The escape velocity from the Earth’s surface is v. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is :

The correct option is 4 v.

To find the escape velocity of the planet, let us establish the relationship between the escape velocity, the radius, and the mass density of a planet.

The escape velocity from the surface of a planet of mass $M$ and radius $R$ is given by the formula:
$v_e=\sqrt{\frac{2GM}{R}}$
where $G$ is the universal gravitational constant.

Since the mass of a spherical planet can be expressed in terms of its uniform density $\rho$ and volume, we have:
$M=\text{Density}\times \text{Volume}=\rho \times \frac{4}{3}\pi R^3$

Now, substituting the expression for mass $M$ into the escape velocity formula, we get:
$v_e=\sqrt{\frac{2G\left(\rho ·\frac{4}{3}\pi R^3\right)}{R}}$
Simplifying the expression under the square root:
$v_e=\sqrt{\frac{8\pi G\rho R^2}{3}}=R\sqrt{\frac{8\pi G\rho }{3}}$

Since the mass density $\rho$ is constant (the same as Earth's density) and $G$ is a constant, the escape velocity is directly proportional to the radius of the planet:
$v_e\propto R$

Let $v$ and $R_E$ be the escape velocity and radius of the Earth, and let $v_p$ and $R_p$ be the escape velocity and radius of the other planet. We can set up the ratio:
$\frac{v_p}{v}=\frac{R_p}{R_E}$

Given that the radius of the planet is four times that of Earth ($R_p=4R_E$):
$\frac{v_p}{v}=\frac{4R_E}{R_E}=4$

Thus, the escape velocity from the surface of the new planet is:
$v_p=4v$