Search any question & find its solution
Question:
Answered & Verified by Expert
Magnetic field of the earth is $H=0.3 \mathrm{~g}$. A magnet is vibrating 5 oscillations per min then the dippreciation required in the magnetic field of the earth of increase time period upto 10 oscillations per minute is
Options:
Solution:
1100 Upvotes
Verified Answer
The correct answer is:
$2.25$ g
We know that $T \propto \frac{1}{\sqrt{H}}$ $\therefore \quad \frac{T_1}{T_2}=\sqrt{\frac{H_2}{H_1}}$,
$\frac{5}{10}=\sqrt{\frac{H_2}{3}} \Rightarrow \frac{1}{2}=\sqrt{\frac{H_2}{3}}$
$\begin{aligned}& & \frac{1}{4} & =\frac{H_2}{3} \\& & H_2 & =0.75\end{aligned}$
Dippreciation in field $=3-0.75=2.25$
$\frac{5}{10}=\sqrt{\frac{H_2}{3}} \Rightarrow \frac{1}{2}=\sqrt{\frac{H_2}{3}}$
$\begin{aligned}& & \frac{1}{4} & =\frac{H_2}{3} \\& & H_2 & =0.75\end{aligned}$
Dippreciation in field $=3-0.75=2.25$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.