NEET Test Series from KOTA - 10 Papers In MS WORD
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Work, Energy and Power
148747
An engine develops $20 \mathrm{~kW}$ of power. How much time will it take to lift a mass of $200 \mathrm{~kg}$ to a height of $40 \mathrm{~m} ?\left(\mathrm{~g}=10 \mathrm{~m} . \mathrm{s}^{-2}\right)$
1 $4 \mathrm{~s}$
2 $5 \mathrm{~s}$
3 $8 \mathrm{~s}$
4 $10 \mathrm{~s}$
Explanation:
A Given, Mass of lift $(\mathrm{m})=200 \mathrm{~kg}$, Height $(\mathrm{h})=40 \mathrm{~m}$, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$, Power $(\mathrm{P})=20 \mathrm{~kW}=20 \times 10^{3} \mathrm{~W}$ $\text { Power }(\mathrm{P}) =\frac{(\mathrm{m} \cdot \mathrm{g} \cdot \mathrm{h})}{\mathrm{t}}$ $\mathrm{t} =\frac{\mathrm{m} \cdot \mathrm{g} \cdot \mathrm{h}}{\mathrm{P}}$ $=\frac{200 \times 10 \times 40}{20 \times 10^{3}}$ $\mathrm{t} =4 \mathrm{sec}$
AP EAMCET-20.08.2021
Work, Energy and Power
148748
A quarter horse power motor runs at a speed of $600 \mathrm{rpm}$. Assuming 60\% efficiency, the work done by the motor in one rotation is -
148749
A lawn roller has been pushed through a distance of $80 \mathrm{~m}$ by applying a force of $50 \mathrm{kgwt}$ in a direction inclined at $60^{\circ}$ with the ground. Calculate the work done on the roller.
148751
A constant force of $5 \mathrm{~N}$ accelerates a stationary particle of mass $500 \mathrm{gm}$ through a displacement of $5 \mathrm{~m}$. The average power delivered is
1 $6.25 \mathrm{~W}$
2 $25 \mathrm{~W}$
3 $62.5 \mathrm{~W}$
4 $50 \mathrm{~W}$
Explanation:
B Given, Force, $\mathrm{F}=5 \mathrm{~N}$, mass of stationary particle $=500 \mathrm{gm}$ Displacement $(\mathrm{s})=5 \mathrm{~m}$ $\therefore \quad \mathrm{F}=\mathrm{ma}$ $5=\frac{500}{1000} \mathrm{a}$ $\mathrm{a}=10 \mathrm{~m} / \mathrm{sec}^{2}$ We know that equation of second law of motion, $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$ $5=\frac{1}{2} \times 10 \mathrm{t}^{2} \quad[\because$ Initial velocity, $\mathrm{u}=0]$ $5=5 \mathrm{t}^{2}$ $\mathrm{t}=1 \mathrm{sec}$ Then, the average power delivered- $\mathrm{P} =\frac{\mathrm{W}}{\mathrm{t}}=\frac{\mathrm{F} . \mathrm{s}}{\mathrm{t}}$ $\mathrm{P} =\frac{5 \times 5}{1}$ $=25 \mathrm{Watt}$
148747
An engine develops $20 \mathrm{~kW}$ of power. How much time will it take to lift a mass of $200 \mathrm{~kg}$ to a height of $40 \mathrm{~m} ?\left(\mathrm{~g}=10 \mathrm{~m} . \mathrm{s}^{-2}\right)$
1 $4 \mathrm{~s}$
2 $5 \mathrm{~s}$
3 $8 \mathrm{~s}$
4 $10 \mathrm{~s}$
Explanation:
A Given, Mass of lift $(\mathrm{m})=200 \mathrm{~kg}$, Height $(\mathrm{h})=40 \mathrm{~m}$, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$, Power $(\mathrm{P})=20 \mathrm{~kW}=20 \times 10^{3} \mathrm{~W}$ $\text { Power }(\mathrm{P}) =\frac{(\mathrm{m} \cdot \mathrm{g} \cdot \mathrm{h})}{\mathrm{t}}$ $\mathrm{t} =\frac{\mathrm{m} \cdot \mathrm{g} \cdot \mathrm{h}}{\mathrm{P}}$ $=\frac{200 \times 10 \times 40}{20 \times 10^{3}}$ $\mathrm{t} =4 \mathrm{sec}$
AP EAMCET-20.08.2021
Work, Energy and Power
148748
A quarter horse power motor runs at a speed of $600 \mathrm{rpm}$. Assuming 60\% efficiency, the work done by the motor in one rotation is -
148749
A lawn roller has been pushed through a distance of $80 \mathrm{~m}$ by applying a force of $50 \mathrm{kgwt}$ in a direction inclined at $60^{\circ}$ with the ground. Calculate the work done on the roller.
148751
A constant force of $5 \mathrm{~N}$ accelerates a stationary particle of mass $500 \mathrm{gm}$ through a displacement of $5 \mathrm{~m}$. The average power delivered is
1 $6.25 \mathrm{~W}$
2 $25 \mathrm{~W}$
3 $62.5 \mathrm{~W}$
4 $50 \mathrm{~W}$
Explanation:
B Given, Force, $\mathrm{F}=5 \mathrm{~N}$, mass of stationary particle $=500 \mathrm{gm}$ Displacement $(\mathrm{s})=5 \mathrm{~m}$ $\therefore \quad \mathrm{F}=\mathrm{ma}$ $5=\frac{500}{1000} \mathrm{a}$ $\mathrm{a}=10 \mathrm{~m} / \mathrm{sec}^{2}$ We know that equation of second law of motion, $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$ $5=\frac{1}{2} \times 10 \mathrm{t}^{2} \quad[\because$ Initial velocity, $\mathrm{u}=0]$ $5=5 \mathrm{t}^{2}$ $\mathrm{t}=1 \mathrm{sec}$ Then, the average power delivered- $\mathrm{P} =\frac{\mathrm{W}}{\mathrm{t}}=\frac{\mathrm{F} . \mathrm{s}}{\mathrm{t}}$ $\mathrm{P} =\frac{5 \times 5}{1}$ $=25 \mathrm{Watt}$
148747
An engine develops $20 \mathrm{~kW}$ of power. How much time will it take to lift a mass of $200 \mathrm{~kg}$ to a height of $40 \mathrm{~m} ?\left(\mathrm{~g}=10 \mathrm{~m} . \mathrm{s}^{-2}\right)$
1 $4 \mathrm{~s}$
2 $5 \mathrm{~s}$
3 $8 \mathrm{~s}$
4 $10 \mathrm{~s}$
Explanation:
A Given, Mass of lift $(\mathrm{m})=200 \mathrm{~kg}$, Height $(\mathrm{h})=40 \mathrm{~m}$, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$, Power $(\mathrm{P})=20 \mathrm{~kW}=20 \times 10^{3} \mathrm{~W}$ $\text { Power }(\mathrm{P}) =\frac{(\mathrm{m} \cdot \mathrm{g} \cdot \mathrm{h})}{\mathrm{t}}$ $\mathrm{t} =\frac{\mathrm{m} \cdot \mathrm{g} \cdot \mathrm{h}}{\mathrm{P}}$ $=\frac{200 \times 10 \times 40}{20 \times 10^{3}}$ $\mathrm{t} =4 \mathrm{sec}$
AP EAMCET-20.08.2021
Work, Energy and Power
148748
A quarter horse power motor runs at a speed of $600 \mathrm{rpm}$. Assuming 60\% efficiency, the work done by the motor in one rotation is -
148749
A lawn roller has been pushed through a distance of $80 \mathrm{~m}$ by applying a force of $50 \mathrm{kgwt}$ in a direction inclined at $60^{\circ}$ with the ground. Calculate the work done on the roller.
148751
A constant force of $5 \mathrm{~N}$ accelerates a stationary particle of mass $500 \mathrm{gm}$ through a displacement of $5 \mathrm{~m}$. The average power delivered is
1 $6.25 \mathrm{~W}$
2 $25 \mathrm{~W}$
3 $62.5 \mathrm{~W}$
4 $50 \mathrm{~W}$
Explanation:
B Given, Force, $\mathrm{F}=5 \mathrm{~N}$, mass of stationary particle $=500 \mathrm{gm}$ Displacement $(\mathrm{s})=5 \mathrm{~m}$ $\therefore \quad \mathrm{F}=\mathrm{ma}$ $5=\frac{500}{1000} \mathrm{a}$ $\mathrm{a}=10 \mathrm{~m} / \mathrm{sec}^{2}$ We know that equation of second law of motion, $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$ $5=\frac{1}{2} \times 10 \mathrm{t}^{2} \quad[\because$ Initial velocity, $\mathrm{u}=0]$ $5=5 \mathrm{t}^{2}$ $\mathrm{t}=1 \mathrm{sec}$ Then, the average power delivered- $\mathrm{P} =\frac{\mathrm{W}}{\mathrm{t}}=\frac{\mathrm{F} . \mathrm{s}}{\mathrm{t}}$ $\mathrm{P} =\frac{5 \times 5}{1}$ $=25 \mathrm{Watt}$
148747
An engine develops $20 \mathrm{~kW}$ of power. How much time will it take to lift a mass of $200 \mathrm{~kg}$ to a height of $40 \mathrm{~m} ?\left(\mathrm{~g}=10 \mathrm{~m} . \mathrm{s}^{-2}\right)$
1 $4 \mathrm{~s}$
2 $5 \mathrm{~s}$
3 $8 \mathrm{~s}$
4 $10 \mathrm{~s}$
Explanation:
A Given, Mass of lift $(\mathrm{m})=200 \mathrm{~kg}$, Height $(\mathrm{h})=40 \mathrm{~m}$, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$, Power $(\mathrm{P})=20 \mathrm{~kW}=20 \times 10^{3} \mathrm{~W}$ $\text { Power }(\mathrm{P}) =\frac{(\mathrm{m} \cdot \mathrm{g} \cdot \mathrm{h})}{\mathrm{t}}$ $\mathrm{t} =\frac{\mathrm{m} \cdot \mathrm{g} \cdot \mathrm{h}}{\mathrm{P}}$ $=\frac{200 \times 10 \times 40}{20 \times 10^{3}}$ $\mathrm{t} =4 \mathrm{sec}$
AP EAMCET-20.08.2021
Work, Energy and Power
148748
A quarter horse power motor runs at a speed of $600 \mathrm{rpm}$. Assuming 60\% efficiency, the work done by the motor in one rotation is -
148749
A lawn roller has been pushed through a distance of $80 \mathrm{~m}$ by applying a force of $50 \mathrm{kgwt}$ in a direction inclined at $60^{\circ}$ with the ground. Calculate the work done on the roller.
148751
A constant force of $5 \mathrm{~N}$ accelerates a stationary particle of mass $500 \mathrm{gm}$ through a displacement of $5 \mathrm{~m}$. The average power delivered is
1 $6.25 \mathrm{~W}$
2 $25 \mathrm{~W}$
3 $62.5 \mathrm{~W}$
4 $50 \mathrm{~W}$
Explanation:
B Given, Force, $\mathrm{F}=5 \mathrm{~N}$, mass of stationary particle $=500 \mathrm{gm}$ Displacement $(\mathrm{s})=5 \mathrm{~m}$ $\therefore \quad \mathrm{F}=\mathrm{ma}$ $5=\frac{500}{1000} \mathrm{a}$ $\mathrm{a}=10 \mathrm{~m} / \mathrm{sec}^{2}$ We know that equation of second law of motion, $\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}$ $5=\frac{1}{2} \times 10 \mathrm{t}^{2} \quad[\because$ Initial velocity, $\mathrm{u}=0]$ $5=5 \mathrm{t}^{2}$ $\mathrm{t}=1 \mathrm{sec}$ Then, the average power delivered- $\mathrm{P} =\frac{\mathrm{W}}{\mathrm{t}}=\frac{\mathrm{F} . \mathrm{s}}{\mathrm{t}}$ $\mathrm{P} =\frac{5 \times 5}{1}$ $=25 \mathrm{Watt}$
