| 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] |
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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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