Q57 / 75 · JEE Main 2025 (29 Jan Shift 2) · Physics

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

  • A
    (A)-(I), (B)-(III), (C)-(II), (D)-(IV)
  • B
    (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  • Correct answer
    C
    (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
  • D
    (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Solution

(A) $[\mathrm{Y}]=\frac{\mathrm{F}}{\mathrm{A}\left(\frac{\Delta \ell}{\ell}\right)} \Rightarrow \frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}$ (B) Torque $(\vec{\tau})=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$ $(\vec{\tau})=\mathrm{L} \times \mathrm{MLT}^{-2}=\mathrm{ML}^2 \mathrm{~T}^{-2}(\mathrm{IV})$ (C) Coefficient of viscosity $\Rightarrow F=\eta A \frac{d V}{d t}$ $\eta \rightarrow \mathrm{Pa} \cdot \mathrm{sec}$ $[\eta]=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2} \times \mathrm{T}=\mathrm{ML}^{-1} \mathrm{~T}^{-1}$ (D) Gravitational constant (G) $\begin{aligned} & \mathrm{F}=\frac{\mathrm{GM}_1 \mathrm{M}_2}{\mathrm{r}^2} \\ & {[\mathrm{G}]=\frac{\mathrm{F} \cdot \mathrm{r}^2}{\mathrm{~m}_1 \mathrm{~m}_2}=\frac{\mathrm{MLT}^{-2} \times \mathrm{L}^2}{\mathrm{M}^2}=\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}} \end{aligned}$

Hence, the correct answer is option: 'C'.

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