A uniform disc of mass 100kg and radius 2m is rotating at 1 rad/ sec about a perpendicular axis passing through its centre. A boy of mass 60 kg standing at the centre of the disc suddenly jumps to a point which is 1m from the centre of the disc. The final angular velocity of the boy (in rad/s) is $\frac{7.7}{x}$ , $x$ is

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 10.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Angular momentum is conserved as there is no external torque. So I1W1 = I2W2
$(\frac{1}{2} M_{d} \times {R^{2}}_{d}) w_{1} = (\frac{1}{2} M_{d} R_{d}^{2} + M_{B} R_{B}^{2}) w_{2}$
$\Rightarrow (\frac{1}{2} \times 100 \times 2^{2}) \times 1 = (\frac{1}{2} \times 100 \times 2^{2} + 60 \times 1^{2}) w_{2} \Rightarrow w_{2} = \frac{200}{200 + 60} = \frac{20}{26} = \frac{10}{13} = 0.77 ∴ w_{2} = 0.77 r a d / s w_{2} = \frac{7.7}{x} (g i v e n) 0.77 = \frac{7.7}{x} = x = \frac{7.7}{0.77} = 10$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE