A bullet is fired vertically upwards with velocity $v$ from the surface of the planet. When it reaches its maximum heights, its acceleration due to the planet's gravity is $\dfrac{1}{4}th$ of $g$. If the escape velocity from the planet is $v\sqrt N $, then $N = $ ?

Hint: The escape velocity is defined as the velocity required escaping from the gravitational force of the Earth. The escape velocity depends on the acceleration due to gravity and radius of the Earth. The acceleration due to gravity decreases as the particle moves far away from the Earth.

Complete step by step answer:
Given: At the maximum height of the bullet the acceleration due to gravity is $\dfrac{1}{4}th$ of $g$.
The formula to calculate the acceleration due to gravity on Earth is $g = \dfrac{{GM}}{{{R^2}}}$ where, $g$ is the acceleration due to gravity, $G$ is the gravitational constant, $M$ is the mass of Earth and $R$ is the radius of Earth.
The acceleration due to gravity decreases at some height. The formula to calculate the acceleration due to gravity on that height is $g' = \dfrac{{GM}}{{{r^2}}}$ , where $g'$ is the acceleration due to gravity at height $r$.
According to the given condition, $g' = \dfrac{1}{4}g$ .
Substitute $\dfrac{{GM}}{{{R^2}}}$ for $g$ and $\dfrac{{GM}}{{{r^2}}}$ for $g'$ in the given condition to calculate the value of height.
$ \dfrac{{GM}}{{{r^2}}} = \dfrac{{GM}}{{4{R^2}}}\\$
$\implies {r^2} = 4{R^2}\\$
$\implies r = 2R$
Now, the mechanical energy of the system is the sum of kinetic energy at potential energy of the system. According to the law of conservation of mechanical energy, the initial mechanical energy is equal to final mechanical energy.
The initial mechanical energy of the bullet is ${U_i} = - \dfrac{{GMm}}{R}$ where, ${U_i}$ is the initial potential energy and $m$ is the mass of the bullet.
The initial kinetic energy of the bullet is ${K_i} = \dfrac{1}{2}m{v^2}$ where, ${K_i}$ is the initial kinetic energy, $v$ is the velocity of the bullet.
The final potential energy of the bullet is ${U_f} = - \dfrac{{GMm}}{r}$ where, ${U_f}$ is the final potential energy.
The final kinetic energy of the bullet is zero because the velocity at the maximum height is always zero.
According to the conservation of mechanical energy, ${U_i} + {K_i} = {U_f} + {K_f}$.
Substitute $ - \dfrac{{GMm}}{R}$ for ${U_i}$ , $\dfrac{1}{2}m{v^2}$ for ${K_i}$, $ - \dfrac{{GMm}}{r}$ for ${U_f}$ and $0$for${K_f}$ in the condition of conservation of energy.
$ - \dfrac{{GMm}}{R} + \dfrac{1}{2}m{v^2} = - \dfrac{{GMm}}{r} + 0\\$
$\implies \dfrac{1}{2}m{v^2} = - \dfrac{{GMm}}{r} + \dfrac{{GMm}}{R}\\$
$\implies \dfrac{1}{2}{v^2} = - \dfrac{{GM}}{r} + \dfrac{{GM}}{R} $
Substitute$2R$ for $r$ in the above equation to calculate the velocity of the bullet.
$\implies \dfrac{1}{2}{v^2} = - \dfrac{{GM}}{{2R}} + \dfrac{{GM}}{R}\\$
$\implies \dfrac{1}{2}{v^2} = \dfrac{{GM}}{{2R}}\\{v^2} = \dfrac{{GM}}{R}\\$
$\implies v = \sqrt {\dfrac{{GM}}{R}} $
The formula to calculate the escape velocity of the planet is ${v_{esc}} = \sqrt {\dfrac{{2GM}}{R}} $ where ${v_{esc}}$ is the escape velocity of the planet.
Substitute $v$for$\sqrt {\dfrac{{GM}}{R}} $ in the formula and compare it with the equation ${v_{esc}} = v\sqrt N $ to calculate the value of $N$.
${v_{esc}} = v\sqrt 2 $
Thus, the value of $N$ is equal to $2$.

Note:
The mechanical energy always consists of both gravitational and kinetic energy of the system. Calculate the velocity of the bullet and substitute its value in escape velocity to calculate the value of $N$