JEE Main 2024 (Online) 9th April Morning Shift | Units & Measurements Question 2 | Physics

One main scale division of a vernier caliper is equal to $$\mathrm{m}$$ units. If $$\mathrm{n}^{\text {th }}$$ division of main scale coincides with $$(n+1)^{\text {th }}$$ division of vernier scale, the least count of the vernier caliper is :

$$\frac{1}{(\mathrm{n}+1)}$$

$$\frac{m}{(n+1)}$$

$$\frac{n}{(n+1)}$$

$$\frac{\mathrm{m}}{\mathrm{n}(\mathrm{n}+1)}$$

JEE Main 2024 (Online) 8th April Evening Shift

MCQ (Single Correct Answer)

-1

If $$\epsilon_{\mathrm{o}}$$ is the permittivity of free space and $$\mathrm{E}$$ is the electric field, then $$\epsilon_{\mathrm{o}} \mathrm{E}^2$$ has the dimensions :

$$[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}]$$

$$[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2}]$$

$$[\mathrm{M}^{\mathrm{0}} \mathrm{L}^{-2} \mathrm{TA}]$$

$$[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2]$$

JEE Main 2024 (Online) 8th April Evening Shift

MCQ (Single Correct Answer)

-1

There are 100 divisions on the circular scale of a screw gauge of pitch $$1 \mathrm{~mm}$$. With no measuring quantity in between the jaws, the zero of the circular scale lies 5 divisions below the reference line. The diameter of a wire is then measured using this screw gauge. It is found that 4 linear scale divisions are clearly visible while 60 divisions on circular scale coincide with the reference line. The diameter of the wire is :

4.65 mm

4.60 mm

4.55 mm

3.35 mm

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