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QJEEM23Ph1e04: JEE Main 2023 Physics - Class 11 JEEM23Ph1e04 Solution: Match List I with List II .tg…

Match List I with List II

List I List II
A. Surface tension I. kg m−1 s−1\mathrm{kg~m^{-1}~s^{-1}}
B. Pressure II. kg ms−1\mathrm{kg~ms^{-1}}
C. Viscosity III. kg m−1 s−2\mathrm{kg~m^{-1}~s^{-2}}
D. Impulse IV. kg s−2\mathrm{kg~s^{-2}}

Choose the correct answer from the options given below :

Multiple Choice Options

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

Step-by-Step Verified Solution

 (A)  Surface Tension =Fℓamp;=MLT−2 L=ML0 T−2amp;=kg s−2(IV)\begin{aligned}\text { (A) } \text { Surface Tension }=\frac{\mathrm{F}}{\ell} & =\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}}=\mathrm{ML}^{0} \mathrm{~T}^{-2} \\\\& =\mathrm{kg\,s}^{-2}(\mathrm{IV})\end{aligned}

amp; (B) Pressure =F A=MLT−2 L2amp;=kg m−1 s−2(III)\begin{aligned}& \text { (B) Pressure }=\frac{F}{\mathrm{~A}}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2} \\\\& =\mathrm{kg} \,\mathrm{m}^{-1} \mathrm{~s}^{-2}(\mathrm{III}) \end{aligned}

 (C) Viscosity amp;=FA(dVdz)=MLT−2 L2(LT−1 L)amp;=ML−1 T−1=kg m−1 s−1(I)\begin{aligned} \text { (C) Viscosity } & =\frac{\mathrm{F}}{\mathrm{A}\left(\frac{\mathrm{dV}}{\mathrm{dz}}\right)}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2\left(\frac{\mathrm{LT}^{-1}}{\mathrm{~L}}\right)} \\\\ & =\mathrm{ML}^{-1} \mathrm{~T}^{-1}=\mathrm{kg} \,\mathrm{m}^{-1} \mathrm{~s}^{-1}(\mathrm{I})\end{aligned}

 (D)  Impulse amp;=∫Fdt=MLT−2×Tamp;=MLT−1=kg ms−1 (II) \begin{aligned}\text { (D) } \text { Impulse } & =\int F d t=\mathrm{MLT}^{-2} \times \mathrm{T} \\\\& =\mathrm{MLT}^{-1}=\mathrm{kg\,ms}^{-1} \text { (II) }\end{aligned}

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