Question

X = a + c /m , find dimension of a and c , where x = displacement and m = mass.

Answer 1

Answer:

[M L T^0]

Explanation:

As the dimension of x is [M^0 L T^0] and dimensional formula of mass is [M L^0 T^0] then a+b have the dimention [M L T^0]. Also a and c have same dimentional formula

Answer 2

Answer:

• Dimension of a = $\sf{[M^0L^1T^0]}$

• Dimension of c = $\sf{[M^1L^1T^0]}$

Explanation:

Given

Given formula is

$\bf{X=a+\dfrac{c}{m}}$

where X = displacement and m = mass

To find

• dimensions of a and c

Solution

Given formula is

$\to \bf{X=a+\dfrac{c}{m}}$

$\to \bf{X=\dfrac{ma+c}{m}}$

$\to \bf{X\;m=m\;a+c}$

Substituting dimensions

$\to \bf{[M^0L^1T^0].[M^1L^0T^0]=[M^1L^0T^0]\;[a]+[c]}$

Let dimension of a be $\sf{[M^pL^qL^r]}$

Substituting

$\to \bf{[M^0L^1T^0].[M^1L^0T^0]}$ = $\;\;\;\;\;\;\;\;\;\;\bf{[M^1L^0T^0].[M^pL^qL^r]+[c]}$

solving dimensionally

$\to \bf{[M^1L^1]}$ = $\bf{[M^{p+1}L^qT^r]+[c]}$

Since, $\sf{[M^{p+1}L^qT^r]+[c]+[c]\:and\:[c]\:are \:in\:sum}$

therefore,

$\to \bf{[M^1L^1]}$ = $\bf{[M^{p+1}L^qT^r]+[M^{p+1}L^qT^r]}$

$\to \bf{[M^1L^1]}$ = $\bf{[M^{p+1}L^qT^r]}$

Using principal of homogeneity

Comparing LHS and RHS

we will get,

• p + 1 = 1 ; p = 0
• q = 1
• t = 0

Therefore,

• Dimensions of a = $\sf{[M^pL^qT^r]}$ = $\sf{[M^0L^1T^0]}$

and

• Dimension of c = $\sf{[M^{p+1}L^qT^r]}$ = $\sf{[M^1L^1T^0]}$.