Question

A particle of mass $m_{1}$ is fastened to one end of a string and another particle of mass $m_{2}$ is attached to the middle point, the other end of the string being fastened to a fixed point on a smooth horizontal table. If the particles are made to revolve in a horizontal circular path as shown in the figure, then the ratio of tension in the part of the string between the centre and $m_{2}$ to the tension in the part of the string between $m_{2}$ and $m_{1}$ is

• A. $\frac{m_{1}}{m_{1} + m_{2}}$
• B. $\frac{m_{1} + m_{2}}{m_{1}}$
• C. $\frac{2 m_{1} + m_{2}}{2 m_{1}}$
• D. $\frac{2 m_{1}}{m_{1} + m_{2}}$

Answer 1

Answer: (C)

for $m_{1},$
$\mathrm{T}_1=\mathrm{m}_1 \omega^2(2 \mathrm{r}) \ldots \ldots$ ......(i)
for $m_{2},$
$T_{2}-T_{1}=m_{2}\omega ^{2}r$ ....(ii)
Dividing both the equations:
$\frac{T_{1}}{T_{2} - T_{1}}=\frac{2 m_{1}}{m_{2}}$
$\Rightarrow \frac{T_{2}}{T_{1}}=\frac{2 m_{1} + m_{2}}{2 m_{1}}$