Question

$AB:P = \sqrt{3}V + C_{1} $$${P_{0} = \sqrt{3}V_{0} + C_{1}(1) }{and3P_{0} = \sqrt{3}V_{B} + C_{1}(2) }{BC:P = - \frac{1}{\sqrt{3}}V + C_{2} }{P_{0} = - \frac{1}{\sqrt{3}}.6V_{0} + C_{2}(3) }{3P_{0} = - \frac{1}{\sqrt{3}}.V_{B} + C_{2}(4)}$$

From equation (1), (2), (3) and (4), $V_{B} = \frac{9}{4}V_{0}$

$\text{Now,}\frac{T_{B}}{T_{A}} = \frac{3P_{0}.\frac{9}{4}V_{0}}{P_{0}.V_{0}} = \frac{27}{4}$