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A heavy wheel of radius `20 cm` and weight `10 kg` is to be dragged over a step of height `10 cm`, by a horizontal force `F` applied at the centre of the wheel. The minimum value of `F` isA. `20 kg w t`B. `1 kg wt`C. `10 sqrt(3) k g w t`D. `10sqrt(2) kg wt` |
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Answer» Correct Answer - C Clockwise torque=anticlockwise torque `mgsqrt((20)^(2)-(10)^(2))=F(20-10)` |
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