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Chapter 4: Problem 492

An electric motor develops \(5 \mathrm{KW}\) of power. How much time will it take to lift a water of mass \(100 \mathrm{~kg}\) to a height of $20 \mathrm{~m}\( ? \)\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$ (A) \(4 \mathrm{sec}\) (B) \(5 \mathrm{sec}\) (C) \(8 \mathrm{sec}\) (D) \(10 \mathrm{sec}\)

Short Answer

Expert verified
The time taken by the electric motor to lift the water of mass 100 kg to a height of 20 m is 4 seconds. The correct option is (A) 4 seconds.

Step by step solution

01

Calculate the potential energy change

In order to calculate the potential energy change of the water, we can use the formula: \[ΔPE = m \cdot g \cdot h,\] where \(ΔPE\) is the potential energy change, \(m\) is the mass of the water (100 kg), \(g\) is the gravitational constant (10 m/s²), and \(h\) is the height the water is raised (20 m).
02

Compute the potential energy change

Plugging in the known values, we get: \[ΔPE = (100\,\text{kg}) (10\,\text{m}/\text{s}^2) (20\,\text{m}).\] \[ΔPE = 20000\, \text{Joules}.\] The potential energy change is 20,000 Joules.
03

Calculate the time taken by the motor to lift the water

Since power is the rate at which work is done (or energy is transferred), we can use the formula: \[P = \dfrac{ΔPE}{t},\] Where \(P\) is the power of the motor (5 KW = 5,000 W), \(ΔPE\) is the potential energy change (20,000 Joules), and \(t\) is the time taken. Solving for \(t\) gives us \[t = \dfrac{ΔPE}{P}.\]
04

Compute the time t

Substituting the known values, we get: \[t = \dfrac{20000\,\text{Joules}}{5000\,\text{Watts}}.\] \[t = 4\,\text{seconds}.\] Thus, the time taken by the electric motor to lift the water of mass 100 kg to a height of 20 m is 4 seconds. The correct option is (A) 4 seconds.

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Most popular questions from this chapter

Two bodies \(\mathrm{P}\) and \(\mathrm{Q}\) have masses \(5 \mathrm{~kg}\) and $20 \mathrm{~kg}\( respectively. Each one is acted upon by a force of \)4 \mathrm{~N}$. If they acquire the same kinetic energy in times \(t_{\mathrm{P}}\) and \(\mathrm{t}_{\mathrm{Q}}\) then the ratio $\left(t_{q} / t_{p}\right)=\ldots \ldots$ \((\mathrm{A})(1 / 2)\) (B) 2 (C) 5 (D) 6

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The potential energy of \(2 \mathrm{~kg}\) particle, free to move along \(\mathrm{x}\) axis is given by $\mathrm{U}(\mathrm{X})=\left[\left\\{\mathrm{x}^{4} / 4\right\\}-\left\\{\mathrm{x}^{2} / 2\right\\}\right] \mathrm{J}$. If its mechanical energy is \(2 \mathrm{~J}\), its maximum speed is $\ldots \mathrm{m} / \mathrm{s}$ (A) \((3 / 2)\) (B) \(\sqrt{2}\) (C) \((1 / \sqrt{2})\) (D) 2
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