Question

The dimensional formula of AB where U =Ax+B/t is

Answer 1

Answer:

Dimensional formula of A = $[T^{-1} ]$

Dimensional formula of B = $[L^{1} ]$

Dimensional formula of AB = $[M^{0} L^{1} T^{-1} ]$

Explanation:

u = Ax +B/t

here,

• x = Displacement : $[M^{0} L^{1} T^{0}]$
• t = Time : $[M^{0} L^{0} T^{1}]$

• u = Velocity : $[M^{0} L^{1} T^{-1}]$

Now,

Dimensions of LHS and RHS should be equal as per the law of homogeneity.

For RHS :

Dimensions of Ax should be equal to u and dimensions of B/t should also be equal to u as per the law of homogeneity.

For A :

⇒ Dimension of Ax = Dimension of u

⇒ A $[M^{0} L^{1} T^{0}]$ = $[M^{0} L^{1} T^{-1}]$

⇒ A = $\frac{[M^{0} L^{1} T^{-1}]}{[M^{0} L^{1} T^{0}]}$ = $[T^{-1} ]$

For B :

⇒ Dimension of B/t = Dimension of u

⇒ $\frac{B}{[M^{0} L^{0} T^{1}]}$ = $[M^{0} L^{1} T^{-1}]$

⇒ B = $[M^{0} L^{0} T^{1}]$ × $[M^{0} L^{1} T^{-1}]$

⇒ B = $[L^{1} ]$

Hope it helps :)

(If you find any mistakes, which I am sure there aren't any, then please contact me within 10 minutes of writing this answer, if not possible, please forgive me)

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