Questions - Answers
1. Show that one kilowatt hour is equal to 36 × 105 J
A: 1 kilowatt hour = 1000 W × 1 hour
= 1000 W × 1 × 60 × 60 s
= 36 × 105 watt second
1 joule = 1 watt second
1 KWh = 36 × 105 J
2. A student remembers Einstein's equation for variation of mass m0 of a body with velocity v as 
He forgets to indicate the term velocity of light C. Where he should put 'C' in the equation?
A: Dimensional formula for m on L.H.S. is [M]
Hence the R.H.S. should also have the same formula. m0 has already the dimensions of M
The denominator should be dimension less, for which C should be kept below V2 as C2 . Such that V2 /C2 is dimension less.
So, the correct equation is 
3. Check the correctness of formula x = a sin (ωt + φ) where x is displacement, a is amplitude, ω is angular velocity, and φ is the angle by dimensional method.
A: Dimensional formula for x is L
dimensional formula for a is L
dimensional formula for 
dimensionless
dimensional formula for φ = dimensionless
Thus dimensions for the equation on L.H.S. is equal to that on R.H.S. Thus equation is dimensionally correct.
4. What are the limitations of dimensional analysis?
A: Dimensional method is a very convenient and simple method of arriving at a physical formula relating to different Physical quantities. But it suffers from the following limitations.
i) The value of the dimensionless constant K in the equation cannot be found by dimensional analysis. It has to be determined experimentally.
ii) In deriving a physical formula, this method used only when the physical quantity varies as the product of other quantities. If the physical quantity is sum or difference of two quantities, this method fails.
e.g.: The expression 
cannot be derived by dimensional method.
iii) The equations containing the trigonometrical terms sinθ, cosθ etc., exponential and log n etc cannot be analysed.
iv) The method fails to derive the exact form of a physical relation which depends more than three fundamental quantities length, mass, time.
Problems
i) Conversion of one system of units into another.
e.g.: Convert the units of density in C.G.S. System into SI
Sol: Density = mass/volume
Dimensional formula for density is M/L3 = ML-3. In C.G.S. and SI let M1, L1, T1 represent gm, cm and second and M2, L2, T2 represent kg, m and second respectively.
So, in these systems dimensional formula of density will be M1 L1-3 and M2L2-3
If numerical values are n1 and n2 in these systems, from the principle of homogeneity of dimensions,
n1 [M1 L1-3] = n2 [M2 L2-3]
Thus 1 gm cm-3 = 103 kg m-3
ii) Checking the correctness of equation.
e.g.: Check dimensionally, the correctness of equation for the distance travelled in the nth second 
where u is the initial velocity of a body, a its acceleration.
Sol: Dimensional formula for distance Dn is 
as Dn is the distance travelled in nthsecond Dimensional formula for velocity u, 
Dimensional formula for acceleration 
; for (2n-1), T
using the given equation, 
The dimensions of each term on L.H.S. are the same as those of every term on the R.H.S.
Hence the given equation is correct.
iii) Deriving a relation between different Physical Quantities.
e.g.: Derive 
, incase of a Simple Pendulum by dimensional method.
Sol: Let the period of oscillation of Simple Pendulum 't' depend on mass (m) of the
bob, length (l) of the pendulum and g the acceleration due to gravity at the place.
Let t ma lbgc .. .... ...... (i)
or t = K ma lb gc
Where K is dimensionless constant and a, b, c are dimensions.
Dimensional formula for 't' is [T]
Dimensional formula for 'l' is [L]
for m, it is [M], for K, nil
for g, [LT-2]
substituting these in equation (i)
[T] = [M]a [L]b [LT-2]c or
M0 L0 T1 = [M]a [L]b+c [T]-2c
According to the principle of homogeneity of dimensions, the powers of M, L, T on both sides should be same
a = 0, b+c = 0 and -2c = 1
substituting the values of a, b, c in equation (i)
t = K m0 l1/2 g-1/2 = 
The value of K is found experimentally = 2π
Since, the dimension of mass is found to be 0, the period of the Simple Pendulum is independent of the mass of the bob.
