Q50 / 110 · NTA Abhyas (NEET) 2020 · Physics · Units and Dimensions

A spherical liquid drop is placed on a horizontal plane. A small disturbance causes the volume of the drop to oscillate. The time period of oscillation T of the liquid drop depends on the radius (r) of the drop, density ρ and surface tension S of the liquid. Which among the following will be a possible expression for the time period? where k is a dimensionless constant.

  • A
    kρrS
  • B
    kρ2rS
  • Correct answer
    C
    kρr3S
  • D
    kρr3S2

Solution

According to the question,

time period, T∝raρbSc

T=kraρbSc ..... (i)

Thus, putting dimension, we get

[ T ]= [ L ] a [ M L −3 ] b [ M T −2 ] c

T=Mb+c⋅La-3b⋅T-2c

Equating the dimensions of both sides, we get

b+c=0, a-3b=0

and -2c=1

∴   c=-12

∴   b=12

and a=3b=32

Putting these value of a, b and c into Equation. (i), we get

T=k⋅r32ρ12⋅S-12 

=kρSr3

Hence, the correct answer is option: 'C'.

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