1. Given data

• Mass of ice $= M_1$ g
• Initial temperature of ice $= -10^\circ C$
• Specific heat of ice $= 0.5\, \text{cal g}^{-1}\, ^\circ C^{-1}$
• Mass of water $= M_2$ g
• Initial temperature of water $= 5^\circ C$
• Final temperature of mixture $= 0^\circ C$
• No ice is left at the end

We need to find latent heat of fusion of ice, say $L$ in $\text{cal g}^{-1}$.

2. Heat required by ice

The ice first warms from $-10^\circ C$ to $0^\circ C$.

Heat needed for this: $Q_1 = M_1 \times 0.5 \times 10 = 5M_1 \text{ cal}$

Then the ice melts completely at $0^\circ C$.

Heat needed for melting: $Q_2 = M_1 L$

So total heat absorbed by ice is $Q_{\text{ice}} = 5M_1 + M_1L$

3. Heat lost by water

Water cools from $5^\circ C$ to $0^\circ C$.

Taking specific heat of water as $1\, \text{cal g}^{-1}\, ^\circ C^{-1}$, $Q_{\text{water}} = M_2 \times 1 \times 5 = 5M_2 \text{ cal}$

4. Apply principle of calorimetry

Since final temperature is $0^\circ C$ and no heat is lost externally,

$\text{Heat lost by water} = \text{Heat gained by ice}$

So, $5M_2 = 5M_1 + M_1L$

5. Solve for $L$

$M_1L = 5M_2 - 5M_1$

$L = \frac{5M_2 - 5M_1}{M_1}$

$L = \frac{5M_2}{M_1} - 5$

6. Match with options

This matches:

Option C: $\frac{5M_2}{M_1} - 5$

7. Comparison with stored correct answer

Stored correct answer is A: $\frac{50M_2}{M_1} - 5$

But this would be correct only if the water were initially at $50^\circ C$. The question text says water is at $5^\circ C$, so using the given statement, the correct expression is $\frac{5M_2}{M_1} - 5$

Hence I disagree with the stored answer. It appears there may be a typo in the question statement or in the stored answer.