Chapter 1 : Dimensions

Important Question and their Solutions :

Question 1 :

 

Define Dimension. Determine dimension of Force, Pressure and Power.

 

Solutions:

Dimensions is a simply a power to be raised in a physical quantities of length, mass and time. The dimension of physical quantities can be written as block letter including large bracket in both sides.
For example:

Area (A) = l x b

A = [L] x [L]

A = [L x L]

A = [L^2]

A = [M^0L^2T^0]

Thus, the dimension of mass in area= 0

The dimension of length in area= 2

The dimension of time in area= 0

To solve next part of question we solve below :

1. Force (F) = Mass (m) x Acceleration (a)

F = Mass (m) x Velocity (v) / Time (t)

F = [M] x [M^0L^1T^-1] / [T]

F = [M^1L^1T^-2]

Thus, the dimension of mass in force= 1

The dimension of length in force= 1

The dimension of time in force= -2

2. Pressure (P) = Force (F)/ Area (A)

P = [M^1L^1T^-2] / [M^0L^2T^0]

P = [M^1L^-1T^-2]

Thus, the dimension of mass in pressure= 1

The dimension of length in pressure= -1

The dimension of time in pressure= -2

3. Power (P) = Work done (W) / Time Taken (t)

P = (Force (F) x Distance traveled (d)) / Time (t)

P = ([M^1L^1T^-2] x [L^1]) / [T]

P = [M^1L^2T^-2] / [T]

P = [M^1L^2T^-3]

Thus, the dimension of mass in power= 1

The dimension of length in power= 2

The dimension of time in power= -3

Question 2 :

 

Write down the application of dimensional equations.

Answer :

The following are the applications of dimensional equations :

1. To check the correctness of equations.

If the formula is correct, it should be dimensional constituent (i.e. equal)

Example:

Let us take an example of v= u+at

The dimension of v= [M^0L^1T^-1]

The dimension of u= [M^0L^1T^-1]

The dimension of a= [M^0L^1T^-2]

The dimension of t=  [M^0L^0T^1]

Then, we have

L.H.S = v = [M^0L^1T^-1]

R.H.S = u+at= [M^0L^1T^-1]+ [M^0L^1T^-2] x [M^0L^0T^1]

= [M^0L^1T^-1] + [M^0L^1T^-1]

= 2[M^0L^1T^-1]

since, '2'is pure number. so, it is dimensionless and neglected in above equation.

thus, L.H.S = R.H.S

v= u+at proved.

2. To convert one system of unit to another system of unit.

Let, a physical quantities has a dimension 'x' in mass, 'y' in length and 'z' in time. Also let, 'n1' and 'n2' be the numerical values of the physical quantities in the first and second system of unit respectively. Again, 'M1','L1' and 'T1' be the fundamental unit of mass, length and time first system of unit respectively. Similarly, 'M2','L2' and 'T2' be the fundamental unit of mass, length and time in second system of unit respectively. 

Then, the physical quantities in the first system of unit can be written as,

Q1 = n1 [M1]^x [L1]^y [T1]^z----------(i)

Similarly, the physical quantities in the second system can be written as,

Q2 = n2 [M2]^x [L2]^y [T]^z------------(ii)

Then, Q1 and Q2 must be equal. Q1=Q2.

or, n1 [M1]^x [L1]^y [T1]^z = n2 [M2]^x [L2]^y [T]^z

or, n2 = n1 [M1/M2]^x [L1/L2]^y [T1/T2]^z

 

Example : Convert 1joule into erg.

Solution:

Since, Joule and erg are the units of work.

The dimension of work = [ML^2T^-2]= [M]^1[L]^2[T]^-2-----------(i)

The numerical values in SI unit (n1) = 1

The numerical values in SI unit (n2) = ? 

we have, 

n2 = n1 [M1/M2]^x [L1/L2]^y [T1/T2]^z---------(ii)

comparing equation(i) to the equation (ii) we get,

x=1,y=2 and z=-2

In SI unit,

M1=1kg,L1=1m,T1=1sec

In CGS unit,

M2=1gram,L2=1cm,T2=1sec

Then,equation (ii) comes as,
n2 = 1x [1kg/1gram]^1 [1m/1cm]^2 [1sec/1sec]^-2

n2 = 1x [1x1000gram/1gram]^1 [1x1 0cm/1cm]^2 [1]^-2

n2 = 1x1000x100^2x1

n2 = 10000000 = 10^7

Hence, 1Joule = 10^7erg

3. To determine Dimension of constant.

According to Newtons Law of Gravitation gives the relation between two  masses 'm1' and 'm2' with distance 'd' and force 'f' is given by:

F = GM1M2 / d^2

or, G = Fd^2 / M1M2

or, G = [MLT^-2] [L^2] / [M^2] 

or, G = [M^-1L^3T^-2]

4. To derive relation between various physical quantities.

Let, take an example of pendulum bob. The time period is,

t = 2π√l/g

where 't' is a time period, 'l' is the length of the string and 'g' is an acceleration due to gravity.

Here, time period (t) depends upon

(i) 't' directly proportional to m^x i.e. t ∝ m^x

(ii)  't directly proportional to l^y i.e. t ∝ l^y

(iii) 't' directly proportional to g^z i.e. t ∝ g^z

combining above relation we get 

t∝m^x l^y g^z

t = k m^x l^y g^z------(i)

Now, 

[M^0L^0T^1] = k [M]^x [L]^y [LT^-2]^z

or, [M]^0 [L]^0 [T]^1 = k [M]^x [L]^y [L]^z [T]^-2z

or, [M]^0 [L]^0 [T]^1 = k [M]^x [L]^y+z [T]^-2z

Applying principle of homogeneity, we get

we get, 

x = 0, y+z = 0 , 1 = -2z

i.e. x = 0, y = 1/2, z = -1/2

Then,equation (i) becomes,

t = k m^0 l^1/2 g^-1/2

or, t = k x 1 ( l^1/2) / (g^1/2)

t = k √l/g

since, k = 2π is an constant term. so, above relation becomes,

t = 2π√l/g proved..

Question 3: 

 

If x = at+bt^2 where x is in meter and t in hours. What is the units of a and b?

solution: 

Here we have,

x = at+bt^2 where x in meter and t in hours and need to find the units of a and b.

Now, x = at+bt^2

Applying principle of homogeneity,

x = at

or, a = x/t

or, a = [M^0LT^0] / [M^0L^0T]

i.e. a = [M^0LT^-1]

The dimension of a in mass = 0

The dimension of a in length = 1

The dimension of a in time = -1

Hence, a is an vector quantity.so, its unit is m/hrs.

Again,

x = bt^2

or, b = x/t^2

or, b = [M^0LT^0] / [M^0L^0T^2]

or, b = [M^0LT^-2]

The dimension of b in mass = 0

The dimension of b in length = 1

The dimension of b in time = -2

Hence,

b is an acceleration.so, its unit is m/hrs^2.