Question

A fixed U-tube shaped smooth wire has a semi-circular bending between A and B as shown in the figures. A bead of mass m moving with uniform speed v through the wire enters the semicircular bed at A and leaves at B. The average force exerted by the bead on the part AB of the wire is

• A. $0$
• B. $\frac{4m{v}^2}{\pi d}$
• C. $\frac{2m{v}^2}{\pi d}$
• D. $\frac{2mv}{\pi d}$

Answer 1

The correct option is A $0$

Since point A and B has same force to be exerted in order to bead to be moved

Average force$=\frac{F_A+F_B}{2}$

But $F_A=F_B=0$

Average force$=\frac{0}{2}=0$