Experimental Physics JEE Advanced previous year questions with solutions

4 solved JEE Advanced questions on Experimental Physics, free to read — no sign-in needed. The full chapter has 14 questions; sign in to attempt the remaining 10 in the exam simulator.

1. Q1JEE Advanced Adv 2022 (Paper 2)
   Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5mm$. The circular scale has $100$ divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below. Measurement condition Main scale reading Circular scale reading Two arms of gauge touching each other without wire $0$ division $4$ divisions Attempt-$1$: With wire $4$ divisions $20$ divisions Attempt-$2$: With wire $4$ divisions $16$ divisions What are the diameter and cross-sectional area of the wire measured using the screw gauge?
   1. A.$2.22\pm 0.02mm,\pi \left(1.23\pm 0.02\right){mm}^{2}$
   2. B.$2.22\pm 0.01mm,\pi \left(1.23\pm 0.01\right){mm}^{2}$
   3. C.$2.14\pm 0.02mm,\pi \left(1.14\pm 0.02\right){mm}^{2}$
   4. D.$2.14\pm 0.01mm,\pi \left(1.14\pm 0.01\right){mm}^{2}$
   Show answer & solution

   Answer: (C)

   Note: This question is given bonus by JEE council. In one rotation 2 divisions of the main scale are crossed. Therefore, the least count of the screw gauge is $LC=\dfrac{2\times 0.5}{100}=0.01mm$ And the zero error is $4\times LC=0.04mm$ Reading-1 ${R}_{1}=MSR+LC\times CSR-\left(Zeroerror\right)$ $=\left(2+0.20-0.04\right)mm$ $=2.16mm$ Reading-2 ${R}_{2}=\left(2+0.16-0.04\right)mm=2.12mm$ Therefore, average reading ${R}_{m}=\dfrac{{R}_{1}+{R}_{2}}{2}=2.14mm$ Average mean error $=\dfrac{\left|{R}_{m}-{R}_{1}\right|+\left|{R}_{m}-{R}_{2}\right|}{2}=0.02mm$ $\Rightarrow$ Diameter $d=\left(2.14\pm 0.02\right)mm$ Area $=\dfrac{\pi {d}^{2}}{4}$ $\Rightarrow ∆Area=\dfrac{2\pi d}{4}\left(∆d\right)$ $\Rightarrow ∆Area\cong \pi \left(0.02\right)$ $\Rightarrow Area=\dfrac{\pi {d}^{2}}{4}\pm ∆Area=\pi \left(1.14\pm 0.02\right){mm}^{2}$
2. Q2JEE Advanced Adv 2015 (Paper 1)
   Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
   1. A.If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm
   2. B.If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm
   3. C.If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm
   4. D.If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm
   Show answer & solution

   Answer: B,C

   1 main scale division (M.S.D) $=\dfrac{1}{8}cm$ 5 Vernier scale division (V.S.D) $=4M.S.D$ $1V.S.D.=\dfrac{4}{5}M.S.D$ Least count of Vernier scale $\left(L.C.\right)=1M.S.D.-1V.S.D.$ $=1M.S.D.-\dfrac{4}{5}M.S.D$ $\left(L.C\right)=\dfrac{1M.S.D}{5}=\dfrac{1}{40}cm$ For option A and B If the pitch of the screw gauge is twice the least count of the Vernier callipers then pitch $=2\times L.C.$ of Vernier scale $=\dfrac{1}{20}cm$ Hence least count of screw gauge $=\dfrac{Pitch}{100}$ $=\dfrac{0.50}{100}$ $=0.005m$ For option C and D Least count of linear scale of screw gauge $=2\times \dfrac{1}{40}=\dfrac{1}{20}cm$ Pitch $=2\times \dfrac{1}{20}=\dfrac{1}{10}cm=1mm$ Least count of screw gauge $=\dfrac{1mm}{100}=0.01mm$ Hence answer is (B, C)
3. Q3JEE Advanced Adv 2013 (Paper 1)
   The diameter of a cylinder is measured using a vernier calipers with no zero error. It is found that the zero of the vernier scale lies between $5.10cm$ and $5.15cm$ of the main scale. The vernier scale has $50$ divisions equivalent to $2.45cm$. The ${24}^{th}$ division of the vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is,
   1. A.$5.112cm$
   2. B.$5.124cm$
   3. C.$5.136cm$
   4. D.$5.148cm$
   Show answer & solution

   Answer: (B)

   $\text{1 MSD}=\text{0.05 ​}$ $\text{1 VSD}=\dfrac{\text{2.45}}{50}=\text{0.04}$ $\text{Diameter of the cylinder}$ $=\text{MSR}+\text{L.C.}\times 24$ $=\text{5.124 cm}$
4. Q4JEE Advanced Adv 2010 (Paper 2)
   A vernier calipers has $1 \mathrm{~mm}$ marks on the main scale. It has 20 equal divisions on the vernier scale which match with 16 main scale divisions. For this vernier calipers, the least count is
   1. A.$0.02 \mathrm{~mm}$
   2. B.$0.05 \mathrm{~mm}$
   3. C.$0.1 \mathrm{~mm}$
   4. D.$0.2 \mathrm{~mm}$
   Show answer & solution

   Answer: (D)

   Least count of vernier calipers $$\begin{aligned} & \text { LC }=1 \mathrm{MSD}-1 \text { VSD } \\ & =\frac{\text { Smallest division on main scale }}{\text { Number of divisions on vernier scale }} \\ & 20 \text { divisions of vernier scale } \\ & =16 \text { divisions of main scale } \\ & \therefore \quad 1 \mathrm{VSD}=\frac{16}{20} \mathrm{~mm}=0.8 \mathrm{~mm} \\ & \therefore \quad \text { LC }=1 \mathrm{MSD}-1 \mathrm{VSD} \\ & =1 \mathrm{~mm}-0.8 \mathrm{~mm} \\ & \therefore \quad=0.2 \mathrm{~mm} \\ & \end{aligned}$$ $\therefore$ The correct option is (d).