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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

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B

50 gm

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C

5 gm

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D

0.5 gm

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Solution

## 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×(10cm)×(25 dyne cm−1) =500 dyne=5×10−3N The force acts vertically upward and should be balanced by the load. Hence the load that should be suspended is5× 10−3N. The mass of the load should be 5×10−3N10ms−2=5×10−4 Kg =0.5g

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