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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2025 · 29 January · Evening Shift · Physics · Units and Dimensions

Match List - I with List - II. Choose the correct answer from the...

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

Options

  1. A(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
  2. B(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  3. C(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
  4. D(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Correct answer

C. (A)-(II), (B)-(IV), (C)-(I), (D)-(III)

Step-by-step solution

(A) [Y]=FA(Δℓℓ)⇒MLT−2 L2=ML−1 T−2[\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 (τ⃗)=r→×F→(τ⃗)=L×MLT−2=ML2 T−2(IV)(\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 ⇒F=ηAdVdtη→Pa⋅sec[η]=MLT−2 L2×\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 T−1^{-1} \mathrm{~T}^{-1} (D) Gravitational constant (G) F=GM1M2r2[G]=F⋅r2 m1 m2=MLT−2×L2M2=M−1 L3 T−2\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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