NEET Solved Paper 2019 Question 42
Question: A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after $ 2\pi $ revolutions is- [NEET 5-5-2019]
Options:
A) $ 12\times 10^{4}Nm $
B) $ 2\times 10^{6}Nm $
C) $ 2\times 10^{6}Nm $
D) $ 2\times 10^{3}Nm $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \omega _2^{2}=\omega _1^{2}+2\alpha \theta $
$ 0={{( 2\pi \frac{3}{60} )}^{2}}-2\times 2\pi (2\pi )\alpha $
$ \alpha =\frac{9}{60\times 60\times 2}=\frac{1}{800} $
$ \tau =I\alpha =\frac{mr^{2}}{2}\alpha $
$ \frac{2\times 16\times 10^{4}}{2}\times \frac{1}{800}=2\times {10^{-6}} $
BETA
