(A) [Y]=A(ℓΔℓ)F⇒ L2MLT−2=ML−1 T−2 (B) Torque (τ)=r×F (τ)=L×MLT−2=ML2 T−2(IV) (C) Coefficient of viscosity ⇒F=ηAdtdV η→Pa⋅sec [η]= L2MLT−2×T=ML−1 T−1 (D) Gravitational constant (G) $\begin{aligned}
& \mathrm{F}=\frac{\mathrm{GM}_1 \mathrm{M}_2}{\mathrm{r}^2} \
& {[\mathrm{G}]=\frac{\mathrm{F} \cdot \mathrm{r}^2}{\mathrm{m}_1 \mathrm{m}_2}=\frac{\mathrm{MLT}^{-2} \times \mathrm{L}^2}{\mathrm{M}^2}=\mathrm{M}^{-1} \mathrm{L}^3 \mathrm{T}^{-2}}
\end{aligned}$
JEE Main 2025 — Physics Mechanics
2025mcqmedium
Match List - I with List - II. 
Choose the correct answer from the options given below :
Official previous-year question
Held on 29 Jan 2025 · Verified 6 Jul 2026.
Options
- A
(A)-(I), (B)-(III), (C)-(II), (D)-(IV)
- B
(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
- C
(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
- D
(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
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