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Q.

In the system of pulleys shown, what should be the value of m1 (in gram) such that 100 g remains at rest ?  (Take g=10m/s2 ) 

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answer is 160.

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Detailed Solution

Since 100 g is at rest,

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lf m1 is going up at acceleration a, we use

2Tm1g=m1a   ……(1)

For 100 g to be at rest, acceleration of 200 g mass is 2a, if m1 is going up with acceleration a.

2T=0.2(2a)21=0.4aa=104=52m/s2

From (1), we have,

2×110m1=m152

2=10m1+5m124=25m1m1=425=0.16kg=160g

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