Match List I with List II for a cyclotronLIST ILIST IIA.Frequency of…

The correct answer is Option (4) → A-IV, B-III, C-II, D-I

Inside a dee the particle moves on a circle because the magnetic force \( qvB \) supplies the centripetal force: \( qvB = mv^{2}/r \), so \( r = mv/(qB) \). Every entry follows from this.

(A) Frequency of revolution of charged particle — this matches (IV) \( Bq/2\pi m \). The time for one circle is \( 2\pi r/v = 2\pi m/(qB) \), independent of speed; this is the cyclotron frequency.

(B) Velocity of charged particle — this matches (III) \( qBR/m \). From \( r = mv/(qB) \) with r = R.

(C) Kinetic energy of charged particle — this matches (II) \( q^2B^2R^2/2m \). \( \frac{1}{2}mv^{2} = \frac{1}{2}m(qBR/m)^{2} \).

(D) Force on charged particle when it is inside dee — this matches (I) \( Bqv \). Inside a dee there is no electric field, only the magnetic force.

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
Cyclotron – “… They move in a semi-circular path in one of the dees and arrive in the gap between the dees in a time interval T/2; where T, the period of revolution, is given by Eq. (4.6),

    \( \displaystyle T = \frac{1}{\nu_c} = \frac{2\pi m}{qB} \quad \text{or} \quad \nu_c = \frac{qB}{2\pi m} \qquad (4.8) \)

This frequency is called the cyclotron frequency for obvious reasons and is denoted by \( \nu_c \).”

Cyclotron – “… From Eq. (4.5) we have,

    \( \displaystyle v = \frac{qBR}{m} \qquad (4.9) \)

where R is the radius of the trajectory at exit, and equals the radius of a dee. Hence, the kinetic energy of the ions is,

    \( \displaystyle \frac{1}{2}mv^2 = \frac{q^2B^2R^2}{2m} \qquad (4.10) \)”