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A bob of mass M is suspended by a massless string of length L. The horizontal velocity v at position A is just sufficient to make it reach the point B. The angle $θ$ at which the speed of the bob is half of that at A, satisfies

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$θ = \frac{π}{4}$

$\frac{3 π}{4} < θ < π$

$\frac{π}{2} < θ < \frac{3 π}{4}$

$\frac{π}{4} < θ < \frac{π}{2}$

answer is C.

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Detailed Solution

When the string rotates by $90 °$ , the speed of ball is

$v^{2} = v^{2} - 2 gL = 3 gL$

$v^{'} = \sqrt{3 gL}$

$v_{1}^{2} = v^{2} - 2 gL (1 + cosα)$

$\frac{v^{2}}{4} = v^{2} - 2 gL (1 + cosα)$

$2 gL (1 + cosα) = \frac{3 v^{2}}{4}$

$= \frac{3}{4} \times 5 gL$

$1 + cosα = \frac{15}{8}$

$\Rightarrow cosα = \frac{7}{8}$

$α = \cos^{- 1} (7 / 8), α < 45 ° = (\frac{π}{4})$

$\frac{π}{2} < θ$

SOLVED] A bob of mass m is suspended by a massless string of le - Self Study 365

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