Question

A toy cart is tied to the end of an unstretched string of length $‘l^{\prime }$, when revolved, the toy cart moves in horizontal circle with radius $‘2l^{\prime }$ and time period $T_1$. If it is speeded until it moves in horizontal circle of radius $‘3l^{\prime }$ with period $T_2$, relation between $T_1$ and $T_2$ is (Hooke’s law is obeyed)

• A. $T_2=\frac{2}{\sqrt{3}}{T}_1$
• B. $T_2=\sqrt{\frac{3}{2}}{T}_1$
• C. $T_2=\sqrt{\frac{2}{3}}{T}_1$
• D. $T_2=\frac{\sqrt{3}}{2}{T}_1$

Answer 1

The correct option is D $T_2=\frac{\sqrt{3}}{2}{T}_1$


$T=\frac{m{v}^2}{r}$

$TimePeriod=\frac{2\pi r}{v}$

$\Delta L=\frac{Fl}{AY}$

$\Rightarrow T=\frac{\Delta lAY}{l}$

$\frac{\Delta lAY}{l}=\frac{m{v}^2}{r}$

$v=\sqrt{\frac{\Delta lAYr}{lm}}$

$TimePeriod\left(T\right)=\frac{2\pi r}{\sqrt{\frac{AYr\Delta l}{lm}}}$

So, $T\propto \sqrt{\frac{r}{\Delta l}}$

$T_1\propto \sqrt{2}$
$T_2\propto \sqrt{\frac{3}{2}}$
$⟹\frac{T_2}{T_1}=\frac{\sqrt{3}}{2}$