A ball weight 1.0 kg is tied to a string 15cm long. Initially the ball is held in position such that the string is horizontal. The ball is now released .A nail N is situated vertically below the support at a distance L. The minimum value of L such that the string will be wound round the nail is

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15 cm

9 cm

none of these

4 cm

answer is C.

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Detailed Solution

Velocity of bob at C should be $\sqrt{5 g (r a d i u s o f c i r c l e)}$

here length of string =R, distance of nail below support =L , required minimum velocity to have vertical circle = $\sqrt{5 g (R - L)}$

velocity of ball at bottom most point of circle is $\sqrt{2 g R}$

equating the above two equations
$\sqrt{5 g (R - L) =} \sqrt{2 g R}$

squaring on both sides
5(R-L) = 2R
3R = 5L
L= $\frac{3}{5}$ R

distance of nail below support is

L= $\frac{3}{5} \times$ 15 = 9cm

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