Question

Two small balls of the same size, having masses $m_{1}$ and $m_{2}$ $\left(\right.m_{1}>m_{2}\left.\right)$ are tied by a thin weightless thread and dropped from a certain height. Taking the force of buoyancy of air into account, the tension $T$ in the thread during the flight, after the motion of the ball becomes uniform, will be

• A. $\left(\right.m_{1}-m_{2}\left.\right)g$
• B. $\left(\right.m_{1}-m_{2}\left.\right)\frac{g}{2}$
• C. $\left(\right.m_{1}+m_{2}\left.\right)g$
• D. $\left(\right.m_{1}+m_{2}\left.\right)\frac{g}{2}$

Answer 1

Answer: (B)

As both the ball are of same size, force of buoyancy on each is same. Therefore, in equilibrium

$F+F= m _{1} g + m _{2} g$ or $F=\left(m_{1}+m_{2}\right) \frac{ g }{2}$
Considering the equilibrium of lower ball,
$T+F=m_{1} g$
$T=m_{1} g -F$
$T=m_{1} g -\left(m_{1}+m_{2}\right) \frac{ g }{2}$
$T=\left(m_{1}-m_{2}\right) \frac{ g }{2}$