Question

Find the dimension of A, B, C, D and E in X=A+Bt+Ct2+ (Dt3/E+t) where X is displacement and it's time
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Answer 1

Explanation:

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Answer 2

The dimensions of A,B, C, D and E are $[L]$, $[LT^{-1} ]$, $[LT^{-2} ]$, $[LT^{-2} ]$ and $[T]$ respectively.

Given,

Equation:$X=A+Bt+Ct^{2} +\frac{Dt^{3} }{E+t}$.

X-displacement and t-time.

To find,

the dimension of A, B, C, D and E.

Solution:

• The principle that will be used here is known as "Principle of homogeneity of dimensions".
• It states that the dimensions of all the terms in an equation must be equal.
• Simply, it states that we add or subtract similar physical quantities.

Dimensions of E=Dimensions of t

Dimensions of E=[T].

Dimensions of A=Dimensions of X

Dimensions of A=[L].

Dimensions of Bt=Dimensions of A

Dimensions of B=Dimensions of A/Dimensions of t

Dimensions of B=[L]/[T]

Dimensions of B=$[LT^{-1} ]$.

Dimensions of $Ct^{2}$=Dimensions of A

Dimensions of C=Dimensions of A/Dimensions of $t^{2}$

Dimensions of C=$\frac{[L]}{[T^{2} ]}$

Dimensions of C=$[LT^{-2} ]$.

Dimensions of $\frac{Dt^{3} }{E+t}$=Dimensions of A

Dimensions of D={Dimensions of A x Dimensions of (E+t)}/Dimensions of $t^{3}$

Dimensions of D=$\frac{[L][T]}{[T]^{3} }$

Dimensions of D=$[LT^{-2} ]$.

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