
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 of the liquid drop depends on radius of the drop, density and surface tension of the liquid. Which among the following will be a possible expression for (where is a dimensionless constant)?
A.
B.
C.
D.
Answer
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Hint:- As time period of oscillation of the liquid drop depends on radius of the drop, density and surface tension of the liquid
So, the time period of oscillation can be given by the expression where k, x, y, z are dimensionless constants.
Now, equate the dimensions according to the expression and find the value of x, y and z.
Complete step-by-step solution:-
As given in the question that the time period of oscillation of the liquid drop depends on radius of the drop, density and surface tension of the liquid
So, the time period of oscillation can be given by the expression where k, x, y, z are dimensionless constants.
Now, we equate the dimensions according to the expression to find the value of x, y and z
We know that the dimension of time period of oscillation is and that of radius is
As we know density , so the dimension of will be
And the surface tension is the force per unit length, so its dimension will be
Now, substituting these dimension in the expression for we have
On simplifying we have
Now, as LHS has on time dimension so the other dimension in RHS must be zero
For M, or …… (1)
For L, …… (2)
For T, or …… (3)
Now, substituting the value of z from equation (3) in equation (1) we get
Now, substituting the value of y in equation (2) we get
Now, substituting the value of x, y and z in the expression of we have
On simplifying we get
Hence, option C is correct.
Note:- Dimensions of any physical quantity are those raised powers on base units to specify its unit. Dimensional formula is the expression which shows how and which of the fundamental quantities represent the dimensions of a physical quantity.
So, the time period of oscillation can be given by the expression
Now, equate the dimensions according to the expression and find the value of x, y and z.
Complete step-by-step solution:-
As given in the question that the time period of oscillation
So, the time period of oscillation can be given by the expression
Now, we equate the dimensions according to the expression to find the value of x, y and z
We know that the dimension of time period of oscillation
As we know density
And the surface tension is the force per unit length, so its dimension will be
Now, substituting these dimension in the expression for
On simplifying we have
Now, as LHS has on time dimension so the other dimension in RHS must be zero
For M,
For L,
For T,
Now, substituting the value of z from equation (3) in equation (1) we get
Now, substituting the value of y in equation (2) we get
Now, substituting the value of x, y and z in the expression of
On simplifying we get
Hence, option C is correct.
Note:- Dimensions of any physical quantity are those raised powers on base units to specify its unit. Dimensional formula is the expression which shows how and which of the fundamental quantities represent the dimensions of a physical quantity.
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