Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A compass needle oscillates 20 times per minute at a place where the dip is and 30 times per minute

A compass needle oscillates 20 times per minute at a place where the dip is 30^{\circ} and 30 times per minute where the dip is 60^{\circ}. The ratio of total magnetic field due to the earth at two places respectively is \frac{4}{\sqrt{\mathrm{x}}}. The value of \mathrm{x} is

Option: 1

243


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

Period of oscillation \alpha \frac{1}{\sqrt{B_{\mathrm{H}}}}
$$ \begin{aligned} & \mathrm{T} \alpha \frac{1}{\sqrt{\mathrm{B} \cos \theta}} \Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\sqrt{\frac{\mathrm{B}_2 \cos \theta_2}{\mathrm{~B}_1 \cos \theta_1}} \\ & \Rightarrow \frac{60 / 20}{60 / 30}=\sqrt{\frac{\mathrm{B}_2}{\mathrm{~B}_1} \frac{\cos 60^{\circ}}{\cos 30^{\circ}}} \Rightarrow \frac{3}{2}=\sqrt{\frac{\mathrm{B}_2}{\sqrt{3} \mathrm{~B}_1}} \\ & \Rightarrow \frac{9}{4}=\frac{\mathrm{B}_2}{\sqrt{3} \mathrm{~B}_1} \Rightarrow \frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{4}{9 \sqrt{3}}=\frac{4}{\sqrt{243}} \\ & \end{aligned}

Posted by

Kshitij

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: NTA exam city intimation slip link available; admit card date
anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads
anam-khan

Bihar Board extends BSEB Super 50 free coaching application last date to March 26

Latest Question