Explanation
$$T=-\alpha V^3+\beta V^2\\ \cfrac{PV}{R}=-\alpha V^2+\beta V^2\\ P=RV(-\alpha V+\beta)$$
For maximum pressure
$$\cfrac{dP}{dV}=0\\ \Rightarrow -2\alpha V+\beta =0\\ \Rightarrow V=\cfrac{\beta }{2\alpha}\\ P=R[-\alpha(\cfrac{\beta }{2\alpha})^2+\cfrac{\beta }{2\alpha}\times \beta]\\P=R[\cfrac{-\beta^2}{4\alpha}+\cfrac{\beta^2}{2\alpha}]\\P=\cfrac{R\beta^2}{4\alpha}$$
For a given volume, among all the three objects, the sphere has the least surface area and the cube will have the maximum surface area. This is also due to the fact that the volume and mass of all three are the same.
According to Heat theory, the rate of loss of heat is proportional to the surface area of the object. Hence, the rate of cooling will be slowest for the sphere and fastest for the cube.
Thus the correct option is A.
Let V is the volume of the vessel.
Now, let $$p_{1}$$ and $$p_{2}$$ be the partial pressure, then using gas law:
$$p_{1}V = \dfrac{m_1}{M_1}RT\\$$
$$p_{2}V = \dfrac{m_2}{M_2}RT,\ p_{0} = p_{1} + p_{2}\\$$
$$p_{0} = \left(\dfrac{m_1}{M_1} + \dfrac{m_2}{M_2}\right)\dfrac{RT}{V}\\$$
$$V = \left(\dfrac{m_1}{M_1} + \dfrac{m_2}{M_2}\right)\dfrac{RT}{p_{0}}\\$$
$$\because \rho_{mix}=\dfrac{(m_{1} + m_{2})}{V}\\$$
$$\rho_{mix}=\dfrac {(m_1 + m_2)M_1 M_2} {(m_1M_2 + m_2M_1)} \times \dfrac{p_0}{RT}\\$$
Substituting values,
$$\rho_{mix}=\dfrac {(7 + 11) \times 28 \times 44\times 10^{-3}} {(7 \times 44 + 11\times 28))} \times \dfrac{10^{5}}{8.3 \times 300}\\$$
$$= 1.446 \ per \ litre$$
Option A is correct.
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