(D) Let the lowest point be $A$ and the horizontal position be $B$. At $A$, velocity is $\vec{u} = u \hat{i}$.Using the law of conservation of energy

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(D) Let the lowest point be $A$ and the horizontal position be $B$. At $A$, velocity is $\vec{u} = u \hat{i}$.Using the law of conservation of energy between $A$ and $B$:$\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgL$$v^2 = u^2 - 2gL$$v = \sqrt{u^2 - 2gL}$.At position $B$, the velocity is directed vertically upward, so $\vec{v} = v \hat{j} = \sqrt{u^2 - 2gL} \hat{j}$.The change in velocity is $\Delta \vec{v} = \vec{v} - \vec{u} = v \hat{j} - u \hat{i}$.The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{v^2 + u^2} = \sqrt{(u^2 - 2gL) + u^2} = \sqrt{2u^2 - 2gL} = \sqrt{2(u^2 - gL)}$.

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

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$A$ stone tied to a string of length $L$ is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is

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A
$\sqrt{u^2 - 2gL}$
B
$\sqrt{2gL}$
C
$\sqrt{u^2 - gL}$
D
$\sqrt{2(u^2 - gL)}$

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$A$ stone is tied to a string of length $L$ and is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed $u$. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is

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$A$ $1\, kg$ stone at the end of a $1\, m$ long string is whirled in a vertical circle at a constant speed of $4\, m/s$. The tension in the string is $6\, N$ when the stone is at $(g = 10\, m/s^2)$.

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In the case of vertical circular motion of a particle attached to a thread of length $r$, if the tension in the thread is zero at an angle of $30^{\circ}$ with the horizontal as shown in the figure, the velocity at the bottom point $(A)$ of the circular path is ($g =$ gravitational acceleration).

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$A$ bucket tied at the end of a $1.6\, m$ long string is whirled in a vertical circle with constant speed. What should be the minimum speed so that the water from the bucket does not spill,when the bucket is at the highest position? (Take $g = 10\, m/s^2$)

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$A$ particle attached to a string of length $r$ is moving in a vertical circular path. If the speed of the particle at the highest point is $\sqrt{7gr}$,then the ratio of the tension in the string at the highest point to the tension at the lowest point is ($g=$ acceleration due to gravity).

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$A$ $1\, kg$ stone at the end of a $1\, m$ long string is whirled in a vertical circle at a constant speed of $4\, m/s$. The tension in the string is $6\, N$ when the stone is at $(g = 10\, m/s^2)$.

In the case of vertical circular motion of a particle attached to a thread of length $r$, if the tension in the thread is zero at an angle of $30^{\circ}$ with the horizontal as shown in the figure, the velocity at the bottom point $(A)$ of the circular path is ($g =$ gravitational acceleration).

$A$ bucket tied at the end of a $1.6\, m$ long string is whirled in a vertical circle with constant speed. What should be the minimum speed so that the water from the bucket does not spill,when the bucket is at the highest position? (Take $g = 10\, m/s^2$)

$A$ particle attached to a string of length $r$ is moving in a vertical circular path. If the speed of the particle at the highest point is $\sqrt{7gr}$,then the ratio of the tension in the string at the highest point to the tension at the lowest point is ($g=$ acceleration due to gravity).

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