Question

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Answer 1

Equate upward and downward pressure at the gas-mercury junction in the two cases. It can be proved that the pressure exerted due  to mercury is proportional to the vertical component of the length of the column $(p=h\rho g)$
$\text{In figure 1 },\\p_{gas}=p_{atm}+p_{Hg}\\\dfrac{nRT}{V_1}=76 cmHg+0\\\Rightarrow \dfrac{nRT}{Al_1}=76cmHg\qquad(1)\\\text{In figure 2},\\p_{gas}=p_{atm}+p_{Hg}\\\dfrac{nRT}{V_2}=76 cmHg+8\sin\theta cmHg\\\Rightarrow \dfrac{nRT}{Al_2}=(76+8\sin\theta)cmHg\qquad(2)\\(2)\div(1)\\\dfrac{l_1}{l_2}=\dfrac{76+8\sin\theta}{76}\\=1+\dfrac{2\sin\theta}{19}$