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Q.

A rolling wheel of 12 kg is on an inclined plane at position P and connected to a mass of 3 kg through a string of fixed length and pulley as shown in figure. Consider PR as friction free surface.

The velocity of center of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be 12xgh m/s. The value of x is __________.

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answer is 2.67.

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Detailed Solution

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Net loss in PE = Gain in KE
12gh-3gh=123v2+1212v2+1212r2vr2 9gh=12[3+12+12]v2 v2=2gh3v=1283gh x=83=2.67

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