JEE Physics Mechanical Properties Of Solids MCQs Set 07

Answer: (c) \( \frac{F\cos^2\theta}{A},\ \frac{F\sin2\theta}{2A};\ \theta=0^\circ,\ \theta=45^\circ \)

Explanation:
1. Area of the inclined plane is \( \frac{A}{\cos\theta} \). Normal component of force \( F\cos\theta \), tangential component \( F\sin\theta \).
2. Tensile stress \( =\frac{F\cos\theta}{A/\cos\theta}=\frac{F\cos^2\theta}{A} \); shear stress \( =\frac{F\sin\theta\cos\theta}{A}=\frac{F\sin2\theta}{2A} \).
3. The first is maximum at \( \theta=0 \), the second at \( \theta=45^\circ \).

Teacher's Note:
1. Resolve the force into components normal and tangential to the plane, and divide by the larger area of the slanted plane.
2. Maximum shear stress is F/(2A) at 45 degrees.

Answer: (a) \( 2.4\times10^{-5} \) J

Explanation:
1. Force \( F=5\times9.8=49 \) N.
2. \( U=\frac{F^2L}{2AY}=\frac{49^2\times0.2}{2\times10^{-4}\times10^{11}}=2.4\times10^{-5} \) J.

Teacher's Note:
1. Elastic energy can be found directly as F squared L over 2AY.
2. Convert kg-weight to newtons with g = 9.8.

Answer: (c) \( \frac{L_a}{L_a-L_w} \)

Explanation:
1. Extension is proportional to the pull: \( L_a\propto W \) and \( L_w\propto W-U \) with U the buoyant force.
2. \( \frac{L_a-L_w}{L_a}=\frac{U}{W}=\frac{\rho_w}{\rho} \).
3. Relative density \( \frac{\rho}{\rho_w}=\frac{L_a}{L_a-L_w} \).

Teacher's Note:
1. Buoyancy reduces the tension in the wire by the weight of displaced water.
2. The ratio of upthrust to weight is the reciprocal of the relative density.

Answer: (d) \( \frac{Y_1+Y_2}{2} \)

Explanation:
1. Equal extension means equal strain; the forces add: \( F=(Y_1+Y_2)A\varepsilon \).
2. The combination has area 2A, so \( Y_{eq}=\frac{F/2A}{\varepsilon}=\frac{Y_1+Y_2}{2} \).

Teacher's Note:
1. Compare with one equivalent wire of the same length and the total area 2A.
2. Parallel wires share the load; their stiffnesses add.

Answer: (c) 25 V

Explanation:
1. Energy \( \tfrac12kx^2\propto x^2 \).
2. Ratio \( (10/2)^2=25 \).

Teacher's Note:
1. Elastic energy varies as the square of the extension.
2. Five times the extension means 25 times the energy.

Answer: (a) 4F

Explanation:
1. Breaking stress is a constant of the material, so breaking force \( \propto A\propto R^2 \).
2. Doubling R gives 4F.

Teacher's Note:
1. Breaking force is stress times area.
2. The length plays no role.

Answer: (d) the volume of water decreases by 46.4 millionths of its original volume for each atmosphere of pressure increase

Explanation:
1. Compressibility is \( \frac1K=\frac{\Delta V/V}{\Delta P} \).
2. So \( \frac{\Delta V}{V}=46.4\times10^{-6} \) for \( \Delta P=1 \) atm.

Teacher's Note:
1. Compressibility is the fractional volume change per unit pressure.
2. The bulk modulus is its reciprocal, about 21,500 atm.

Answer: (a) \( 2\times10^9 \) N/m²

Explanation:
1. Pressure increase \( \rho gh=10^3\times10\times200=2\times10^6 \) Pa.
2. \( K=\frac{2\times10^6}{10^{-3}}=2\times10^9 \) N/m².

Teacher's Note:
1. Use only the excess pressure due to the water column.
2. 0.1 percent equals 10^-3.

Answer: (c) 40 cm

Explanation:
1. Maximum stress occurs in the 5 cm² part: \( F=5\sigma \) (areas in cm²). Stepped bar: \( U=\frac{F^2}{2Y}\left(\frac{x}{10}+\frac{1-x}{5}\right) \).
2. Uniform bar at the same maximum stress carries \( F'=10\sigma \): \( U'=\frac{(10\sigma)^2}{2Y}\cdot\frac{1}{10} \).
3. \( \frac{U}{U'}=\frac{25(0.2-0.1x)}{10}=0.5-0.25x=0.4\Rightarrow x=0.4 \) m.

Teacher's Note:
1. Strain energy of a bar in series pieces is the sum of F squared L over 2AY.
2. The thinner part fixes the maximum stress for the given load.

Answer: (a) 100 rad/s

Explanation:
1. \( k=\frac{4\times10^{-4}\times2\times10^{10}}{4}=2\times10^6 \) N/m.
2. \( \omega=\sqrt{\frac km}=\sqrt{\frac{2\times10^6}{200}}=100 \) rad/s.

Teacher's Note:
1. Treat the rod as a spring.
2. Use omega = sqrt(k/m).

Answer: (b) 1 cm

Explanation:
1. Energy conservation: \( mg(h+x)=\tfrac12kx^2 \): \( 100(0.99+x)=10^6x^2 \).
2. \( 10^6x^2-100x-99=0\Rightarrow x=0.01 \) m = 1 cm.

Teacher's Note:
1. The mass loses potential energy over both the free fall and the extension.
2. The quadratic has the positive root 0.01 m.

Answer: (d) \( \frac{100}{3} \) rad/s

Explanation:
1. Stiffnesses: \( k_1=\frac{2\times10^{-4}\times2\times10^{10}}{4}=10^6 \) N/m, \( k_2=2\times10^6 \) N/m.
2. Series: \( k=\frac{k_1k_2}{k_1+k_2}=\frac23\times10^6 \) N/m.
3. \( \omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{2\times10^6}{3\times600}}=\frac{100}{3} \) rad/s.

Teacher's Note:
1. Springs in series add flexibilities.
2. Calculate each rod's constant from AY/L.

Answer: (b) \( \frac{YA\delta^2}{2L} \) and \( \frac{YA\delta^2}{2L} \)

Explanation:
1. Since F is only slightly above the elastic force there is no acceleration and no heat is produced.
2. Work \( =\int_0^\delta\frac{YA}{L}x\,dx=\frac{YA\delta^2}{2L} \), all stored as elastic energy.

Teacher's Note:
1. For a quasi-static process all the work becomes stored energy.
2. The force-extension graph is a straight line through the origin.

Answer: (c) \( \tfrac12Mg\delta \) and \( \tfrac12Mg\delta \)

Explanation:
1. Final equilibrium: \( k\delta=Mg \), so the stored energy is \( \tfrac12k\delta^2=\tfrac12Mg\delta \).
2. Loss of gravitational potential energy is \( Mg\delta \); the rest, \( \tfrac12Mg\delta \), is dissipated as heat.

Teacher's Note:
1. Energy conservation: the bob oscillates and finally comes to rest.
2. Half the lost potential energy is stored and half is dissipated.

Answer: (a) 2.00

Explanation:
1. Same tension in both. \( \Delta L=\frac{FL}{\pi r^2Y}\propto\frac{L}{r^2} \).
2. \( \frac{\Delta L_{thin}}{\Delta L_{thick}}=\frac{L/R^2}{2L/4R^2}=2 \).

Teacher's Note:
1. Equal force, equal material: compare L divided by radius squared.
2. The thin wire stretches more.

Answer: (d) 9F

Explanation:
1. Volume equal: wire 2 has length \( L/3 \) if wire 1 has L.
2. Stiffness \( k=\frac{YA}{L} \): \( k_2=\frac{Y\cdot3A}{L/3}=9k_1 \).
3. Same extension needs \( 9F \).

Teacher's Note:
1. Equal volume means a threefold area comes with one-third of the length.
2. Force for a given extension is proportional to stiffness.

Answer: (a) \( 2.2\times10^8 \) Pa

Explanation:
1. Thermal strain \( =\alpha\Delta T=1.1\times10^{-3} \).
2. \( \text{Stress}=Y\alpha\Delta T=2\times10^{11}\times1.1\times10^{-3}=2.2\times10^8 \) Pa.

Teacher's Note:
1. The applied pressure must cancel the thermal expansion strain.
2. The length of the wire does not appear.

Answer: (c) \( \frac{mg}{3Ka} \)

Explanation:
1. Extra pressure \( \Delta P=\frac{mg}{a} \) is transmitted equally to the sphere.
2. Volume strain \( \frac{\Delta V}{V}=\frac{\Delta P}{K}=\frac{mg}{Ka} \) and \( \frac{\Delta V}{V}=3\frac{\Delta r}{r} \).
3. So \( \frac{\Delta r}{r}=\frac{mg}{3Ka} \).

Teacher's Note:
1. For a sphere V is proportional to r cubed, so the volume strain is three times the radial strain.
2. Pressure in a liquid is transmitted undiminished.

Answer: (a) \( 3.77\times10^4 \) N

Explanation:
1. Shear force per rivet \( =\sigma\pi r^2=1.5\times10^9\times\pi\times(2\times10^{-3})^2=1.885\times10^4 \) N.
2. Two rivets: \( 3.77\times10^4 \) N.

Teacher's Note:
1. Each rivet is sheared across its circular cross-section.
2. Multiply by the number of rivets.

Answer: (c) 1.33 cm

Explanation:
1. Maximum stress is in the thinner part: \( F=50\times1300=65000 \) N.
2. Strain energy \( U=\frac{F^2}{2Y}\left(\frac{1000}{1300}+\frac{1000}{2000}\right)\approx1.34\times10^4 \) N mm; extension \( x=\frac{F}{Y}(1.269)=0.41 \) mm.
3. \( mg(h+x)=U\Rightarrow h=\frac{13406}{980}-0.41\approx13.3 \) mm = 1.33 cm.

Teacher's Note:
1. Energy released by the falling load, including the extra extension, goes into strain energy.
2. The thinner section is the one that reaches 50 N/mm² first.

Answer: (a) \( 10^{-3} \) J

Explanation:
1. Force \( F=40 \) N and the stretch needed \( \delta=\alpha\Delta T\,L=10^{-5}\times10\times0.5=5\times10^{-5} \) m.
2. \( U=\tfrac12F\delta=\tfrac12\times40\times5\times10^{-5}=10^{-3} \) J.

Teacher's Note:
1. Elastic energy is half of force times extension.
2. The extension equals the free thermal contraction.