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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 = $$\frac{3}{30}kg/cm=0.1kg/cm$$
Mass of $$10cm$$ part = $$m_1$$ = $$1kg$$
Mass of $$20cm$$ part = $$m_2$$ = $$2kg$$
Let $$F$$ be the force of contact betwween them
From free body diagram
$$F-20-a=0$$........(1)
$$32-F-2a=0$$..........(2)
$$32-20-2a-a=0$$
$$12-3a=0$$
$$a=\frac{12}{3}=4ms^{-2}$$
Contact force $$F=20+a=20+4=24N$$
Force exerted = $$24N$$

1879235_1901392_ans_e019b67e68104289b9766939c2f31320.png

Physics

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