Magnetism Question 258
Question: A compass needle placed at a distance r from a short magnet in Tan A position shows a deflection of $ 60{}^\circ $ . If the distance is increased to r $ {{(3)}^{1/3}} $ , then deflection of compass needle is
Options:
A) $ 30^{o} $
B) $ 60\times {{3}^{\frac{1}{3}}} $
C) $ 60\times {{3}^{\frac{2}{3}}} $
D) $ 60\times {{3}^{\frac{3}{3}}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ \frac{\tan ,{\theta _{2}}}{\tan ,{\theta _{1}}}=\frac{d _{1}^{3}}{d _{2}^{3}}=\frac{r^{3}}{{{[r{{(3)}^{1/3}}]}^{3}}}=\frac{1}{3} $
$ \tan ,{\theta _{2}}=\frac{1}{3}\tan ,{\theta _{1}}=\frac{\tan 60}{3}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}} $
$ \
Therefore ,,,,,{\theta _{2}}=30{}^\circ $
BETA
