Question

In the following equations, the distance $x$ is in $meters$ , the time $t$ in $seconds$ and the velocity $v$ in $m s^{- 1}$ . What is the SI unit of the constant $C_{1}$ and $C_{2}$ $?$

(a) $v^{2} = 2 C_{1} x$

(b) $x = C_{1} \cos C_{2} t$

(c) $v = C_{1} e^{- C_{2} t}$

Accepted Answer

(a) Given expression:

$meters$

As we know that the unit on both sides should be the same. Below is the unit on the left-side expression,

$\Rightarrow {[\frac{m}{s}]}^{2} = m^{2} s^{- 2}$ .

Also, unit on the right-side expression should also be the same as above. So,

$\Rightarrow C_{1} \times m = m^{2} s^{- 2}$

$\Rightarrow C_{1} = \frac{m^{}}{s^{2}}$

(b) Given expression:

$\Rightarrow x = C_{1} \cos (C_{2} t)$

As we know that the unit on both sides should be same. Below is the unit on the left-side expression:

$\Rightarrow m$

On right-side of expression $\cos C_{2} t$ has no unit and $C_{2} t$ should be unitless too.

So, $C_{2} t = 1$

$\Rightarrow C_{2} = \frac{1}{t} = s^{- 1}$

Hence, the unit of $C_{2}$ would be $s^{- 1}$ .

Again, the unit of $C_{1}$ should also be $m$ as it should be equal to the right-hand side unit.

(c) By using the principle of homogeneity, the dimensions of left hand side should be equal to the dimensions of right-hand side.

$[v] = [C_{1} e^{- C_{2} t}] as e^{- C_{2} t} term should be unitless so, ∵ [e^{- C_{2} t}] = [M^{0} L^{0} T^{0}] and within that the term - C_{2} t should be also unitless so, [C_{2}] = [T^{- 1}] Again, as e^{- C_{2} t} term is unitless so, [v] = [C_{1}] \Rightarrow [C_{1}] = [{LT}^{- 1}]$
So, the unit of $C_{1}$ is $m s^{- 1}$ and the unit of $C_{2}$ is $s^{- 1}$ .

BM Sharma Solutions for Chapter: Unit Dimension and Measurement, Exercise 1: DPP

BM Sharma Physics Solutions for Exercise - BM Sharma Solutions for Chapter: Unit Dimension and Measurement, Exercise 1: DPP

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