A particle of mass ' $m$ ' performs uniform circular motion of radius ' $r$ ' with linear speed ' $v$ '…
- $125 \%$
- $150 \%$
- $100 \%$
- $225 \%$
Solution
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
$\frac{F^{\prime}-F}{F} \times 100=\left(\frac{9}{4}-1\right) \times 100=125 \%$
.Asked in: MHT CET 2022 (07 Aug Shift 2)
Practice more Motion In Two Dimensions questions on Aicharya