datavorous/entrance-exam-dataset huggingface dataset:
Waste water treatment generates a large quantity of sludge, which can be treated by

Options:
A
anaerobic digesters

B
activated sludge

C
chemicals

D
oxidation pond

The correct answer is:
anaerobic digesters

Solution:
$\ldots$, Sludge is further treated by anaerobic digesters. These are large heated tanks in which the microorganism catalyses the chemical decomposition of sludge.
After treating it by bacteria that breakdown the organic matter remaining in solution. It is then sent to oxidation ponds where heterotrophic bacteria continue the breakdown of the organics and solar UV light destroys the harmful bacteria. Chemicals are not used in these treatments.

Topic: Biology, Microbes in Human Welfare
Water are biodegradable and non-biodegradable pollutants?

Options:

None

Solution:
Biodegradable pollutants are those which can be decomposed by bacteria or any other environmental factors like fruits, sewages etc. Non-biodegradable pollutants are not easily decomposed by bacteria. It remains as such in the environment. Examples like DDT, mercury etc.

Topic: Chemistry, Environmental Chemistry
Water boils in an electric kettle in 20 minutes after being switched on. Using the same main supply, the length of the heating element should be _____ to _____ times of its initial length if the water is to be boiled in 15 minutes.

Options:
A decreased, $3 / 4$
B increased, $4 / 3$
C decreased, 4/3
D increased, $3 / 4$

The correct answer is:
decreased, $3 / 4$

Solution:
$\begin{aligned} & \mathrm{P}=\frac{\mathrm{v}^2}{\mathrm{R}}, \mathrm{R}=\frac{\rho \ell}{\mathrm{A}} \\ & \mathrm{P} \propto \frac{1}{\ell} \\ & \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{t}_2}{\mathrm{t}_1}=\frac{15}{20}=\frac{\ell_2}{\ell_1} \\ & \ell_2=\frac{3}{4} \ell_1\end{aligned}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (08 Apr Shift 2)
Water coming with velocity v drops onto a mesh through which 50% of the water passes by, 25% loses its entire momentum and 25% bounces back. The pressure on the mesh is

Options:
A 12ρv2
B 32ρv2
C 3ρv2
D 34ρv2

The correct answer is:
34ρv2

Solution:
Mass per unit time =ρAv
Force due to momentum loss =14ρAv×v
Force due to bounce back =14ρAv×2v
Pressure =ρAv24+ρAv22A=34ρv2

Topic: Physics, Laws of Motion, JEE Main
Water decomposes at 2300 K

H2Og→H2g+12O2g

The percent of water decomposing at 2300 K and 1 bar is _____ (Nearest integer). Equilibrium constant for the reaction is 2×10-3 at 2300 K

Options:

The correct answer is:
2

Solution:
Decomposition of water gives:
H2Og ⇌ H2g + 12O2g

P01-α P0α P0α2 partial pr. at eq.

Thus, P01+α2=1 …1

Kp=PH2PO21/2PH2O

P0αP0α21/2P01-α=2×10-3

since α is negligible w.r.t 1 thus, 1-α≈1

⇒αα2=2×10-3

⇒α3/2=23/2×10-3

⇒α=23/2×2/3×10-3×2/3

⇒α=2×10-2, %α=2%

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Water does not produce CO on reacting with :

Options:
A CO2
B CH4
C C3H8
D C

The correct answer is:
CO2

Solution:
Carbon monoxide does not react with water to form an acid.

The carbon monoxide molecule is one that won't react with water at all. As the acidic oxide has a higher oxidation state, it reacts with water to produce acid, and reacts with base to produce salt.

CO2+H2O→H2CO3

C+H2Osteam→CO+H2

CH4+H2O→CO+3H2C3H8+H2O→3CO+H2 both reactions are carried out at 1270 K temp. with Ni catalyst

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Water droplets are coming from an open tap at a particular rate. The spacing between a droplet observed at 4th second after its fall to the next droplet is 34.3 m. At what rate the droplets are coming from the tap ? (Take g=9.8 m s-2)

Options:
A 3 drops/2 seconds
B 2 drops/second
C 1 drop/second
D 1 drop/7 seconds

The correct answer is:
1 drop/second

Solution:
In 4th sec, 1st drop will travel

⇒12×9.8×42=78.4 m

∴ 2nd drop would have travelled

⇒78.4-34.3=44.1 m

Time for 2nd drop

129.8t2=44.1

t=3 sec

∴ Each drop have time gap of 1 sec.

∴ 1 drop per second.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Water drops are falling from a nozzle of a shower onto the floor from a height of 9.8 m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.

Options:
A 2.94 m
B 4.18 m
C 2.45 m
D 7.35 m

The correct answer is:
7.35 m

Solution:
For first drop.

h=12g(2n)2

For 2nd drop

h'=12g(n)2

hh'=41

h'=h4=9.84

so height of 2nd drop

H=h-h'=9.8-9.84=34×9.8=7.35 m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (27 Aug Shift 2)
Water emerges from a tap vertically downwards with an initial speed of 1.0 m/s. The cross-sectional area of the tap is 10â4 m2. Assume that the pressure is constant throughout and that the flow is steady, the cross-sectional area of the stream 0.15 m below the tap is

Options:
A 5.0Ã10-4 m2
B 1.0Ã10-4Â m2
C 5.0Ã10-5 m2
D 2.0Ã10-5Â m2

The correct answer is:
5.0Ã10-5 m2

Solution:
From conservation of energy
v22=v12+2gh... (1)
[can also be found by applying Bernouilli's theorem between 1 and 2]
From continuity equation
A1 v1=A2 v2
v2=A1A2v1 (2)
Substituting value of v2 from Eq. (2) in Eq. (1)
A12A22.v12=v12+2gh
or A22=A12 v12v12+2gh
â A 2 = A 1 v 1 v 1 2 +2gh
Substituting the given values
A2=10-4m21.0 m/s1.0 m/s 2+210 m/s 2 0.15 m
A2=5.0Ã10-5 m2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Water enters through end $A$ with speed $v_1$ and leaves through end $B$ with speed $V_2$ of a cylindrical tube $A B$. The tube is always completely filled with water. In case I tube is horizontal and in case II it is vertical with end A upwards and in case III it is vertical with end $B$ upwards. We have $v_1=v_2$ for

Options:
A Case I
B Case II
C Case III
D Each case

The correct answer is:
Each case

Solution:
This happens in accordance with equation of continuity and this equation was derived on the principle of conservation of mass and it is true in every case, either tube remain horizontal or vertical.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Water enters through end A with speed v1 and leaves through end B with speed v2 of a cylindrical tube AB. The tube is always completely filled with water. In case I, the tube is horizontal and in case II, it is vertical with end A upwards and in case III, it is vertical with end B upwards. We have v1=v2 for

Options:
A case I
B case II
C case III
D each case

The correct answer is:
each case

Solution:
This happens in accordance with equation of continuity and this equation was derived on the principle of conservation of mass and it is true in every case, either tube remain horizontal or vertical.

Topic: Physics, Mechanical Properties of Fluids, NEET
Water enters through end A with speed v 1 and leaves through end B with speed v 2 of a cylindrical tube AB. The tube is always completely filled with water. In case I tube is horizontal and in case II it is vertical with end A upwards and in case III it is vertical with end B upwards. We have v 1 = v 2 for

Options:
A case I
B case II
C case III
D each case

The correct answer is:
each case

Solution:
This happens in accordance with the equation of continuity and this equation was derived on the principle of conservation of mass and it is true in every case, whether the tube remains horizontal or vertical.

Topic: Physics, Mechanical Properties of Fluids, NEET
Water falls down a $500.0 \mathrm{~m}$ shaft to reach a turbine which generates electricity. How much water must fall per second in order to generate $1.00 \times 10^{9}$ Watts of power? (Assume $50 \%$ efficiency of conversion and $\left.\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$

Options:
A $250 \mathrm{~m}^{3}$
B $400 \mathrm{~m}^{3}$
C $500 \mathrm{~m}^{3}$
D $200 \mathrm{~m}^{3}$

The correct answer is:
$400 \mathrm{~m}^{3}$

Solution:
$500 \mathrm{~m}\left[\begin{array}{c}\mathrm{P}=\frac{\mathrm{mah}}{\operatorname{time}} \\ \eta=\frac{\mathrm{P}_{\text {cut }}}{\mathrm{P}_{\mathrm{input}}} \\ \mathrm{P}_{\mathrm{in}}=\frac{\mathrm{P}_{\mathrm{cut}}}{\eta}\end{array}\right.$
$=\frac{10^{9}}{0.5}$
$P_{\text {in }}=2 \times 10^{9}$
$\frac{\text { mah }}{\text { time }}=2 \times 10^{9}$
$\mathrm{~m} / \mathrm{t}=\frac{2 \times 10^{9}}{10 \times 500}=\frac{2}{5} \times 10^{6}$
$=4 \times 10^{5}$
$=400 \mathrm{~m}^{3}$

Topic: Physics, Work Power Energy, JEE Main
Water falls from a 40 m high dam at the rate of 9×104 kg per hour. Fifty percentage of gravitational potential energy can be converted into electrical energy. Using this hydro electric energy number of 100 W lamps, that can be lit, is
(Take g=10 ms-2)

Options:
A 25
B 50
C 100
D 18

The correct answer is:
50

Solution:
Since 50% efficiency is given, half of the energy of the falling water will be used by the bulbs. Therefore,

12∆mghΔt=n×P⇒12×9×1043600×10×40=100n⇒n=50

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (28 Jun Shift 2)
Water falls from a height 500 m. The rise in temperature of water at the bottom if the whole of energy remains in the water, will be (specific heat of water is C=4.2 KJ Kg-1)

Options:
A 0.23 °C
B 1.16 °C
C 0.96 °C
D 1.02 °C

The correct answer is:
1.16 °C

Solution:
When water falls from a height, it has potential energy (mgh),
this is used in heating up the water (mc∆θ).
Hence, we have
mgh=mc∆θ
⇒ ∆θ=ghc
=9.8×5004.2×103=1.16 °C

Topic: Physics, Thermal Properties of Matter, NEET
Water falls from a height of 60 m at the rate of 15 kg s-1 to operate a turbine. The losses due to frictional force are 10% of the input energy. How much power is generated by the turbine? g=10 m s-2)

Options:
A 10.2 kW
B 12.3 kW
C 8.1 kW
D 7.0 kW

The correct answer is:
8.1 kW

Solution:
Change in potential energy of water per second =15×g×60

=9000 J s-1

Energy remaining per second after loss =90% of 9000 J s-1

=90100×9000

=8100 J s-1

=8.1 kW

Topic: Physics, Work Power Energy, NEET, NEET 2021
Water falls from a height of $60 \mathrm{~m}$ at the rate of $15 \mathrm{~kg} / \mathrm{s}$ to operate a turbine. The losses due to frictional forces are $10 \%$ of energy. How much power is generated by the turbine? $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$

Options:
A $8.1 \mathrm{~kW}$
B $10.2 \mathrm{~kW}$
C $12.3 \mathrm{~kW}$
D $7.0 \mathrm{~kW}$

The correct answer is:
$8.1 \mathrm{~kW}$

Solution:
Power generated by the turbine is,
$\begin{aligned}
P_{\text {gene rated }} & =P_{\text {input }} \times \frac{90}{100} \\
& =\frac{M g h}{t} \times \frac{90}{100}
\end{aligned}$
Putting the given values
$\begin{aligned}
& \frac{M}{t}=15 \mathrm{~kg} / \mathrm{s}, g=10 \mathrm{~m} / \mathrm{s}^2, h=60 \mathrm{~m} \\
& P_{\text {generated }}=(15 \times 10 \times 60) \times \frac{90}{100} \\
&=8.1 \mathrm{~kW}
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
Water falls from a height of $60 \mathrm{~m}$ at the rate of $15 \mathrm{~kg} / \mathrm{s}$ to operate a turbine. The losses due to frictional forces are $10 \%$ of energy. How much power is generated by the turbine? $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right)$

Options:
A $7.0 \mathrm{~kW}$
B $8.1 \mathrm{~kW}$
C $10.2 \mathrm{~kW}$
D $12.3 \mathrm{~kW}$

The correct answer is:
$8.1 \mathrm{~kW}$

Solution:
$$
P=0.9 \mathrm{gH} \frac{d m}{d t}
$$

Topic: Physics, Work Power Energy, NEET, NEET 2008 (Mains)
Water filled in two glasses A and B gave BOD values of 10 and 20, respectively. The correct statement regarding them is

Options:
A A is suitable for drinking, whereas B is not
B B is more polluted than A
C Both A and B are suitable for drinking
D A is more polluted than B

The correct answer is:
B is more polluted than A

Solution:
A water supply with a BOD level of 3-5 ppm is considered moderately clean. In water with a BOD level of 6-9 ppm, the water is considered somewhat polluted because there is usually organic matter present and bacteria are decomposing this waste. Hence, Both A and B are not suitable for drinking. B with a higher value of BOD level is more polluted.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Water flows into a large tank with flat bottom at the rate of 10-4 m3s-1. Water is also leaking out of a hole of area 1 cm2 at its bottom. If the height of the water in the tank remains steady then this height is:

Options:
A 5.1 cm
B 1.7 cm
C 2.9 cm
D 4 cm

The correct answer is:
5.1 cm

Solution:
Amount of water going out = amount of water coming in.
2gh×A=V
h=10-422×9.8×10-42 =5.1 cm

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Water flows m a horizontal tube (see figure). The pressure of water changes by 700 Nm-2 between A and B where the area of cross section are 40 cm2 and 20 cm2, respectively. Find the rate of flow of water through the tube. (density of water =1000 kgm-3 )

Options:
A 3020cm3/s

B 2720cm3/s

C 2420cm3/s

D 1810cm3/s

The correct answer is:
2720cm3/s

Solution:
Using equation of continuity
40VA=20VB
⇒2VA=VB
Using Bernoulli’s equation
PA+12ρVA2=PB+12ρVB2
⇒PA-PB=12VB2-VA2
⇒ΔP=121000VB2-VB24
⇒ΔP=500×3VB24
⇒VB=ΔP×41500=700×41500ms
Volume flow rate =20×100×VB=2732 cm3/s

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2020 (09 Jan Shift 1)
Water flows out of a big tank along a tube bent at right angles; the inside radius of the tube is equal to r = 0.50 cm. The length of the horizontal section of the tube is equal to l = 22 cm. The water flow rate is Q = 0.50 L s-1. Find the moment of reaction forces of flowing water, acting on the tube's walls, relative to the point Q.

Options:
A Ï Qâ¡ 2 lâ¡ Ï râ¡ 2
B ρ Q&ApplyFunction; 2 l&ApplyFunction; π r&ApplyFunction;
C - ρ Q&ApplyFunction; 2 l&ApplyFunction; π r&ApplyFunction; 2
D ρ Q&ApplyFunction; l&ApplyFunction; π r&ApplyFunction; 2

The correct answer is:
ρ Q⁡ 2 l⁡ π r⁡ 2

Solution:
Consider only bend part of the tube.
The change in momentum of mass Δm of liquid as it passes through the bend,
ΔP→=P→f⁡-P→i⁡=Δm⁡v⁡i^-Δm⁡v⁡-j^=Δm⁡v⁡i^+Δm⁡v⁡j^
∴ F → = Δ P⁡ Δ t = v⁡ Δ m⁡ Δ t i ^ + v⁡ Δ m⁡ Δ t j ^
=v⁡ρS⁡ΔL⁡Δti^+v⁡ρS⁡ΔL⁡Δtj⁡^=ρS⁡v⁡2i^+ρS⁡v⁡2j^
∴ Reaction force on the tube at bend point is F → 0 = - F →
∴ F→0=-ρS⁡v⁡2i⁡^-ρS⁡v⁡2j^=-ρπr⁡2v⁡2i^-ρπr⁡2v⁡2j^
Moment of the force F→0about O⁡ is
N → = l → × F → 0 = l⁡ i ^ × - ρ π r⁡ 2 v⁡ 2 i ^ - ρ π r⁡ 2 v⁡ 2 j ^
N → = - ρ π r⁡ 2 v⁡ 2 l⁡ k ^
∴ N → = ρ π r⁡ 2 v⁡ 2 l⁡
ButQ=Sv
∴ v⁡=Q⁡S⁡=Q⁡πr⁡2

∴N→=ρπr2l⁡Q⁡πr⁡22=ρQ⁡2l⁡πr⁡2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Water from a hosepipe of radius 5 cm strikes a wall at a speed of 5 m s-1 normally and stops. The force exerted on the wall in newton is

Options:
A 13.5π
B 6.25π
C 62.5π
D 27π

The correct answer is:
62.5π

Solution:
Given, Radius of hose pipe r=5 cm
=5×10-2 m
Speed of water v=5 m s-1
We know that, F=Av2ρ
F=πr2v2ρ
F=π5×10-2252×1000
F=62.5π N

Topic: Physics, Laws of Motion, JEE Main
Water from a pipe is coming at a rate of 100 liters per minute. If the radius of the pipe is 5 cm, the Reynolds number for the flow is of the order of: (density of water = 100 kg/m3, coefficient of viscosity of water =1 mPa s)

Options:
A 102
B 104
C 103
D 106

The correct answer is:
104

Solution:
Volume flow rate =πr2v=100 lmin
⇒πr2v=1600 m3/sec
⇒v=1600 πr2 m/sec
Reynolds number Re=ρvDη
=10001×10-3×1600 πr2×2r
=20000.6π5×10-2=21220.66=2.1×104

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2019 (08 Apr Shift 1)
Water from a tap emerges vertically downwards with an initial speed of 1.0 m s-1. The cross-sectional area of the tap is 10-4 m2. If we assume that the pressure is constant throughout and that the flow is steady, then the cross-sectional area of the stream, 0.15 m below the tap, is nÃ10-5 m2. What is the value of n?

Options:

The correct answer is:
5

Solution:
From conservation of energy
v 2 2 = v 1 2 + 2 gh ...(i)
[can also be found by applying Bernouilli's theorem between 1 and 2]
From continuity equation
A 1 v 1 = A 2 v 2
v 2 = A 1 A 2 v 1 ...(ii)
Substituting value of v2 from Eq. (ii) in Eq. (i)

A 1 2 A 2 2 Â· v 1 2 = v 1 2 + 2 gh
or A 2 2 = A 1 2 v 1 2 v 1 2 + 2 gh
â´ A 2 = A 1 v 1 v 1 2 + 2 gh
Substituting the given values
A 2 = 1 0 - 4 m 2 1.0 m/s 1.0 m/s 2 + 2 1 0 0.15
A 2 = 5 Ã 1 0 - 5 m 2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Water from a tap emerges vertically downwards with an initial speed of 1.0 ms-1 . The cross - sectional area of the tap is 10-4 m2 . Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross - sectional area of the stream, 0.15 m below the tap would be:
(Take g=10 ms-2 )

Options:
A 1×10-5 m2
B 5×10-4 m2
C 2×10-5 m2
D 5×10-5 m2

The correct answer is:
5×10-5 m2

Solution:
Velocity of water after falling through distance 0.15 m is
v2=u2+2as
=1+2×10×0.15
v2=4
v=2 m/s
From equation of continuity
A1v1=A2v2
A2=10-4×12=5×10-5 m2

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2019 (10 Apr Shift 2)
Water from a tap emerges vertically downwards with an initial velocity V0. Assuming pressure is constant throughout the stream of water and the flow is steady, find the distance from the tap at which cross-sectional area of the stream is half of the cross-sectional area of the stream at the tap.

Options:
A V022g
B 3V022g
C 2V02g
D 5V022g

The correct answer is:
3V022g

Solution:
Using the 3rd equation of motion
V2=u2+2gh
V2=V02+2gh
Using the equation of continuity
A1V1=A2V2
we get A2A1=V0V02+2gh
Required condition A1A2=12=V0V02+2gh
⇒4V02=V02+2gh
h=3V022g

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Water has maximum density at

Options:
A $0^{\circ} \mathrm{C}$
B $32^{\circ} \mathrm{F}$
C $-4^{\circ} \mathrm{C}$
D $4^{\circ} \mathrm{C}$

The correct answer is:
$4^{\circ} \mathrm{C}$

Solution:
Water has maximum density at $4^{\circ} \mathrm{C}$.

Topic: Physics, Thermal Properties of Matter, JEE Main
Water is a

Options:
A Amphoteric acid
B Aprotic solvent
C Protophobic solvent
D None of these

The correct answer is:
Amphoteric acid

Solution:
Amphoteric acid: This is true. Water is amphoteric, meaning it can act as both an acid and a base

Topic: Chemistry, Ionic Equilibrium, JEE Main
Water is being filled at the rate of 1cm3sec-1 in a right circular conical vessel (vertex downwards) of height 35cm and diameter 14cm. When the height of the water level is 10cm, the rate (in cm2 sec-1) at which the wet conical surface area of the vessel increases is

Options:
A 5
B 215
C 265
D 2610

The correct answer is:
265

Solution:
Let the volume of the cone be V c.c.

Given dvdt=1cc/sec, h=35cm, r=7cm

i.e. hr=5

We know for a cone l2=r2+h2

Lateral surface area, S=πrr2+h2
S=πh5h225+h2=π2625h2

Also V=13πr2h=13πh52h=π75h3

⇒dVdt=π25h2dhdt⇒π25h2dhdt=1

⇒dhdt=25πh2 …i

Now dSdt=π2625×2hdhdt=226h (from i)

dSdth=10=265

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2022 (25 Jun Shift 2)
Water is being poured in a vessel at a constant rate Î± m3 s-1. There is a small hole of the area a at the bottom of the tank. The maximum level of water in the vessel is

Options:
A h=Î±22ga2
B h=α24ga2
C h=α2ga2
D h=3α2ga2

The correct answer is:
h=α22ga2

Solution:
Level of water in the vessel will be maximum when rate of inflow of water=rate of outflow of water.
or Î±=av
or Î±=a2gh
or h=Î±22ga2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Water is conveyed through a uniform tube of 8 $\mathrm{cm}$ in diameter and $3140 \mathrm{~m}$ in length at the rate $2 \times 10^{-3} \mathrm{~m}^3$ per second. The pressure required in maintain the flow is (Viscosity of water $=10^{-3}$ SI units)

Options:
A $6.25 \times 10^3 \mathrm{Nm}^{-3}$
B $0.625 \mathrm{Nm}^{-2}$
C $0.0625 \mathrm{Nm}^{-2}$
D $0.00625 \mathrm{Nm}^{-2}$

The correct answer is:
$6.25 \times 10^3 \mathrm{Nm}^{-3}$

Solution:
Radius $r=\frac{8}{2} \mathrm{~cm}=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}$,
Rate of flow of water
$\begin{aligned} & Q=2 \times 10^{-3} \mathrm{~m}^3 / \mathrm{s} \\ & Q=\frac{\pi P r^2}{8 \eta l}\end{aligned}$
$\Rightarrow \quad P=\frac{Q(8 \eta l)}{\pi r^4}$
$\eta=$ coefficient of viscosity of water $=10^{-3}$ SI units
$\begin{aligned} \therefore \quad P & =\frac{2 \times 10^{-3} \times 8 \times 10^{-3} \times 3140}{3.14 \times\left(4 \times 10^{-2}\right)^4} \\ & =\frac{2 \times 8 \times 3140 \times 10^{-6}}{3.14 \times 256 \times 10^{-8}} \\ & =\frac{3140 \times 10^2}{3.14 \times 16} \\ & \left.=\frac{100 \times 10^3}{16}\right] \\ & =6.25 \times 10^3 \mathrm{~N} / \mathrm{m}^2\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, AP EAMCET, AP EAMCET 2001
Water is conveyed through a uniform tube of 8 $\mathrm{cm}$ in diameter and $3140 \mathrm{~m}$ in length at the rate $2 \times 10^{-3} \mathrm{~m}^3$ per second. The pressure required in maintain the flow is (Viscosity of water $=10^{-3}$ SI units)

Options:
A $6.25 \times 10^3 \mathrm{Nm}^{-3}$
B $0.625 \mathrm{Nm}^{-2}$
C $0.0625 \mathrm{Nm}^{-2}$
D $0.00625 \mathrm{Nm}^{-2}$

The correct answer is:
$6.25 \times 10^3 \mathrm{Nm}^{-3}$

Solution:
Radius $r=\frac{8}{2} \mathrm{~cm}=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}$,
Rate of flow of water
$\begin{aligned} & Q=2 \times 10^{-3} \mathrm{~m}^3 / \mathrm{s} \\ & Q=\frac{\pi P r^2}{8 \eta l}\end{aligned}$
$\Rightarrow \quad P=\frac{Q(8 \eta l)}{\pi r^4}$
$\eta=$ coefficient of viscosity of water $=10^{-3}$ SI units
$\begin{aligned} \therefore \quad P & =\frac{2 \times 10^{-3} \times 8 \times 10^{-3} \times 3140}{3.14 \times\left(4 \times 10^{-2}\right)^4} \\ & =\frac{2 \times 8 \times 3140 \times 10^{-6}}{3.14 \times 256 \times 10^{-8}} \\ & =\frac{3140 \times 10^2}{3.14 \times 16} \\ & \left.=\frac{100 \times 10^3}{16}\right] \\ & =6.25 \times 10^3 \mathrm{~N} / \mathrm{m}^2\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Water is conveyed through a uniform tube of 8 $\mathrm{cm}$ in diameter and $3140 \mathrm{~m}$ in length at the rate $2 \times 10^{-3} \mathrm{~m}^3$ per second. The pressure required in maintain the flow is (Viscosity of water $=10^{-3}$ SI units)

Options:
A $6.25 \times 10^3 \mathrm{Nm}^{-3}$
B $0.625 \mathrm{Nm}^{-2}$
C $0.0625 \mathrm{Nm}^{-2}$
D $0.00625 \mathrm{Nm}^{-2}$

The correct answer is:
$6.25 \times 10^3 \mathrm{Nm}^{-3}$

Solution:
Radius $r=\frac{8}{2} \mathrm{~cm}=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}$,
Rate of flow of water
$\begin{aligned} & Q=2 \times 10^{-3} \mathrm{~m}^3 / \mathrm{s} \\ & Q=\frac{\pi P r^2}{8 \eta l}\end{aligned}$
$\Rightarrow \quad P=\frac{Q(8 \eta l)}{\pi r^4}$
$\eta=$ coefficient of viscosity of water $=10^{-3}$ SI units
$\begin{aligned} \therefore \quad P & =\frac{2 \times 10^{-3} \times 8 \times 10^{-3} \times 3140}{3.14 \times\left(4 \times 10^{-2}\right)^4} \\ & =\frac{2 \times 8 \times 3140 \times 10^{-6}}{3.14 \times 256 \times 10^{-8}} \\ & =\frac{3140 \times 10^2}{3.14 \times 16} \\ & \left.=\frac{100 \times 10^3}{16}\right] \\ & =6.25 \times 10^3 \mathrm{~N} / \mathrm{m}^2\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, TS EAMCET, TS EAMCET 2001
Water is filled up to a height h in a beaker of radius R as shown in the figure. The density of water is ρ , the surface tension of water is T and the atmospheric pressure is p o . Consider a vertical section ABCD of the water column through a diameter of the beaker. The force on water on one side of this section by water on the other side of this section has magnitude

Options:
A |2p0Rh+πR2ρgh-2RT|
B |2p0Rh+Rρgh2-2RT|
C |p0πR2+Rρgh2-2RT|
D |p0πR2+Rρgh2+2RT|

The correct answer is:
|2p0Rh+Rρgh2-2RT|

Solution:
Force from right hand side liquid on left hand side liquid,
(i) Due to surface tension force
= 2RT (towards right)
(ii) Due to liquid pressure force
= p 0 +Ïgx 2Rdx
=2R Ï 0 h+RÏg h 2
Net force is 2 p o Rh+Rpg h 2 â2RT

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Water is filled up to a height h in a cylindrical vessel. It takes time t to completely drain the vessel by means of a small hole at the bottom. If water is filled up to a height 4h then the time it takes to completely drain the vessel is

Options:
A t
B 4t
C 2t
D t4

The correct answer is:
2t

Solution:
Time required to empty the tank
t = A A 0 2 H g â t 2 t 1 = H 2 H 1 = 8 h 2 h = 2
â´ t 2 = 2 t

Topic: Physics, Mechanical Properties of Fluids, JEE Main