Question

A bird of mass 1.23 kg is able to hover by imparting a downward velocity of $10 ms^{-1}$ uniformly to air of density $\rho kg-m^{-3}$ over an effective area $0.1 m^2$. If the acceleration due to gravity is $10 ms^{-2}$ then the magnitude of $\rho$ is in $kg-m^{-3}$

• A. 0.0123
• B. 0.123
• C. 1.23
• D. 1.32

Answer 1

Answer: (C)

1.23

Answer 2

The density is given by

$\rho=\frac{Mass}{Volume}=\frac{M}{V}=\frac{M}{Av}$

$=\frac{100}{0.1\times10}=1.23 kg-m^{-3}$

Therefore, the correct option is (C): 1.23