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A small steel ball of mass `m` and radius `r` is falling under gravity through a viscous liquid of coefficient of viscosity `eta`. If `g` is the value of acceleration due to gravity. Then the terminal velocity of the ball is proportional to (ignore buoyancy)A. `(mg(eta))/(r )`B. `mg(eta)r`C. `(mgr)/(eta)`D. `(mg)/(reta)` |
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Answer» Correct Answer - D Since the steedl ball is given to be small therefore Now, `6pietarv_0=mg` or `v_0pro(mg)/(etar)` |
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