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A particle of mass ' $m$ ' performs uniform circular motion of radius ' $r$ ' with linear speed ' $v$ ' under the application of force ' $F$ '. If $\mathrm{m}, \mathrm{v}$ and $\mathrm{r}$ are all increased by $50 \%$, the necessary change in force required to maintain the particle in uniform circular motion is
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$125 \%$
Given that, Mass $=\mathrm{m}$, Velocity $=\mathrm{v}$, Radius $=\mathrm{r}$
We know that, The force is
Now, the increment is $50 \%$ in mass, velocity and radius
Now, new force is
$F^{\prime}=\frac{\left(m+\frac{m}{2}\right)\left(v+\frac{v}{2}\right)^2}{\left(r+\frac{r}{2}\right)}=\frac{\frac{3}{2} m \times \frac{9}{4} v^2}{\frac{3}{2} r}=F \times \frac{9}{4}$
Now, the \% change in force
$\frac{F^{\prime}-F}{F} \times 100=\left(\frac{9}{4}-1\right) \times 100=125 \%$
We know that, The force is
Now, the increment is $50 \%$ in mass, velocity and radius
Now, new force is
$F^{\prime}=\frac{\left(m+\frac{m}{2}\right)\left(v+\frac{v}{2}\right)^2}{\left(r+\frac{r}{2}\right)}=\frac{\frac{3}{2} m \times \frac{9}{4} v^2}{\frac{3}{2} r}=F \times \frac{9}{4}$
Now, the \% change in force
$\frac{F^{\prime}-F}{F} \times 100=\left(\frac{9}{4}-1\right) \times 100=125 \%$
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