Question

Find the dimensions of a/b in the relation v=a+bt where v is velocity t is time and a and b are constants

Answer 1

Remember according to dimension conservative , if a equation have dimension then right side of the equation are also be give same dimension formula.
i.e. dimension of l.h.s= dimension of r.h.s
2nd thinks is that we can add or substract two quantity if both are same dimension .
so in above question v have dimension =[LT^-1]
so a+Bt have also [LT^1]
according to 2nd thinks A and BT have same dimension
so A=[LT^-1]
bt=[LT^-1]
so b=[LT^-2]
because [LT^-2][T]=[LT^-1]
hope u got ur answer


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Answer 2

Answer:

The dimensions of a/b is [T].

Step-by-step explanation:

Given:

• The velocity of a body $\mathrm{v}=\mathrm{a}+\mathrm{bt}$
• t= Time

According to the principle of homogeneity,

• Homogeneity Principle of Dimensional Analysis Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same.
• The dimensional formula of $\mathrm{A}= Dimensions of \mathrm{v} .$

                                                  $=\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-1}\right]$

• Similarly, the dimensions of Bt = Dimensions of $\mathrm{v}$
• Dimensions of  $\mathrm{B}= dimensions of \frac{\mathrm{v}}{\mathrm{t}}$

                             $\begin{aligned}&=\frac{\left[\mathrm{L}^{1} \mathrm{~T}^{-1}\right]}{\left[\mathrm{T}^{-1}\right]}=\left[\mathrm{L}^{1} \mathrm{~T}^{-2}\right] \\&=\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right]\end{aligned}$

• Find  the dimensions of a/b =

                                           $\frac{a}{b} =\frac{M^{0}LT^{-1} }{M^{0}LT^{-2} } =[T]$