Question

21. In a new system of units energy (E), density (d)and power (P) are taken as fundamental units, thenthe dimensional formula of universal gravitationalconstant G will be(1) \lbrack Eld2p21(2) \lbrack E-d-1P2\rbrack(3) \lbrack E2d-1P-11(4) \lbrack E'd- 2P-21

Answer 1

Step 1: Given that:

In a new system of units,

Fundamental units = Energy(E), density(d) and power(P)

Step 2: Finding the dimension of universal gravitational constant in the new system of unit:

According to the universal law of gravitation, the force between two masses($m_1$ and $m_2$) kept at a distance r from each other is given by;

$F_g=\frac{G{m}_1{m}_2}{r^2}$

Where G is the universal gravitational constant.

From the above relation, the universal gravitational constant is given by;

$G=\frac{F_g{r}^2}{m_1{m}_2}$

In the original system of units, the dimensional formula of G is;

$=\frac{\left[ML{T}^{-2}\right]\left[L^2\right]}{\left[M\right]\left[M\right]}$

$=\frac{\left[M{L}^3{T}^{-2}\right]}{\left[M^2\right]}$

$=\left[M^{-1}{L}^3{T}^{-2}\right]$

Let the dimensional formula of G in the new system of units is given as $\left[E^a{d}^b{P}^c\right]$ , then according to the property of unit of a physical quantity in different systems

$\left[M^{-1}{L}^3{T}^{-2}\right]$ = $\left[E^a{d}^b{P}^c\right]$ ...............(1)

Now,

The dimensional formula of energy is $\left[M{L}^2{T}^{-2}\right]$ , the dimensional formula of density is $\left[M{L}^{-3}\right]$ and the dimensional formula of power is $\left[M{L}^2{T}^{-3}\right]$ .

Putting these values at the place of E, d and P respectively in equation (1), we get;

$\left[M^{-1}{L}^3{T}^{-2}\right]$ = $\left\{{\left[M{L}^2{T}^{-2}\right]}^a{\left[M{L}^{-3}\right]}^b{\left[M{L}^2{T}^{-3}\right]}^c\right\}$

$\left[M^{-1}{L}^3{T}^{-2}\right]=\left\{{\left[M{L}^2{T}^{-2}\right]}^a{\left[M{L}^{-3}\right]}^b{\left[M{L}^2{T}^{-3}\right]}^c\right\}$

$\left[M^{-1}{L}^3{T}^{-2}\right]=\left[M^a{L}^{2a}{T}^{-2a}\right]\left[M^b{L}^{-3b}\right]\left[M^c{L}^{2c}{T}^{-3c}\right]$

$\left[M^{-1}{L}^3{T}^{-2}\right]=\left[M^{a+b+c}{L}^{2a-3b+2c}{T}^{-2a-3c}\right]$

Comparing the power of both sides, we get

$a+b+c=-1$.............(2)

$2a-3b+2c=3$..........(3)

$-2a-3c=-2$.............(4)

Adding equations (2), (3) and (4) we get

$a-2b=0$

$a=2b$

From equation (1),

$3b+c=-1$ ............(5)

And from equation (4)

$-4b-3c=-2$ ........(6)

From equation (5)$\times 3+$ equation (6)

$5b=-5$

$b=-1$

Thus,

$a=-2$

And

$c=2$

Putting the values of a, b and c in equation (1) we get,

The dimensional formula of G in new system of unit is $\left[E^{-2}{d}^{-1}{P}^2\right]$

.Thus,

The dimensional formula of G in the new system of unit is $\left[E^{-2}{d}^{-1}{P}^2\right]$