JEE Physics 2023 Question with Solution

Given: Match the physical quantities in List I with their dimensional formulas in List II.

Find: The correct correspondence and hence the correct option.

For Young's Modulus $Y$,

$$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{(F/A)}{(\Delta L/L)}$$

Since strain is dimensionless, the dimensions of $Y$ are the same as stress:

$$[Y] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$$

So, A (\rightarrow) III.

For co-efficient of viscosity $\eta$, using

$$F = 6\pi \eta r v$$

we get

$$\eta = \frac{F}{6\pi r v}$$

Hence,

$$[\eta] = \frac{[MLT^{-2}]}{[L][LT^{-1}]} = [ML^{-1}T^{-1}]$$

So, B (\rightarrow) I.

For Planck's constant $h$, from

$$E = h\nu$$

we have

$$h = \frac{E}{\nu}$$

Therefore,

$$[h] = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$$

So, C (\rightarrow) II.

For work function $\phi$, it has the same dimensions as energy:

$$[\phi] = [ML^2T^{-2}]$$

So, D (\rightarrow) IV.

Thus the matching is A-III, B-I, C-II, D-IV. Therefore, the correct option is B.

The solution labels the option as D, but its own written matching is A-III, B-I, C-II, D-IV, which corresponds to option B in the given choices.

Given: Four physical quantities and four dimensional formulas.

Find: Which dimensional formula belongs to each quantity.

1. Young's Modulus is stress divided by strain. Since strain has no dimensions,

$$[Y] = [\text{stress}] = \frac{[F]}{[A]} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$$

Hence A (\rightarrow) III.

2. For co-efficient of viscosity,

$$F = 6\pi \eta r v$$

So,

$$[\eta] = \frac{[F]}{[r][v]} = \frac{[MLT^{-2}]}{[L][LT^{-1}]} = [ML^{-1}T^{-1}]$$

Hence B (\rightarrow) I.

3. For Planck's constant,

$$E = h\nu$$

Therefore,

$$[h] = \frac{[E]}{[\nu]} = \frac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$$

Hence C (\rightarrow) II.

4. Work function is an energy quantity, so

$$[\phi] = [ML^2T^{-2}]$$

Hence D (\rightarrow) IV.

So the final matching is A-III, B-I, C-II, D-IV, giving option B.