171

You visited us 171 times! Enjoying our articles? Unlock Full Access!

Instantaneous Force and Impulse

A bird of mas...

Question

A bird of mass 1.23kg is able to hover by imparting a downward velocity of 10 m/s uniformly to air of density 'p' kg/ $m^{3}$ over an effective area 0.1 $m^{2}$ . The acceleration due to gravity is 10m/ $s^{2}$ . Then the magnitude of 'p' in kg/ $m^{3}$

0.34

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.89

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.23

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4.8

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $1.23$
$m = 1.23 k g A = 0.1 m^{2} v = 10 m / s g = 10 m s^{- 2} e = ?$
Weight=Force by wings
$m g = e A u^{2}$
$e = \frac{m g}{A u^{2}} = \frac{1.23 \times 10}{0.1 \times 10 \times 10} = 1.23 k g / m^{3}$

Suggest Corrections

Similar questions

m / s^{2}

k g / m^{3}

k g / m^{3}

d > p

m^{3}

m^{3}

s^{2}

0.3 m m

1.8 \times 10^{- 5} N s / m^{2}

(

10^{3} k g / m^{3},

g = 9.8 m / s^{2}

)

1.8 x 10^{- 5} N - s / m^{2}

m^{3}

m^{3}

T a k e g = 10 m / s^{- 2} (F o r c e o f b u o y a n c y d u e t o a i r i s n e g l e c t e d)

0.04 mm

1.9 \times 10^{- 5} {Ns/m}^{2}

1.2 {kg/m}^{3}

1000 {kg/m}^{3}

g = 10 {m/s}^{2}

Join BYJU'S Learning Program