Question 71:
moderateAn engine has an efficiency of $1/6$. When the temperature of sink is reduced by $62\text{ }^\circ\text{C}$, its efficiency is doubled. Temperature of the source is (2007)
Initial efficiency $\eta_1 = 1 - \frac{T_2}{T_1} = \frac{1}{6}$. After reduction, $\eta_2 = 2(\frac{1}{6}) = \frac{1}{3} = 1 - \frac{T_2-62}{T_1}$. Solving these equations yields source temperature $T_1 = 372\text{ K} = 99\text{ }^\circ\text{C}$.
Question 72:
moderateA particle executes linear simple harmonic motion with an amplitude of $ 3 \text{ cm} $. When the particle is at $ 2 \text{ cm} $ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is: (2017-Delhi)
Given $ |v| = |a| $, we have $ \omega \sqrt{A^2 - x^2} = \omega^2 x $. Substituting $ A = 3 $ and $ x = 2 $, we get $ \sqrt{3^2 - 2^2} = \omega (2) \implies \omega = \frac{\sqrt{5}}{2} $. The time period is $ T = \frac{2\pi}{\omega} = \frac{4\pi}{\sqrt{5}} $.