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Question

All the surfaces are smooth and the strings and pulley's are light. The force exerted by the 20cm part of the rod on the 10cm part is

1901392_5d6a57a721814e98b36b2f6b651b2eed.png

A
6N
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B
12N
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C
24N
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D
36N
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Solution

The correct option is C 24N
Mass per unit length = 330kg/cm=0.1kg/cm
Mass of 10cm part = m1 = 1kg
Mass of 20cm part = m2 = 2kg
Let F be the force of contact betwween them
From free body diagram
F20a=0........(1)
32F2a=0..........(2)
32202aa=0
123a=0
a=123=4ms2
Contact force F=20+a=20+4=24N
Force exerted = 24N

1879235_1901392_ans_e019b67e68104289b9766939c2f31320.png

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