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MHT CET 202625 AprilEvening Shift
Exam
MHT CET
Year
2026
Date
25 April
Shift
Evening Shift
Subject
Physics
Chapter
Mathematics in Physics

MHT CET 2026 · 25 April · Evening Shift · Physics · Mathematics in Physics

The heat generated in a circuit is given by Q = I^2 R t . If the...

The heat generated in a circuit is given by Q=I2RtQ = I^2 R t. If the percentage error in the measurement of QQ is 1%1\%, RR is 2%2\%, and tt is 1%1\%, then the percentage error in the measurement of current II is

Options

  1. A3%3\%
  2. B2%2\%
  3. C4%4\%
  4. D5%5\%

Correct answer

B. 2%2\%

Step-by-step solution

Given Q=I2RtQ = I^2 R t

Rearranging the formula for II, we get:

I=(QRt)1/2I = \left(\dfrac{Q}{R t}\right)^{1/2}

The maximum percentage error in II is given by:

ΔII×100=12(ΔQQ×100)+12(ΔRR×100)+12(Δtt×100)\dfrac{\Delta I}{I} \times 100 = \dfrac{1}{2} \left( \dfrac{\Delta Q}{Q} \times 100 \right) + \dfrac{1}{2} \left( \dfrac{\Delta R}{R} \times100\right) + \dfrac{1}{2} \left( \dfrac{\Delta t}{t} \times100\right)

Substituting the given percentage errors:

ΔII×100=12(1%)+12(2%)+12(1%)\dfrac{\Delta I}{I} \times100 = \dfrac{1}{2}(1\%) + \dfrac{1}{2}(2\%) + \dfrac{1}{2}(1\%)

ΔII×100=0.5%+1%+0.5%=2%\dfrac{\Delta I}{I} \times100 = 0.5\% + 1\% + 0.5\% = 2\%

Answer: 2%2\%
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