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A block of mass 4kg is attached to one end of a wire of cross sectional area $1 m m^{2}$ and is whirled in a vertical circle of radius 90 cm. The speed of the block at the bottom of the circle $6 m s^{- 1}$ . Find the elongation of the wire when the block is at the bottom. $(Y = 2 \times 10^{11} N m^{- 2})$ $(g = 10 m s^{- 2})$

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0.5 mm

0.7 mm

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0.9 mm

answer is B.

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Detailed Solution

$T = F = \frac{m v^{2}}{r} + m g$

$= \frac{4 \times 36}{90 \times 10^{- 2}} + 4 \times 10$

$= 160 + 40 = 200 N$ and

$e = \frac{F L}{A y}$

$= \frac{200 \times 90 \times 10^{- 2}}{10^{- 6} \times 2 \times 10^{11}}$

$= 9 \times 10^{- 4} m = 0.9 \times 10^{- 3} m = 0.9 m m$

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