CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Water from a tap emerges vertically down with an initial speed of $$1.0\ ms^{-1}$$. The cross sectional area of tap is $$10\ cm^{2}$$. Assume that the pressure is constant through out the stream of water,and that the flow is steady, the cross sectional area of the steam $$0.15\ m$$ below the tap is


A
5×104 m2
loader
B
5×105 m2
loader
C
5.0×105 m2
loader
D
50×104 m2
loader

Solution

The correct options are
A $$5\times 10^{-4}\ m^{2}$$
C $$5.0 \times 10^{-5}\ m^{2}$$
Given initial velocity $$u = 1  m s^{-1}$$
Area $$A = 10 c m^{2} = 10 \times 10^{-4} m^{2}$$
Velocity at height  0.15 m below  the  tap is: $$v^{2}  =  u^{2} + 2 gh$$
                                                                              $$= (1)^{2} + 2\times 10\times 0.1 s$$
                                                                              $$=  4$$
$$\Rightarrow  v = 2 m s^{-1}$$
We  know  that  by  continuity  Theoram  $$A_{1}v_{1} = A_{2}v_{2}$$
         $$1\times 10\times 10^{-4} = A^{2}\times  2$$
                 $$5\times 10^{-4}  = A^{2}$$

Physics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image