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

\(\begin{aligned} & \text { Centripetal force, } \mathrm{F}=\frac{\mathrm{mv}^2}{\mathrm{r}} \\ & \mathrm{F}_1=\frac{\mathrm{mv}_1^2}{\mathrm{r}_1} \text { and } \mathrm{F}_2=\frac{\mathrm{mv}_2^2}{\mathrm{r}_2} \\ & \frac{\mathrm{F}_2}{\mathrm{~F}_1}=\frac{\mathrm{v}_2^2}{\mathrm{v}_1^2} \cdot \frac{\mathrm{r}_1}{\mathrm{r}_2} \\ & \mathrm{v}_2=\frac{\mathrm{v}_1}{2} \text { and } \mathrm{r}_2=3 \mathrm{r}_1 \\ & \frac{\mathrm{F}_2}{\mathrm{~F}_1}=\left(\frac{1}{2}\right)^2 \cdot \frac{1}{3}=\frac{1}{12} \\ & \mathrm{v}_2=\frac{\mathrm{v}_1}{2} \text { and } \mathrm{r}_2=3 \mathrm{r}_1 \\ & \frac{\mathrm{F}_2}{\mathrm{~F}_1}=\left(\frac{1}{2}\right)^2 \cdot \frac{1}{3}=\frac{1}{12} \\ & \mathrm{~F}_2=\frac{\mathrm{F}_1}{12}\end{aligned}\)
Change in force \(=F_2-F_1=\frac{F_1}{12}-F_1=-\frac{11}{12} F_1\) Hence force is decreased by \(\frac{11}{12}\) times.