If for a given angle of projection, the horizontal range is doubled, the time of flight becomes

The correct option is √2 times.

To understand why, let us write down the mathematical expressions for the horizontal range and the time of flight of a projectile. Let the initial velocity of projection be u and the angle of projection be $\theta$.

The formula for the horizontal range (R) of a projectile is given by:
$R=\frac{u^{2}sin\left(2\theta \right)}{g}$
where g is the acceleration due to gravity.

The formula for the time of flight (T) of a projectile is given by:
$T=\frac{2usin\theta }{g}$

For a given (constant) angle of projection $\theta$ and constant acceleration due to gravity g, we can establish the proportionalities of both variables with respect to the initial velocity u:

From the horizontal range formula:
$R\propto u^2$
This implies that:
$u\propto \sqrt{R}$

From the time of flight formula:
$T\propto u$

Substituting the relationship between u and R into the proportionality for the time of flight T, we get:
$T\propto \sqrt{R}$

This means we can write the ratio of the new time of flight (T₂) to the initial time of flight (T₁) as:
$\frac{T_2}{T_1}=\sqrt{\frac{R_2}{R_1}}$

According to the problem, the horizontal range is doubled, which means:
$R_2=2R_1$

Substituting this value into the ratio equation:
$\frac{T_2}{T_1}=\sqrt{\frac{2R_1}{R_1}}=\sqrt{2}$

Thus, the new time of flight becomes:
$T_2=\sqrt{2}{T}_1$
So, the time of flight becomes √2 times the original value.