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 ρ kg/m3 over an effective area 0.1m2 If the acceleration due to gravity is 10.m/s2, then the magnitude of ρ is in kg/m3 :

Question

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 ρ kg/m3 over an effective area 0.1m2 If the acceleration due to gravity is 10.m/s2, then the magnitude of ρ is in kg/m3 : (A) 0.0123 (B) 0.123 (C) 1.23 (D) 1.32

Tardigrade Admin · Accepted Answer

Given: Mass of bird m = 1.25 kg Velocity v =10m/s Area, A=0.1 m2 The density is given by ρ =(Mass/Volume)=(M/V)=(M/Av) (∵ V=Av) =(12.3/0.1× 10)=1.23 kg/m3

Q. 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 $ρ k g / m^{3}$ over an effective area $0.1 m^{2}$ If the acceleration due to gravity is $10. m / s^{2},$ then the magnitude of $ρ$ is in $k g / m^{3}$ :

0.0123

0.123

1.23

1.32

Given: Mass of bird m = 1.25 kg Velocity v =10m/s Area, $A = 0.1 m^{2}$ The density is given by $ρ = \frac{M a ss}{V o l u m e} = \frac{M}{V} = \frac{M}{A v}$ $(∵ V = A v)$ $= \frac{12.3}{0.1 \times 10} = 1.23 k g / m^{3}$

Q. 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 ρkg/m3 over an effective area 0.1m2 If the acceleration due to gravity is 10.m/s2, then the magnitude of ρ is in kg/m3 :