$(a)\;\large\frac{3d}{2}\qquad(b)\;\large\frac{2d}{3}\qquad(c)\;\large\frac{5d}{3}\qquad(d)\;\large\frac{3d}{5}$

Thermodynamics

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Answer : $\;\large\frac{3d}{2}$

Explanation :

$PV=nRT \quad $ n=no.of moles = mass / molar mass

Therefore , $\;P=\large\frac{mRT}{VM}=\large\frac{\rho RT}{M}$

$\rho = \large\frac{MP}{RT}$

$d_{A}=\large\frac{M}{R}\;\large\frac{P_{0}}{T_{0}}$

$d_{B}=\large\frac{M}{R}\;\large\frac{3P_{0}}{2T_{0}}$

$d_{B}=\large\frac{3}{2}\;d_{A}$

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