Class 11 Physics | Mechanical Properties of Fluids MCQs with Answers – Part 3

Explanation: Bernoulli’s equation includes \(P\), \(\frac{1}{2}\rho v^2\), and \(\rho gh\). The term \(\rho gh\) represents the gravitational potential energy per unit volume of the fluid. If two points are at different heights, this term changes and must be included. For horizontal flow, the height term cancels, but that simplification is not valid when elevation changes. Ignoring height can give the wrong pressure or speed relation.

Explanation: Statement I is true because for horizontal ideal flow, \(P+\frac{1}{2}\rho v^2=\text{constant}\). If speed increases, the dynamic term increases and static pressure decreases. Statement II is true because the usual Bernoulli relation is applied along a streamline for steady incompressible non-viscous flow. Statement III is also true because \(\frac{1}{2}\rho v^2\) has unit \(\text{kg m}^{-1}\text{s}^{-2}\), which is equivalent to \(\text{Pa}\). These statements combine the condition, meaning, and unit consistency of Bernoulli’s equation.

Explanation: Bernoulli’s equation can be obtained by applying the work-energy idea to a moving fluid element. Pressure forces at the two ends of a small fluid volume do work as the fluid moves through the pipe. This work can change the kinetic energy and gravitational potential energy of the fluid element. The pressure term \(P\), kinetic term \(\frac{1}{2}\rho v^2\), and height term \(\rho gh\) are all written per unit volume. The derivation is meaningful for steady, incompressible, non-viscous flow along a streamline.

Explanation: The pressure term \(P\) has unit \(\text{Pa}\), which is the same as \(\text{J m}^{-3}\). The kinetic term \(\frac{1}{2}\rho v^2\) also has unit \(\text{J m}^{-3}\). The gravitational term \(\rho gh\) has the same unit as well. Therefore, Bernoulli’s equation in this form compares mechanical energy per unit volume. This is why pressure can be added directly to the other two terms.

Explanation: Rows P, Q, and R correctly describe the three Bernoulli terms as energy per unit volume. The pressure term, kinetic term, and gravitational term can be added because they share the same unit. Row S is incorrect because Bernoulli’s equation in the usual form needs suitable conditions such as steady, incompressible, non-viscous flow along a streamline. In strongly viscous or turbulent flow, mechanical energy may be dissipated. The constant-sum idea is not a free rule for every possible fluid motion.

Explanation: \( \textbf{Bernoulli relation:} \) \[ P+\frac{1}{2}\rho v^2+\rho gh=\text{constant} \] The speeds at P and Q are equal, so the kinetic terms are equal. Let \(h_Q\gt h_P\), since Q is higher. Bernoulli’s equation gives: \[ P_P+\rho gh_P=P_Q+\rho gh_Q \] Rearranging: \[ P_P-P_Q=\rho g(h_Q-h_P) \] Since \(h_Q-h_P\) is positive, pressure is larger at the lower point P. \( \textbf{Final answer:} \) Pressure at P is larger by \(\rho g(h_Q-h_P)\).

Explanation: For horizontal ideal flow, Bernoulli’s equation reduces to \(P+\frac{1}{2}\rho v^2=\text{constant}\). If speed is larger at section Q, the dynamic term \(\frac{1}{2}\rho v^2\) is larger there. To keep the sum constant, the static pressure \(P\) must be smaller at Q. This result is often seen in constricted parts of pipes. The faster region has lower static pressure when height and ideal-flow conditions are the same.

Explanation: \( \textbf{Horizontal Bernoulli relation:} \) \[ P+\frac{1}{2}\rho v^2=\text{constant} \] Rearranging: \[ P=\text{constant}-\frac{1}{2}\rho v^2 \] This has the form: \[ y=c-mx \] Here \(P\) is plotted against \(v^2\). Therefore, the slope is: \[ -\frac{1}{2}\rho \] The magnitude of the slope is \(\frac{1}{2}\rho\), while the negative sign shows pressure decreases as \(v^2\) increases. \( \textbf{Final answer:} \) The magnitude of the slope is \(\frac{1}{2}\rho\).

Explanation: In horizontal ideal flow, \(P+\frac{1}{2}\rho v^2\) remains constant along a streamline. When the speed increases, the kinetic energy per unit volume term \(\frac{1}{2}\rho v^2\) increases. Since the total must remain constant, the static pressure \(P\) decreases. The argument that both kinetic term and pressure must increase ignores the trade-off built into Bernoulli’s equation. The pressure decrease is for static pressure under the stated ideal horizontal-flow conditions.

Explanation: \( \textbf{Given:} \) \(A_1=12\,\text{cm}^2\), \(v_1=1.0\,\text{m s}^{-1}\), \(A_2=3\,\text{cm}^2\), and \(\rho=1000\,\text{kg m}^{-3}\). \( \textbf{First use continuity:} \) \[ A_1v_1=A_2v_2 \] So: \[ v_2=\frac{A_1v_1}{A_2} \] Substitute the area ratio: \[ v_2=\frac{12}{3}(1.0)=4.0\,\text{m s}^{-1} \] \( \textbf{Horizontal Bernoulli relation:} \) \[ P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2 \] Pressure drop: \[ P_1-P_2=\frac{1}{2}\rho(v_2^2-v_1^2) \] Substitution: \[ P_1-P_2=\frac{1}{2}(1000)(4.0^2-1.0^2) \] \[ P_1-P_2=500(16-1) \] \[ P_1-P_2=7.5\times10^{3}\,\text{Pa} \] \( \textbf{Final answer:} \) The pressure drop is \(7.5\times10^{3}\,\text{Pa}\).

Explanation: In a Venturi tube, continuity makes the fluid speed increase in the narrow throat. For horizontal ideal flow, Bernoulli’s principle then gives lower static pressure where speed is higher. The pressure drop between the wide part and the throat is the working idea of a Venturi meter. Density does not become zero, and the throat still carries the flow. The pressure change comes from the speed-pressure trade-off, not from the absence of fluid.

Explanation: In the narrow throat Q, the flow speed is greater than in the wide section P. For horizontal ideal flow, greater speed corresponds to lower static pressure. A pressure tube rises to a height depending on the local static pressure. Since pressure at Q is lower, the water level in the tube connected at Q is lower than the level at P. Equal pipe height does not mean equal pressure when speeds differ in Bernoulli flow.

Explanation: The Venturi meter involves two sections of different cross-sectional areas. Continuity gives \(A_1v_1=A_2v_2\) for steady incompressible flow. This relates the speed in the wide section to the speed in the narrow section. Bernoulli’s equation then connects the speed difference with pressure difference. The two relations together allow flow rate to be found from a measured pressure difference.

Explanation: \( \textbf{Given:} \) \(A_1=4A_2\) and \(v_1=v\). For steady incompressible flow: \[ A_1v_1=A_2v_2 \] Substitute \(A_1=4A_2\): \[ (4A_2)v=A_2v_2 \] Cancel \(A_2\): \[ 4v=v_2 \] Therefore: \[ v_2=4v \] The throat speed is larger because the same volume flow rate must pass through a smaller area. \( \textbf{Final answer:} \) The speed at the throat is \(4v\).

Explanation: \( \textbf{Given:} \) \(A_1=5.0\,\text{cm}^2\), \(A_2=2.0\,\text{cm}^2\), \(v_1=2.0\,\text{m s}^{-1}\), and \(\rho=1000\,\text{kg m}^{-3}\). \( \textbf{Use continuity first:} \) \[ A_1v_1=A_2v_2 \] Rearranging: \[ v_2=\frac{A_1v_1}{A_2} \] Substitute: \[ v_2=\frac{5.0}{2.0}(2.0) \] \[ v_2=5.0\,\text{m s}^{-1} \] \( \textbf{Use horizontal Bernoulli equation:} \) \[ P_1-P_2=\frac{1}{2}\rho(v_2^2-v_1^2) \] Substitute: \[ P_1-P_2=\frac{1}{2}(1000)(5.0^2-2.0^2) \] Calculate: \[ 5.0^2-2.0^2=25-4=21 \] Therefore: \[ P_1-P_2=500(21)=1.05\times10^{4}\,\text{Pa} \] \( \textbf{Final answer:} \) The pressure difference is \(1.05\times10^{4}\,\text{Pa}\).

Explanation: A manometer attached to a Venturi meter measures pressure difference between two sections. A larger vertical height difference in the manometer means a larger pressure difference, according to \(\Delta P=\rho_m g\Delta h\) for the manometer liquid. In a horizontal Venturi tube, this pressure difference is connected with the difference in speeds between the wide section and the throat. A larger pressure difference usually indicates a stronger speed contrast and hence a larger flow rate under the same geometry. The height difference is a pressure signal, not a sign that the flow has stopped.

Explanation: Statement I is true because the throat is the narrow part of a Venturi tube. Statement II is true because continuity requires higher speed where the area is smaller. Statement III is false for horizontal ideal flow, because Bernoulli’s principle gives lower static pressure where the speed is higher. The word squeezed can be misleading if it is used to suggest higher pressure at the throat. The Venturi effect depends on pressure dropping in the faster narrow section.

Explanation: Since the fluid is incompressible and the flow is steady, continuity gives \(A_1v_1=A_2v_2\). If \(A_2\lt A_1\), then \(v_2\gt v_1\). For horizontal ideal flow, Bernoulli’s equation gives \(P+\frac{1}{2}\rho v^2=\text{constant}\). The higher speed in the narrow section means a larger kinetic term there. Therefore, the static pressure in the narrow section is lower, so \(P_2\lt P_1\).

Explanation: The Venturi tube is horizontal. So, the height term \(\rho gh\) is the same at the wide section and the throat. Given: \(v_{\text{wide}}=2.0\,\text{m s}^{-1}\) \(v_{\text{throat}}=3v_{\text{wide}}\) \(\rho=1000\,\text{kg m}^{-3}\) First find the throat speed: \(v_{\text{throat}}=3(2.0)\) \(v_{\text{throat}}=6.0\,\text{m s}^{-1}\) For horizontal ideal flow, Bernoulli’s equation becomes: \(P+\frac{1}{2}\rho v^2=\text{constant}\) Apply it between the wide section and the throat: \(P_{\text{wide}}+\frac{1}{2}\rho v_{\text{wide}}^2=P_{\text{throat}}+\frac{1}{2}\rho v_{\text{throat}}^2\) Rearrange: \(P_{\text{wide}}-P_{\text{throat}}=\frac{1}{2}\rho(v_{\text{throat}}^2-v_{\text{wide}}^2)\) The throat has greater speed, so its pressure is lower. Now substitute: \(P_{\text{wide}}-P_{\text{throat}}=\frac{1}{2}(1000)(6.0^2-2.0^2)\) Square the speeds: \(6.0^2=36\) \(2.0^2=4\) So: \(6.0^2-2.0^2=36-4=32\) Now calculate: \(P_{\text{wide}}-P_{\text{throat}}=500(32)\) \(P_{\text{wide}}-P_{\text{throat}}=16000\,\text{Pa}\) \(P_{\text{wide}}-P_{\text{throat}}=1.6\times10^{4}\,\text{Pa}\) \( \textbf{Final answer:} \) The pressure difference is \(1.6\times10^{4}\,\text{Pa}\).

Explanation: A Venturi meter calculation uses continuity together with Bernoulli’s equation. The Bernoulli part assumes that energy losses due to viscosity are negligible. If the flow is strongly viscous, mechanical energy is dissipated as internal energy, so the simple ideal pressure-speed relation is no longer exact. The measured pressure difference may then not correspond to the ideal flow-rate formula. This is why real flow meters often need calibration factors or corrections.