Earth Magnetism — NEET Physics PYQ MCQs with Solutions
Free NEET Physics PYQ Earth Magnetism MCQs with step-by-step solutions (7 questions). Part of Magnetism and Matter. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Earth Magnetism · easy · theory
The relations amongst the three elements of earth's magnetic field, namely horizontal component $H$, vertical component $V$ and dip $\delta$, are ($B_E=$ total magnetic field)
A. $V=B_E\tan\delta,\,H=B_E$
B. $V=B_E\sin\delta,\,H=B_E\cos\delta$ ✓ Correct
C. $V=B_E\cos\delta,\,H=B_E\sin\delta$
D. $V=B_E,\,H=B_E\tan\delta$
Solution: From the vector triangle of the earth's magnetic field, the horizontal component is $H=B_E\cos\delta$ and the vertical component is $V=B_E\sin\delta$.
Q2 — Earth Magnetism · medium · theory
At a point $A$ on the earth's surface the angle of dip is $\delta=+25^\circ$. At a point $B$ on the earth's surface the angle of dip is $\delta=-25^\circ$. We can interpret that
A. $A$ is located in the southern hemisphere and $B$ is located in the northern hemisphere
B. $A$ is located in the northern hemisphere and $B$ is located in the southern hemisphere ✓ Correct
C. $A$ and $B$ are both located in the southern hemisphere
D. $A$ and $B$ are both located in the northern hemisphere
Solution: The angle of dip is taken positive in the northern hemisphere and negative in the southern hemisphere, so $A$ (dip $+25^\circ$) lies in the northern hemisphere and $B$ (dip $-25^\circ$) lies in the southern hemisphere.
Q3 — Earth Magnetism · medium · theory
If $\delta_1$ and $\delta_2$ be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip $\delta$ is given by
A. $\cot^2\delta=\cot^2\delta_1+\cot^2\delta_2$ ✓ Correct
B. $\tan^2\delta=\tan^2\delta_1+\tan^2\delta_2$
C. $\cot^2\delta=\cot^2\delta_1-\cot^2\delta_2$
D. $\tan^2\delta=\tan^2\delta_1-\tan^2\delta_2$
Solution: With $\cot\delta_1=\dfrac{B_H\cos\theta}{B_V}$ and $\cot\delta_2=\dfrac{B_H\sin\theta}{B_V}$, squaring and adding gives $\cot^2\delta_1+\cot^2\delta_2=\dfrac{B_H^2}{B_V^2}=\cot^2\delta$.
Q4 — Earth Magnetism · medium · numerical
A vibration magnetometer placed in the magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2\,\text{s}$ in the earth's horizontal magnetic field of $24\,\mu\text{T}$. When a horizontal field of $18\,\mu\text{T}$ is produced opposite to the earth's field by placing a current carrying wire, the new time period of the magnet will be
A. $1\,\text{s}$
B. $2\,\text{s}$ ✓ Correct
C. $3\,\text{s}$
D. $4\,\text{s}$
Solution: Time period of a vibration magnetometer is $T=2\pi\sqrt{\dfrac{I}{MB_H}}$, so $T\propto \dfrac{1}{\sqrt{B_H}}$. Thus $\dfrac{T_1}{T_2}=\sqrt{\dfrac{(B_H)_2}{(B_H)_1}}\Rightarrow \dfrac{2}{T_2}=\sqrt{\dfrac{18}{24}}\Rightarrow T_2\approx 2\,\text{s}$.
Q5 — Earth Magnetism · medium · theory
Two bar magnets having the same geometry with magnetic moments $M$ and $2M$ are firstly placed in such a way that their similar poles are on the same side, then its period of oscillation is $T_1$. Now, the polarity of one of the magnets is reversed and the time period of oscillations becomes $T_2$. Then,
A. $T_1<T_2$ ✓ Correct
B. $T_1=T_2$
C. $T_1>T_2$
D. $T_2=\infty$
Solution: With similar poles together, net moment $M_1=M+2M=3M$ and $T_1=2\pi\sqrt{\dfrac{I}{3MH}}$. With one polarity reversed, net moment $M_2=2M-M=M$ and $T_2=2\pi\sqrt{\dfrac{I}{MH}}$. Since $M_1>M_2$, $T_1<T_2$.
Q6 — Earth Magnetism · medium · theory
Due to the earth's magnetic field, charged cosmic ray particles
A. can never reach the poles
B. can never reach the equator
C. require less kinetic energy to reach the equator than the poles
D. require greater kinetic energy to reach the equator than the poles ✓ Correct
Solution: At the poles the earth's field is vertical and parallel to the particle's velocity, so the magnetic force $qv B\sin\theta=0$ and particles reach easily. At the equator the field is horizontal (perpendicular to $v$), so the deflecting force $qvB$ is maximum; only particles with greater kinetic energy can overcome this and reach the equator.
Q7 — Earth Magnetism · medium · theory
A bar magnet is oscillating in the earth's magnetic field with a period $T$. What happens to its period of motion if its mass is quadrupled?
A. Motion remains simple harmonic with new period $\dfrac{T}{2}$
B. Motion remains simple harmonic with new period $2T$ ✓ Correct
C. Motion remains simple harmonic with new period $4T$
D. Motion remains simple harmonic and the period stays nearly constant
Solution: $T=2\pi\sqrt{\dfrac{I}{MB}}$, so $T\propto\sqrt{I}$. Since $I\propto m$ (for the same geometry), quadrupling the mass quadruples $I$, so $T$ becomes $\sqrt{4}=2$ times, i.e. new period $=2T$.