Question

A body is made to revolve along a circle of radius $r$ in a horizontal plane with a time period $T$ with the help of a light horizontal string such that the tension in the string is $F$. If the radius of the circle is increased to $2r$ and time period of revolution is decreased to $\frac{T}{2}$, then the tension in the string will be:

• A. $\frac{F}{2}$
• B. $2F$
• C. $4F$
• D. $8F$

Answer 1

The correct option is D $8F$

We know, Tension, $F=\frac{m{v}^2}{r}=\frac{m{r}^2{\omega }^2}{r}=mr{\omega }^2$
But we know that, ${\omega }^2={\left(\frac{2\pi }{T}\right)}^2$
So, $F\propto \frac{r}{T^2}$

If, $F_1=$ Tension in the first case, and
$F_2=$ Tension in the second case, then

$\frac{F_1}{F_2}=\frac{r_1}{T_1^2}\times \frac{T_2^2}{r_2}=\frac{r}{T^2}\times \frac{T^2}{4}\times \frac{1}{2r}=\frac{1}{8}$

So, $F_2=8F_1=8F$