A block of mass 2 kg is attached to one end of a wire of cross-sectional area 1 mm² and is whirled in a vertical circle of radius 40 cm. The speed of the block at the bottom of the circle is 5 ms^-1. Elongation of the wire when the block is at the top of the circle in is $x \times 10^{- 2}$ mm. The value of $x$ is ______ (Y = 2 × 10¹¹Nm^-2)

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Detailed Solution

Velocity at the topmost point ( $v$ ) in terms of velocity at bottom most point ( $u$ ) is given by
$v^{2} = u^{2} - 4 g R = 5^{2} - 4 \times 10 \times 0.4 = 9 \Rightarrow v = 3 m / s$
Tension at the top most point is then given by
$T + m g = \frac{m v^{2}}{R} \Rightarrow T = \frac{m v^{2}}{R} - m g = \frac{2 \times 3^{2}}{0.4} - 2 \times 10 = 25 N$
Elongation in the wire is then given by
$Δ l = \frac{T l}{A Y} = \frac{25 \times 0.4}{1 \times 10^{- 6} \times 2 \times 10^{11}} = 5 \times 10^{- 5} m = 5 \times 10^{- 2} mm$