Question

In the system shown in the figure, blocks $A$ and $B$ have mass $m_1=2kg$ and $m_2=\frac{26}{7}kg$ respectively. A pulley having moment of inertia $I=0.11kg{m}^2$ can rotate without friction about a fixed axis. Inner and outer radii of the pulley are $a=10cm$ and $b=15cm$ respectively. $B$ is hanging with the thread wrapped around the pulley, while $A$ lies on a rough inclined plane . Coefficient of friction is $\mu =\frac{\sqrt{3}}{10}(g=10m{s}^{-2})$


Column-I Column-II

i.	Tension in the thread connecting block A and the pulley (in $N$)	p.	17
ii.	Tension in the thread connecting block B and pulley (in $N$)	q.	26
iii.	Angular acceleration of the pulley(in $rad{s}^{-2}$)	r.	20
iv.	Acceleration of the block B (in $m{s}^{-2}$)	s.	3

• A. i-p ; ii-q ; iii-r ; iv-s
• B. i-q ; ii-r ; iii-s ; iv-p
• C. i-s ; ii-p ; iii-q ; iv-r
• D. i-p ; ii-r ; iii-q ; iv-s

Answer 1

The correct option is A i-p ; ii-q ; iii-r ; iv-s

When the system is released, weight $m_{2}g$ tries to rotate the pulley in clockwise direction while the down-plane component $m_{1}gsin30$ of weight of block $A$ tries to rotate it anticlockwise. But the 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 $b\alpha$ (vertically 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 $A$,

$N=m_{1}g\cos {30}^{\circ }$
$T_1-m_{1}g\sin {30}^{\circ }-\mu N=m_1\left(a\alpha \right)$

For forces on block $B$,
$m_{2}g-T_2=m_2\left(b\alpha \right)$

Taking moments of forces acting on pulley, about axis of rotation,

$T_{2}b-T_{1}a=I\alpha$

From the above equations,

$N=10\sqrt{3}N$
$T_1=17N,T_2=26N,$
$\alpha =20rad{s}^{-2}$

Acceleration of block $A=a\alpha =2m{s}^{-2}$ (up the plane)
Acceleration of block $B=b\alpha =3m{s}^{-2}$ (downward)