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Question

# Two masses m and M are connected by a light string passing over a smooth pulley. When set free, m moves up by 1.4 m in 2 s. The ratio mM is (g=9.8 ms−2)

A
1315
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B
1513
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C
97
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D
79
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Solution

## The correct option is A 1315 Given s=1.4 m t=2s s=12at2 1.4=12×a×4 a=0.7 m/s2 Applying the result of newton 2nd law F=ma Mg−T=M(0.7)... i T−mg=m(0.7)...ii We get (M−mM+m)g=0.7 (M−mM+m)9.8=710 14M−14m=M+m 13M=14m mM=1315 Alternative; For block having mass m as it is going upward T=m(g+a)(a) For block having mass M as it is going downward T=M(g−a)(b) As Tension is same equating both the equations m(g+a)=M(g−a) mM=(9.8−0.7)(9.8+0.7)=9.110.5=1.31.5=1315

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