Question

# Two blocks of masses 400 g and 200 g are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has a moment of inertia 1.6×10−4kg−m2 and a radius 2.0 cm. Find 'x', which is the kinetic energy of the system as the 400 g block falls through 50 cm and also find 'y', the speed of the blocks at this instant.

A

x = 10.5 J, y = 2.6 m/s

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B

x = 9.8 J, y = 2.6 m/s

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C

x = 9.8 J, y = 1.4 m/s

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D

x = 0.98 J, y = 1.4 m/s

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Solution

## The correct option is D x = 0.98 J, y = 1.4 m/s According to the question 0.4g−T1=0.4 a ...(1) T2−0.2g=0.2 a ....(2) (T1−T2)r=Iar ...(3) From equation 1,2 and 3 ⇒ a=(0.4−0.2)g(0.4+0.2+1.60.4)=g5 Therefore (y) V=√2ah=√(2×gl5×0.5) ⇒√(g5)=√(9.85)=1.4ms. (x) total kinectic energy of the system =12m1V2+12m2V2+12182 =(12×0.4×1.42)+(12×0.2×1.42)+(12×(1.64)×1.42)=0.98 joule.

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