Consider a cylindrical tank of radius 1m is filled with water. The top surface of water is at 15m front the bottom of the cylinder. There is a hole on the wall of a cylinder at a height of 5m from the bottom. A force of 5 × 105N is applied an the top surface of water using a piston. The speed of ifflux from the hole will be:

(given atmospheric pressure PA = 1.01 × 105Pa, density of water ρ w = 1 0 0 0 k g / m 3  and gravitational acceleration g = 10m/s2)

Option 1 -

11.6 m/s

Option 2 -

10.8 m/s

Option 3 -

17.8 m/s

Option 4 -

14.4 m/s

0 10 Views | Posted 4 months ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago
    Correct Option - 3


    Detailed Solution:

    ρω=1000kg/m3

    G = 10 m/s2

    Using bernaulli’s eqn

    5×105π+ρg×10=ρ0+12ρVe2

    =318? 17.8m/s

Similar Questions for you

A
alok kumar singh

NLM 1 

ρω43πr3g=6πηrVT

ρω43πr2g= 6πηvT

VT=43ρωπr2g6πη

=29×103× (106)2×101.8×105

=0.1234×103

vT = 123.4 × 10-6m/sec

 

V
Vishal Baghel

Kindly consider the following answer

 Re=ρvdη

P
Payal Gupta

After switch ‘S’ is closed

Q1+Q2=C1V - (1)

Using KVL

Q1C1Q2C2=0

Q1=Q2C1C2 - (2)

from (1) & (2)

Q2 [C1+C2C2]=C1VQ2= (C1C2C1+C2)V

V
Vishal Baghel

 43πR3=72943πr3

R3 = 729r3

R= (729)13 (r) (13)

R = 9r

Δu=T (4πr2)×729T×4πR2

=T×4π×8R2

=75.39×105JΔu=7.5×104J

V
Vishal Baghel

A = 10 cm2, V = 20 m/s

F=dpdt=ρAV2=103×10×104×400

= 400 N.

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