Question

A block of mass 2 Kg is attached to one end of a wire of cross sectional area $1m{m}^2$ and is whirled in a vertical circle of radius 40 cm. The speed of the block at the bottom of the circle $5m{s}^{-1}$. Find the elongation of the wire when the block is at the bottom $(Y=2\times {10}^{11}N{M}^2)$

• A. 0.2892 mm
• B. 0.054 mm
• C. 0.1446 mm
• D. 0.0017 mm

Answer 1

The correct option is A 0.2892 mm

Given,

Mass of the block, $m=2kg$.

Length of wire, $L=0.4m$

Velocity of block at bottom $v=5m{s}^{-1}$

Young modulus of wire $Y=2\times {10}^{11}N{m}^{-2}$

When block at bottom, the tension$\left(T\right)$ in the wire is due to centrifugal force and weight of block.

$T=mg+\frac{m{v}^2}{L}=2\times 9.8+\frac{2\times 5^2}{0.4}=144.6N$

Stress = young modulus x strain

$\frac{T}{A}=Y\times \frac{\Delta L}{L}$

$\Delta L=\frac{TL}{YA}=\frac{144.6\times 0.4}{2\times {10}^{11}\times {10}^{-6}}=2.892\times {10}^{-4}m$

$\Delta L=0.2892mm$

Extension in wire is $0.2892mm$