When $M_1$ gram of ice at $-10\,^{\circ}C$ (specific heat $= 0.5\, cal\, g^{-1}\,^{\circ}C^{-1}$) is added to $M_2$ gram of water at $50\,^{\circ}C$,finally no ice is left and the water is at $0\,^{\circ}C$. The value of latent heat of ice,in $cal\, g^{-1}$ is

  • A
    $\frac{50M_2}{M_1} - 5$
  • B
    $\frac{5M_2}{M_1} - 5$
  • C
    $\frac{50M_2}{M_1}$
  • D
    $\frac{5M_1}{M_2} - 50$

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