Home
Class 11
PHYSICS
A bird of mass 1.23 kg is able to hover ...

A bird of mass 1.23 kg is able to hover by imparting a downward velocity of `10 m//s` uniformly to air of density `rho kg//m^(3)` over an effective area `0.1m^(2)` the acceleration due to gravity is `10m//s^(2)` then the magnitude of `rho` in `kg//m^(3)`

A

0.34

B

0.89

C

1.23

D

4.8

Text Solution

Verified by Experts

The correct Answer is:
C
`"Weight of bird"="force exerted by the bird by moving its wings. "rho Au^(2)=mg" "(F=Au^(2)rho)`
Promotional Banner

Topper's Solved these Questions

  • MECHANICAL PROPERTIES OF FLUIDS

    NARAYNA|Exercise EXERCISE - II (C.W) (VARIATION OF PRESSURE, UPTHRUST)|8 Videos

  • MECHANICAL PROPERTIES OF FLUIDS

    NARAYNA|Exercise EXERCISE - II (C.W) (EQUATION OF CONTINUITY, BERNOULLI.S THEOREM AND ITS APPLICATIONS)|1 Videos

  • MECHANICAL PROPERTIES OF FLUIDS

    NARAYNA|Exercise EXERCISE - I - (H.W) (COMBINATION OF DROPS & BUBBLES)|4 Videos

  • MATHEMATICAL REVIEW & PHYSICAL WORLD

    NARAYNA|Exercise C.U.Q|13 Videos

  • MECHANICAL PROPERTIES OF SOLIDS

    NARAYNA|Exercise LEVEL-II (H.W)|24 Videos

Similar Questions

Explore conceptually related problems

In the figure ( g = 10 m//s^(2) ) .Acceleration of 2kg block is :

If acceleration due to gravity is 10 m//s^(2) , what will be the potential energy of a body of mass 1 kg kept at a height of 5 m?

A body of mass 1.0 kg is falling with an acceleration of 10 m//s^(2) . Its apparent weight will be (g=10m//sec^(2))

In the system shown in the adjoining figure, the acceleration of the 1 kg mass is m//s^(2) . (take g = 10 m//s^(2) )

NARAYNA-MECHANICAL PROPERTIES OF FLUIDS-EXERCISE - II (C.W) (PRESSURE AND PASCAL.S LAW)
  1. A bird of mass 1.23 kg is able to hover by imparting a downward veloci...

    Text Solution

    |