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JEE Main 20258 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2025 · 8 April · Evening Shift · Physics · Rotational Motion

A thin solid disk of 1 kg is rotating along its diameter axis at...

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 25πNm25 \pi \mathrm{Nm} for 40 s, the speed increases to 2100 rpm. The diameter of the disk is ________ m.

Correct answer

0

Step-by-step solution

Given, m=1 kg\mathrm{m}=1 \mathrm{~kg}
<br/>ωi=1800rpm=1800×2π60=60πradsecωf=2100rpm=2100×2π60=70πradsecτext=25πNmt=40sec<br/>\begin{aligned}<br/>& \omega_{\mathrm{i}}=1800 \mathrm{rpm}=1800 \times \frac{2 \pi}{60}=60 \pi \frac{\mathrm{rad}}{\mathrm{sec}} \\ & \omega_{\mathrm{f}}=2100 \mathrm{rpm}=2100 \times \frac{2 \pi}{60}=70 \pi \frac{\mathrm{rad}}{\mathrm{sec}} \\ & \tau_{e x t}=25 \pi \mathrm{Nm} \\ & \mathrm{t}=40 \mathrm{sec}<br/>\end{aligned}
Using equation of motion
<br/>ωf=ωi+αt70π=60π+α(40)α=π4rad/sec2<br/>\begin{aligned}<br/>& \omega_{\mathrm{f}}=\omega_{\mathrm{i}}+\alpha \mathrm{t} \\ & 70 \pi=60 \pi+\alpha(40) \\ & \alpha=\frac{\pi}{4} \mathrm{rad} / \mathrm{sec}^2<br/>\end{aligned}
Also, τ=\tau=\mathrm{I}α\alpha
τ=mR24α\tau=\frac{\mathrm{mR}^2}{4} \alpha
<br/>25π=1×R24×π4R=20 m<br/>\begin{aligned}<br/>& 25 \pi=\frac{1 \times \mathrm{R}^2}{4} \times \frac{\pi}{4} \\ & \mathrm{R}=20 \mathrm{~m}<br/>\end{aligned}
Hence, diameter of disk =2R=2×20=40 m=2 \mathrm{R}=2 \times 20=40 \mathrm{~m}
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