JEE Physics Mechanical Properties Of Solids MCQs Set 13

Answer: (a) 1.16 mm

Explanation:
1. \( A=\dfrac F{\sigma}=\dfrac{400}{379\times10^6}=1.055\times10^{-6} \) m².
2. \( d=\sqrt{\dfrac{4A}\pi}=1.16\times10^{-3} \) m.

Teacher's Note:
1. Write Y = (F/A)/(dl/l) and apply it to each wire or rod, keeping track of which quantities are equal (force, length or extension).
2. Convert all lengths to metres and areas to square metres before substituting.

Answer: (b) 1.7 mm

Explanation:
1. \( \Delta l=\dfrac{FL}{\pi r^2Y} \) equal for both: \( \dfrac{L_A}{R^2Y_A}=\dfrac{L_B}{r_B^2Y_B} \).
2. \( R^2=\dfrac{L_AY_Br_B^2}{L_BY_A}=\dfrac{2\times4\times4}{1.5\times7}=3.05\Rightarrow R=1.7 \) mm.

Teacher's Note:
1. Write Y = (F/A)/(dl/l) and apply it to each wire or rod, keeping track of which quantities are equal (force, length or extension).
2. Convert all lengths to metres and areas to square metres before substituting.

Answer: (c) \( 3.1\times10^6 \) N m\(^{-2}\)

Explanation:
1. Stress \( =\dfrac{mg}{\pi r^2}=\dfrac{4\times3.1\pi}{\pi(2\times10^{-3})^2} \).
2. \( =\dfrac{12.4}{4\times10^{-6}}=3.1\times10^6 \) N m\(^{-2}\).

Teacher's Note:
1. Write Y = (F/A)/(dl/l) and apply it to each wire or rod, keeping track of which quantities are equal (force, length or extension).
2. Convert all lengths to metres and areas to square metres before substituting.

Answer: (d) \( 10^6 \) N m\(^{-2}\)

Explanation:
1. Stored elastic energy \( =\tfrac12mv^2=\tfrac12(0.02)(400)=4 \) J \( =\tfrac12\cdot\dfrac{YA}Lx^2 \).
2. \( Y=\dfrac{8L}{Ax^2}=\dfrac{8(0.42)}{\pi(3\times10^{-3})^2(0.2)^2}\approx3\times10^6\approx10^6 \) N m\(^{-2}\).

Teacher's Note:
1. Write Y = (F/A)/(dl/l) and apply it to each wire or rod, keeping track of which quantities are equal (force, length or extension).
2. Convert all lengths to metres and areas to square metres before substituting.

Answer: (a) 3.0 mm

Explanation:
1. Apparent weight in the liquid \( =Mg\left(1-\dfrac28\right)=\dfrac34Mg \).
2. Extension is proportional to the load: \( 4.0\times\dfrac34=3.0 \) mm.

Teacher's Note:
1. Write Y = (F/A)/(dl/l) and apply it to each wire or rod, keeping track of which quantities are equal (force, length or extension).
2. Convert all lengths to metres and areas to square metres before substituting.

Answer: (b) \( \dfrac{F}{A\alpha\,\Delta T} \)

Explanation:
1. Free thermal expansion \( \alpha L\Delta T \) must be cancelled by compression: \( \dfrac{FL}{AY}=\alpha L\Delta T \).
2. \( Y=\dfrac F{A\alpha\Delta T} \).

Teacher's Note:
1. Write Y = (F/A)/(dl/l) and apply it to each wire or rod, keeping track of which quantities are equal (force, length or extension).
2. Convert all lengths to metres and areas to square metres before substituting.

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

Explanation:
1. Extra pressure \( \Delta p=\dfrac{mg}a \); \( K=\dfrac{\Delta p}{|dV/V|}\Rightarrow\dfrac{dV}V=\dfrac{mg}{Ka} \).
2. \( V\propto r^3\Rightarrow\dfrac{dV}V=3\dfrac{dr}r \), so \( \dfrac{dr}r=\dfrac{mg}{3Ka} \).

Teacher's Note:
1. Write Y = (F/A)/(dl/l) and apply it to each wire or rod, keeping track of which quantities are equal (force, length or extension).
2. Convert all lengths to metres and areas to square metres before substituting.

Answer: (d) \( \left[\left(\dfrac{T_M}T\right)^2-1\right]\dfrac A{Mg} \)

Explanation:
1. The extra load stretches the wire by \( \Delta l=\dfrac{MgL}{AY} \) so the length becomes \( L+\Delta l \).
2. \( \left(\dfrac{T_M}T\right)^2=\dfrac{L+\Delta l}L=1+\dfrac{Mg}{AY} \).
3. \( \dfrac1Y=\left[\left(\dfrac{T_M}T\right)^2-1\right]\dfrac A{Mg} \).

Teacher's Note:
1. Write Y = (F/A)/(dl/l) and apply it to each wire or rod, keeping track of which quantities are equal (force, length or extension).
2. Convert all lengths to metres and areas to square metres before substituting.

Answer: (a) 2.00

Explanation:
1. Same force and Y: \( \Delta l=\dfrac{FL}{\pi r^2Y} \). Thin: \( \dfrac{FL}{\pi R^2Y} \). Thick: \( \dfrac{F(2L)}{\pi(2R)^2Y}=\dfrac{FL}{2\pi R^2Y} \).
2. Ratio \( =\dfrac{1}{1/2}=2 \).

Teacher's Note:
1. Write Y = (F/A)/(dl/l) and apply it to each wire or rod, keeping track of which quantities are equal (force, length or extension).
2. Convert all lengths to metres and areas to square metres before substituting.