Match List-I with List-II.
Choose the correct answer from the options given below:

Question

Match List-I with List-II. Choose the correct answer from the options given below:

Accepted Answer

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

Find: The correct matching and hence the correct option.

Use dimensional analysis for each quantity.

For Young's Modulus:

\text{Young's Modulus} = \frac{\text{stress}}{\text{strain}}

Since strain is dimensionless,

[\text{Young's Modulus}] = [\text{stress}] = \frac{[\text{Force}]}{[\text{Area}]} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}

So, (A) $\to$ (I).

For Torque:

[\tau] = [\text{Force}] \times [\text{distance}] = (MLT^{-2})(L) = ML^2T^{-2}

So, (B) $\to$ (II).

For Coefficient of Viscosity: Using

\eta = \frac{\text{Force/Area}}{\text{velocity gradient}}

we get

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

So, (C) $\to$ (I).

For Gravitational Constant: From Newton's law of gravitation,

F = G\frac{m_1m_2}{r^2}

Hence,

G = \frac{Fr^2}{m_1m_2}

and therefore

[G] = \frac{(MLT^{-2})L^2}{M^2} = M^{-1}L^3T^{-2}

So, (D) $\to$ (IV).

Therefore, the final matching is (A)-(I), (B)-(II), (C)-(I), (D)-(IV). The correct option is D.

Given: Four physical quantities are to be matched with their dimensional formulas.

Find: The correct correspondence.

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

Hence, (A) $\to$ (I).

[\text{torque}] = (MLT^{-2})(L) = ML^2T^{-2}

Hence, (B) $\to$ (II).

\text{Force per unit area} = \eta \frac{dv}{dy}

\left[\frac{dv}{dy}\right] = T^{-1}

we get

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

Hence, (C) $\to$ (I).

F = G\frac{m_1m_2}{r^2}

G = \frac{Fr^2}{m_1m_2}

[G] = \frac{(MLT^{-2})L^2}{M^2} = M^{-1}L^3T^{-2}

Hence, (D) $\to$ (IV).

So the required match is (A)-(I), (B)-(II), (C)-(I), (D)-(IV), which corresponds to option D.

Answer