$\begin{array}{1 1}(A)\;1cm\\(B)\;2cm\\(C)\;3cm\\(D)\;4cm\end{array} $

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$P_0V_0=P_1V_1$

$\Rightarrow P_1=\large\frac{P_0V_0}{V_1}=\frac{P_0(45)}{45+x}$

$P_0V_0=P_2V_2$

$\Rightarrow P_2=\large\frac{P_045}{45-x}$

$P_2=P_1+10$

$P_0=76cm$ Hg

$\large\frac{45P_0}{45+x}$$+10=\large\frac{45P_0}{45-x}$

$\Rightarrow x^2+684x-2025=0$

$x=2.95\approx 3$

Hence (C) is the correct answer.

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