Question

If energy, $E=G^p{h}^q{c}^r$, where $G$ is the universal gravitational constant, $h$ is the Planck's constant and $c$ is the velocity of light, then the values of $p,q$ and $r$ are, respectively:

• A. $-1/2,1/2$ and $5/2$
• B. $-1/2,1/2$ and $3/2$
• C. $1/2,-1/2$ and $-5/2$
• D. $1/2,-1/2$ and $-3/2$

Answer 1

Answer: (A)

Dimension of Energy, $E=\left[M{L}^2{T}^{-2}\right]$

Dimension of gravitational constant, $G=\left[M^{-1}{L}^3{T}^{-2}\right]$

Dimension of Planck's constant, $h=\left[M{L}^2{T}^{-1}\right]$

Dimension of speed of light, $c=\left[M^{0}L{T}^{-1}\right]$

Given :

$E=G^p{h}^q{c}^r$

$\therefore$ $\left[M{L}^2{T}^{-2}\right]=[M^{-1}{L}^3{T}^{-2}{]}^p\times$ $[M{L}^2{T}^{-1}{]}^q$ $\times [M^{0}L{T}^{-1}{]}^r$

$⟹$ $\left[M{L}^2{T}^{-2}\right]$$=\left[M^{(-p+q)}{L}^{(3p+2q+r)}{T}^{(-2p-q-r)}\right]$

Equating both sides, we get:

$-p+q=1$ .............(1)

$3p+2q+r=2$ .............(2)

$-2p-q-r=-2$ .............(3)

Adding (2) and (3), we get: $p+q=0$ .......(4)

Solving (1) and (4),

$⟹p=-\frac{1}{2}$ and $q=\frac{1}{2}$

Now from (2),

$3\times (-\frac{1}{2})+2\times \frac{1}{2}+r=2$

$⟹r=\frac{5}{2}$