10^{11}}=2.5\times10^{-4}=25\times10^{-5}$.

Answer. $25$.

Solution by Sreeraj P, M.Sc Physics

Answer: 11

Concept. Tensile stress is the force per unit cross-sectional area: $\sigma=\dfrac FA$. In equilibrium the wire's tension equals the weight it holds, so $mg=\sigma A$.

Step 1. Area of the wire. Radius $=7$ mm $=7\times10^{-3}$ m:
$$A=\pi r^2=\frac{22}{7}\times(7\times10^{-3})^2=154\times10^{-6}\ \text{m}^2$$

Step 2. Tension. $F=\sigma A=7\times10^5\times154\times10^{-6}=107.8$ N.

Step 3. Mass. $m=\dfrac{107.8}{9.8}=11$ kg.

Answer. $11$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) 5600

Concept. Energy per unit volume $u=\tfrac12Y\varepsilon^2$ for strain $\varepsilon$.

Step 1. Strain. $0.04\%=4\times10^{-4}$.

Step 2. Energy density.
$$u=\frac12\times7\times10^{10}\times(4\times10^{-4})^2=\frac12\times7\times10^{10}\times1.6\times10^{-7}=5600\ \text{J m}^{-3}$$

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: 20

Concept. Read the figure: the steel wire is on top, so it carries both blocks. Then use $\Delta l=\dfrac{TL}{AY}$ with the steel wire's own length.

Step 1. Tension in the steel wire. $T=(2+1.14)\times10=31.4$ N.

Step 2. Area. $A=\pi(2\times10^{-3})^2=1.256\times10^{-5}$ m$^2$.

Step 3. Extension.
$$\Delta l=\frac{31.4\times1.6}{1.256\times10^{-5}\times2\times10^{11}}=\frac{50.24}{2.512\times10^6}=2\times10^{-5}\ \text{m}$$

Answer. $20\times10^{-6}$ m.

Solution by Sreeraj P, M.Sc Physics

Answer: 1

Concept. $B = \dfrac{P}{\Delta V/V}$. The same pressure acts on both, so $B \propto \dfrac{1}{\Delta V/V}$; the actual volumes do not matter.

Step 1. Ratio.
$$\frac{B_w}{B_l} = \frac{0.03\%}{0.01\%} = 3 = \frac31$$

Answer. $1$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $0.1\ \text{mm}$

Concept. $\Delta L=\dfrac{FL}{AY}$ with the load's weight on that planet.

Step 1. Tension. $4\times\dfrac{10}{4}=10$ N.

Step 2. Extension. $\dfrac{10\times6}{3\times10^{-6}\times2\times10^{11}}=10^{-4}$ m $=0.1$ mm.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $6\times10^{8}\ \text{N m}^{-2}$

Concept. $\Delta P=K\dfrac{\Delta V}{V}$.

Step 1. Substitute. $3\times10^{10}\times0.02=6\times10^8$ N m$^{-2}$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 6

Concept. $\Delta l = \dfrac{mgL}{AY}$; the same wire and load, so $\Delta l \propto g$.

Step 1. Ratio.
$$g' = 10 \times \frac{6 \times 10^{-5}}{10^{-4}} = 6\ \text{m s}^{-2}$$

Answer. $6$.

Solution by Sreeraj P, M.Sc Physics

Answer: 25

Concept. Energy density $u=\tfrac12Y\varepsilon^2$. Get $Y$ from the line: here strain is plotted against stress, so $Y=\dfrac{\text{stress}}{\text{strain}}=$ 1/slope.

Step 1. Young's modulus. $\dfrac{20}{1\times10^{-10}}=2\times10^{11}$ Pa.

Step 2. Energy density. $u=\tfrac12(2\times10^{11})(5\times10^{-4})^2=2.5\times10^4$ J m$^{-3}=25$ kJ m$^{-3}$.

Answer. $25$.

Solution by Sreeraj P, M.Sc Physics

Answer: 5

Concept. $\Delta L\propto\dfrac{FL}{r^2}$.

Step 1. Scale. $\dfrac{4\times4}{16}=1$: $5$ cm.

Answer. $5$.

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. $Y=\dfrac{4FL}{\pi d^2\,\Delta l}$. Relative errors add, each multiplied by its power: $\dfrac{\Delta Y}{Y}=2\dfrac{\Delta d}{d}+\dfrac{\Delta(\Delta l)}{\Delta l}$ ($F$ and $L$ are exact).

Step 1. Value of $Y$ ($F=10$ N, $L=1$ m, $d=0.4\times10^{-3}$ m, $\Delta l=0.4\times10^{-3}$ m):
$$Y=\frac{4\times10\times1}{\pi(0.4\times10^{-3})^2\times0.4\times10^{-3}}=\frac{40}{2.01\times10^{-10}}=2\times10^{11}\ \text{N m}^{-2}$$

Step 2. Relative error.
$$\frac{\Delta Y}{Y}=2\times\frac{0.01}{0.4}+\frac{0.02}{0.4}=0.05+0.05=0.1$$

Step 3. Absolute error. $\Delta Y=0.1\times2\times10^{11}=2\times10^{10}$ N m$^{-2}$.

Answer. $x=2$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $6.25\times10^{-4}\ \text{m}^2$

Concept. The rope breaks at a fixed breaking stress, so the load it can carry is proportional to its area.

Step 1. Scale. $A_2=2.5\times10^{-4}\times\dfrac{25}{10}=6.25\times10^{-4}$ m$^2$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: 25

Concept. In a rod hanging under its own weight the tension falls from $Mg$ at the top to zero at the bottom. Adding the stretch of each small piece gives half of what the full weight would give at the end: $\Delta l=\dfrac{MgL}{2AY}$.

Step 1. Why the factor $\tfrac12$. A piece $dx$ at distance $x$ from the bottom carries the weight $\dfrac{Mg}{L}x$ below it, so it stretches by $\dfrac{Mgx\,dx}{LAY}$. Integrating $x$ from $0$ to $L$ gives $\dfrac{MgL}{2AY}$.

Step 2. Numbers.
$$\Delta l=\frac{20\times10\times20}{2\times0.4\times2\times10^{11}}=\frac{4000}{1.6\times10^{11}}=2.5\times10^{-8}\ \text{m}$$

Answer. $25\times10^{-9}$ m, so $x=25$.

Solution by Sreeraj P, M.Sc Physics

Answer: 48

Concept. Shear modulus $\eta=\dfrac{F/A}{\Delta x/h}$, where $A$ is the face the force acts on and $h$ is the height perpendicular to that face. So $\Delta x=\dfrac{Fh}{\eta A}$.

Step 1. Identify the face and the height. The force acts on the narrow face, $60\times15$ cm$^2$: $A=0.6\times0.15=0.09$ m$^2$. The height from the fixed edge to that face is the side, $h=0.6$ m.

Step 2. Displacement.
$$\Delta x=\frac{1.8\times10^5\times0.6}{25\times10^9\times0.09}=\frac{1.08\times10^5}{2.25\times10^9}=4.8\times10^{-5}\ \text{m}$$

Answer. $48\ \mu$m.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $154$

Concept. Wires joined end to end (in series) carry the same tension. The total extension is the sum: $\Delta l=\dfrac FA\left(\dfrac{L_1}{Y_1}+\dfrac{L_2}{Y_2}\right)$.

Step 1. Bracket.
$$\frac{3.2}{2\times10^{11}}+\frac{4.4}{1.1\times10^{11}}=1.6\times10^{-11}+4\times10^{-11}=5.6\times10^{-11}$$

Step 2. Area. $A=\dfrac{22}{7}(1.4\times10^{-3})^2=6.16\times10^{-6}$ m$^2$.

Step 3. Load.
$$1.4\times10^{-3}=\frac{F}{6.16\times10^{-6}}\times5.6\times10^{-11}\ \Rightarrow\ F=\frac{1.4\times10^{-3}\times6.16\times10^{-6}}{5.6\times10^{-11}}=154\ \text{N}$$

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. The slope of each extension–load line is $\dfrac{L}{AY}$. Plotted against $L$, those slopes form a line of slope $\dfrac{1}{AY}$.

Step 1. Slope of the graph. $0.25\times10^{-5}$ m/N per metre of length.

Step 2. Young's modulus. $AY=\dfrac{1}{0.25\times10^{-5}}=4\times10^5$ N; $Y=\dfrac{4\times10^5}{2\times10^{-6}}=2\times10^{11}$ N m$^{-2}$.

Answer. $x=2$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $2\times10^7$ N

Concept. $F = YA \times \text{strain}$. Doubling the length means a strain of $1$ (assuming Hooke's law still held).

Step 1. Substitute. $A = 1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$:
$$F = 2 \times 10^{11} \times 10^{-4} \times 1 = 2 \times 10^7\ \text{N}$$

Answer. $2 \times 10^7$ N (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: 20

Concept. The stretched rubber acts as a spring of constant $k=\dfrac{YA}{L}$. Its stored energy $\tfrac12kx^2$ all becomes the stone's kinetic energy.

Step 1. Spring constant. $k=\dfrac{0.5\times10^9\times10^{-6}}{0.1}=5000$ N m$^{-1}$.

Step 2. Stored energy. $U=\tfrac12\times5000\times(0.04)^2=4$ J.

Step 3. Speed. $\tfrac12(0.02)v^2=4\Rightarrow v^2=400\Rightarrow v=20$ m s$^{-1}$.

Answer. $20$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $5\times10^{-10}$ m

Concept. Under its own weight, the tension falls linearly from $W$ at the top to $0$ at the bottom; the average tension $W/2$ gives $\Delta L=\dfrac{WL}{2AY}$.

Step 1. Substitute. $\Delta L=\dfrac{10\times0.2}{2\times10^{-2}\times2\times10^{11}}=5\times10^{-10}$ m.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: 500

Concept. The extra pressure needed is $\Delta P = B\dfrac{\Delta V}{V}$, supplied by depth: $\Delta P = \rho gh$.

Step 1. Pressure.
$$\Delta P = 9.8 \times 10^8 \times 0.005 = 4.9 \times 10^6\ \text{Pa}$$

Step 2. Depth.
$$h = \frac{4.9 \times 10^6}{10^3 \times 9.8} = 500\ \text{m}$$

Answer. $500$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $2.60\times10^{-7}$

Concept. Each column carries a quarter of the weight over an annular area; strain $= \dfrac{F}{AY}$.

Step 1. Load per column.
$$F = \frac{50 \times 10^3 \times 9.8}{4} = 1.225 \times 10^5\ \text{N}$$

Step 2. Area.
$$A = \pi(1^2 - 0.5^2) = 0.75\pi \approx 2.356\ \text{m}^2$$

Step 3. Strain.
$$\text{strain} = \frac{1.225 \times 10^5}{2.356 \times 2 \times 10^{11}} \approx 2.60 \times 10^{-7}$$

Answer. $2.60 \times 10^{-7}$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $K = \frac{Y\eta}{9\eta-3Y}\ \text{N m}^{-2}$

Concept. The three moduli are linked through Poisson's ratio $\sigma$:
$$Y=2\eta(1+\sigma),\qquad Y=3K(1-2\sigma)$$
Eliminate $\sigma$ to get one relation, then rearrange it for $K$.

Step 1. Express $\sigma$ from each. $1+\sigma=\dfrac{Y}{2\eta}$ and $1-2\sigma=\dfrac{Y}{3K}$.

Step 2. Eliminate $\sigma$. Double the first and add the second: $3=\dfrac{Y}{\eta}+\dfrac{Y}{3K}$, i.e.
$$\frac{9}{Y}=\frac{3}{\eta}+\frac1K$$

Step 3. Solve for $K$.
$$\frac1K=\frac9Y-\frac3\eta=\frac{9\eta-3Y}{Y\eta}\ \Rightarrow\ K=\frac{Y\eta}{9\eta-3Y}$$

Answer. Option C. Check option A against this: it has the wrong sign; the correct form is $Y=\dfrac{9K\eta}{3K+\eta}$.

Solution by Sreeraj P, M.Sc Physics

Answer: 2

Concept. $\Delta L\propto\dfrac{L}{d^2}$ at fixed force and material.

Step 1. Scale. $\dfrac{2}{4}=\dfrac12$: $0.02$ m $=2$ cm.

Answer. $2$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\frac{T_1\ell_2 - T_2\ell_1}{T_1 - T_2}$

Concept. Within the elastic limit the extension is proportional to the tension: $\ell = \ell_0 + cT$.

Step 1. Equations. $\ell_1 - \ell_0 = cT_1$ and $\ell_2 - \ell_0 = cT_2$.

Step 2. Eliminate $c$.
$$T_2(\ell_1 - \ell_0) = T_1(\ell_2 - \ell_0) \;\Rightarrow\; \ell_0(T_1 - T_2) = T_1\ell_2 - T_2\ell_1$$

Step 3. Solve.
$$\ell_0 = \frac{T_1\ell_2 - T_2\ell_1}{T_1 - T_2}$$

Answer. $\dfrac{T_1\ell_2 - T_2\ell_1}{T_1 - T_2}$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\frac{\ell_1T_2 - \ell_2T_1}{T_2 - T_1}$

Concept. Within Hooke's law the extension is proportional to the tension: $l=L+kT$, where $L$ is the natural length and $k$ a constant of the wire. Two readings give two equations in $L$ and $k$.

Step 1. Two equations. $l_1=L+kT_1$ and $l_2=L+kT_2$.

Step 2. Eliminate $k$. Multiply the first by $T_2$ and the second by $T_1$, then subtract:
$$l_1T_2-l_2T_1=L(T_2-T_1)$$

Step 3. Natural length. $L=\dfrac{l_1T_2-l_2T_1}{T_2-T_1}$.

Answer. Option B. Check: $T_1=0$ gives $L=l_1$, as it should.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\frac{\rho P}{K}$

Concept. Mass stays the same while volume shrinks, so density rises: $\rho V=$ constant gives $\dfrac{\Delta\rho}{\rho}=-\dfrac{\Delta V}{V}=\dfrac PK$.

Step 1. Volume strain. $\dfrac{|\Delta V|}{V}=\dfrac PK$.

Step 2. Density change. $\Delta\rho=\rho\dfrac PK$.

Answer. $\dfrac{\rho P}{K}$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\frac{T_2l_1-T_1l_2}{T_2-T_1}$

Concept. $l_1=L_0+\dfrac{T_1}{k}$, $l_2=L_0+\dfrac{T_2}{k}$. Eliminate $k$.

Step 1. Multiply. $T_2l_1-T_1l_2=L_0(T_2-T_1)$.

Step 2. Solve. $L_0=\dfrac{T_2l_1-T_1l_2}{T_2-T_1}$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $12.5$ cm

Concept. First find the tension in the wire from the Atwood machine. The wire must not break, so $\sigma_b\ge\dfrac{T}{\pi r^2}$, which gives the least radius.

Step 1. Tension in an Atwood machine.
$$T=\frac{2m_1m_2g}{m_1+m_2}=\frac{2\times3\times5\times10}{8}=37.5\ \text{N}$$

Step 2. Least area.
$$\pi r^2=\frac{T}{\sigma_b}=\frac{37.5}{\frac{24}{\pi}\times10^2}=\frac{37.5\pi}{2400}\ \Rightarrow\ r^2=\frac{37.5}{2400}=0.015625$$

Step 3. Radius. $r=0.125$ m $=12.5$ cm.

Answer. Option D. The tension is not $5g$ or $3g$; the blocks accelerate, so use the Atwood formula.

Solution by Sreeraj P, M.Sc Physics

Answer: 40

Concept. Each wire carries the weight of everything hanging below it. Find the largest load each wire can take ($\sigma_bA$), then find the pan mass that keeps both wires safe. The smaller limit wins.

Step 1. Breaking tensions.
$$W_1:\ 1.25\times10^9\times8\times10^{-7}=1000\ \text{N},\qquad W_2:\ 1.25\times10^9\times4\times10^{-7}=500\ \text{N}$$

Step 2. Lower wire $W_2$ holds the $10$ kg block and the pan load $M$:
$$(10+M)\times10\le500\ \Rightarrow\ M\le40\ \text{kg}$$

Step 3. Upper wire $W_1$ holds everything: $20+10+M$:
$$(30+M)\times10\le1000\ \Rightarrow\ M\le70\ \text{kg}$$

Step 4. Both must hold. The lower wire limits: $M_{\max}=40$ kg.

Answer. $40$.

Solution by Sreeraj P, M.Sc Physics

Answer: 4

Concept. In a space station there is no effective gravity, so the wire's tension alone supplies the centripetal force: $T=m\omega^2l$. The largest allowed tension is $\sigma_bA$.

Step 1. Largest tension. $A=10^{-2}$ cm$^2=10^{-6}$ m$^2$:
$$T_{\max}=4.8\times10^7\times10^{-6}=48\ \text{N}$$

Step 2. Angular speed.
$$m\omega^2l=48\ \Rightarrow\ \omega^2=\frac{48}{10\times0.3}=16\ \Rightarrow\ \omega=4\ \text{rad s}^{-1}$$

Answer. $4$.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $1.67$

Concept. Volume strain $\dfrac{\Delta V}{V}=\dfrac PB$. For a cube $V=L^3$, so for small changes $\dfrac{\Delta V}{V}=3\dfrac{\Delta L}{L}$.

Step 1. Volume strain. $\dfrac{4\times10^9}{8\times10^{10}}=0.05=5\%$.

Step 2. Length strain. $\dfrac{\Delta L}{L}=\dfrac{5\%}{3}=1.67\%$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $1.16\ \text{mm}$

Concept. Stress must not exceed the elastic limit: $\dfrac{F}{\pi d^2/4}\le\sigma_e$.

Step 1. Least area. $A=\dfrac{400}{379\times10^6}=1.055\times10^{-6}$ m$^2$.

Step 2. Diameter. $d=\sqrt{\dfrac{4A}{\pi}}=\sqrt{1.344\times10^{-6}}\approx1.16\times10^{-3}$ m.

Answer. $1.16$ mm (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $8.0\times10^6\ \text{N/m}^2$

Concept. Wires in series carry the same tension; with equal areas they have the same stress $\sigma$. Their extensions add.

Step 1. Total extension. $\sigma\left(\dfrac{L}{Y_s}+\dfrac{L}{Y_b}\right)=\sigma\left(\dfrac1{120\times10^9}+\dfrac1{60\times10^9}\right)=\dfrac{\sigma}{40\times10^9}$.

Step 2. Solve. $\dfrac{\sigma}{40\times10^9}=0.2\times10^{-3}\Rightarrow\sigma=8\times10^6$ N/m$^2$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{F}{A\alpha\Delta T}$

Concept. The compressive force must produce a compression equal to the free thermal expansion: $\dfrac{FL}{AY}=L\alpha\Delta T$.

Step 1. Solve. $Y=\dfrac{F}{A\alpha\Delta T}$. (A temperature change in kelvin equals the change in °C.)

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $10^6\ \text{N m}^{-2}$

Concept. The cord's stored energy becomes the stone's kinetic energy: $\dfrac{YAx^2}{2L}=\tfrac12mv^2$.

Step 1. Area. $A=\pi(3\times10^{-3})^2=2.83\times10^{-5}$ m$^2$.

Step 2. Solve for $Y$.
$$Y=\frac{mv^2L}{Ax^2}=\frac{0.02\times400\times0.42}{2.83\times10^{-5}\times0.04}\approx3\times10^6\ \text{N m}^{-2}$$

Step 3. Closest. $10^6$ N m$^{-2}$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $3.1\times10^6\ \text{N m}^{-2}$

Concept. Tensile stress $=\dfrac{\text{tension}}{\text{area}}$; a hanging load gives tension $mg$.

Step 1. Area. $A=\pi(2\times10^{-3})^2=4\pi\times10^{-6}$ m$^2$.

Step 2. Stress. $\sigma=\dfrac{4\times3.1\pi}{4\pi\times10^{-6}}=3.1\times10^6$ N m$^{-2}$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $1.7\ \text{mm}$

Concept. $\Delta l = \dfrac{FL}{\pi r^2Y}$. Equal extensions under the same load means $\dfrac{L}{r^2Y}$ is the same for both wires.

Step 1. Set up.
$$\frac{L_A}{R^2Y_A} = \frac{L_B}{r_B^2Y_B} \;\Rightarrow\; R^2 = r_B^2 \cdot \frac{L_A}{L_B} \cdot \frac{Y_B}{Y_A}$$

Step 2. Substitute. $r_B = 2$ mm:
$$R^2 = 4 \times \frac{2}{1.5} \times \frac47 = 3.05\ \text{mm}^2 \;\Rightarrow\; R \approx 1.7\ \text{mm}$$

Answer. $1.7$ mm (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $3.0$ mm

Concept. Extension is proportional to the tension. Immersing the load reduces the tension by the buoyant force, so the extension falls in the same ratio.

Step 1. Tension in the liquid. Buoyancy $=\dfrac{\rho_{\text{liq}}}{\rho_{\text{load}}}W=\dfrac28W$, so $T=W\left(1-\dfrac14\right)=\dfrac34W$.

Step 2. New extension. $\dfrac34\times4.0=3.0$ mm.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $3F/(\pi r^2YT)$

Concept. Length unchanged means the compression undoes the thermal expansion: $\dfrac{F}{AY}=\alpha T$. Then the volume coefficient is $\gamma=3\alpha$.

Step 1. Linear coefficient.
$$\frac{F}{\pi r^2Y}=\alpha T\ \Rightarrow\ \alpha=\frac{F}{\pi r^2YT}$$

Step 2. Volume coefficient. $\gamma=3\alpha=\dfrac{3F}{\pi r^2YT}$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $0.25\ \text{cm}$

Concept. For a cube of side $L$ in shear: $\eta=\dfrac{F/L^2}{\Delta x/L}$, so $\Delta x=\dfrac{F}{\eta L}$. Same material and same force: $\Delta x\propto\dfrac1L$.

Step 1. Ratio. Side doubles from $10$ cm to $20$ cm, so $\Delta x$ halves.

Step 2. New displacement. $\dfrac{0.5}{2}=0.25$ cm.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $9$

Concept. Stress in the leg is weight divided by the leg's cross-sectional area. Weight scales with volume ($L^3$) at fixed density; area scales with $L^2$.

Step 1. Scaling. Weight $\times9^3$, area $\times9^2$.

Step 2. Stress. $\dfrac{9^3}{9^2}=9$.

Answer. Stress increases by a factor $9$ (Option B). This is why very large animals need disproportionately thick legs.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $AY\alpha\,\Delta T$

Concept. Zero net change: compression from $F$ cancels thermal expansion, $\dfrac{Fl}{AY}=l\alpha\Delta T$.

Step 1. Solve. $F=AY\alpha\Delta T$, independent of the length.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $1\times10^5$ N

Concept. If a rod is stopped from expanding, the walls compress it by exactly the length it wanted to gain. The thermal strain $\alpha\Delta T$ becomes a mechanical strain, so the stress is $Y\alpha\Delta T$ and the force is $YA\alpha\Delta T$, independent of the length.

Step 1. Force.
$$F=YA\alpha\Delta T=2\times10^{11}\times40\times10^{-4}\times1.2\times10^{-5}\times10$$
$$=2\times10^{11}\times4\times10^{-3}\times1.2\times10^{-4}=9.6\times10^4\ \text{N}$$

Answer. About $1\times10^5$ N (Option D). The rail's length ($5$ m) does not enter.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{mg}{3Ka}$

Concept. The mass on the piston raises the liquid's pressure by $\dfrac{mg}{a}$ everywhere (Pascal's law), so the sphere is squeezed equally from all sides. Volume strain $=\dfrac{\Delta P}{K}$, and for a sphere $\dfrac{dV}{V}=3\dfrac{dr}{r}$.

Step 1. Extra pressure. $\Delta P=\dfrac{mg}{a}$.

Step 2. Volume strain. $\dfrac{dV}{V}=\dfrac{mg}{Ka}$.

Step 3. Radius strain. $\dfrac{dr}{r}=\dfrac13\cdot\dfrac{mg}{Ka}$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) **(A)** is true but **(R)** is false

Concept. When a coil spring is stretched, its wire hardly changes length; it is twisted.

Step 1. Assertion. Twisting is a shear deformation, so the stretching of the spring is governed by the shear modulus → true.

Step 2. Reason. Steel has much larger elastic moduli and tensile strength than copper, so the steel spring is the stronger one → false.

Answer. (A) is true but (R) is false (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: 25

Concept. Poisson's ratio links the sideways strain to the lengthwise strain: $\sigma=\dfrac{\text{lateral strain}}{\text{longitudinal strain}}$. The stored energy uses the longitudinal strain: $u=\tfrac12Y\varepsilon^2$.

Step 1. Longitudinal strain. $\varepsilon=\dfrac{10^{-3}}{0.2}=5\times10^{-3}$.

Step 2. Energy density.
$$u=\frac12\times2\times10^{11}\times(5\times10^{-3})^2=10^{11}\times2.5\times10^{-5}=25\times10^5\ \text{J m}^{-3}$$

Answer. $25$.

Solution by Sreeraj P, M.Sc Physics

Answer: 1

Concept. Hooke's law: $T = k(l - l_0)$. Two readings fix $l_0$.

Step 1. Equations.
$$5 = k(1.4 - l_0),\qquad 7 = k(1.56 - l_0)$$

Step 2. Divide and solve.
$$5(1.56 - l_0) = 7(1.4 - l_0) \;\Rightarrow\; 7.8 - 5l_0 = 9.8 - 7l_0 \;\Rightarrow\; l_0 = 1\ \text{m}$$

Answer. $1$.

Solution by Sreeraj P, M.Sc Physics

Answer: 1

Concept. $\eta=\dfrac{F}{A\theta}$, with $A$ the face the force acts on. Here the force acts on the narrow faces, of area $l\times d$.

Step 1. Areas. $A_2=ld_2=2A_1$.

Step 2. Ratio. $\dfrac{\eta_2}{\eta_1}=\dfrac{A_1\theta_1}{A_2\theta_2}=\dfrac{1}{2}\times\dfrac12=\dfrac14$.

Step 3. Value. $\eta_2=1\times10^9$ N/m$^2$.

Answer. $x=1$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{1}{160\pi}$

Concept. For a small displacement the angular displacement equals the shear strain: $\theta = \dfrac{F}{AG}$, with the force spread over the top face $A = \pi r^2$.

Step 1. Area. $A = \pi(0.04)^2 = 16\pi \times 10^{-4}\ \text{m}^2$.

Step 2. Angle.
$$\theta = \frac{10^5}{16\pi \times 10^{-4} \times 10^{10}} = \frac{1}{160\pi}\ \text{rad}$$

Answer. $\dfrac{1}{160\pi}$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1 : 12$

Concept. $\Delta L\propto\dfrac{L}{d^2}$.

Step 1. Ratio. $\dfrac{\Delta L_A}{\Delta L_B}=\dfrac{1/3}{2^2}=\dfrac1{12}$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $5$

Concept. $Y=\dfrac{FL}{A\,\Delta L}$.

Step 1. Substitute.
$$Y=\frac{50\times3\pi\times3}{\pi(3\times10^{-3})^2\times10^{-4}}=\frac{450}{9\times10^{-10}}=5\times10^{11}\ \text{N m}^{-2}$$

Answer. $P=5$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: 4

Concept. $\dfrac{\Delta V}{V}=\dfrac{\Delta P}{K}$, so $V=\dfrac{K\,\Delta V}{\Delta P}$.

Step 1. Pressure rise. $5-1=4$ atm $=4\times10^5$ Pa.

Step 2. Volume. $V=\dfrac{2\times10^9\times0.8\times10^{-6}}{4\times10^5}=4\times10^{-3}$ m$^3=4$ L.

Answer. $4$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $1 : 8$

Concept. The slope of a stress–strain line is Young's modulus $Y$. Extension $\Delta L = \dfrac{FL}{AY}$.

Step 1. Young's moduli from the graph.
$$Y_P = \frac{4 \times 10^8}{2 \times 10^{-3}} = 2 \times 10^{11}\ \text{N/m}^2,\qquad Y_Q = \frac{2 \times 10^8}{4 \times 10^{-3}} = 5 \times 10^{10}\ \text{N/m}^2$$
So $Y_P = 4Y_Q$.

Step 2. Ratio of extensions. Same $F$ and $L$:
$$\frac{\Delta L_P}{\Delta L_Q} = \frac{A_QY_Q}{A_PY_P} = \frac{1}{2 \times 4} = \frac18$$

Answer. $1 : 8$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $9 : 11$

Concept. The tension varies along the wire because each part also supports the wire below it. The extension of a small element $dx$ is $\dfrac{T\,dx}{AY}$, so the extension of a section is $\dfrac{1}{AY}\int T\,dx$.

Step 1. Tension. At height $x$ above the lower end:
$$T(x) = 2mg + \frac{mg}{L}x$$

Step 2. Lower half ($0$ to $\frac L2$).
$$\int_0^{L/2}T\,dx = 2mg\cdot\frac L2 + \frac{mg}{L}\cdot\frac{L^2}{8} = \frac98mgL$$

Step 3. Upper half ($\frac L2$ to $L$).
$$\int_{L/2}^{L}T\,dx = 2mg\cdot\frac L2 + \frac{mg}{2L}\left(L^2 - \frac{L^2}{4}\right) = \frac{11}{8}mgL$$

Step 4. Ratio. $\dfrac98 : \dfrac{11}{8} = 9 : 11$.

Answer. $9 : 11$ (Option D). Without the block the ratio would be $1 : 3$; with a block of mass $m$ it would be $5 : 7$.

Solution by Sreeraj P, M.Sc Physics

Answer: 90

Concept. Bulk modulus $B = \dfrac{\Delta P}{|\Delta V/V|}$. For a sphere $V \propto R^3$, so $\dfrac{\Delta V}{V} = 3\dfrac{\Delta R}{R}$. The extra pressure at depth $h$ is $\rho gh$.

Step 1. Volume strain.
$$\frac{\Delta V}{V} = 3 \times 0.02\% = 6 \times 10^{-4}$$

Step 2. Pressure needed.
$$\Delta P = B\frac{\Delta V}{V} = 1.5 \times 10^9 \times 6 \times 10^{-4} = 9 \times 10^5\ \text{Pa}$$

Step 3. Depth.
$$h = \frac{\Delta P}{\rho g} = \frac{9 \times 10^5}{10^3 \times 10} = 90\ \text{m}$$

Answer. $90$.

Solution by Sreeraj P, M.Sc Physics