A bird of mass 1.23 kg is able to hover by imparting t downward velocity of 10 ms-1 uniformly to air o density ρ kg-m-3 over an effective area 0.1 m2. If the acceleration due to gravity is 10 ms-2 then the magnitude of ρ is in kg-m-3

Question

A bird of mass 1.23 kg is able to hover by imparting t downward velocity of 10 ms-1 uniformly to air o density ρ kg-m-3 over an effective area 0.1 m2. If the acceleration due to gravity is 10 ms-2 then the magnitude of ρ is in kg-m-3 (A) 0.0123 (B) 0.123 (C) 1.23 (D) 1.32

Tardigrade Admin · Accepted Answer

The density is given by ρ=(Mass/Volume)=(M/V)=(M/Av) =(100/0.1×10)=1.23 kg-m-3

Q. A bird of mass 1.23 kg is able to hover by imparting t downward velocity of $10 m s^{- 1}$ uniformly to air o 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 $ρ i s in k g - m^{- 3}$

0.0123

0.123

1.23

1.32

The density is given by
$ρ = \frac{M a ss}{V o l u m e} = \frac{M}{V} = \frac{M}{A v}$
$= \frac{100}{0.1 \times 10} = 1.23 k g - m^{- 3}$

Q. A bird of mass 1.23 kg is able to hover by imparting t downward velocity of 10ms−1 uniformly to air o density ρkg−m−3 over an effective area 0.1m2. If the acceleration due to gravity is 10ms−2 then the magnitude of ρisinkg−m−3