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

  • [JEE MAIN 2019]
  • A

    $\frac{{50{M_2}}}{{{M_1}}} - 5$

  • B

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

  • C

    $\frac{{50{M_2}}}{{{M_1}}}$

  • D

    $\frac{{5{M_1}}}{{{M_2}}} - 50$

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