# 1)a water tank which is on ground has an arrangement to maintain a constant water levvel of depth 60cm.thru a hole on its vertical wall at a depth of 20cm from the free surface water comes out and reaches the ground at a certain distance.to have the same horizontal range anothe r hole can be made at a depth of? 2)a tank full of water has a small at its bottom.if one fourth of the tank is emptied in t1 sec and the remaining three-fourths of the tank is emptied in t2 sec.the ratio t1/t2= 3)a large tank filled with water to a height h is to be emptied thru a small hole at the bottom.the ratio of the time taken for the level to fall from h to h/2 and that taken for the level to fall from h/2 to 0 is

1
by aimanakhtar799
ohhh my head aches after reading it. you just fried my mind
i have done this. i will write answer after some time please
i want it asap
you should write one problem in one question.. it is not easy to write answers for so many questions.. it is not convenient. the question is not formatted too.
i will not answer if u write multiple problems in one question. it takes too much time.

2014-11-20T05:12:47+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Bernoulli's energy conservation principle for the fluids tells the relation between pressures and kinetic energy at two locations in a fluid.

Total Energy at the surface of water = Total energy at the hole at 20 cm depth
P₁ + 1/2 ρ v₁² + ρ g h₁    =    P₂ + 1/2 ρ v₂² + ρ g h₂

P₁ = P₂ = atmospheric pressure,        g = 10 m/sec²
v₁ = 0 as the water at the surface is maintained stationary.
h₁ - h₂ = h = 20 cm = 0.20 m

So v₂ = √ (2gh)  ---  equation 1

Hence,    v₂² = 2 * 0.20 g  = 4    =>  v₂ = 2 m/sec

We find the range for the water particles with this speed horizontally directed and at a height of 0.40 meters above ground.

t = time to reach ground = √ (2 * 0.40 / g ) = 0.2 *√2  sec
Range = v₂ t = 0.4 √2 meters

If h is the depth of hole from water surface, then v = √(2gh)
time to reach ground = t = √ [2*(0.60-h)/g ]
Then range R = v₂ t = 2 √[ h (60-h) ] =  0.4√2    --- equation 2

Range will be equal for  h = 20cm, and  h = 40 cm, as the product of h and (60-h) will be same.   If you solve quadratic equation 2, we get the same.
==============================================
2 )
velocity v₂ of water coming out of hole at the bottom of tank = √(2 g h)  from equation 1 above.  Here we neglect v₁ at the water surface as it is very small as compared to v₂. Let A be the area of cross section of the tank.

Volume of water in tank = V = A h
water flow rate out of tank = decrease rate in volume inside tank
A v₂  =  d V/dt = - A dh/dt
=>     v₂ = √(2 g h )  =  - dh/dt

=>       - dh/√h = √(2g) dt        ------- equation 3
Integrating the expressions on both sides ,

Time for water level to go from h₀ to 3 h₀/4,  is

Time for water level to go from 3 h₀ /4 to 0  is

Ratio t1/t2 =

=================================================

3)   Using equation 4,  above   we get the values for t1 and t2.