Class 11 Physics Units and Measurements | Applications of Dimensional Analysis |

__Applications of Dimensional Analysis__

__Checking Dimensional Consistency of equations__

- A
**dimensionally correct equation**must have same dimensions on both sides of the equation. - A dimensionally correct equation need not be a correct equation but a dimensionally incorrect equation is always wrong. It can test dimensional validity but not find exact relationship between the physical quantities.

Example, x = x_{0} + v_{0}t + (1/2) at^{2}Or Dimensionally, [L] = [L] + [LT^{-1}][T] + [LT^{-2}][T^{2}]

x – Distance travelled in time t, x_{0} – starting position, v_{0} - initial velocity, a – uniform acceleration.

Dimensions on both sides will be [L] as [T] gets cancelled out. Hence this is dimensionally correct equation.

__Deducing relation among physical quantities__

- To deduce relation among physical quantities, we should know the dependence of one quantity over others (or independent variables) and consider it as product type of dependence.
- Dimensionless constants cannot be obtained using this method.

Example, T = k l^{x}g^{y}m^{z}

Or [L^{0}M^{0}T^{1}] = [L^{1}]^{x} [L^{1}T^{-2}]^{y} [M^{1}]^{z}= [L^{x+y}T^{-2y }M^{z}]

Means, x+y = 0, -2y = 1 and z = 0. So, x = ½, y = -½ and z = 0

So the original equation reduces to T = k √l/g

.