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NEET-UG Biology

Earth's magnetic field and magnetic elements — practice questions

21 questions in the bank on this idea. Below are 10 of them, exactly as they appear in a test.

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  1. Question 1 · difficulty L1 · recall

    The angle between geographic north and magnetic north in the horizontal plane is called

    • A. angle of dip
    • B. magnetic inclination only
    • C. magnetic declination
    • D. latitude
  2. Question 2 · difficulty L1 · understanding

    The angle made by Earth's magnetic field with the horizontal is called

    • A. declination
    • B. azimuth only
    • C. colatitude
    • D. angle of dip or inclination
  3. Question 3 · difficulty L2 · understanding

    If Earth's field magnitude is B and dip is I, the horizontal component is

    • A. B cos I
    • B. B sin I
    • C. B tan I
    • D. B/cos I
  4. Question 4 · difficulty L2 · application

    If Earth's field magnitude is B and dip is I, the vertical component magnitude is

    • A. B cos I
    • B. B sin I
    • C. B cot I
    • D. B/sin I
  5. Question 5 · difficulty L2 · understanding

    The vertical component of the earth's magnetic field is 6×105 T6 \times 10^{-5} \mathrm{~T} at any place where the angle of dip is 3737^{\circ}. The earth's resultant magnetic field at that place will be (\left(\right.Given tan37=34)\left.\tan 37^{\circ}=\frac{3}{4}\right)

    • A. 8×105 T8 \times 10^{-5} \mathrm{~T}
    • B. 6×105 T6 \times 10^{-5} \mathrm{~T}
    • C. 5×104 T5 \times 10^{-4} \mathrm{~T}
    • D. 1×104 T1 \times 10^{-4} \mathrm{~T}
  6. Question 6 · difficulty L2 · understanding

    An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1 ×\times 10 -4 Wbm -2 . The frequency of revolution of the electron will be : (Take mass of electron = 9.0 ×\times 10 -31 kg)

    • A. 1.6×105 Hz1.6 \times 10^{5} \mathrm{~Hz}
    • B. 5.6×105 Hz5.6 \times 10^{5} \mathrm{~Hz}
    • C. 2.8×106 Hz2.8 \times 10^{6} \mathrm{~Hz}
    • D. 1.8×106 Hz1.8 \times 10^{6} \mathrm{~Hz}
  7. Question 7 · difficulty L2 · understanding

    At a certain place the angle of dip is 30^\circ and the horizontal component of earth's magnetic field is 0.5 G. The earth's total magnetic field (in G), at that certain place, is :

    • A. 13{1 \over {\sqrt 3 }}
    • B. 12{1 \over 2}
    • C. 3\sqrt 3
    • D. 1
  8. Question 8 · difficulty L3 · application

    At a place where the dip is 60° and total field is B, the horizontal component equals

    • A. sqrt(3)B/2
    • B. 2B
    • C. B/2
    • D. B/sqrt(2)
  9. Question 9 · difficulty L3 · application

    If BH and BV are the horizontal and vertical field components, tan I equals

    • A. BH/BV
    • B. B/BH
    • C. BH·BV
    • D. BV/BH
  10. Question 10 · difficulty L3 · understanding

    A compass needle of oscillation magnetometer oscillates 20 times per minute at a place P\mathrm{P} of dip30\operatorname{dip} 30^{\circ}. The number of oscillations per minute become 10 at another place Q\mathrm{Q} of 6060^{\circ} dip. The ratio of the total magnetic field at the two places (BQ:BP)\left(B_{Q}: B_{P}\right) is :

    • A. 3:4\sqrt{3}: 4
    • B. 4:34: \sqrt{3}
    • C. 3:2\sqrt{3}: 2
    • D. 2:32: \sqrt{3}

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