What is the average power required to lift a mass of 100kg to a height of 50m in 50 seconds?

• Correct Answer: D

  Solution :

              \[P=\frac{mgh}{t}\]=\[\frac{100\times 9.8\times 50}{50}=980J/s\]

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Page 2

• Correct Answer: A

  Solution :

              \[P=\left( \frac{m}{t} \right)gh=100\times 10\times 100={{10}^{5}}W=100\ kW\]

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Page 3

• Correct Answer: A

  Solution :

              \[p=\frac{mgh}{t}=\frac{200\times 10\times 200}{10}=40\,kW\]

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Page 4

• Correct Answer: C

  Solution :

              Volume of water to raise = 22380 l = 22380×10?3m3             \[P=\frac{mgh}{t}=\frac{V\rho gh}{t}\] Þ \[t=\frac{V\rho gh}{P}\]             t\[=\frac{22380\times {{10}^{-3}}\times {{10}^{3}}\times 10\times 10}{10\times 746}=15\]min

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Page 5

• Correct Answer: C

  Solution :

              Force produced by the engine\[F=\frac{P}{v}\,=\,\frac{30\times {{10}^{3}}}{30}\]=103N Acceleration=\[\frac{\text{Forward force by engine--resistive force}}{\text{mass of car}}\] \[=\frac{1000-750}{1250}=\frac{250}{1250}\]=\[\frac{1}{5}m/{{s}^{2}}\]

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Page 6

• Correct Answer: B

  Solution :

              Power\[=\frac{\text{Work done}}{\text{time}}=\frac{\frac{1}{2}m({{v}^{2}}-{{u}^{2}})}{t}\]             P\[=\frac{1}{2}\times \frac{2.05\times {{10}^{6}}\times [{{(25)}^{2}}-({{5}^{2}})]}{5\times 60}\]             \[P=2.05\times {{10}^{6}}\,W=2.05\,MW\]

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Page 7

• Correct Answer: A

  Solution :

              As truck is moving on an incline plane therefore only component of weight \[(mg\sin \theta )\] will oppose the upward motion Power = force × velocity = \[mg\sin \theta \times v\] \[=30000\times 10\times \left( \frac{1}{100} \right)\times \frac{30\times 5}{18}=25\,kW\]

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Page 8

• Correct Answer: C

  Solution :

              \[P=\frac{mgh}{t}\] Þ \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{m}_{1}}}{{{m}_{2}}}\times \frac{{{t}_{2}}}{{{t}_{1}}}\]                     (As h = constant)             \\[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{60}{50}\times \frac{11}{12}=\frac{11}{10}\]

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Page 9

• Correct Answer: C

  Solution :

              Power of a pump\[=\frac{1}{2}\rho A{{v}^{3}}\] To get twice amount of water from same pipe v has to be made twice. So power is to be made 8 times.

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Page 10

• Correct Answer: B

  Solution :

              Power\[=\frac{\text{Work done}}{\text{time}}=\frac{\text{Increase in K}\text{.E}\text{.}}{\text{time}}\] \[P=\frac{\frac{1}{2}m{{v}^{2}}}{t}=\frac{\frac{1}{2}\times {{10}^{3}}\times {{(15)}^{2}}}{5}=22500\,W\]

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Page 11

• Correct Answer: A

  Solution :

              Motor makes 600 revolution per minute             \ n = \[600\frac{\text{revolution}}{\text{minute}}=10\frac{rev}{\sec }\]             \ Time required for one revolution \[=\frac{1}{10}\]sec             Energy required for one revolution = power × time                                     =\[\frac{1}{4}\times 746\times \frac{1}{10}=\frac{746}{40}J\]             But work done = 40% of input                                     \[=40%\times \frac{746}{40}=\frac{40}{100}\times \frac{746}{40}=7.46\,J\]

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Page 12

• Correct Answer: A

  Solution :

              Work output of engine = mgh = \[100\times 10\times 10={{10}^{4}}J\] Efficiency (h) = \[\frac{\text{output}}{\text{input}}\] \ Input energy = \[\frac{\text{outupt}}{\eta }\] \[=\frac{{{10}^{4}}}{60}\times 100=\frac{{{10}^{5}}}{6}J\] \ Power = \[\frac{\text{input energy}}{\text{time}}\]= \[\frac{{{10}^{5}}/6}{5}=\frac{{{10}^{5}}}{30}=3.3\ kW\]

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Page 13

• Correct Answer: A

  Solution :

  \[P=\frac{\overrightarrow{F}.\overrightarrow{s}}{t}=\frac{(2\hat{i}+3\hat{j}+4\hat{k}).(3\hat{i}+4\hat{j}+5\hat{k})}{4}\]\[=\frac{38}{4}=9.5\ W\]

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Page 14

• Correct Answer: A

  Solution :

                          \[P=\frac{W}{t}=\frac{mgh}{t}=\frac{200\times 10\times 50}{10}\]\[=10\times {{10}^{3}}\,W\]            

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Page 15

• Correct Answer: A

  Solution :

                          Power of gun \[=\frac{\text{Total K}\text{.E}\text{. of fired bullet}}{\text{time}}\]             \[=\frac{n\times \frac{1}{2}m{{v}^{2}}}{t}=\frac{360}{60}\times \frac{1}{2}\times 2\times {{10}^{-2}}\times {{(100)}^{2}}=600\,W\]            

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Page 16

• Correct Answer: A

  Solution :

              Energy supplied to liquid per second by the pump = \[\frac{1}{2}\frac{m{{v}^{2}}}{t}\]= \[\frac{1}{2}\frac{V\rho {{v}^{2}}}{t}\]= \[\frac{1}{2}A\times \left( \frac{l}{t} \right)\times \rho \times {{v}^{2}}\] \[\left[ \frac{l}{t}=v \right]\] \[=\frac{1}{2}A\times v\times \rho \times {{v}^{2}}\]= \[\frac{1}{2}A\rho {{v}^{3}}\]

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Page 17

• Correct Answer: A

  Solution :

                          \[\text{Power}=\frac{\text{workdone}}{\text{time}}=\frac{\text{pressure }\times \text{ change in volume}}{\text{time}}\]             = \[\frac{20000\times 1\times {{10}^{-6}}}{1}\]\[=2\times {{10}^{-2}}=0.02\ W\]          

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Page 18

• Correct Answer: C

  Solution :

                          \[\text{Power}=\frac{W}{t}.\] If W is constant then \[P\propto \frac{1}{t}\]             i.e. \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{t}_{2}}}{{{t}_{1}}}=\frac{20}{10}=\frac{2}{1}\]            

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