At $40^{\circ} \mathrm{C}$, a brass wire of 1 mm diameter is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from $40^{\circ} \mathrm{C}$ to $20^{\circ} \mathrm{C}$ it regains its original length of 0.2 m . The value of M is close to : (Coefficient of linear expansion and Young's modulus of brass are $10^{-5} /{ }^{\circ} \mathrm{C}$ and $10^{11} \mathrm{N} / \mathrm{m}^{2}$, respectively; $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Elasticity & SHM

4 0^{\circ} C

, a brass wire of 1 mm diameter is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from

4 0^{\circ} C

2 0^{\circ} C

it regains its original length of 0.2 m . The value of M is close to :
(Coefficient of linear expansion and Young's modulus of brass are

1 0^{- 5} /^{\circ} C

and

1 0^{11} N / m^{2}

, respectively;

g = 10 ms^{- 2}

)

Answer: (D)

On cooling, the wire tends to contract by the thermal strain

α Δ T

, while the hanging mass stretches it by the elastic strain

\frac{M g}{A Y}

. It regains its original length when these balance:

\frac{M g}{A Y} = α Δ T \Rightarrow M g = A Y α Δ T

(length cancels). With diameter

1 mm

A = π (0.5 \times 1 0^{- 3})^{2} = 7.85 \times 1 0^{- 7} m^{2}

and

Δ T = 2 0^{\circ} C

M g = 7.85 \times 1 0^{- 7} \times 1 0^{11} \times 1 0^{- 5} \times 20 = 15.7 N

, so

M = 1.57 kg \approx 1.5 kg

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option