The escape velocity of an object on a planet whose $g$ value is $9times$ on earth and whose radius is $4times$ that of the earth in km/s is

The correct option is A

$67.2$

Step 1: Given data

Let the value of acceleration due to the gravity of another planet is $g\text{'}$

The value of acceleration due to the gravity of $g$ of another planet is $g\text{'}=9g$

Let the radius of the earth is $R$.

The radius of another planet $R\text{'}=4R$

Step 2: Formula used

The escaped velocity is computed using the formula $v=\sqrt{2gR}$

where $v=velocity,g=accelerationduetogravityandR=radius$

The escape velocity of earth is $11.2km/s$

Step 3: Compute the escape velocity of an object

So, the velocity of the new object is,

$$\begin{gathered} v_n=\sqrt{2\times g\text{'}\times R\text{'}} \\ {v}_n=\sqrt{2\times 9g\times 4R} \\ {v}_n=6\times \sqrt{2gR}...............\left(1\right) \end{gathered}$$

It is understood that the escape velocity of the earth is $11.2km/s$. So, it can be written that,

$\sqrt{2gR}=11.2..............\left(2\right)$

Substitute equation $\left(2\right)$ in equation $\left(1\right)$,

$$\begin{gathered} v_n=6\times 11.2 \\ {v}_n=67.2km/s \end{gathered}$$

Hence, option A is the correct answer.