Question

A light wire AB of length 10 cm can slide on a vertical frame as shown in figure. There is a film of soap solution trapped between the frame and the wire. Find the load W that should be suspended from the wire to keep it in equilibrium. Neglect friction. Surface tension of soap solution = 25 dyne cm^-1. Take g = 10 m s^-2

• A. 0.5 kg
• B. 50 gm
• C. 5 gm
• D. 0.5 gm

Answer 1

The correct option is D

0.5 gm

Soap solution film will be formed on both sides of the frame. Each film is in contact with the wire along a distance of 10 cm. the force exerted by the film on the wire
=$2\times \left(10cm\right)\times \left(25dynec{m}^{-1}\right)$
=$500dyne=5\times {10}^{-3}N$
The force acts vertically upward and should be balanced by the load. Hence the load that should be suspended is5$\times$
${10}^{-3}N$. The mass of the load should be $\frac{5\times {10}^{-3}N}{10m{s}^{-2}}=5\times {10}^{-4}$ Kg
=0.5g