Mathematics in Physics JEE Advanced previous year questions with solutions

4 solved JEE Advanced questions on Mathematics in Physics, free to read — no sign-in needed. The full chapter has 13 questions; sign in to attempt the remaining 9 in the exam simulator.

Answers checked against the official answer keys.

1. Q1JEE Advanced Adv 2025 (Paper 1)
   Length, breadth and thickness of a strip having a uniform cross section are measured to be 10.5 cm , 0.05 mm , and $6.0 \mu \mathrm{~m}$, respectively. Which of the following option(s) give(s) the volume of the strip in $\mathrm{cm}^3$ with correct significant figures:
   1. A.$3.2 \times 10^{-5}$
   2. B.$32.0 \times 10^{-6}$
   3. C.$3.0 \times 10^{-5}$
   4. D.$3 \times 10^{-5}$
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   Answer: (D)

   $\mathrm{L}=10.5 \mathrm{~cm} \rightarrow 3$ significant digits $\mathrm{b}=0.05 \mathrm{~cm} \rightarrow 1$ significant digit $\mathrm{t}=6.0 \mu \mathrm{~m} \rightarrow 2$ significant digits Volume, $\mathrm{V}=$ Lbt must have only 1 significant digit $\begin{aligned} & \Rightarrow \mathrm{V}=10.5 \times 0.05 \times 10^{-1} \times 6.0 \times 10^{-4} \mathrm{~cm}^3 \\ & =3 \times 10^{-5} \mathrm{cc} \end{aligned}$
2. Q2JEE Advanced Adv 2020 (Paper 2)
   Two capacitors with capacitance values ${C}_{1}=\left(2000\pm 10\right)pF$ and ${C}_{2}=\left(3000\pm 15\right)pF$ are connected in series. The voltage applied across this combination is $V=\left(5.00\pm 0.02\right)V$. The percentage error in the calculation of the energy stored in this combination of capacitors is __________.
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   Answer: 1.3

   For the purpose of calculation of error, fundamental formula is considered $\dfrac{1}{C}=\dfrac{1}{{C}_{1}}+\dfrac{1}{{C}_{2}}\Rightarrow C=1200pF$ $-\dfrac{dC}{{C}_{1}^{2}}=-\dfrac{d{C}_{1}}{{C}_{1}^{2}}-\dfrac{d{C}_{2}}{{C}_{2}^{2}}$ $dC=6pF$ Equivalent capacitance $=\left(1200\pm 6\right)pF$ $E=1/2{CV}^{2}$ $(dE/E=dC/C+2dV/V)\times 100$ $=1.3$
3. Q3JEE Advanced Adv 2017 (Paper 2)
   A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is $\delta T=0.01$ seconds and he measures the depth of the well to be L = 20 meters. Take the acceleration due to gravity $g=10m{s}^{-2}$ and the velocity of sound is $300m{s}^{-1}$ . Then the fractional error in the measurement, $\dfrac{\delta L}{L}$ , is closest to
   1. A.0.2%
   2. B.5%
   3. C.3%
   4. D.1%
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   Answer: (D)

   Total time taken, $T=\sqrt{\dfrac{2L}{g}}+\dfrac{L}{c}$, $c$ is the sound speed in air. Now, for an error $\delta L$ in L, We have an error $\delta T$ in T So, $T+\delta T=\sqrt{\dfrac{2\left(L+\delta L\right)}{g}}+\dfrac{\left(L+\delta L\right)}{c}$ $=\sqrt{\dfrac{2L}{g}\left(1+\dfrac{\delta L}{L}\right)}+\dfrac{L}{c}\left(1+\dfrac{\delta L}{L}\right)$ Since, $\dfrac{\delta T}{T}$ is very small, hence $\dfrac{\delta L}{L}$ is also small, so taking binomial approximation $T+\delta T=\sqrt{\dfrac{2L}{g}}\left(1+\dfrac{1}{2}\dfrac{\delta L}{L}\right)+\dfrac{L}{c}\left(1+\dfrac{\delta L}{L}\right)$ $\Rightarrow T+\delta T=\left(\sqrt{\dfrac{2L}{g}}\right)+\sqrt{\dfrac{2L}{g}}\left(\dfrac{1}{2}\dfrac{\delta L}{L}\right)+\left(\dfrac{L}{c}\right)+\dfrac{L}{c}\left(\dfrac{\delta L}{L}\right)$ $\Rightarrow \delta T=\sqrt{\dfrac{2L}{g}}\left(\dfrac{1}{2}\dfrac{\delta L}{L}\right)+\dfrac{L}{c}\left(\dfrac{\delta L}{L}\right)$ $\Rightarrow \delta T=\left[\sqrt{\dfrac{L}{2g}}+\left(\dfrac{L}{c}\right)\right]\dfrac{\delta L}{L}$ $\Rightarrow 0.01=\left[\sqrt{\dfrac{20}{2\times 10}}+\left(\dfrac{20}{300}\right)\right]\dfrac{\delta L}{L}$ $\Rightarrow 0.01=\left[1+\dfrac{1}{15}\right]\dfrac{\delta L}{L} \Rightarrow \dfrac{\delta L}{L}=\dfrac{0.15}{16}$ $$ $=\dfrac{15}{16}$ $\approx 1$
4. Q4JEE Advanced Adv 2015 (Paper 2)
   The energy of a system as a function of time $t$ is given as $E\left(t\right)={A}^{2}\left(-\alpha t\right)$ , where $\alpha =0.2{s}^{-1}$ . The measurement of $A$ has an error of 1.25%. If the error in the measurement of time is 1.50%, the percentage error in the value of $E(t)$ at $t=5$ s is
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   Answer: 4

   Energy $E={A}^{2}{e}^{-\alpha t}$ For small % errors, we can, do differentiation $dE=2A\left(dA\right){e}^{-\alpha t}+{A}^{2}\left(-\alpha {e}^{-\alpha t}dt\right)$ Fractional error $=\dfrac{dE}{E}=\dfrac{2A{e}^{-\alpha t}\left(dA\right)+\left(-\alpha {A}^{2}{e}^{-\alpha t}\right)dt}{{A}^{2}{e}^{-\alpha t}}=2\left(\dfrac{dA}{A}\right)+\left(-\alpha \dfrac{dt}{t}\right)t$ % error $=2\left(1.25\%\right)+\left(0.2\times 1.5\%\right)\times 5$ $=4\%$ (errors always add up) Alternate solution: $E={A}^{2}{e}^{-\alpha t}$ Taking natural logarithm on both sides, $ln\left(E\right)=ln\left({A}^{2}\right)+\left(-\alpha t\right)$ Differentiating $\dfrac{dE}{E}=2\left(\dfrac{dA}{A}\right)+\left(-\alpha dt\right)$ For small fractional erros, errors always add up $\left|\dfrac{dE}{E}\right|=2\left|\dfrac{dA}{A}\right|+\alpha \left(\dfrac{dt}{t}\right)\times t$ $=2\left(1.25\%\right)+\left(0.2\right)\left(1.5\%\right)5$ $=4\%$