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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A pendulum is released from $\theta_0=60 ^{\circ}$. The rate of change of speed of bob at $\theta=30 ^{\circ}$ is $(g=10 m/s^2)$

\[(a)\;5 \sqrt 3\; m/s^2 \quad (b)\;5\; m/s^2 \quad (c)\;10\; m/s^2 \quad (d)\;2.5\; m/s^2\]

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Answer: 5 m/s$^2$
Rate change of speed, $\large\frac{dv}{dt}$$=$ tangential acceleration $=\large\frac{\text{tangential force}}{\text{mass}}$
When $\theta=30^{\circ}$, tangential force is $mg\; \sin \theta$
Therefore $a=\large\frac{mg \sin 30}{m}$
$\qquad\qquad=10 \sin 30$
$\qquad\qquad=10 \times \frac{1}{2}$
$\qquad\qquad=5 m/s^2$
answered Jul 2, 2013 by meena.p
edited Aug 19, 2014 by balaji.thirumalai

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