A block of mass 2 kg hangs from the rim of a wheel of radius 0.5m . On releasing from rest the block falls through 5m height in 2 s . The moment of inertia of the wheel will be

The correct option is 1.5 kg-m².

To find the moment of inertia of the wheel, we can analyze the system using the equations of linear motion and rotational dynamics.

Step 1: Calculate the linear acceleration of the falling block
The block starts from rest (initial velocity $u=0$), falls through a height $h=5\mathrm{m}$ in time $t=2\mathrm{s}$. We can find its constant linear acceleration $a$ using the second equation of motion:

$h=ut+\frac{1}{2}$

Substituting the given values into the equation:

$5=0+\frac{1}{2}$

$5=2a$

$a=2.5\mathrm{m}/{\mathrm{s}}^2$

Step 2: Determine the tension in the string
Let $T$ be the tension in the string and $m=2\mathrm{kg}$ be the mass of the block. Taking the acceleration due to gravity $g=10\mathrm{m}/{\mathrm{s}}^2$, the equation of motion for the downward-moving block is:

$mg-T=ma$

Substituting the values of $m$, $g$, and $a$:

$2(10)-T=2(2.5)$

$20-T=5$

$T=15\mathrm{N}$

Step 3: Relate tension to the rotational motion of the wheel
The tension in the string exerts a torque on the wheel of radius $R=0.5\mathrm{m}$. The relation between torque $\tau$ and angular acceleration $\alpha$ is given by:

$\tau =TR=I\alpha$

Since the string does not slip, the angular acceleration $\alpha$ of the wheel is related to the linear acceleration $a$ of the block by:

$\alpha =\frac{a}{R}$

Substituting this relationship into the torque equation gives:

$TR=I\left(\frac{a}{R}\right)$

Rearranging the equation to solve for the moment of inertia $I$:

$I=\frac{T{R}^2}{a}$

Step 4: Calculate the moment of inertia
Substitute the values of $T$, $R$, and $a$ into the formula:

$I=\frac{15\times (0.5{)}^2}{2.5}$

$I=\frac{15\times 0.25}{2.5}$

$I=15\times 0.1=1.5\mathrm{kg}\cdot {\mathrm{m}}^2$

Therefore, the moment of inertia of the wheel is 1.5 kg-m².