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**Hint:**Recall that momentum is the product of mass and velocity of the bodies or the objects. The external force on a system is directly proportional to the rate of change of momentum. The change in momentum takes place in the direction of applied force.

**Complete step by step solution:**

Step I: Since the balloon is suddenly released and gas starts leaking out, it is clear that there is no external force applied on the system. This means the external force is zero.

Also $F = \dfrac{{dp}}{{dt}}$

Where $F$is the force

$\dfrac{{dp}}{{dt}}$is the rate of change of momentum

Step II: Since external force on the balloon is zero, so

$\dfrac{{dp}}{{dt}} = 0$

$\dfrac{{d(mv)}}{{dt}} = 0$

Step III:

$v\dfrac{{dm}}{{dt}} + m\dfrac{{dv}}{{dt}} = 0$

$m\dfrac{{dv}}{{dt}} = - v\dfrac{{dm}}{{dt}}$

$mdv = - vdm$

$\dfrac{{dm}}{m} = - \dfrac{{dv}}{v}$

Step IV: When the gas is filled then the mass of the balloon is $M$ and velocity will be $0$.

But when the gas starts leaking the mass is $\dfrac{M}{2}$and velocity will be $M$

Integrating above equation with the given conditions,

$\int\limits_M^{\dfrac{M}{2}} {\dfrac{{dm}}{m}} = - \int\limits_0^M {\dfrac{{dv}}{v}} $

$\int\limits_M^{\dfrac{M}{2}} {\dfrac{{dm}}{m}} = - \dfrac{1}{v}\int\limits_0^M {dv} $

The derivative of the final velocity of the object gives its initial velocity that is ‘$u$’.

$\ln \dfrac{m}{{2m}} = - \dfrac{1}{v}(u)$

$u = - v\ln \dfrac{1}{2}$

$u = v\ln 2$

$u = 2\ln 2$

Step V: So when the gas leaks out and the mass of the balloon is reduced to half, then the velocity of the balloon will be $2\ln 2$.

**Therefore Option (C) is the right answer.**

**Note:**It is to be noted that any object that has mass and is moving has momentum. But the change of momentum of an object is equal to its impulse. Also impulse is a quantity related to force but it is not equal to momentum. Impulse of a body is the increase or decrease in momentum.

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