The moment of inertia of HCl molecule about an axes passing through its centre of mass perpendicular to the bond is (Given inter nuclear distance $d = 1.3$ angstrom, mass of proton $1.7 \times 10^{- 27} K g . A t . W t of Cl is 35) .$

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$2.79 \times 10^{- 44} K g m^{2}$

$2.79 \times 10^{- 49} K g m^{2}$

$2.79 \times 10^{- 42} K g m^{2}$

$2.79 \times 10^{- 47} K g m^{2}$

answer is C.

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Detailed Solution

$I = (\frac{m_{1} m_{2}}{m + m_{2}}) r^{2} = (\frac{1 \times 35}{1 + 35}) {(1.3)}^{2} a m u A^{0} = \frac{35}{36} \times 1.7 \times 10^{- 27} \times {(1.3 \times 10^{- 10})}^{2} = 2.79 \times 10^{- 47} k g m^{2}$

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