NEET Solved Paper 2016 Question 24
Question: A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tang entail acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to $ 8\times 10^{4}\text{ J} $ by the end of the second revolution after the beginning of the motion?
Options:
A) $ 0.1m/s^{2} $
B) $ 0.15m/s^{2} $
C) $ 0.18m/s^{2} $
D) $ 0.2m/s^{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \frac{1}{2}mv^{2}=E\Rightarrow \frac{1}{2}( \frac{10}{1000} )v^{2}=8\times {10^{-4}} $
$ \Rightarrow $ $ v^{2}=16\times {10^{-2}}\Rightarrow v=4\times {10^{-1}}=0.4m/s $ Now, $ v^{2}=u^{2}+2a _{t}s $ $ (s=4\pi R) $
$ \Rightarrow $ $ \frac{16}{100}=0^{2}+2a _{t}( 4\times \frac{22}{7}\times \frac{6.4}{100} ) $
$ \Rightarrow $ $ a _{t}=\frac{16}{100}\times \frac{7\times 100}{8\times 22\times 6.4}=0.1m/s^{2} $
BETA
