Question

Write the dimensions of a and b in the relation e=b-x^2/at where e=energy , x = distance , t= time

Answer 1

Answer:

The dimension of a and b are $[M^{-1}T^{1}]$ and $[ML^{2}T^{-2}]$.

Explanation:

Given the relation as:

$e=b-\dfrac{x^2}{at}$

Here, e = energy

x = distance

t = time

We know that,

The dimension formula

$e = [ML^2T^{-2}]$

$x = [L]$

$t =[T]$

Using principle of homogeneity

e = b...(I)

$[ML^2T^{-2}]=b$

Using principle of homogeneity

$e= \dfrac{x^2}{at}$...(II)

$at=\dfrac{x^2}{e}$

$at=\dfrac{[L^2]}{[ML^2T^{-2}]}$

$at=[M^{-1}T^{2}]$

$a= \dfrac{[M^{-1}T^{2}}{[T]}$

$a=[M^{-1}T^{1}]$

Hence. The dimension of a and b are $[M^{-1}T^{1}]$ and $[ML^{2}T^{-2}]$.

Answer 2

Answer:

a= [M^-1 T^-1] , b=[L^2]

Explanation:

E=b-x^2/at

using principle of homogeneity

b= x^2 = [L^2]

a= b-x^2/Et

a= [L^2]/[ML^2][T]

a= [MT^-1]