Verified Solution • 2025–26 Session moderate

QJEEM24Phc804: JEE Main 2024 Physics - Class 11 JEEM24Phc804 Solution: Match List I with List II. .tg…

Match List I with List II.

List I List II
(A) Coefficient of viscosity (I) [ML2 T−2]\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2}\right]
(B) Surface tension (II) [ML2 T−1]\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-1}\right]
(C) Angular momentum (III) [ML−1 T−1]\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-1}\right]
(D) Rotational kinetic energy (IV) [ML0 T−2]\left[\mathrm{M} \mathrm{L}^0 \mathrm{~T}^{-2}\right]

Choose the correct answer from the options given below :

Multiple Choice Options

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

Step-by-Step Verified Solution

amp;F=ηAdvdyamp;[MLT−2]=η[L2][T−1]amp;η=[ML−1T−1]amp;S.T=Fℓ=[MLT−2][L]=[ML0T−2]amp;L=mvr=[ML2T−1]amp;K.E=12Iω2=[ML2T−2]\begin{aligned}& F=\eta A \frac{d v}{d y} \\& {\left[M L T^{-2}\right]=\eta\left[L^2\right]\left[T^{-1}\right]} \\& \eta=\left[M L^{-1} T^{-1}\right] \\& S . T=\frac{F}{\ell}=\frac{\left[M L T^{-2}\right]}{[L]}=\left[M L^0 T^{-2}\right] \\& L=m v r=\left[M L^2 T^{-1}\right] \\& K . E=\frac{1}{2} I \omega^2=\left[M L^2 T^{-2}\right]\end{aligned}

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring