Question

In the formula, X=3YZ2,X and Z have dimensions of capacitance and magnetic field respectively. What are the dimensions of Y in the MKSQ system?

• A. (A) [M−3 L−1 T3Q4]
• B. (B) [M−3 L−2 T4Q4
• C. (C) [M−2L−2 T4Q4]
• D. (D) [M−3 L−2 T4Q]

Answer 1

Answer: (B)

(B) [M−3 L−2 T4Q4

Answer 2

Explanation:
The magnetic force (F) experienced by a moving charged particle(q) in a uniform magnetic field (B) is given byF=qvBsin⁡θ⇒B=Fqvsin⁡θwhere, θ is angle between the magnetic field and the velocity of the particle.The dimensions of charge, force and velocity are[q]=[Q][F]=[MLT−2][v]=[LT−1]sin⁡θ is a dimensionless quantity. Now, the dimension of magnetic field is[B]=[F]|q|v∣sin⁡θ]=[MLT−2][Q][LT−1]=[MT−1Q−1]The energy stored in a capacitor is given byE=q22Cwhere, C is capacitance.The dimension of energy is[E]=[ML2T−2]Thus, the dimension of capacitance is[C]=[q2][E]⇒[C]=[Q2][ML2T−2]⇒[C]=[M−1L−2T2Q2]Given: X=3YZ2 where, X and Z have the dimensions of capacitance and magnetic field respectively. Thus, [X]=[C]=[M−1L−2T2Q2[z]=[B]=[MT−1Q−1]Therefore, the dimension of Y is [Y]=[X][Z]2⇒[Y]=[M−1 L−2 T2Q2][MT−1Q−1]2⇒[Y]=[M−1L−2T2Q2]|M2T−2Q−2|⇒[Y]=[M−3L−2T4Q4]Hence, the correct option is (B).