Question

A 15 kg mass fastened to the end of a steel wire of unstretched length 1.0 m is whirled in a vertical circle with an angular velocity of 2 rev$s^{- 1}$ at the bottom of the circle. The cross-section of the wire is$0.05cm^{2}$ . The elongation of the wire when the mass is at the lowest point of its path is

$(\text{Take, }g\ = 1\text{0 m }\text{s}^{- 2}\text{, }Y_{steel} = 2 \times 10^{11}\text{ N }\text{m}^{- 2})$

• A. 0.52 mm
• B. 1.52 mm
• C. 2.52 mm
• D. 3.52 mm

Answer 1

Answer: (C)

2.52 mm

Answer 2

: Here, m = 15 kg, r=L=1 m

$\upsilon = 2revs^{- 1},A = 0.05cm^{2} = 0.5 \times 10^{- 4}m^{2}$

When the mass is at the lowest point of the vertical circle, the stretching force is

$F = mg + mr\omega^{2} = mg + mr(2\pi\upsilon)^{2}$

$= 15 \times 10 + 15 \times (1) \times (2\pi \times 2)^{2} = 2516N$

As $Y = \frac{F/A}{\Delta L/L}$

$\therefore\Delta L = \frac{FL}{AY}$

$\Delta L = \frac{2516N \times 1m}{0.05 \times 10^{- 4}m^{2} \times 2 \times 10^{11}}$

$= 2.52 \times 10^{- 3}m = 2.52mm$