Question

Magnitude of acceleration of the two blocks is :
(Radius of pulley is $1\text{m}$)

• A. $3{\text{m/s}}^2$
• B. $1.5{\text{m/s}}^2$
• C. $2{\text{m/s}}^2$
• D. $2.5{\text{m/s}}^2$

Answer 1

The correct option is D $2.5{\text{m/s}}^2$

Let the acceleration of blocks be $a$ and tension in the string be $T_1$ and $T_2$ on left and right side respectively.

FBD:


For $2\text{kg}$ block on applying Newton's second law in the direction of acceleration,
$20-T_1=2a...\left(1\right)$
For $1\text{kg}$ block on applying Newton's second law in the direction of acceleration,
$T_2-10=a...\left(2\right)$
On applying equation of torque for the pulley about the axis passing through its centre.
$\sum \tau =I\alpha$
[consider anti-clockwise sense of rotation as $+ve$]
$(1\times T_1)-(1\times T_2)=1\times \alpha$
$\because \tau =r_{\perp }F$
$\Rightarrow T_1-T_2=\alpha$
For no slipping of string over the pulley, the tangential acceleration of pulley must be equal to the linear acceleration of blocks.
$a_t=\alpha R$
$\Rightarrow a=\alpha R$
or $\alpha =\frac{a}{R}=a$ $(\because R=1\text{m})$
$\Rightarrow T_1-T_2=a...\left(3\right)$
On adding Eq.$\left(1\right),\left(2\right)$ and $\left(3\right)$ we get,
$\Rightarrow (20-T_1)+(T_2-10)+(T_1-T_2)=4a$
or $4a=10$
$\therefore a=2.5{\text{m/s}}^2$
Magnitude of acceleration of the two blocks is $2.5{\text{m/s}}^2$