Units & Dimensions

JEE Main level, single correct, medium

43 of 149

Question 43 of 149

Match List - I with List - II.


Choose the correct answer from the options given below :

Show answer and solution

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

(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→(\vec{\tau})=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}
(τ⃗)=L×MLT−2=ML2 T−2(IV)(\vec{\tau})=\mathrm{L} \times \mathrm{MLT}^{-2}=\mathrm{ML}^2 \mathrm{~T}^{-2}(\mathrm{IV})
(C) Coefficient of viscosity ⇒F=ηAdVdt\Rightarrow F=\eta A \frac{d V}{d t}
η→Pa⋅sec\eta \rightarrow \mathrm{Pa} \cdot \mathrm{sec}
[η]=MLT−2 L2×T=ML−1 T−1[\eta]=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2} \times \mathrm{T}=\mathrm{ML}^{-1} \mathrm{~T}^{-1}
(D) Gravitational constant (G)
$\begin{aligned}
& \mathrm{F}=\frac{\mathrm{GM}_1 \mathrm{M}_2}{\mathrm{r}^2} \\
& {[\mathrm{G}]=\frac{\mathrm{F} \cdot \mathrm{r}^2}{\mathrm{~m}_1 \mathrm{~m}_2}=\frac{\mathrm{MLT}^{-2} \times \mathrm{L}^2}{\mathrm{M}^2}=\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}}
\end{aligned}$

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Units & Dimensions (Dimensions), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.