JEE Physics Mechanical Properties Of Solids MCQs Set 08

Answer: (b) (a) and (c)

Explanation:
1. Elasticity is judged by the modulus: the larger the modulus, the larger the force needed for a given strain.
2. Young's modulus of steel and glass is of the order of \( 10^{11}\ \text{N/m}^2 \) while that of rubber is only about \( 10^{6}\ \text{N/m}^2 \).
3. So glass and steel are both more elastic than rubber: statements (a) and (c).

Teacher's Note:
1. Do not confuse 'easily stretched' with 'more elastic'. A material that is hard to deform and returns to its shape is more elastic.
2. Remember steel as the standard example of a highly elastic material.

Answer: (c) the ivory ball bounces to a greater height

Explanation:
1. Ivory is highly elastic, so it regains its shape quickly and returns most of the kinetic energy it had at impact.
2. Clay is nearly plastic, so most of the energy goes into permanent deformation and heat.
3. Hence the ivory ball rises higher.

Teacher's Note:
1. Bounce height measures how much energy is stored elastically and returned.
2. Nearly plastic bodies such as putty do not rebound.

Answer: (a) perfectly plastic

Explanation:
1. A perfectly plastic body does not tend to recover its original shape after the deforming force is removed.
2. A permanent elongation is therefore the signature of plastic behaviour (loading beyond the elastic limit).

Teacher's Note:
1. Beyond the elastic limit the material shows permanent set.
2. A perfectly elastic body would recover completely.

Answer: (c) The shear modulus of a liquid is infinite.

Explanation:
1. A liquid cannot sustain shear, so its shear modulus is zero, not infinite. Statement (a) is wrong.
2. A perfectly rigid body shows no strain for any stress, so all its moduli are infinite: (b) and (d) are right.
3. Hooke's law states that stress/strain is constant within the elastic limit: (c) is right.

Teacher's Note:
1. Gases and liquids have zero shear modulus but a finite bulk modulus.
2. 'Not correct' questions need every option checked.

Answer: (a) independent of \( l \)

Explanation:
1. Thermal strain prevented by the supports: \( \Delta l/l=\alpha\,\Delta T \).
2. Tension \( F=YA\,\alpha\,\Delta T \).
3. This contains no \( l \), so the tension is independent of the length (but it does depend on \( A \)).

Teacher's Note:
1. Thermal stress \( Y\alpha\Delta T \) depends only on material and temperature change.
2. Force is stress times area, so it depends on the area.

Answer: (c) \( Y_1\alpha_1=Y_2\alpha_2 \)

Explanation:
1. The junction stays at the middle, so each rod is held at its original length.
2. Thermal stress in rod 1 is \( Y_1\alpha_1\Delta T \), in rod 2 it is \( Y_2\alpha_2\Delta T \).
3. The cross-sections are equal, so the tensions must be equal for equilibrium of the junction: \( Y_1\alpha_1=Y_2\alpha_2 \).

Teacher's Note:
1. The junction is in equilibrium only if the two forces on it are equal.
2. Same area means equal stresses.

Answer: (d) remain unchanged

Explanation:
1. Young's modulus is a property of the material, not of the dimensions of the wire.
2. Doubling the radius reduces the stress and the strain, but their ratio stays the same.

Teacher's Note:
1. Modulus values depend on the material and temperature only.
2. Resistance to stretching (force constant) does change with area.

Answer: (c) contract, but its final length will be greater than its initial length

Explanation:
1. Beyond the elastic limit some permanent set is produced.
2. On release the wire partially recovers, leaving a residual elongation, so its final length exceeds the original length.

Teacher's Note:
1. Permanent set is the leftover deformation beyond the elastic limit.
2. The unloading curve is roughly parallel to the initial straight part of the loading curve.

Answer: (c) \( YA/L \)

Explanation:
1. \( F=YA\dfrac{\Delta L}{L}=\left(\dfrac{YA}{L}\right)\Delta L \).
2. Comparing with \( F=k\,\Delta L \) gives \( k=YA/L \).

Teacher's Note:
1. A longer wire is softer, a thicker wire is stiffer.
2. Cutting a wire in half doubles its force constant.

Answer: (d) \( 4F \)

Explanation:
1. Breaking stress is a property of the material, so breaking force \( =\text{breaking stress}\times A \).
2. Area \( \propto R^2 \): doubling the radius multiplies the area by 4.
3. Required force \( =4F \).

Teacher's Note:
1. Breaking force scales with cross-sectional area, not with length.
2. Doubling the diameter quadruples the load a rope can carry.

Answer: (d) the extension of A is twice that of B

Explanation:
1. \( \Delta L=\dfrac{FL}{AY}\propto L \) for the same \( F,A,Y \).
2. So the longer wire A stretches twice as much.
3. Strain \( =\dfrac{F}{AY} \) is the same for both, so (c) and (d) are wrong.

Teacher's Note:
1. Extension depends on length; strain does not.
2. Same stress and same material give the same strain.

Answer: (d) \( Y\alpha^2/2 \)

Explanation:
1. Energy per unit volume \( =\tfrac12\times\text{stress}\times\text{strain} \).
2. Stress \( =Y\alpha \), strain \( =\alpha \).
3. So \( u=\tfrac12 Y\alpha^2 \).

Teacher's Note:
1. Equivalent forms: \( \tfrac12 Y\varepsilon^2=\tfrac12\sigma\varepsilon=\sigma^2/2Y \).
2. The factor 1/2 arises because the force grows from zero to its final value.

Answer: (d) 0.24%

Explanation:
1. Lateral strain in each of two directions \( =-\sigma\varepsilon \).
2. Volume strain \( =\varepsilon-2\sigma\varepsilon=\varepsilon(1-2\sigma) \).
3. \( =4\times10^{-3}\times(1-0.4)=2.4\times10^{-3} \), i.e. 0.24%.

Teacher's Note:
1. For small strains the volume strain is the sum of the three linear strains.
2. Incompressible materials have \( \sigma=0.5 \) and no volume change.

Answer: (b) For a large deformation of a material, the ratio stress/strain decreases.

Explanation:
1. For small deformations (below the proportional limit) stress/strain is the constant Young's modulus, so (a) is wrong.
2. Beyond the proportional limit the stress-strain curve bends, and the ratio stress/strain falls: (b) is correct.
3. In (c) the wires carry the same force and have the same area, so the stress is the same; their strains differ because the moduli differ. (c) is wrong.

Teacher's Note:
1. Hooke's law holds only up to the proportional limit.
2. Wires joined in series carry equal force, hence equal stress.

Answer: (a) (A) and (D) only

Explanation:
1. (A) is true: steel has the larger modulus.
2. (B) is false: stretching a coil spring twists the wire, so the shear modulus governs it.
3. (C) is false: a tuning fork's prong vibrates by bending, which depends on Young's modulus.
4. (D) is true: with no stress the atoms sit at their equilibrium separations.
5. Hence (A) and (D).

Teacher's Note:
1. Torsion and shear are linked to the modulus of rigidity.
2. Bending of a bar involves Young's modulus.

Answer: (b) (A), (C) and (D)

Explanation:
1. The mass descends by \( l \), so it loses \( Mgl \) of gravitational potential energy: (A) is true.
2. The wire stores \( \tfrac12 F l=\tfrac12 Mgl \): (C) is true and (B) is false.
3. The remaining \( \tfrac12 Mgl \) appears as heat (oscillations die down): (D) is true.

Teacher's Note:
1. The mass is let down without a final kinetic energy, so energy conservation needs the heat term.
2. If the load were applied gradually by hand, no heat would be produced.

Answer: (a) (1), (2) and (3)

Explanation:
1. Tension in B \( =F=mg/3 \); tension in A \( =mg+mg/3=4mg/3 \).
2. Stress in A \( =\dfrac{4mg}{3\pi r_A^2} \), in B \( =\dfrac{mg}{3\pi r_B^2} \).
3. Ratio \( \dfrac{\sigma_A}{\sigma_B}=\dfrac{4r_B^2}{r_A^2} \). It exceeds 1 when \( r_A<2r_B \) (so A breaks first for equal radii too) and equals 1 when \( r_A=2r_B \).
4. All three statements are correct.

Teacher's Note:
1. The upper wire carries the load of everything below it.
2. Equal stress means either wire can fail first.

Answer: (a) (1), (3) and (4)

Explanation:
1. Restoring force at extension \( x \): \( \dfrac{YA}{L}x \).
2. Work \( =\int_0^l \dfrac{YA}{L}x\,dx=\dfrac{YAl^2}{2L} \), so (1) is true and (2) false.
3. The stretch is quasi-static, so all this work is stored as elastic energy: (3) is true and no heat is produced: (4) is true.

Teacher's Note:
1. Work equals the area under the force-extension graph.
2. A quasi-static process has no kinetic energy and no dissipation.

Answer: (a) \( 12\times10^{7}\ \text{N/m}^2 \)

Explanation:
1. Strain \( =0.1\%=10^{-3} \).
2. Stress \( =Y\times\text{strain}=12\times10^{10}\times10^{-3}=12\times10^{7}\ \text{N/m}^2 \).

Teacher's Note:
1. One-tenth of one per cent means 0.1 / 100.
2. Stress = modulus times strain.

Answer: (a) \( \dfrac{l_1T_2-l_2T_1}{T_2-T_1} \)

Explanation:
1. Let the natural length be \( L \) and \( k=YA/L \). Then \( T_1=k(l_1-L) \), \( T_2=k(l_2-L) \).
2. Dividing: \( \dfrac{T_1}{T_2}=\dfrac{l_1-L}{l_2-L} \).
3. Solving: \( L=\dfrac{l_1T_2-l_2T_1}{T_2-T_1} \).

Teacher's Note:
1. Eliminate the unknown force constant by taking a ratio.
2. Check: if \( T_1=0 \) then \( L=l_1 \).

Answer: (b) 3600

Explanation:
1. Strain \( =6\times10^{-4} \).
2. Energy density \( u=\tfrac12 Y\varepsilon^2=\tfrac12(2\times10^{10})(6\times10^{-4})^2 \).
3. \( =\tfrac12(2\times10^{10})(3.6\times10^{-7})=3600\ \text{J/m}^3 \).

Teacher's Note:
1. Do not forget the factor 1/2 in \( \tfrac12Y\varepsilon^2 \).
2. Square the percentage after converting it to a fraction.

Answer: (b) \( W \)

Explanation:
1. The breaking load depends on cross-section and material, not on length.
2. Each half has the same area and material as the full cable, so it also supports \( W \).

Teacher's Note:
1. Cutting along the length does not change the breaking stress or the area.
2. Cutting across the area would halve the load.