JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2025
- Medium
- MCQ
Match List - I with List - II.
| List - I | List - II |
|---|---|
| (A) Young’s Modulus | (I) \(M L^{-1} T^{-1}\) |
| (B) Torque | (II) \(M L^{-1} T^{-2}\) |
| (C) Coefficient of Viscosity | (III) \(M^{-1} L^3 T^{-2}\) |
| (D) Gravitational Constant | (IV) \(M L^2 T^{-2}\) |
Choose the correct answer from the options given below :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
Step-by-step Solution
Detailed explanation
(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)
\(\begin{aligned}
& \mathrm{F}=\frac{\mathrm{GM}_1 \mathrm{M}_2}{\mathrm{r}^2} \\
& {[\mathrm{G}]=\frac{\mathrm{F} \cdot \mathrm{r}^2}{\mathrm{~m}_1 \mathrm{~m}_2}=\frac{\mathrm{MLT}^{-2} \times \mathrm{L}^2}{\mathrm{M}^2}=\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}}
\end{aligned}\)
(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)
\(\begin{aligned}
& \mathrm{F}=\frac{\mathrm{GM}_1 \mathrm{M}_2}{\mathrm{r}^2} \\
& {[\mathrm{G}]=\frac{\mathrm{F} \cdot \mathrm{r}^2}{\mathrm{~m}_1 \mathrm{~m}_2}=\frac{\mathrm{MLT}^{-2} \times \mathrm{L}^2}{\mathrm{M}^2}=\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}}
\end{aligned}\)
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