NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI06:WORK ENERGY AND POWER
355476
Assertion : The power of a pump which raises \(100\;kg\) of water in \(10\,{\rm{sec}}\) to a height of \(100\;m\) is \(10\;kW\) Reason : A practical unit of power in British system is horse power (HP).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The power of a pump can be calculated using the formula \(\therefore\) Power \(=\dfrac{\text { work }}{\text { time }}=\dfrac{m g h}{t}=\dfrac{100 \times 10 \times 100}{10}\) \(P=t m g h\), where ' \(m\) ' is the mass, ' \(g\) 'is the acceleration due to gravity, ' \(h\) ' is the height, and ' \(t\) ' is the time. Using this formula, the power is indeed \(10\;kW\) in this case. It is a separate fact that British system of units uses 'Horse Power(HP)' as practical unit of power. So correct option is (2)
PHXI06:WORK ENERGY AND POWER
355477
For power to be constant, the force has to vary with speed as:
1 \(F \propto v^{2}\)
2 \(F \propto \dfrac{1}{v}\)
3 \(F \propto \dfrac{1}{\sqrt{v}}\)
4 \(F \propto v\)
Explanation:
Power: \(P = Fv\) For power to be constant: \(F \propto \dfrac{1}{v}\)
PHXI06:WORK ENERGY AND POWER
355478
A truck of mass 30000 \(kg\) moves up an inclined plane of slope 1 in 100 at speed of 30 \(kmph\). The power of the truck is (Given \(g = 10\;m{s^{ - 2}}\) )
1 10 \(kW\)
2 5 \(kW\)
3 2.5 \(kW\)
4 25 \(kW\)
Explanation:
A truck is moving on an inclined plane therefore only component of weight \((m g \sin \theta)\) will oppose the upward motion Power \(=\) force \(\times\) velocity \(=m g \sin \theta \times v\) \( = 30000 \times 10 \times \left( {\frac{1}{{100}}} \right) \times \frac{{30 \times 5}}{{18}} = 25\;kW.\)
PHXI06:WORK ENERGY AND POWER
355479
A 60 \(kg\) man runs up a staircase in 12 seconds while a 50 \(kg\) man runs up the same staircase in 11 seconds. The ratio of the rate of doing their work is
1 \(11: 10\)
2 \(6: 5\)
3 \(10: 11\)
4 \(12: 11\)
Explanation:
\(P=\dfrac{m g h}{t} \Rightarrow \dfrac{P_{1}}{P_{2}}=\dfrac{m_{1}}{m_{2}} \times \dfrac{t_{2}}{t_{1}}\) [ as \(h=\) constant \(]\) \(\therefore \dfrac{P_{1}}{P_{2}}=\dfrac{60}{50} \times \dfrac{11}{12}=\dfrac{11}{10} \text {. }\)
355476
Assertion : The power of a pump which raises \(100\;kg\) of water in \(10\,{\rm{sec}}\) to a height of \(100\;m\) is \(10\;kW\) Reason : A practical unit of power in British system is horse power (HP).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The power of a pump can be calculated using the formula \(\therefore\) Power \(=\dfrac{\text { work }}{\text { time }}=\dfrac{m g h}{t}=\dfrac{100 \times 10 \times 100}{10}\) \(P=t m g h\), where ' \(m\) ' is the mass, ' \(g\) 'is the acceleration due to gravity, ' \(h\) ' is the height, and ' \(t\) ' is the time. Using this formula, the power is indeed \(10\;kW\) in this case. It is a separate fact that British system of units uses 'Horse Power(HP)' as practical unit of power. So correct option is (2)
PHXI06:WORK ENERGY AND POWER
355477
For power to be constant, the force has to vary with speed as:
1 \(F \propto v^{2}\)
2 \(F \propto \dfrac{1}{v}\)
3 \(F \propto \dfrac{1}{\sqrt{v}}\)
4 \(F \propto v\)
Explanation:
Power: \(P = Fv\) For power to be constant: \(F \propto \dfrac{1}{v}\)
PHXI06:WORK ENERGY AND POWER
355478
A truck of mass 30000 \(kg\) moves up an inclined plane of slope 1 in 100 at speed of 30 \(kmph\). The power of the truck is (Given \(g = 10\;m{s^{ - 2}}\) )
1 10 \(kW\)
2 5 \(kW\)
3 2.5 \(kW\)
4 25 \(kW\)
Explanation:
A truck is moving on an inclined plane therefore only component of weight \((m g \sin \theta)\) will oppose the upward motion Power \(=\) force \(\times\) velocity \(=m g \sin \theta \times v\) \( = 30000 \times 10 \times \left( {\frac{1}{{100}}} \right) \times \frac{{30 \times 5}}{{18}} = 25\;kW.\)
PHXI06:WORK ENERGY AND POWER
355479
A 60 \(kg\) man runs up a staircase in 12 seconds while a 50 \(kg\) man runs up the same staircase in 11 seconds. The ratio of the rate of doing their work is
1 \(11: 10\)
2 \(6: 5\)
3 \(10: 11\)
4 \(12: 11\)
Explanation:
\(P=\dfrac{m g h}{t} \Rightarrow \dfrac{P_{1}}{P_{2}}=\dfrac{m_{1}}{m_{2}} \times \dfrac{t_{2}}{t_{1}}\) [ as \(h=\) constant \(]\) \(\therefore \dfrac{P_{1}}{P_{2}}=\dfrac{60}{50} \times \dfrac{11}{12}=\dfrac{11}{10} \text {. }\)
355476
Assertion : The power of a pump which raises \(100\;kg\) of water in \(10\,{\rm{sec}}\) to a height of \(100\;m\) is \(10\;kW\) Reason : A practical unit of power in British system is horse power (HP).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The power of a pump can be calculated using the formula \(\therefore\) Power \(=\dfrac{\text { work }}{\text { time }}=\dfrac{m g h}{t}=\dfrac{100 \times 10 \times 100}{10}\) \(P=t m g h\), where ' \(m\) ' is the mass, ' \(g\) 'is the acceleration due to gravity, ' \(h\) ' is the height, and ' \(t\) ' is the time. Using this formula, the power is indeed \(10\;kW\) in this case. It is a separate fact that British system of units uses 'Horse Power(HP)' as practical unit of power. So correct option is (2)
PHXI06:WORK ENERGY AND POWER
355477
For power to be constant, the force has to vary with speed as:
1 \(F \propto v^{2}\)
2 \(F \propto \dfrac{1}{v}\)
3 \(F \propto \dfrac{1}{\sqrt{v}}\)
4 \(F \propto v\)
Explanation:
Power: \(P = Fv\) For power to be constant: \(F \propto \dfrac{1}{v}\)
PHXI06:WORK ENERGY AND POWER
355478
A truck of mass 30000 \(kg\) moves up an inclined plane of slope 1 in 100 at speed of 30 \(kmph\). The power of the truck is (Given \(g = 10\;m{s^{ - 2}}\) )
1 10 \(kW\)
2 5 \(kW\)
3 2.5 \(kW\)
4 25 \(kW\)
Explanation:
A truck is moving on an inclined plane therefore only component of weight \((m g \sin \theta)\) will oppose the upward motion Power \(=\) force \(\times\) velocity \(=m g \sin \theta \times v\) \( = 30000 \times 10 \times \left( {\frac{1}{{100}}} \right) \times \frac{{30 \times 5}}{{18}} = 25\;kW.\)
PHXI06:WORK ENERGY AND POWER
355479
A 60 \(kg\) man runs up a staircase in 12 seconds while a 50 \(kg\) man runs up the same staircase in 11 seconds. The ratio of the rate of doing their work is
1 \(11: 10\)
2 \(6: 5\)
3 \(10: 11\)
4 \(12: 11\)
Explanation:
\(P=\dfrac{m g h}{t} \Rightarrow \dfrac{P_{1}}{P_{2}}=\dfrac{m_{1}}{m_{2}} \times \dfrac{t_{2}}{t_{1}}\) [ as \(h=\) constant \(]\) \(\therefore \dfrac{P_{1}}{P_{2}}=\dfrac{60}{50} \times \dfrac{11}{12}=\dfrac{11}{10} \text {. }\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI06:WORK ENERGY AND POWER
355476
Assertion : The power of a pump which raises \(100\;kg\) of water in \(10\,{\rm{sec}}\) to a height of \(100\;m\) is \(10\;kW\) Reason : A practical unit of power in British system is horse power (HP).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
The power of a pump can be calculated using the formula \(\therefore\) Power \(=\dfrac{\text { work }}{\text { time }}=\dfrac{m g h}{t}=\dfrac{100 \times 10 \times 100}{10}\) \(P=t m g h\), where ' \(m\) ' is the mass, ' \(g\) 'is the acceleration due to gravity, ' \(h\) ' is the height, and ' \(t\) ' is the time. Using this formula, the power is indeed \(10\;kW\) in this case. It is a separate fact that British system of units uses 'Horse Power(HP)' as practical unit of power. So correct option is (2)
PHXI06:WORK ENERGY AND POWER
355477
For power to be constant, the force has to vary with speed as:
1 \(F \propto v^{2}\)
2 \(F \propto \dfrac{1}{v}\)
3 \(F \propto \dfrac{1}{\sqrt{v}}\)
4 \(F \propto v\)
Explanation:
Power: \(P = Fv\) For power to be constant: \(F \propto \dfrac{1}{v}\)
PHXI06:WORK ENERGY AND POWER
355478
A truck of mass 30000 \(kg\) moves up an inclined plane of slope 1 in 100 at speed of 30 \(kmph\). The power of the truck is (Given \(g = 10\;m{s^{ - 2}}\) )
1 10 \(kW\)
2 5 \(kW\)
3 2.5 \(kW\)
4 25 \(kW\)
Explanation:
A truck is moving on an inclined plane therefore only component of weight \((m g \sin \theta)\) will oppose the upward motion Power \(=\) force \(\times\) velocity \(=m g \sin \theta \times v\) \( = 30000 \times 10 \times \left( {\frac{1}{{100}}} \right) \times \frac{{30 \times 5}}{{18}} = 25\;kW.\)
PHXI06:WORK ENERGY AND POWER
355479
A 60 \(kg\) man runs up a staircase in 12 seconds while a 50 \(kg\) man runs up the same staircase in 11 seconds. The ratio of the rate of doing their work is
1 \(11: 10\)
2 \(6: 5\)
3 \(10: 11\)
4 \(12: 11\)
Explanation:
\(P=\dfrac{m g h}{t} \Rightarrow \dfrac{P_{1}}{P_{2}}=\dfrac{m_{1}}{m_{2}} \times \dfrac{t_{2}}{t_{1}}\) [ as \(h=\) constant \(]\) \(\therefore \dfrac{P_{1}}{P_{2}}=\dfrac{60}{50} \times \dfrac{11}{12}=\dfrac{11}{10} \text {. }\)