JEE Main Solved Paper 2017 Question 21
Question: A magnetic needle of magnetic moment $ 6.7\times {10^{-2}},Am^{2} $ and moment of inertia $ 7.5\times {10^{-6}},kg,m^{2} $ is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is: [JEE Main Solved Paper-2017]
Options:
A) 6.98 s
B) 8.76 s
C) 6.65 s
D) 8.89 s
Show Answer
Answer:
Correct Answer: C
Solution:
- $ T=2\pi \sqrt{\frac{I}{MB}} $ $ I=7.5\times {10^{-6}}kg-m^{2} $ $ M=6.7\times {10^{-2}}Am^{2} $ By substituting value in the formula $ T=.665,\sec $ for 10 oscillation, time taken will be Time = 10 T = 6.65 sec Answer option 3
