Question

Two bodies A, B of masses $m_1,m_2$ are knotted to a mass-less string at different points rotated along concentric circles in horizontal plane. The distances of A, B from common centre are 50cm, 1m. If the tensions in the string between centre to A and A to B are in the ratio 5:4, then the ratio of $m_1$ to $m_2$ is:

• A. 2 : 3
• B. 3 : 2
• C. 1 : 1
• D. 1 : 2

Answer 1

The correct option is D 1 : 2

The tension will be more between center to A as it has to bear the centripetal force of 2 bodies.

applying force balance on the masses,

$T_1-T_2=m_1{\omega }^2{r}_1$.........(1)

$T_2=m_2{\omega }^2{r}_2$.........(1)

$\Rightarrow \frac{T_2}{T_1}=\frac{4}{5}$

Now: $r_1=0.5$, $r_2=1$

=> $\frac{5}{4}=\frac{m_1\left(0.5\right)}{m_2+1}$

=> $\frac{m_1}{m_2}=\frac{2\times 1}{4}=\frac{1}{2}$