Question

The dimensions of gravitational constant $G$ and the moment of inertia are, respectively

• A. $\left[M{L}^3{T}^{-2}\right],\left[M{L}^2{T}^0\right]$
• B. $\left[M^{-1}{L}^3{T}^{-2}\right],\left[M{L}^2{T}^0\right]$
• C. $\left[M^{-1}{L}^3{T}^{-2}\right],\left[M^{-1}{L}^{2}T\right]$
• D. $\left[M{L}^3{T}^{-2}\right],\left[M^{-1}{L}^{2}T\right]$

Answer 1

The correct option is B

$\left[M^{-1}{L}^3{T}^{-2}\right],\left[M{L}^2{T}^0\right]$

Step 1: Use the gravitational formula,

Know that the gravitational formula,

$F=\frac{G{m}_1{m}_2}{r^2}$, where $G=$Gravitational Constant, $m_1\&m_2=$mass of the first body and the mass of the second body, and $r=$distance between both bodies.

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

The formula for the moment of inertia is,

$I=m{k}^2$, where $m=$mass, and $k=$length.

Step 2: Calculate the dimensions of gravitational constant and moment of inertia,

For the dimension of force, we know that

$\begin{array}{rcl}Force\left(F\right)& =& mass\left(m\right)\times acceleration\left(a\right)\\ & =& \left[M\right]\left[L{T}^{-2}\right]\\ & =& \left[ML{T}^{-2}\right]\end{array}$

The dimensional formula for gravitation will be

$\begin{array}{rcl}G& =& \frac{F{r}^2}{m_1{m}_2}\\ \left[G\right]& =& \frac{\left[ML{T}^{-2}\right]\left[L^2\right]}{\left[M\right]\left[M\right]}\\ & =& \frac{\left[ML{T}^{-2}\right]\left[L^2\right]}{\left[M^2\right]}\\ & =& \left[M^{-1}{L}^3{T}^{-2}\right]\end{array}$

Now find the dimensional formula of the inertia,

$\begin{array}{rcl}I& =& \left[M\right]\left[L^2\right]\\ & =& \left[M{L}^2\right]\end{array}$

Therefore dimensions of gravitational constant $G$ and the moment of inertia are, respectively,$\left[M^{-1}{L}^3{T}^{-2}\right]\mathrm{and}\left[M{L}^2{T}^0\right]$.

Hence, the correct option is B.