Units & Dimensions

JEE Main level, single correct, medium

10 of 149

Question 10 of 149

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:

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

(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)\eta =\dfrac{F}{A\left(\dfrac{dv}{dx}\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=\int 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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