A boy is pushing a ring of mass \(2 \mathrm{~kg}\) and radius \(0.5 \mathrm{~m}\) with a stick as shown in the figure. The stick applies a force of \(2 \mathrm{~N}\) on the ring and rolls it without slipping with an acceleration of \(0.3 \mathrm{~m} / \mathrm{s}^2\). The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is \(\frac{P}{10}\). The value of \(P\) is

There is no slipping between ring and ground. Hence, \(f_2\) is not maximum. But there is slipping between ring and stick. Therefore, \(f_1\) is maximum. Now, let us write the equations.
\(I =m R^2=(2)(0.5)^2 \)
\( =\frac{1}{2} \mathrm{kgm}^{-2} \)
\( N_1-F_2 =m a \)
\( \text { or } N_1-F_2 =(2)(0.3)=0.6 \mathrm{~N} \)
\( a =R \alpha=\frac{R \tau}{I} \)
\( =\frac{R\left(f_2-f_1\right) R}{I}=\frac{R^2\left(f_2-f_1\right)}{I} \)
\( \therefore 0.3 =\frac{(0.5)^2\left(f_2-f_1\right)}{(1 / 2)} \)
\( \text { or } \quad f_2-f_1 =0.6 \mathrm{~N} \)
\( N_1^2+f_1^2 =(2)^2=4\)
Further \(\quad F_1=\mu N_1=\left(\frac{P}{10}\right) N_1\)
Solving above four equation, we get \(P \simeq 3.6\)
Therefore, the correct answer should be 4 .
Analysis of Question
(i) Question is moderately tough from concept point of view. But calculations are lengthy.
(ii) One has to think about the two components of the force applied by the stick.
(iii) Answer comes an integer when you consider only \(N_1\).