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JEE Main 202325 JanuaryMorning Shift
Exam
JEE Main
Year
2023
Date
25 January
Shift
Morning Shift
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2023 · 25 January · Morning Shift · Physics · Units and Dimensions

Match List I with List II List - I List - II A Surface tension I....

Match List I with List II

List - I List - II
A Surface tension I. kgm−1s−1kg{m}^{- 1}{s}^{- 1}
B Pressure II. kgms−1kgm{s}^{- 1}
C Viscosity III. kgm−1s−2kg{m}^{- 1}{s}^{- 2}
D Impulse IV. kgs−2kg{s}^{- 2}

Choose the correct answer from the options given below:

Options

  1. AA-IV, B-III, C-II, D-I
  2. BA-IV, B-III, C-I, D-II
  3. CA-III, B-IV, C-I, D-II
  4. DA-II, B-I, C-III, D-IV

Correct answer

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

Step-by-step solution

(A) Surface tension is related to force and length as S=Fl=MLT−2L=MT−2S = \dfrac{F}{l} = \dfrac{ML{T}^{- 2}}{L} = M{T}^{- 2}

=kgs−2= kg{s}^{- 2} (IV)

(B) Pressure is related to force and area as P=FAP = \dfrac{F}{A}

=MLT−2L2=kgm−1s−2= \dfrac{ML{T}^{- 2}}{{L}^{2}} = kg{m}^{- 1}{s}^{- 2} (III)

(C) Viscosity in terms of force, area and velocity gradient is given by η=FA(dvdx)=MLT−2L2(LT−1L)η = \dfrac{F}{A\left(\dfrac{d v}{d x}\right)} = \dfrac{ML{T}^{- 2}}{{L}^{2}\left(\dfrac{L{T}^{- 1}}{L}\right)}

=ML−1T−1=kgm−1s−1= M{L}^{- 1}{T}^{- 1} = kg{m}^{- 1}{s}^{- 1} (I)

(D) Impulse is related to force and time of application as J=∫Fdt=MLT−2×TJ = ∫ Fdt = ML{T}^{- 2} \times T

=MLT−1=kgms−1= ML{T}^{- 1} = kgm{s}^{- 1} (II).

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

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