A railway wagon (open at the top) of mass M₁ is moving with speed v₁ along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M₂ and speed becomes v₂. Taking the rain to be falling vertically and water stationary inside the wagon, the relation between the two speeds v₁ and v₂ is :

GS&EI ElectEngg 14 2024 1601-2025 GS&EI ElectEngg 14 2024 1601-2025 This question is identical to ID 80852 and ID 80861. In a Schering bridge balance, the unknown capacitance $C_1$ and its loss angle $\delta$ are determined by the following formulas: $C_1 = C_2 \frac{R_4}{R_3}$ $\tan \delta = \omega C_4 R_4$Given values: $C_2 = 100 \mu \text{F}$ $R_3 = 100 \Omega$ $R_4 = 300 \Omega$ $C_4 = 0.5 \mu \text{F}$ $f = 50 \text{ Hz}$Calculation for $C_1$: $C_1 = 100 \mu \text{F} \times \frac{300 \Omega}{100 \Omega} = 100 \mu \text{F} \times 3 = 300 \mu \text{F}$.Calculation for $\delta$: Angular frequency, $\omega = 2\pi f = 2\pi \times 50 = 100\pi \text{ rad/s}$. $\tan \delta = (100\pi \text{ rad/s}) \times (0.5 \times 10^{-6} \text{ F}) \times (300 \Omega)$ $\tan \delta = 100\pi \times 0.04712$.The loss angle $\delta$ is the arctangent of the dissipation factor: $\delta = \arctan(0.04712) \approx 2.698^{\circ}$. Rounding to one decimal place, $\delta \approx 2.7^{\circ}$.Thus, the unknown capacitance is $300 \mu \text{F}$ and its loss angle is $2.7^{\circ}$. This matches option C. The initial horizontal momentum of the wagon is $P_1 = M_1 v_1$. After the rain has collected, the total mass is $M_2$ and the speed is $v_2$. The final horizontal momentum is $P_2 = M_2 v_2$. By conservation of horizontal momentum, $P_1 = P_2$, so $M_1 v_1 = M_2 v_2$.