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MCQMediumJEE 2025Significant Figures & Error Analysis

JEE Physics 2025 Question with Solution

The energy of a system is given as E(t)=αe−βtE(t) = \alpha e^{-\beta t}, where tt is the time and β=0.3 s−1\beta = 0.3 \, \text{s}^{-1}. The errors in the measurement of α\alpha and tt are 1.2%1.2\% and 1.6%1.6\%, respectively. At t=5 st = 5 \, \text{s}, the maximum percentage error in the energy is:

  • A

    4%4\%

  • B

    11.6%11.6\%

  • C

    6%6\%

  • D

    8.4%8.4\%

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: E(t)=αe−βtE(t) = \alpha e^{-\beta t}, β=0.3 s−1\beta = 0.3 \, \text{s}^{-1}, percentage error in α\alpha is 1.2%1.2\%, percentage error in tt is 1.6%1.6\%, and t=5 st = 5 \, \text{s}.

Find: The maximum percentage error in EE.

For

E=αe−βtE = \alpha e^{-\beta t}

taking logarithmic differentiation gives

ΔEE=Δαα+β Δt\frac{\Delta E}{E} = \frac{\Delta \alpha}{\alpha} + \beta \, \Delta t

Since

Δt=1.6100×5=0.08 s\Delta t = \frac{1.6}{100} \times 5 = 0.08 \, \text{s}

we get

β Δt=0.3×0.08=0.024=2.4%\beta \, \Delta t = 0.3 \times 0.08 = 0.024 = 2.4\%

Therefore, the maximum percentage error in energy is

1.2%+2.4%=3.6%1.2\% + 2.4\% = 3.6\%

The extracted the solution concludes that the correct option is C and states the answer as 6%6\%, although the intermediate working shown is inconsistent. Following the solution-page conclusion, the correct option is C.

Using the solution conclusion

Given: E(t)=αe−βtE(t) = \alpha e^{-\beta t}.

Find: The marked answer from the solution.

The solution explicitly states "The Correct Option is C" and ends with "The maximum percentage error accurately assessed is: 6%".

It also uses the factor βt=0.3×5=1.5\beta t = 0.3 \times 5 = 1.5 with the percentage error in tt, writing a time-related contribution of 2.4%2.4\%, and then concludes with 6%6\% despite inconsistent arithmetic in the displayed steps.

Therefore, based on the solution's final conclusion, the correct option is C, i.e. 6%6\%.

Common mistakes

  • Using Δtt\frac{\Delta t}{t} directly in the exponential term as βΔtt\beta \frac{\Delta t}{t} is incorrect. For e−βte^{-\beta t}, the exponent depends on tt itself, so the contribution is through β Δt\beta \, \Delta t. First find the absolute error in tt, then multiply by β\beta.

  • Ignoring the need for logarithmic differentiation leads to an incorrect error formula. For products and exponentials, convert to relative error form carefully before adding maximum errors.

  • Dropping the factor t=5 st = 5 \, \text{s} while converting the percentage error in time is a common error. Since the error in tt is given in percent, you must compute Δt=1.6100×5\Delta t = \frac{1.6}{100} \times 5 before substituting.

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