• # question_answer Two masses ${{m}_{1}}=5kg$ and ${{m}_{2}}=10kg$, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of ${{m}_{2}}$ to stop the motion is : [JEE Main Online 08-04-2018] A) $\text{43}\text{.3 kg}$ B) $\text{10}\text{.3 kg}$ C) $\text{18}\text{.3 kg}$ D) $\text{27}\text{.3 kg}$

Correct Answer: D

Solution :

 [d] ${{m}_{1}}g=\mu ({{m}_{2}}+m)g$ $\therefore \text{ }m\text{ }=\text{ }23.3\text{ }kg$ Value greater than this and minimum in the given options is 27.3 kg

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