Solved Examples - Units and Dimension, JEE Main Notes JEE Notes | EduRev

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The document Solved Examples - Units and Dimension, JEE Main Notes JEE Notes | EduRev is a part of the JEE Course JEE Main Mock Test Series 2020 & Previous Year Papers.
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Problem 1:-

The Speed of Sound V in a Gas might plausibly depend on the Pressure P, the Density Ρ, and the Volume V of the Gas. Use Dimensional Analysis To Determine The Exponents X, Y, And Z in the Formula

V = Cpxρyvz

where C is a dimensionless constant. Incidentally, The mks units of pressure are kilograms per meter per second squared.
Solution:-
Equating the dimensions of both sides of the above equation, we obtain

Solved Examples - Units and Dimension, JEE Main Notes JEE Notes | EduRevA comparison of the exponents of [L], [M], and [T] on either side of the above expression yields,
1 = -x – 3y +3z
0 = x+y
-1 = -2x
The third equation immediately gives x = ½ ; the second equation then yields y = – ½ ; finally, the first equation gives z = 0. Hence,
Solved Examples - Units and Dimension, JEE Main Notes JEE Notes | EduRev


Problem 2:-
Milk is flowing through a full pipe whose diameter is known to be 1.8 cm. The only measure available is a tank calibrated in cubic feet, and it is found that it takes 1 h to fill 12.4 ft3.
What is the velocity of flow of the liquid in the pipe'?
Solution:-
velocity is [L]/[t] and the units in the SI system for velocity are therefore m s-1:
v = L/t where v is the velocity.
Now    V = AL where V is the volume of a length of pipe L of cross-sectional area A
i.e. L = V/A.
Therefore v = V/At
Checking this dimensionally
[L][t]-1 = [L]3[L]-2[t]-1 = [L][t]-1
which is correct.
Since the required velocity is in m s-1, volume must be in m3, time in s and area in m2.
From the volume measurement
V/t = 12.4ft3 h-1
We know that,
1 ft= 0.0283 m3
1 = (0.0283 m/1 ft3)
1 h = 60 x 60 s
So, (1 h/3600 s) = 1
Therefore V/t = 12.4 ft3/h x (0.0283 m3/1 ft3) x (1 h/3600 s)
= 9.75 x 10-5 m3 s-1.
Also the area of the pipe A = πD2/4
= π(0.018)2 /4 m2
= 2.54 x 10-4 m2
v = V/t x 1/A
= 9.75 x 10-5/2.54 x 10-4
= 0.38 m s-1.

Problem 3:-
Force of viscosity F acting on a spherical body moving through a fluid depends upon its velocity (v), radius (r) and co-efficient of viscosity ‘η’ of the fluid. Using method of dimensions obtain an expression for ‘F’.
Solution:-
Let F ∝ va, F ∝ rb and F ∝ η       …... (1)
So, F = Kvarbηc
where ‘K’ is a dimensional constant.
Dimensional formula of F = [M1L1T-2]
Dimensional formula of v = [L1T-1]
Dimensional formula of  r = [L1]
Dimensional formula of η = [M1L-1T-1]
Substitute for the dimensional formulae in equation (1),
[M1L1T-2] = [L1T-1]a [L1]b [M1L-1T-1]c
[M1L1T-2] = [Mc La+b+c T-a-c]        …... (2)
In accordance to principle of homogeneity, the dimensions of the two sides of relation (2) should be same.
So, c = 1     …... (3)
a+b-c = 1     …... (4)
-a – c = -2    …... (5)
Putting  c = 1 in (5), we get a = 1
Putting a = 1 and c = 1 in (4), we get b = 1
Substituting for a, b and c in (1), F = kηrv, which is required relation.

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