If the radius of earth reduces by 4% and density remains same then escape velocity will

The correct option is Reduce by 4%.

To understand why, we can derive the relationship between the escape velocity, the radius of the Earth, and its density.

The escape velocity ($v_e$) from the surface of the Earth is given by the formula:

$v_e=\sqrt{\frac{2GM}{R}}$

where:
$G$ is the universal gravitational constant,
$M$ is the mass of the Earth, and
$R$ is the radius of the Earth.

We are given that the density ($\rho$) of the Earth remains the same. The mass ($M$) of a spherical body can be expressed in terms of its density and volume ($V$) as:

$M=\rho \times V=\rho \times \frac{4}{3}\pi R^3$

Substituting this expression for $M$ back into the escape velocity formula:

$v_e=\sqrt{\frac{2G\left(\rho \times \frac{4}{3}\pi R^3\right)}{R}}$

Simplifying the terms inside the square root:

$v_e=\sqrt{\frac{8}{3}\pi G\rho R^2}$

Taking the radius $R$ out of the square root gives:

$v_e=R\sqrt{\frac{8}{3}\pi G\rho }$

Since the density $\rho$ and the gravitational constant $G$ remain constant, the entire term under the square root is constant. Therefore, the escape velocity is directly proportional to the radius:

$v_e\propto R$

Because of this direct linear proportionality, any percentage change in the radius of the Earth will result in an identical percentage change in the escape velocity.

Therefore, if the radius of the Earth reduces by 4%, the escape velocity will also reduce by 4%.