A projectile of mass M is fired so that the horizontal range is 4 km. At the highest point the projectile explodes in two parts of masses M/4 and 3M/4 respectively and the heavier part starts falling down vertically with zero initial speed. The horizontal range (distance from point of firing) of the lighter part is :

A projectile of mass M has a horizontal range of 4 km. The range formula for a projectile is given by $$ R = \frac{u^2 \sin 2\theta}{g} $$, where u is the initial velocity and θ is the angle of projection. So, we have:

$$ R = \frac{u^2 \sin 2\theta}{g} = 4 \text{km} $$

At the highest point of its trajectory, the projectile explodes into two parts: one of mass $$ \frac{M}{4} $$ and the other of mass $$ \frac{3M}{4} $$. The heavier part (mass $$ \frac{3M}{4} $$) falls vertically downward with zero initial speed, meaning its velocity immediately after the explosion is zero in both horizontal and vertical directions.

Since the explosion is an internal force, it does not affect the center of mass motion. Before the explosion, at the highest point, the entire projectile has only horizontal velocity $$ u_x = u \cos \theta $$ and zero vertical velocity. Therefore, the center of mass continues with velocity $$ (u_x, 0) $$ after the explosion.

Let $$ \vec{v_1} = (v_{1x}, v_{1y}) $$ be the velocity of the lighter part (mass $$ \frac{M}{4} $$) and $$ \vec{v_2} = (0, 0) $$ be the velocity of the heavier part (mass $$ \frac{3M}{4} $$) after the explosion. The center of mass velocity is:

$$ \vec{v_{cm}} = \frac{ \left( \frac{M}{4} \right) \vec{v_1} + \left( \frac{3M}{4} \right) \vec{v_2} }{ M } = \frac{1}{4} \vec{v_1} + \frac{3}{4} \vec{v_2} $$

Setting $$ \vec{v_{cm}} = (u_x, 0) $$ and substituting $$ \vec{v_2} = (0, 0) $$:

$$ (u_x, 0) = \frac{1}{4} (v_{1x}, v_{1y}) + \frac{3}{4} (0, 0) $$

This simplifies to:

$$ (u_x, 0) = \left( \frac{v_{1x}}{4}, \frac{v_{1y}}{4} \right) $$

Equating components:

$$ \frac{v_{1x}}{4} = u_x \quad \Rightarrow \quad v_{1x} = 4u_x $$

$$ \frac{v_{1y}}{4} = 0 \quad \Rightarrow \quad v_{1y} = 0 $$

So, the lighter part has horizontal velocity $$ 4u_x $$ and zero vertical velocity immediately after the explosion.

The explosion occurs at the highest point, which is at a horizontal distance of half the range from the firing point. Since the range is 4 km, this distance is:

$$ \frac{R}{2} = \frac{4}{2} = 2 \text{km} $$

The lighter part is now projected horizontally from this point with initial velocity $$ 4u_x $$ and from a height H, where H is the maximum height of the original projectile. The maximum height H is given by:

$$ H = \frac{u^2 \sin^2 \theta}{2g} $$

The time taken for the lighter part to fall to the ground from height H with zero initial vertical velocity is found using the equation of motion:

$$ H = \frac{1}{2} g t^2 \quad \Rightarrow \quad t = \sqrt{\frac{2H}{g}} $$

In this time, the horizontal distance covered by the lighter part from the explosion point is:

$$ d = v_{1x} \cdot t = 4u_x \cdot \sqrt{\frac{2H}{g}} $$

Substituting $$ H = \frac{u^2 \sin^2 \theta}{2g} $$:

$$ d = 4u_x \cdot \sqrt{ \frac{2}{g} \cdot \frac{u^2 \sin^2 \theta}{2g} } = 4u_x \cdot \sqrt{ \frac{u^2 \sin^2 \theta}{g^2} } = 4u_x \cdot \frac{u \sin \theta}{g} $$

Let $$ u_y = u \sin \theta $$ (the initial vertical component of velocity for the original projectile). Then:

$$ d = 4u_x \cdot \frac{u_y}{g} $$

From the original range equation:

$$ R = \frac{2 u_x u_y}{g} = 4 \text{km} $$

So:

$$ \frac{u_x u_y}{g} = \frac{R}{2} = \frac{4}{2} = 2 \text{km} $$

Substituting this into the expression for d:

$$ d = 4 \cdot 2 = 8 \text{km} $$

This is the horizontal distance traveled by the lighter part from the explosion point. The total horizontal range from the firing point is the distance to the explosion point plus d:

$$ \text{Total range} = 2 \text{km} + 8 \text{km} = 10 \text{km} $$

Hence, the horizontal range of the lighter part is 10 km.

So, the answer is Option C.