Solution to JEE Mains 2018 Physics Q3

Q3: Two masses m 1 = 5 kg and m 2 = 10 kg, 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 :

(1) 18.3 g
(2) 27.3 kg
(3) 43.3 kg
(4) 10.3 kg
Ans:
Motion will stop when pull due to m1 will become equal to the force of friction due to mass (m1+m) and surface.
$$m_{1}g=\mu (m_{2}+m)g$$
$$m_{1}=\mu (m_{2}+m)$$
$$5=0.15(10+m)$$
m=23.3kg
Hence the nearest solution is (2).
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