(A) The formula for centripetal force is $F = \frac{mv^2}{r}$.Initial force $F_1 = \frac{mV^2}{r}$.New speed $v_2 = \frac{V}{2}$ and new radius $r_2 =

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decreased by $\frac{11}{12}$ times.

(A) The formula for centripetal force is $F = \frac{mv^2}{r}$.Initial force $F_1 = \frac{mV^2}{r}$.New speed $v_2 = \frac{V}{2}$ and new radius $r_2 = 3r$.New force $F_2 = \frac{m(V/2)^2}{3r} = \frac{mV^2/4}{3r} = \frac{mV^2}{12r} = \frac{F_1}{12}$.The change in force is $\Delta F = F_2 - F_1 = \frac{F_1}{12} - F_1 = -\frac{11}{12}F_1$.The negative sign indicates a decrease in the magnitude of the force. Thus,the force is decreased by $\frac{11}{12}$ times.

Class 11 Physics 3-2.Motion in Plane Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

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$A$ body of mass $m$ is moving with speed $V$ along a circular path of radius $r$. Now,the speed is reduced to $\frac{V}{2}$ and the radius is increased to $3r$. For this change,the initial centripetal force needs to be:

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A
decreased by $\frac{11}{12}$ times.
B
increased by $\frac{11}{12}$ times.
C
decreased by $\frac{1}{12}$ times.
D
increased by $\frac{1}{12}$ times.

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3-2.Motion in Plane

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Dynamics of circular Motion (Centrifugal force) and Pendulum and Motion on Curved path

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$A$ body of mass $m$ is moving with speed $V$ along a circular path of radius $r$. Now,the speed is reduced to $\frac{V}{2}$ and the radius is increased to $3r$. For this change,the initial centripetal force needs to be:

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$A$ stone of mass $16 \, kg$ is attached to a string $144 \, m$ long and is whirled in a horizontal circle. The maximum tension the string can withstand is $16 \, N$. The maximum velocity of revolution that can be given to the stone without breaking it will be ....... $m/s$.

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