JEE Main Solved Paper 2017 Question 14
Question: A body of mass $ m={10^{-2}}kg $ is moving in a medium and experiences a frictional force $ m={10^{-2}}kg $ Its intial speed is $ v _0=10m{s^{-1}}. $ If, after 10 s, its energy is $ \frac{1}{8}mv_0^{2}, $ [JEE Main Solved Paper-2017]
Options:
A) $ {10^{-4}}kg,{m^{-1}} $
B) $ {10^{-1}}kg,{m^{-1}}{S^{-1}} $
C) $ {10^{-3}}kg,{m^{-1}} $
D) $ {10^{-3}}kg,{s^{-1}} $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \frac{1}{2}mv_f^{2}=\frac{1}{8}mv_0^{2} $ $ v _{f}=\frac{v _0}{2}=5,m/s $ $ ({10^{-2}})\frac{dv}{dt}=-kv^{2} $ $ \int _10^{5}{\frac{dv}{v^{2}}}=-1000k\int_0^{10}{dt} $ $ \frac{1}{5}-\frac{1}{10}=100,k(10) $ $ k={10^{-4}} $
