Two small balls of the same size, having masses m1 and m2 (.m1m2.) 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

Question

Two small balls of the same size, having masses m1 and m2 (.m1>m2.) 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) (.m1-m2.)g (B) (.m1-m2.)(g/2) (C) (.m1+m2.)g (D) (.m1+m2.)(g/2)

Tardigrade Admin · Accepted Answer

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 = (m_{1} + m_{2}) \frac{g}{2}

Considering the equilibrium of lower ball,

T + F = m_{1} g

T = m_{1} g - F

T = m_{1} g - (m_{1} + m_{2}) \frac{g}{2}

T = (m_{1} - m_{2}) \frac{g}{2}

$(m_{1} - m_{2}) g$

$(m_{1} - m_{2}) \frac{g}{2}$

$(m_{1} + m_{2}) g$

$(m_{1} + m_{2}) \frac{g}{2}$

(m1​−m2​)g

(m1​−m2​)2g​

(m1​+m2​)g

(m1​+m2​)2g​

As both the ball are of same size, force of buoyancy on each is same. Therefore, in equilibrium F+F=m1​g+m2​g or F=(m1​+m2​)2g​ Considering the equilibrium of lower ball, T+F=m1​g T=m1​g−F T=m1​g−(m1​+m2​)2g​ T=(m1​−m2​)2g​

$(m_{1} - m_{2}) g$

$(m_{1} - m_{2}) \frac{g}{2}$

$(m_{1} + m_{2}) g$

$(m_{1} + m_{2}) \frac{g}{2}$