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Chapter 4: Problem 560

An engine pump is used to pump a liquid of density \(\rho\) continuously through a pipe of cross-sectional area \(\mathrm{A}\). If the speed of flow of the liquid in the pipe is \(\mathrm{v}\), then the rate at which kinetic energy is being imparted to the liquid is (A) \((1 / 2) \mathrm{A} \rho \mathrm{V}^{3}\) (B) \((1 / 2) \mathrm{A} \rho \mathrm{V}^{2}\) (C) \((1 / 2) \mathrm{A} \rho \mathrm{V}\) (B) \(\mathrm{ApV}\)

Short Answer

Expert verified
The correct answer is (A) \(\frac{1}{2} \rho A v^3\).

Step by step solution

01

Determining the mass flow rate

In this case, the mass flow rate can be found by multiplying the density (ρ) of the liquid by the cross-sectional area (A) of the pipe and the flow velocity (v). The formula for mass flow rate is given by: Mass flow rate = ρAv
02

Applying the kinetic energy formula

Now, let's substitute the mass flow rate into the kinetic energy formula to determine the rate of imparted kinetic energy. Rate of imparted kinetic energy = (1/2)(Mass flow rate)v^2 Substitute the mass flow rate (ρAv) into the formula: Rate of imparted kinetic energy = (1/2)(ρAv)v^2
03

Simplifying the expression

Now, we'll simplify the expression: Rate of imparted kinetic energy = (1/2)ρAv^3
04

Comparing the expression with given options

After simplifying the expression, we have found that the rate of imparted kinetic energy to the liquid is (1/2)ρAv^3. Comparing this expression with the given options, we can see that the correct answer is: (A) (1 / 2) ρAV^3

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Most popular questions from this chapter

The force constant of a wire is \(\mathrm{K}\) and that of the another wire is \(3 \mathrm{k}\) when both the wires are stretched through same distance, if work done are \(\mathrm{W}_{1}\) and \(\mathrm{W}_{2}\), then... (A) \(\mathrm{w}_{2}=3 \mathrm{w}_{1}^{2}\) (B) \(\mathrm{W}_{2}=0.33 \mathrm{~W}_{1}\) (C) \(\mathrm{W}_{2}=\mathrm{W}_{1}\) (D) \(\mathrm{W}_{2}=3 \mathrm{~W}_{1}\)

A bomb of mass \(3.0 \mathrm{~kg}\) explodes in air into two pieces of masses \(2.0 \mathrm{~kg}\) and \(1.0 \mathrm{~kg}\). The smaller mass goes at a speed of \(80 \mathrm{~m} / \mathrm{s}\). The total energy imparted to the two fragments is (A) \(1.07 \mathrm{KJ}\) (B) \(2.14 \mathrm{KJ}\) (C) \(2.4 \mathrm{KJ}\) (D) \(4.8 \mathrm{KJ}\)

Three objects \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) are kept in a straight line on a frictionless horizontal surface. These have masses $\mathrm{m}, 2 \mathrm{~m}\( and \)\mathrm{m}\( respectively. The object \)\mathrm{A}$ moves towards \(\mathrm{B}\) with a speed \(9 \mathrm{~m} / \mathrm{s}\) and makes an elastic collision with it. Thereafter, B makes completely inelastic collision with \(\mathrm{C}\). All motion occur on the same straight line. Find final speed of the object \(\mathrm{C}\) (A) \(3 \mathrm{~m} / \mathrm{s}\) (B) \(4 \mathrm{~m} / \mathrm{s}\) (C) \(5 \mathrm{~m} / \mathrm{s}\) (D) \(1 \mathrm{~m} / \mathrm{s}\)

Two bodies of masses \(\mathrm{m}\) and \(3 \mathrm{~m}\) have same momentum. their respective kinetic energies \(E_{1}\) and \(E_{2}\) are in the ratio..... (A) \(1: 3\) (B) \(3: 1\) (C) \(1: 3\) (D) \(1: 6\)

A single conservative force \(\mathrm{F}(\mathrm{x})\) acts on a $2.5 \mathrm{~kg}\( particle that moves along the \)\mathrm{x}$ -axis. The potential energy \(\mathrm{U}(\mathrm{x})\) is given by \(\mathrm{U}(\mathrm{x})=\left[10+(\mathrm{x}-4)^{2}\right]\) where \(\mathrm{x}\) is in meter. \(\mathrm{At} \mathrm{x}=6.0 \mathrm{~m}\) the particle has kinetic energy of \(20 \mathrm{~J}\). what is the mechanical energy of the system? (A) \(34 \mathrm{~J}\) (B) \(45 \mathrm{~J}\) (C) \(48 \mathrm{~J}\) (D) \(49 \mathrm{~J}\)
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