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Three perfect gases at absolute temperatures $T_{1}, T_{2} a n d T_{3}$ are mixed. The masses of the molecules are $m_{1}, m_{2} a n d m_{3}$ and number of molecules are $n_{1}, n_{2} a n d n_{3}$ respectively. Assuming no loss of energy, the final temperature of the mixture is

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$\frac{T_{1} + T_{2} + T_{3}}{3}$

$\frac{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}{n_{1} + n_{2} + n_{3}}$

$\frac{n_{1} T_{1}^{2} + n_{2} T_{2}^{2} + n_{3} T_{3}^{2}}{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}$

$\frac{n_{1}^{2} T_{1}^{2} + n_{2}^{2} T_{2}^{2} + n_{3}^{2} T_{3}^{2}}{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}$

answer is B.

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Detailed Solution

Let $T_{3} > T_{2} > T_{1}$ and final temperature be T such that $T_{3} > T > T_{2} > T_{1}$ .
Heat lost by first two gases is equal to heat gained by third gas

$n_{1} C (T - T_{1}) + n_{2} C (T - T_{2}) = n_{1} C (T_{3} - T)$

$\Rightarrow T = \frac{n_{1} T_{1} + n_{2} T_{2} + n_{3} T_{3}}{n_{1} + n_{2} + n_{3}}$

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