(B) By the law of conservation of energy,the potential energy lost equals the kinetic energy gained as the pendulum moves from the $90^{\circ}$ positi

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(B) By the law of conservation of energy,the potential energy lost equals the kinetic energy gained as the pendulum moves from the $90^{\circ}$ position to the mean position:$mgl = \frac{1}{2} mv^2$$v^2 = 2gl$At the mean position,the forces acting on the pendulum are the tension $T$ (upwards) and the weight $mg$ (downwards). The net force provides the necessary centripetal force:$T - mg = \frac{mv^2}{L}$Substituting $v^2 = 2gl$ into the equation:$T - mg = \frac{m(2gl)}{L}$$T - mg = 2mg$$T = 3mg$Thus,the maximum tension in the string is $3mg$.

Class 11 Physics Work, Energy, Power and Collision Vertical Circular Motion

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The string of a pendulum of length $L$ is displaced through $90^{\circ}$ from the vertical and released. The maximum tension the string must withstand as the pendulum passes through the mean position is ($m =$ mass of the pendulum,$g =$ acceleration due to gravity).

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A
$mg$
B
$3 mg$
C
$5 mg$
D
$6 mg$

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Work, Energy, Power and Collision

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Vertical Circular Motion

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$A$ particle is moving in a vertical circle. The tensions in the string when passing through two positions at angles $30^\circ$ and $60^\circ$ from the vertical (lowest position) are $T_1$ and $T_2$ respectively. Then:

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$A$ particle of mass $200 \,g$ is whirled in a vertical circle of radius $80 \,cm$ using a massless string. The speed of the particle when the string makes an angle of $60^{\circ}$ with the vertical line is $1.5 \,m/s$. The tension in the string at this position is ........... $N$.

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$A$ body crosses the topmost point of a vertical circle with critical speed. Its centripetal acceleration,when the string is horizontal,will be $....g$

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$A$ mass $m$ is released from point $A$ as shown in the figure. The tension in the string at point $B$ will be:

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The string of a pendulum of length $L$ is displaced through $90^{\circ}$ from the vertical and released. The maximum tension the string must withstand as the pendulum passes through the mean position is ($m =$ mass of the pendulum,$g =$ acceleration due to gravity).

Similar Questions

$A$ weightless thread can bear a tension of up to $3.7 \, kg \, wt$. $A$ stone of mass $500 \, g$ is tied to it and revolved in a circular path of radius $4 \, m$ in a vertical plane. If $g = 10 \, m/s^2$,then the maximum angular velocity of the stone will be ........ $rad/s$.

$A$ particle is moving in a vertical circle. The tensions in the string when passing through two positions at angles $30^\circ$ and $60^\circ$ from the vertical (lowest position) are $T_1$ and $T_2$ respectively. Then:

$A$ particle of mass $200 \,g$ is whirled in a vertical circle of radius $80 \,cm$ using a massless string. The speed of the particle when the string makes an angle of $60^{\circ}$ with the vertical line is $1.5 \,m/s$. The tension in the string at this position is ........... $N$.

$A$ body crosses the topmost point of a vertical circle with critical speed. Its centripetal acceleration,when the string is horizontal,will be $....g$

$A$ mass $m$ is released from point $A$ as shown in the figure. The tension in the string at point $B$ will be:

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