Q.31. A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the

Q: What is the correct answer for JEE Main Physics Q31?

The correct answer is D. Sol. Radius of sphere $R = 4 \text{ cm} = 0.04 \text{ m}$. Mass $m = 5 \text{ kg}$. Initial angular velocity $\omega_0 = 1200 \text{ rpm} = 1200 \times \frac{2\pi}{60} \text{ rad/s} = 40\pi \text{ rad/s}$. Final angular velocity $\omega = 0$. Time $t = 10 \text{ s}$. For a solid sphere, the moment of inertia $I = \frac{2}{5}mR^2 = \frac{2}{5}(5 \text{ kg})(0.04 \text{ m})^2 = 2(0.0016) = 0.0032 \t

Q: Q.31. A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of 1200 rpm. It is brought to rest in 10 s by…

The correct answer is D. Sol. Radius of sphere $R = 4 \text{ cm} = 0.04 \text{ m}$. Mass $m = 5 \text{ kg}$. Initial angular velocity $\omega_0 = 1200 \text{ rpm} = 1200 \times \frac{2\pi}{60} \text{ rad/s} = 40\pi \text{ rad/s}$. Final angular velocity $\omega = 0$. Time $t = 10 \text{ s}$. For a solid sphere, the moment of inertia $I = \frac{2}{5}mR^2 = \frac{2}{5}(5 \text{ kg})(0.04 \text{ m})^2 = 2(0.0016) = 0.0032 \t

Q: What are the options for this JEE Main Physics question?

A. 128 \pi Nm, 100 · B. 0128 \pi Nm, 50 · C. 128 \pi Nm, 50 · D. 0128 \pi Nm, 100

Question

Q.31. A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of 1200 rpm. It is brought to rest in 10 s by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are ________ and ________ respectively.

FirstInTest · Accepted Answer

Sol. Radius of sphere $R = 4 	ext{ cm} = 0.04 	ext{ m}$. Mass $m = 5 	ext{ kg}$. Initial angular velocity $\omega_0 = 1200 	ext{ rpm} = 1200 	imes \frac{2\pi}{60} 	ext{ rad/s} = 40\pi 	ext{ rad/s}$. Final angular velocity $\omega = 0$. Time $t = 10 	ext{ s}$. For a solid sphere, the moment of inertia $I = \frac{2}{5}mR^2 = \frac{2}{5}(5 	ext{ kg})(0.04 	ext{ m})^2 = 2(0.0016) = 0.0032 	ext{ kg m}^2$. Using the first equation of rotational motion: $\omega = \omega_0 + \alpha t$. $0 = 40\pi + \alpha(10) \Rightarrow \alpha = -4\pi 	ext{ rad/s}^2$. The torque applied is $	au = I\alpha = (0.0032 	ext{ kg m}^2)(-4\pi 	ext{ rad/s}^2) = -0.0128\pi 	ext{ N m}$. The negative sign indicates that the torque opposes the motion. The magnitude of the torque is $0.0128\pi 	ext{ N m}$. Now, for the number of rotations. Using the equation $	heta = \omega_0 t + \frac{1}{2}\alpha t^2$. $	heta = (40\pi)(10) + \frac{1}{2}(-4\pi)(10)^2 = 400\pi - \frac{1}{2}(4\pi)(100) = 400\pi - 200\pi = 200\pi 	ext{ radians}$. Number of rotations $N = \frac{	heta}{2\pi} = \frac{200\pi}{2\pi} = 100$. So, the torque applied is $0.0128\pi 	ext{ N m}$ and the number of rotations is 100.

Answer

128 \pi Nm, 100

Answer

0128 \pi Nm, 50

Answer

128 \pi Nm, 50

Answer

0128 \pi Nm, 100

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