Units & Dimensions

JEE Main level, single correct, medium

76 of 149

Question 76 of 149

Match List I with List II List I List II A Angular momentum I [ML2T−2]\left[{ML}^{2}{T}^{-2}\right] B Torque II [ML−2T−2]\left[{ML}^{-2}{T}^{-2}\right] C Stress III [ML2T−1]\left[{ML}^{2}{T}^{-1}\right] D Pressure gradient IV [ML−1T−2]\left[{ML}^{-1}{T}^{-2}\right] Choose the correct answer from the options given below :

Show answer and solution

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

As we know, angular momentum is given by, L⃗=r⃗×p⃗⇒[L]=[M0L1T0][M1L1T−1]\vec{L}=\vec{r}\times \vec{p} \Rightarrow \left[L\right]=\left[{M}^{0}{L}^{1}{T}^{0}\right]\left[{M}^{1}{L}^{1}{T}^{-1}\right]
=[M1L2T−1]=\left[{M}^{1}{L}^{2}{T}^{-1}\right]
Torque is given by,
τ⃗=r⃗×F⃗⇒[τ]=[L1][MLT−2]\vec{\tau }=\vec{r}\times \vec{F} \Rightarrow \left[\tau \right]=\left[{L}^{1}\right]\left[{MLT}^{-2}\right]
=[ML2T−2]=\left[{ML}^{2}{T}^{-2}\right]
Also, Stress ≡\equiv Pressure =FA=\dfrac{F}{A}
Hence, [Stress] =[ML−1T−2]=\left[{ML}^{-1}{T}^{-2}\right]
Pressure Gradient =dPdx=\dfrac{dP}{dx}
Therefore, [Pressure Gradient]=[ML−2T−2]=\left[{ML}^{-2}{T}^{-2}\right]

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.