Question

Steel wire of length '$L$' at ${40}^{o}C$ is suspended from the ceiling and then a mass '$m$' is hung from its free end. The wire is cooled down from ${40}^{o}C$ to ${30}^{o}C$ to regain its original length ${}^{\prime }{L}^{\prime }$. The coefficient of linear thermal expansion of the steel is ${10}^{-5}{/}^{o}C$, Young's modulus of steel is ${10}^{11}N/m^2$ and radius of the wire is 1 mm. Assume that $L>>$ diameter of the wire. Then the value of '$m$' in kg is nearly

• A. $3$
• B. $2$
• C. $9$
• D. $5$

Answer 1

The correct option is A $3$

We know that

$E=\frac{\frac{F}{A}}{\frac{△L}{L}}$⇒$\frac{\Delta L}{L}=\frac{F}{AE}$......(i)

Also

$\frac{\Delta L}{L}\propto \alpha \Delta T$.....(ii)

from (i) and (ii)

$\frac{F}{AE}=\alpha \Delta T$

⇒$mg=\left(\alpha \Delta T\right)AE$

⇒$m=g\alpha \Delta TAE$

=$\frac{{10}^{-5}\times 10\times \pi \times {10}^{-6}\times {10}^{11}}{10}$

=$\pi \approx 3$

So correct answer is a