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← JEE Advanced 2015 · Paper 2
PhysicsInteger+4Question 2 of 60

JEE Advanced 2015 · Paper 2, Physics question 2

The energy of a system as a function of time tt is given as E(t)=A2exp⁡(−αt)E(t)=A^2\exp(-\alpha t), where α=0.2 s−1\alpha=0.2\ \mathrm s^{-1}. The measurement of AA has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of E(t)E(t) at t=5t=5 s is

Answer and worked solution

Official answer: 4

Relative error. ln⁡E=2ln⁡A−αt\ln E=2\ln A-\alpha t, so
ΔEE=2ΔAA+αtΔtt.\frac{\Delta E}E=2\frac{\Delta A}A+\alpha t\frac{\Delta t}t.

At t = 5 s.
2(1.25%)+0.2×5×1.5%=2.5%+1.5%=4%.2(1.25\%)+0.2\times5\times1.5\%=2.5\%+1.5\%=4\%.

Answer: 4