Physics Q47 – JEE-MAIN 2025

Home Physics Rotational MotionQuestion #455

A thin solid disk of 1 kg is rotating along its diameter axis at the speed of 1800 rpm. By applying an external torque of

25 π

Nm for 40s, the speed increases to 2100 rpm. The diameter of the disk is _______ m.

Enter Calculated Numerical Value:

Clue 1 of 3: Rotational Dynamics

This problem involves the relationship between torque, moment of inertia, and angular acceleration, as well as rotational kinematics.

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Step 1: Convert angular speeds and calculate angular acceleration✦ Active

Convert the initial and final angular speeds from revolutions per minute (rpm) to radians per second (rad/s):

ω_{1} = 1800 rpm = 1800 \times \frac{2 π}{60} rad/s = 60 π rad/s

ω_{2} = 2100 rpm = 2100 \times \frac{2 π}{60} rad/s = 70 π rad/s

Now, calculate the angular acceleration $α$ using the kinematic equation $ω_{2} = ω_{1} + α t$ :

α = \frac{ω_{2} - ω_{1}}{t} = \frac{70 π - 60 π}{40} = \frac{10 π}{40} = \frac{π}{4} {rad/s}^{2}

Step 2: Calculate the moment of inertia○ Expand

Use the relationship between torque, moment of inertia, and angular acceleration, $τ = I α$ , to find the moment of inertia $I$ :

I = \frac{τ}{α} = \frac{25 π}{π / 4} = 25 π \times \frac{4}{π} = 100 {kg m}^{2}

Step 3: Determine the diameter of the disk○ Expand

For a thin solid disk rotating about its diameter, the moment of inertia is given by $I = \frac{1}{4} M R^{2}$ , where $M$ is the mass and $R$ is the radius. We can now solve for $R$ :

100 = \frac{1}{4} (1 kg) R^{2}

R^{2} = 400 m^{2}

R = \sqrt{400} = 20 m

Finally, calculate the diameter $D = 2 R$ :

D = 2 \times 20 = 40 m

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Physics Question 47 – JEE-MAIN 2025

Clue 1 of 3: Rotational Dynamics

🧠 Full Solution Path