Two bar magnets A and B are placed one over t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The frequency of oscillation in a vibration magnetometer is given by $f = \frac{1}{2\pi} \sqrt{\frac{MB}{I}}$.When magnets are placed together, th

Vedclass · Accepted Answer

$25:7$

Vedclass · Answer

(B) The frequency of oscillation in a vibration magnetometer is given by $f = \frac{1}{2\pi} \sqrt{\frac{MB}{I}}$.When magnets are placed together, the effective magnetic moment is $M_{eff} = M_A + M_B$ (for similar poles) and $M_{eff} = M_A - M_B$ (for opposite poles).Let $f_s = 20 	ext{ oscillations/min}$ and $f_d = 15 	ext{ oscillations/min}$.Since $f \propto \sqrt{M_{eff}}$, we have $\frac{f_s}{f_d} = \sqrt{\frac{M_A + M_B}{M_A - M_B}}$.Squaring both sides: $\left(\frac{20}{15}ight)^2 = \frac{M_A + M_B}{M_A - M_B} \Rightarrow \left(\frac{4}{3}ight)^2 = \frac{M_A + M_B}{M_A - M_B} \Rightarrow \frac{16}{9} = \frac{M_A + M_B}{M_A - M_B}$.Using componendo and dividendo: $\frac{M_A}{M_B} = \frac{16+9}{16-9} = \frac{25}{7}$.

Similar Questions

Two magnets of same size and mass make respectively $10$ and $15$ oscillations per minute at a certain place. The ratio of their magnetic moments is

The time period of a thin bar magnet in Earth's magnetic field is $T$. If the magnet is cut into two equal parts perpendicular to its length,the time period of each part in the same field will be

The time period of a magnet is $T$. If it is divided into four equal parts along its axis and perpendicular to its axis as shown,then the time period for each part will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module