In a perfectly inelastic collision, two spheres made of the same material with masses 15 kg and 25 kg, moving in opposite directions with speeds of 10 m/s and 30 m/s, respectively, strike each other and stick together. The rise in temperature (in $$^{\circ}C$$), if all the heat produced during the collision is retained by these spheres, is :
(specific heat of sphere material 31 cal/kg.$$^{\circ}C$$ and 1 cal =4.2 J)

Let the masses be $$m_1 = 15\ \text{kg}$$ and $$m_2 = 25\ \text{kg}$$, and their velocities before collision be $$v_1 = +10\ \text{m/s}$$ and $$v_2 = -30\ \text{m/s}$$ (opposite directions).

In a perfectly inelastic collision, momentum is conserved. The formula is:

$$m_1v_1 + m_2v_2 = (m_1 + m_2)\,v_f$$ $$-(1)$$

Substitute the given values into $$(1)$$:

$$15\times 10 + 25\times(-30) = 40\,v_f$$
$$150 - 750 = 40\,v_f$$
$$-600 = 40\,v_f$$
$$v_f = -\frac{600}{40} = -15\ \text{m/s}$$ $$-(2)$$

The heat produced equals the loss in kinetic energy. First, compute the initial kinetic energy:

$$K_i = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$$ $$-(3)$$

Substitute values:

$$K_i = \frac{1}{2}\times 15\times 10^2 + \frac{1}{2}\times 25\times 30^2$$
$$K_i = 750 + 11250 = 12000\ \text{J}$$ $$-(4)$$

Next, compute the final kinetic energy of the combined mass:

$$K_f = \frac{1}{2}(m_1 + m_2)\,v_f^2$$ $$-(5)$$

Substitute values:

$$K_f = \frac{1}{2}\times 40\times (-15)^2 = 20\times 225 = 4500\ \text{J}$$ $$-(6)$$

Thus, the heat generated is:

$$Q = K_i - K_f = 12000 - 4500 = 7500\ \text{J}$$ $$-(7)$$

Given specific heat in calories: $$c = 31\ \text{cal/kg}^{\circ}\text{C}$$ and $$1\ \text{cal} = 4.2\ \text{J}$$. Converting to SI units:

$$c = 31\times 4.2 = 130.2\ \text{J/kg}^{\circ}\text{C}$$ $$-(8)$$

The total mass is:

$$M = m_1 + m_2 = 15 + 25 = 40\ \text{kg}$$ $$-(9)$$

The temperature rise is:

$$\Delta T = \frac{Q}{M\,c} = \frac{7500}{40\times 130.2} \approx 1.44\ ^{\circ}\text{C}$$ $$-(10)$$

Final Answer: The rise in temperature is $$1.44\ ^{\circ}\text{C}$$ (Option D).