The magnetic needle of a vibration magnetomet | vedclass

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

Question

(B) The frequency of oscillation $n$ of a vibration magnetometer is given by $n = \frac{1}{2 \pi} \sqrt{\frac{MH}{I}}$,where $M$ is the magnetic momen

Vedclass · Accepted Answer

$\sqrt{63}$

Vedclass · Answer

(B) The frequency of oscillation $n$ of a vibration magnetometer is given by $n = \frac{1}{2 \pi} \sqrt{\frac{MH}{I}}$,where $M$ is the magnetic moment,$H$ is the horizontal component of the earth's magnetic field,and $I$ is the moment of inertia. Thus,$n \propto \sqrt{B_{net}}$,where $B_{net}$ is the net magnetic field.Case $1$: Initially,$n_1 = 12$ oscillations/min with field $H$.Case $2$: With the bar magnet,$n_2 = 15$ oscillations/min. The net field is $H + H_1$,where $H_1$ is the field due to the bar magnet. $\frac{n_2}{n_1} = \sqrt{\frac{H + H_1}{H}} \Rightarrow \frac{15}{12} = \sqrt{1 + \frac{H_1}{H}} \Rightarrow \frac{25}{16} = 1 + \frac{H_1}{H} \Rightarrow \frac{H_1}{H} = \frac{9}{16}$.Case $3$: When poles are interchanged,the field becomes $H - H_1$. Let the new frequency be $n_3$.$\frac{n_3}{n_1} = \sqrt{\frac{H - H_1}{H}} = \sqrt{1 - \frac{H_1}{H}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.$n_3 = 12 	imes \frac{\sqrt{7}}{4} = 3\sqrt{7} = \sqrt{9 	imes 7} = \sqrt{63}$ oscillations per minute.

Similar Questions

$\cos \left[\cos ^{-1}\left(-\frac{1}{7}\right)+\sin ^{-1}\left(-\frac{1}{7}\right)\right]$ is equal to

What is the structure and hybridization of $Si(CH_3)_4$?

In the equation $\log S = \log C + Z \log A$,what does $S$ represent?

The temperature in $K$ at which $\Delta G = 0$ for a given reaction with $\Delta H = -20.5 \ kJ \ mol^{-1}$ and $\Delta S = -50.0 \ J \ K^{-1} \ mol^{-1}$ is:

Explain why $BeH_{2}$ molecule has a zero dipole moment although the $Be-H$ bonds are polar.

Vedclass Test Series

Exam Paper Generator

Online Exam Module