Question

In the system shown in figure blocks A and B have mass $m_1=2kgand{m}_2=\frac{26}{7}$ kg respectively. Pulley having moment of inertia I = 0.11 kg $m^2$ can rotate without friction about a fixed axis. Inner and outer radii of pulley are a = 10 cm and b = 15 cm respectively. B is hanging with the thread wrapped around the pulley, while A lies on a rough inclined plane.


Coefficient of friction being $\mu =\frac{\sqrt{3}}{10}$

$\begin{array}{cc}ColumnI& ColumnII\\ P.\text{Tension in the thread connecting block A}& w.26N\\ Q.\text{Tension in the thread connecting block B}& x.2\frac{m}{s^2}\\ RaccelerationofA& \\ R.\text{Acceleration of A}& y.3\frac{m}{s^2}\\ S.\text{Acceleration of B}& z.17N\end{array}$

• A. P-w, Q-z, R-x, S-y
• B. P-z, Q-w, R-y, S-x
• C. P-z, Q-w, R-x, S-y
• D. P-z, Q-z, R-y, S-y

Answer 1

The correct option is C

P-z, Q-w, R-x, S-y

When the system is released, weight $m_{2}g$ tries to rotate the pulley in clockwise direction while down the plane component $m_{1}g$ sin ${30}^{\circ }$ of weight of block A tries to rotate it anticlockwise. But moment produced by $m_{2}g$ is greater, therefore, the pulley has tendency to rotate clockwise. Let its angular acceleration be '$\alpha$'. Then acceleration of blocks A and B will be equal to a $\alpha$ (up the plane) and b $\alpha$ (ertically downwards) respectively. Let tension in threads connected with blocks A and B be $T_{1}and{T}_2$ respectively.

Considering free body diagrams

For forces on block A, $N=m_{1}gcos{30}^{\circ }$ , ...(i)
$T_1-m_{1}gsin{30}^{\circ }-\mu N=m_1\left(a\alpha \right)$...(ii)
For forces on block $B,m_{2}g-T_2=m_2\left(b\alpha \right)$...(iii)

Taking moments of forces acting on pulley, about axis of rotation, $T_{2}b-T_{1}a=1\alpha$ ...(iv)

From above equation $N=10\sqrt{3}N$

$T_1=17N,T_2=26N\alpha =20\frac{rad}{se{c}^2}$

Acceleration of block, $A=a\alpha =2m{s}^{-2}$ (up the plane)

Acceleration of block, $B=b\alpha =3m{s}^{-2}$ (downward)