A steel wire of length L at 40^{circ}C is sus | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The elongation produced by the mass $m$ is $\Delta \ell = \frac{mgL}{AY}$.The contraction due to cooling is $\Delta \ell = L\alpha \Delta T$.Since

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The elongation produced by the mass $m$ is $\Delta \ell = \frac{mgL}{AY}$.The contraction due to cooling is $\Delta \ell = L\alpha \Delta T$.Since the wire regains its original length,the elongation due to the mass must equal the contraction due to cooling:$\frac{mgL}{AY} = L\alpha \Delta T$Solving for $m$:$m = \frac{YA\alpha \Delta T}{g}$Given:$Y = 10^{11} N/m^2$$A = \pi r^2 = \pi (10^{-3} m)^2 = \pi 	imes 10^{-6} m^2$$\alpha = 10^{-5} /^{\circ}C$$\Delta T = 40^{\circ}C - 30^{\circ}C = 10^{\circ}C$$g = 10 m/s^2$Substituting the values:$m = \frac{10^{11} 	imes (\pi 	imes 10^{-6}) 	imes 10^{-5} 	imes 10}{10}$$m = \pi 	imes 10^{11-6-5+1-1} = \pi \approx 3.14 kg$Thus,the value of $m$ is nearly $3 kg$.

Similar Questions

The mean distance between the atoms of iron is $3 \times 10^{-10} \ m$ and the interatomic force constant for iron is $7 \ N/m$. The Young's modulus of elasticity for iron is:

When a force of $4 \, N$ is applied to a wire,its length is $a \, m$. When a force of $5 \, N$ is applied,its length is $b \, m$. What will be its length when a force of $9 \, N$ is applied?

In steel,the Young's modulus and the strain at the breaking point are $2 \times 10^{11} \, N/m^2$ and $0.15$ respectively. The stress at the breaking point for steel is therefore:

Vedclass Test Series

Exam Paper Generator

Online Exam Module