Q. In a simple pendulum, the breaking strength of the string is double the weight of the bob. The bob is released from rest when the string is horizontal. The string breaks when it makes an angle $\theta$ with the vertical :

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A

$\theta=\cos ^{-1}(1 / 3)$

B

$\theta=60^{\circ}$

C

$\theta=\cos ^{-1}(2 / 3)$

D

$\theta=0$

Solution:

Let speed of bob at $B$ is $v$
Tension at $\theta$ angle is

$T =m g \cos \theta+\frac{m v^{2}}{l} $
$[$ here $v=\sqrt{2 g l \cos \theta}]$
$2 m g =m g \cos \theta+\frac{m v^{2}}{l} $
$m v^{2} =2 m g l-m g l \cos \theta $
$3 \cos \theta =2$
$\theta =\cos ^{-1}\left(\frac{2}{3}\right)$

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Q. In a simple pendulum, the breaking strength of the string is double the weight of the bob. The bob is released from rest when the string is horizontal. The string breaks when it makes an angle $\theta$ with the vertical :

$\theta=\cos ^{-1}(1 / 3)$

$\theta=60^{\circ}$

$\theta=\cos ^{-1}(2 / 3)$

$\theta=0$

Let speed of bob at $B$ is $v$ Tension at $\theta$ angle is $T =m g \cos \theta+\frac{m v^{2}}{l} $ $[$ here $v=\sqrt{2 g l \cos \theta}]$ $2 m g =m g \cos \theta+\frac{m v^{2}}{l} $ $m v^{2} =2 m g l-m g l \cos \theta $ $3 \cos \theta =2$ $\theta =\cos ^{-1}\left(\frac{2}{3}\right)$

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1. Identify the false statement from the following

2. The machine gun fires $240 \, $ bullets per minute. If the mass of each bullet is $10 \, g$ and the velocity of the bullets is $600 \, m \, s^{- 1}$ ,the power (in $kW$ )of the gun is

4. A particle of mass $m$ is moving in a circular path of constant radius $r$ such that its centripetal acceleration $a_{c}$ is varying with time $t$ as $a_{c}=k^{2} \, rt^{2}$ , where $k$ is a constant. The power delivered to the particle by the force acting on it is

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1. A car starts from rest and accelerates at $5\, m / s ^{2}$ At $t=4\, s$, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at $t =6\, s$ ? (Take $\left. g =10\, m / s ^{2}\right)$

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Latest Updates

Let speed of bob at $B$ is $v$
Tension at $\theta$ angle is

$T =m g \cos \theta+\frac{m v^{2}}{l} $
$[$ here $v=\sqrt{2 g l \cos \theta}]$
$2 m g =m g \cos \theta+\frac{m v^{2}}{l} $
$m v^{2} =2 m g l-m g l \cos \theta $
$3 \cos \theta =2$
$\theta =\cos ^{-1}\left(\frac{2}{3}\right)$