A rocket is launched with velocity 10 km/s. If radius of earth is R, then maximum height attained by it will be

The correct option is 4R.

To find the maximum height attained by the rocket, we can apply the principle of conservation of mechanical energy.

Let:
- $m$ be the mass of the rocket.
- $M$ be the mass of the Earth.
- $R$ be the radius of the Earth.
- $v$ be the launch velocity of the rocket, which is given as 10 km/s.
- $h$ be the maximum height attained by the rocket from the surface of the Earth.
- $G$ be the universal gravitational constant.
- $g$ be the acceleration due to gravity on the surface of the Earth, where $g=\frac{GM}{R^2}$ or $GM=g{R}^2$.

The escape velocity from the surface of the Earth is given by:
$v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}$

Taking the acceleration due to gravity $g\approx 9.8{\text{m/s}}^2$ and the radius of the Earth $R\approx 6400\text{km}$:
$v_e\approx 11.2\text{km/s}$

Since the launch velocity $v=10\text{km/s}$ is less than the escape velocity ($11.2\text{km/s}$), the rocket will reach a maximum height $h$ where its final velocity becomes zero.

According to the law of conservation of energy:
Total energy at the surface of the Earth = Total energy at the maximum height

$K_i+U_i=K_f+U_f$

Where:
- Initial kinetic energy: $K_i=\frac{1}{2}m{v}^2$
- Initial potential energy on the Earth's surface: $U_i=-\frac{GMm}{R}$
- Final kinetic energy at maximum height: $K_f=0$
- Final potential energy at maximum height (distance $R+h$ from the center of the Earth): $U_f=-\frac{GMm}{R+h}$

Substituting these values into the conservation of energy equation:
$\frac{1}{2}m{v}^2-\frac{GMm}{R}=0-\frac{GMm}{R+h}$

We can cancel the mass of the rocket $m$ from both sides:
$\frac{1}{2}{v}^2-\frac{GM}{R}=-\frac{GM}{R+h}$

Rearranging the equation to solve for $h$:
$\frac{GM}{R+h}=\frac{GM}{R}-\frac{1}{2}$

Divide both sides by $GM$:
$\frac{1}{R+h}=\frac{1}{R}-\frac{v^2}{2GM}$

Since $v_e=\sqrt{\frac{2GM}{R}}$, we have $2GM=R{v}_e^2$. Substituting this in:
$\frac{1}{R+h}=\frac{1}{R}-\frac{v^2}{R{v}_e^2}=\frac{1}{R}\left(1-\frac{v^2}{v_e^2}\right)$

Using the values:
- $v=10\text{km/s}$
- $v_e\approx 11.2\text{km/s}$ (where $v_e^2=2gR\approx 2\times 9.8\times {10}^{-3}\times 6400\approx 125.44{\text{km}}^2{\text{/s}}^2$, and more precisely with standard textbook approximations, $v_e=11.2\text{km/s}$, meaning $v_e^2\approx 125{\text{km}}^2{\text{/s}}^2$).
Let's plug in the ratio:
$\frac{v^2}{v_e^2}=\frac{{10}^2}{125}=\frac{100}{125}=\frac{4}{5}=0.8$

Now, substitute this ratio back into the equation:
$\frac{1}{R+h}=\frac{1}{R}\left(1-\frac{4}{5}\right)$
$\frac{1}{R+h}=\frac{1}{R}\left(\frac{1}{5}\right)$
$\frac{1}{R+h}=\frac{1}{5R}$

Taking the reciprocal of both sides:
$R+h=5R$
$h=5R-R$
$h=4R$

Thus, the maximum height attained by the rocket is 4R.