Home
Class 12
PHYSICS
A lift weighing 480 kg is to be lifted...

A lift weighing 480 kg is to be
lifted up at a constant velocity of
0.40 m `s^(-1)` . The minimum horse
power of the motor to be used is
_____ hp.(Take g=10 m `s^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
2.57
F = mg = 480 xx 10=4800 N
Now, `P = overset rarr F.overset rarr V=F vcos theta`
In this case, `theta=0^(@)` ,
`therefore P=4800 xx 0.4=1920W`
Now, 1hp=746 W
`therefore P=(1920)/(746)=2.57 hp`
Promotional Banner

Topper's Solved these Questions

  • NTA TPC JEE MAIN TEST 30

    NTA MOCK TESTS|Exercise PHYSICS (SUBJECTIVE NUMERICAL)|10 Videos

  • NTA TPC JEE MAIN TEST 124

    NTA MOCK TESTS|Exercise PHYSICS |30 Videos

  • NTA TPC JEE MAIN TEST 36

    NTA MOCK TESTS|Exercise PHYSICS |30 Videos

Similar Questions

Explore conceptually related problems

A lift weighing 250 kg is to be lifted up at a constant velocity of 0.20 m. What would be the minimum horse power of the motor to be used ?

An elevator weighing 500 kg is to be lifted up at a constant velociyt of 0.20 m/s. What would be the minimum horsepower of the motor to be used?

An elevatore weighinig 500kg is to be lifted up at a constant velocity of 0*4m//s . What should be the minimum horse power of the motor to be used ?

A lift weighing 900 kg is to be lifted at a constant speed of 0.5 m/s. What shoud be the horse power of the motor used for running the lift? [Use g=10m//s^(2) and 1 Horse power =750 watt]

A crane lifts a mass of 100 kg to a height of 10 m in 20 s. The power of the crane is (Take g = 10 m s^(-2) )

A crane lifts a mass of 100 kg to a height of 10m in 20s . The power of the crane is (Take g = 10 ms^(-2))

Ten litre of water per second is lifted from well through 20m and delivered with a velocity of 10 m/s, then the power of the motor is :

When a man holding spring balance in stationary lift then it's reading is 60 kg . Now if lift starts descends with constant acceleration 1.8 m/ s^(2) then what is the new reading of spring balance in newton (take g =10 m/ s^(2) )

Mass of a person sitting in a lift is 50 kg . If lift is coming down with a constant acceleration of 10 m//sec^(2) . Then the reading of spring balance will be (g=10m//sec^(2))