Question

The flow of oil of density $800\text{ kgm}^{-3}$ through a horizontal pipe is streamlined. If the pressure between two points separated by 1m in the tube falls by $1.6\text{ Nm}^{-2}$, then the change in kinetic energy per kg of the liquid between these points is

• A. $16\times 10^{-4}\text{ J kg}^{-1}$
• B. $2\times 10^{-3}\text{ J kg}^{-1}$
• C. $4\times 10^{-3}\text{ J kg}^{-1}$
• D. $8\times 10^{-4}\text{ J kg}^{-1}$
• E. $32\times 10^{-2}\text{ J kg}^{-1}$

Hint:

Logic Tip: Pay close attention to the requested units. "Energy per kg" means dividing the standard energy expression by mass. Since pressure is energy per unit volume, dividing pressure by density (mass per unit volume) directly yields energy per unit mass.

Answer 1

The Correct Option is B

Concept:
According to Bernoulli's equation for a horizontal pipe (where the height $h_1 = h_2$), the sum of pressure energy and kinetic energy per unit volume remains constant: $$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$ Rearranging this relates the pressure drop to the change in kinetic energy per unit volume: $$P_1 - P_2 = \frac{1}{2}\rho v_2^2 - \frac{1}{2}\rho v_1^2 = \Delta KE_{vol}$$ To find the change in kinetic energy \textit{per unit mass} ($\Delta KE_m$), we divide the volume-based energy by density ($\rho$).
Step 1: Identify the given parameters.
Density of oil, $\rho = 800\text{ kgm}^{-3}$ Pressure drop, $\Delta P = P_1 - P_2 = 1.6\text{ Nm}^{-2}$
Step 2: Formulate the required expression.
The change in kinetic energy per unit mass is: $$\Delta KE_m = \frac{\frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2}{m} = \frac{1}{2}v_2^2 - \frac{1}{2}v_1^2$$ From Bernoulli's equation: $$P_1 - P_2 = \rho \left( \frac{1}{2}v_2^2 - \frac{1}{2}v_1^2 \right)$$ $$\Delta P = \rho \cdot \Delta KE_m$$ $$\Delta KE_m = \frac{\Delta P}{\rho}$$
Step 3: Calculate the value.
Substitute the given values into the formula: $$\Delta KE_m = \frac{1.6}{800}$$ To simplify, rewrite 800 in scientific notation: $$\Delta KE_m = \frac{1.6}{8 \times 10^2}$$ $$\Delta KE_m = 0.2 \times 10^{-2}$$ Adjust the decimal to match standard scientific options: $$\Delta KE_m = 2 \times 10^{-3}\text{ J kg}^{-1}$$