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Q.The magnetic flux ϕ\phi (in Wb) linked with a coil is related to time tt (in s) as ϕ=5At2+Bt−2C\phi = 5At^2 + Bt - 2C. The SI units of AA and BB are respectively (A) Wb s2, Wb s\text{Wb s}^2,\ \text{Wb s} (B) Wb s−1, Wb\text{Wb s}^{-1},\ \text{Wb} (C) Wb s−2, Wb s−1\text{Wb s}^{-2},\ \text{Wb s}^{-1} (D) Wb s−1, Wb s−2\text{Wb s}^{-1},\ \text{Wb s}^{-2}

CBSECBSE Class XII Board 2026MCQ· 1mImportance★★★★★
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The key idea is dimensional consistency: each term in ϕ=5At2+Bt−2C\phi = 5At^2 + Bt - 2C must have the same unit as ϕ\phi (Wb). This forces AA to have units Wb s−2\text{Wb s}^{-2} and BB to have units Wb s−1\text{Wb s}^{-1}, so the correct option is (C).

The problem gives you a relation between magnetic flux ϕ\phi (in webers) and time tt (in seconds):

ϕ=5At2+Bt−2C.\phi = 5At^2 + Bt - 2C.

You’re asked for the SI units of AA and BB. The constants 55 and 22 are pure numbers — they have no units. So the only way this equation makes physical sense is if every term on the right-hand side has the same unit as ϕ\phi, which is the weber (Wb). This is the principle of dimensional homogeneity, and it’s the entire foundation of the solution.

Let’s apply it term by term.

  1. First term: 5At25At^2 Since 55 is dimensionless, the unit of 5At25At^2 is the unit of AA multiplied by the unit of t2t^2. Time tt is in seconds, so t2t^2 has unit s2\text{s}^2. For this term to equal a flux in Wb, we need:

unit of A×s2=Wb.\text{unit of }A \times \text{s}^2 = \text{Wb}.

Therefore:

unit of A=Wbs2=Wb s−2.\text{unit of }A = \frac{\text{Wb}}{\text{s}^2} = \text{Wb s}^{-2}.

  1. Second term: BtBt Here BB is multiplied by tt (unit s). So:

unit of B×s=Wb.\text{unit of }B \times \text{s} = \text{Wb}.

Hence:

unit of B=Wbs=Wb s−1.\text{unit of }B = \frac{\text{Wb}}{\text{s}} = \text{Wb s}^{-1}.

  1. Third term: −2C-2C …
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