Question 12 Marks
Rectangular film of soap solution is 6 cm long and 4 cm wide. To make its size $8 cm \times 5 cm$, how much work would have to be done against surface tension? If surface tension of soap solution is $40 \times 10^{-3} N / m$
Answer$
\begin{aligned}
A & =2 \times 6 \times 10^{-2} \times 4 \times 10^{-2} \\
& =48 \times 10^{-4} m^2
\end{aligned}
$
$
\begin{aligned}
A+\Delta A & =2 \times 8 \times 10^{-2} \times 5 \times 10^{-2} \\
& =80 \times 10^{-4} m^2 \\
\therefore \quad \Delta A & =(A+\Delta A)-(A) \\
& =80 \times 10^{-4}=48 \times 10^{-4} \\
& =32 \times 10^{-4} m^2
\end{aligned}
$
We know that $W = T \Delta A$
$\begin{array}{l}=40 \times 10^{-3} \times 32 \times 10^{-4} \\ =1280 \times 10^{-7} \\ =1.2 \times 10^{-4} \text { Joule }\end{array}$
View full question & answer→Question 22 Marks
Water rises to a height of 10 cm in a capillary tube. If the surface tension of water is $73 \times$ $10^{-3} N / m$ and density is $1 \times 10^3 kg / m ^3$ and $g=9.8 m /$ $s^2$ then find the radius of capillary tube.
AnswerIf a liquid of density $\rho$ rises to a height $h$ in a capillary tube of radius $r$, then
Surface tension $T =\frac{h r \rho g}{2}$ where $g$ is the acceleration due to gravity.
Here,
$
\begin{array}{l}
h=10 cm=0.1 m \\
T=73 \times 10^{-3} kg / m^3 \\
g=9.8 m / sec^2
\end{array}
$
$
\therefore \quad 73 \times 10^{-3}=\frac{0.1 \times\left(1 \times 10^3\right) \times 9.8 \times r}{2}
$
$
2 \times 73 \times 10^{-3}=980 r
$
$
r=\frac{146 \times 10^{-3}}{980} M
$
$
=0.1489 \times 10^{-3} M
$
$
r=0.1489 \times 10^{-3} \times 10^2 cm=0.1489 \times 10^{-1} cm
$
$
r=0.015 cm \text { (Round off) }
$
View full question & answer→Question 32 Marks
Between a flat plate of $100 sq . cm$ area a bigger plate is a layer of 1 mm thick glycerine. If the coefficient of viscosity of glycerine is $1.0 kg / m$ sec then how much force is required to move the plate with a velocity of $7 cm / sec$ ?
AnswerThe force required to move the plate with a definite velocity will be equal to the viscous force $F$ acting on the plate. According to the formula,
$
F=-\eta A \frac{d v}{d x}=-\eta A \frac{\Delta v}{\Delta x}
$
Given :
$ \begin{aligned}
\eta & =1.0 kg / m \times s \\
A & =100 cm^2=10^{-2} m^2 \\
\Delta v & =-7 \times 10^{-2} m / s \\
\Delta x & =1 mm=10^{-3} m \\
\therefore \quad \quad F & =\frac{1 \times 10^{-2} \times 7 \times 10^{-2}}{10^{-3}}=0.7 N
\end{aligned}
$
View full question & answer→Question 42 Marks
Water flows in a horizontal pipe one end of which is closed by a value and the reading in the pressure meter on the pipe is $3.5 \times 10^5$ Newton $/ m ^2$.
When the value on the pipe is opened the reading on the pressure meter becomes $3.0 \times 10^5 Newton / m ^2$. Calculate the velocity of water flowing in the pipe.
AnswerFrom Bernoulli's principle
$
\begin{aligned}
P_1+\frac{1}{2} \rho v_1^2 & =P_2+\frac{1}{2} \rho v_2^2 \\
\text { or } \frac{1}{2} \rho\left(v_2^2-v_1^2\right) & =P_1-P_2
\end{aligned}
$
Here, $v _1=0$ (Initially the value is closed so velocity of water will be zero.)
$
v_2^2=\frac{2}{\rho}\left(P_1-P_2\right)
$
For water
$
\begin{aligned}
\rho & =1 \times 10^3 kg / m^3 \\
P_1 & =3.5 \times 10^5 N / m^2 \\
P_2 & =3.0 \times 10^5 N / m^2 \\
v_2^2 & =\frac{2}{1 \times 10^3}(3.5-3.0) \times 10^5 \\
& =2 \times 0.5 \times 10^2=100 \\
v_2 & =\sqrt{100}=10 m / sec
\end{aligned}
$
View full question & answer→Question 52 Marks
2 cm from a horizontal pipe of unequal bore. At a point of pressure equal to Hg column, water is flowing out at a velocity of $40 cm / sec$. Calculate the pressure at the point where the velocity of water flow is $60 cm / sec$.
AnswerPipe is horizontal.
From Bernoulli's equation
$
P_1+\frac{1}{2}~ \text {pv}_1^2=P_2+\frac{1}{2} \text {pv}_2^2
$
Here, $\quad P _1=2 \times 13.6 \times 980=26656$
$P _1=26656$ dyne $/ cm ^2$
$v _1=40 cm / sec$
$v _2=60 cm / sec$
$P _2=$ ?
$
\begin{aligned}
26656+\frac{1}{2} \times 1 & \times(40)^2=P_2+\frac{1}{2} \times 1 \times(60)^2 \\
26656+800 & =P_2+1800 \\
P_2 & =25656 \text { dyne } / cm^2 \\
& hgd=25656 \text { dyne } / cm^2
\end{aligned}
$
Changing in Hg column
$
\begin{array}{l}
=\frac{25656}{dg}=\frac{25656}{13.6 \times 980} \\
=1.92 cm \text { equal to } Hg \text { column }
\end{array}
$
View full question & answer→Question 62 Marks
Water is flowing in a pipe of uneuqal cross-section at the place where the radius of the pipe is 1 cm whereas the velocity of water is 10 $cm / sec$. At any other place where the radius of the pipe is 2 cm then find the velocity of water.
AnswerGiven :
$ r_1=1 cm$
$
v_1=10 cm / sec
$
$r_2=2 cm$
Let velocity of water $\left( v _2\right)=$ ?
We know that
$\begin{aligned}A_1 v_1 & =A_2 v_2 \\
\pi r_1^2 v_1 & =\pi r_2^2 v_2 \\r_1^2 v_1 & =r_2^2 v_2 \\
(1)^2 \times 10 & =(2)^2 \times v_2 \\
10 & =4 V_2 \\
v_2 & =\frac{10}{4}=2.5 cm / sec\end{aligned}$
View full question & answer→Question 72 Marks
The velocity of flow of water in a horizontal pipe is $10 m / sec$. Find the velocity of water at the top? $\left( g =9.8 m / s ^2\right)$
AnswerIf the velocity of liquid in the pipe is v ,
Velocity at top $=\frac{1}{2} \frac{ v ^2}{g}$
Here, $v =10 m / sec$ and $g =9.8 m / sec ^2$
Velocity of water at the top $=\frac{1}{2} \frac{(10)^2}{9.8}$
$
\begin{array}{l}
=\frac{100}{19.6}=\frac{1000}{196}=5.102 \\
\cong 5 m
\end{array}
$
View full question & answer→Question 82 Marks
The pressure of water at one end of a horizontal pipe is equal to the pressure of one atmosphere. Find the pressure head of water. $\left( g =9.8 m / sec ^2\right)$
AnswerLet the pressure of water at the top $= hm$
Atmosphere pressure $=1 \times 10^5 N / m ^2$
We know that $P = h \rho g$
$
\begin{array}{l}
h=\frac{P}{\rho g} \\
h=\frac{1 \times 10^5}{10^3 \times 10}
\end{array}$
Since density of water $\rho=10^3 kg / m ^3$
$h=\frac{10^5}{10^4}=10 m$
Therefore, the pressure of water at top 10 is meters.
View full question & answer→Question 92 Marks
Fast air is flowing near a horizontal aircraft wing such a way that its velocity on the upper surface of the wing is $120 m / sec$ and on the lower surface it is $90 m / sec$. If the density of air is $1.3 kg /$ m , then find the pressure difference between the upper and lower of the wing. If the wing is 10 m long and average width is 2 metres, then calculate the total buoyant force on it.
AnswerVelocity of air flowing under the aircraft
$v_1=90 m / sec$
and here the pressure is considered to be $P _1$.
Velocity of air flowing over the wing
$v_2=120 m / sec$
and due to this the pressure is considered to be $P _2$.
Then from Bernoulli's theorem
$P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2$
Here, $\rho=$ Density of air
$\begin{aligned}\Rightarrow \quad & P_1-P_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right) \\& P_1-P_2=\frac{1}{2} \rho\left(v_2-v_1\right)\left(v_2+v_1\right)\end{aligned}$
$\begin{aligned} & =\frac{1}{2} \rho(120+90)(120-90) \\ & =\frac{1}{2} \times \rho \times 210 \times30=3150 \times \rho \\ P_1-P_2 & =3150 \times 1.3=4095 N / m ^2\end{aligned}$
Hence, the value of the pressure difference between upper and lower of the wing will be $4095 N / m ^2$.
Area of the wing $= L \times B =10 \times 2=20 m^2$
$\therefore$ The value of total buoyant force on the aircraft $=20 \times 4095=8.19 \times 10^4 N$
View full question & answer→Question 102 Marks
Water of pressure $4 \times 10^4 N / m ^2$ is flowing through a pipe of cross-sectional area 0.02 $m ^2$ with a velocity of $2 m / sec$. If the cross-sectional area of pipe in reduced to $0.01 m^2$ then what will be the value of pressure in the pipe?
AnswerGiven :
$\begin{aligned}P_1 & =4 \times 10^{+} N / m^2 \\v_1 & =2 m / sec \\A_1 & =0.02 m^2 \\A_2 & =0.01 m^2 \\P_2 & =? \text { and } v_2=?\end{aligned}$
From continuity equation
$\begin{aligned}A_1 v_1 & =A_2 v_2 \\v_2 & =\frac{A_1 v_1}{A_2}=\frac{0.02 \times 2}{0.01}=4 m / sec\end{aligned}$
From Bernoulli's principle
$\begin{array}{c}P_1+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2{ }^2 \\4 \times 10^4+\frac{1}{2} \times 10^3 \times(2)^2=P_2+\frac{1}{2} \times 10^3 \times(4)^2 \\P_2=4 \times 10^4+2 \times 10^3-8 \times 10^3\end{array}$
$\begin{array}{l} P _2^2=34 \times 10^3 \\ P _2=3.4 \times 10^4 N / m ^2\end{array}$
Pressure in the small cross-sectional area of pipe is $3.4 \times 10^4 N / m^2$
View full question & answer→Question 112 Marks
A glass ball of radius 4 mm is released into the sea. It starts moving with a velocity of 52.3 $m / sec$. If the density of sea water is $1.015 \times 10^3 kg /$ $m ^3$ then what will be the value of the viscous force acting on the bullet? Find out coefficient of viscosity of sea water $=0.01$ Poise.
AnswerGiven : $r=4 mm=4 \times 10^{-3} m$
$\begin{aligned}v & =52.3 m / sec \\\eta & =0.01 \text { Poise }=\frac{0.01}{10} kg / m-sec \\& =10^{-3} kg / m-sec\end{aligned}$
Therefore, viscous force
$F=6 \pi \eta r v$
Putting values,
$\begin{array}{l}F=6 \times 3.14 \times 10^{-3} \times 4 \times 10^{-3} \times 52.3 \\F=9.84 \times 10^{-4} \text { Newton }\end{array}$
View full question & answer→Question 122 Marks
Blood is flowing 0.1 cm long and $2 \times 10^{-4}$ cm radius in the tube. The pressure difference at the ends of this tube is 20 mm mercury. Coefficient of viscosity of blood is $0.5 \times 10^3 kg m ^{-1} sec ^{-1}$. Find the rate of blood flow in the tube.
AnswerGiven :
$l=0.1 cm=10^{-3} m,$
radius $r=2 \times 10^{-4} cm=2 \times 10^{-6} m$
Pressure difference $P =20 mm$ mercury
$\begin{aligned}& =20 \times 10^{-3} m \text { mercury } \\& =20 \times 10^{-3} \times\left(13.6 \times 10^3\right) \times 9.8 N / m^2 \\\eta & =0.5 \times 10^{-3} kg / m-sec\end{aligned}$
Hence, rate of blood flow in the tube
$\begin{array}{l}Q=\frac{\pi Pr^4}{8 \eta l} \\Q=\frac{3.14 \times\left(20 \times 10^{-3} \times 13.6 \times10^3 \times 9.8\right) \times\left(2 \times 10^{-6}\right)^4}{8 \times 0.5 \times 10^{-3} \times10^{-3}} \\Q=3.42 \times 10^{-15} m^3 / sec .\end{array}$
View full question & answer→Question 132 Marks
When a 200 gm piece of metal is weighed in pure water, its weight is 160 gm. Find the relative density of the metal and the volume of the piece.
AnswerRelative density of metal
$\begin{array}{l}=\frac{\text { Weight of metal in air }}{\begin{array}{c}\text { Reduction in weight when } \\\text { weighed in water }\end{array}} \\=\frac{200}{200-160}=\frac{200}{40}=5\end{array}$
Density of metal $=$ Relative density of metal $\times$ density of water
$=5 \times 1=5 gm / cm^3$
$\therefore$ Volume of piece $=\frac{\text { Mass }}{\text { Density }}=\frac{200 gm }{5 gm / cm ^3}$
View full question & answer→Question 142 Marks
How much weight will be required to lift a stone weighting 800 kg in a hydraulic lift. Given that the ratio of the cross-sectional areas of the two piston is 4 then is the work done by the machine more than the work done on the machine? Explain.
AnswerOn the basis of Pascal's law, from the principle of hydraulic lift
$\begin{aligned}& \frac{F_2}{F_1}=\frac{A_2}{A_1} \Rightarrow \frac{W_2}{W_1}=\frac{A_2}{A_1} \\\Rightarrow \quad & W_1=W_2\left(\frac{A_2}{A_1}\right)\end{aligned}$
Here, given $W _2=800 kg$
$\begin{array}{l}\frac{A_2}{A_1}=4 \text { or } \frac{A_1}{A_2}=\frac{1}{4} \\W_1=\text { ? }\end{array}$
Therefore, required weight
$w_1=800 \times \frac{1}{4}=200 kg$
But the mechanical adrantage
$\frac{W_2}{W_1}=\frac{A_2}{A_1}=4$
i.e. $\frac{ W _2}{W_1}>1 \Rightarrow W_2> W _1$
i.e. work done by the machine > work done on the machine.
View full question & answer→Question 152 Marks
A two metre long tube contains $0.8 \times 10^3$ $kg / m ^3$ the tube is filled with a liquid of high density and is tilted and rested against a wall so that the vertical height of the upper surface of the liquid from the bottom of the tube remains 1.8 m . If the atmospheric pressure $1.01 \times 10^5 N / m ^2$ then find the total pressure at the bottom of the tube.
View full question & answer→Question 162 Marks
A 1 meter thick cement wall can withstand a pressure of $10^5 N / m ^2$. What should be the thickness of the wall at the base of a 200 m depth water of dam?
(Density of water $=10^3 kg / m ^3$ and acceleration due to gravity $=9.8 m / s ^2$ )
AnswerThe value of pressure (P) on the wall near the dam
$\begin{array}{l}= h \rho g \\ =200 \times 10^3 \times 9.8 \\ =19.6 \times 10^5 N / m ^2\end{array}$
Since the required thickness of the wall to bear the pressure of $10^5 N / m =1 m$
Required thickness of the wall to withstand a pressure of $19.6 \times 10^5 N / m ^2$
$=\left(\frac{1}{10^5}\right) \times 19.6 \times 10^5=19.6 m$
View full question & answer→Question 172 Marks
An object whose volume is $100 cm^3$ is completely immersed in water. Find the buoyant force on the object.
AnswerSol. Buoyancy force on the object $=$ Weight of water of equal volume displaced by the object
$
\begin{array}{l}
=V \times \rho \times g \\
=200 cm^3 \times\left(1 gm / cm^3\right) \times 9.8 m / s^2 \\
=200 gm \times 9.8 m / s^2 \\
=200 \times 10^{-3} kg \times 9.8 m / s^2 \\
=1.96 N
\end{array}
$
View full question & answer→Question 182 Marks
If the atmospheric pressure on Earth is 75.0 cm of mercury. If it is equal to the column, then what will be the measurement of the monometer on the top of a 250 meter high mountain? (Average density of air $=1.29 kg / m ^3$ )
AnswerSol. Let the value of pressure at the top of the tower be equal to the value of height h of the mercury column. Therefore, the value of atmospheric pressure at the base of the mountain will be equal to the sum of the pressure of the 250 meter high air column and the pressure of the mercury column.
$0.75 \times 13.6 \times 10^3 \times g =1.29 \times g \times 250+ h \times 13.6 \times 10^3 \times g$
$\Rightarrow 0.75 \times 13.6 \times 10^3=1.29 \times 250+ h \times 13.6 \times 10^3$
$\begin{array}{l}\Rightarrow \frac{0.75 \times 13.6 \times 10^3}{13.6 \times 10^3}-\frac{1.29 \times 250}{13.6 \times 10^3}=h \\ \Rightarrow 0.75-0.02371= h \\ \Rightarrow 0.72629= h \\ \text { Therefore, } h =0.72629 m=72.63 cm\end{array}$
View full question & answer→Question 192 Marks
The length of a rectangular tank is 4.0 m and breadth is 3.0 m. It is filled with water up to a height of 2.5 m. Find the total force and water pressure exerted by the water on the bottom of the tank.
Answer(Density of water $=1000 kg / m ^3$ and $g =10 m / s ^2$ )
Volume of water filled in the tank
$
\begin{array}{l}
=\text { Length } \times \text { breadth } \times \text { height } \\
=4 m \times 3 m \times 2.5 m \\
=30.0 m^3
\end{array}
$
Density of water $=10^3 kg / m ^3$
Mass of water $M =$ Volume $\times$ density
$
\begin{array}{l}
=\left(30.0 m^3\right) \times\left(10^3 kg / m^3\right) \\
=30000 kg=3 \times 10^4 kg
\end{array}
$
Weight of water in the tank
$
\begin{array}{l}
=\text { Mass } \times \text { acceleration due to } \\
\text { gravity }(g) \\
=3 \times 10^4 \times 10=3 \times 10^5 \text { Newton }
\end{array}
$
$\therefore$ Total force on the bottom of the tank
$
\begin{aligned}
F & =\text { Weight of water } \\
& =3 \times 10^5 \text { Newton }
\end{aligned}
$
Area of the bottom
$
\begin{aligned}
A & =\text { Length } \times \text { breadth } \\
& =4 m \times 3 m=12 m^2
\end{aligned}
$
Pressure of water on the bottom
$
\begin{array}{l}
=\frac{\text { Total force on bottom }}{\text { Area of tank }} \\
=\frac{3 \times 10^5}{12}=2.5 \times 10^4 N / m^2 \end{array}
$
View full question & answer→Question 202 Marks
The mass of a person is 30 kg and the area of the sole of his foot is $0.01 m^2$. If $g =10 m /$ $s^2$ then calculate the pressure exerted by the person on the floor.
(i) When the person is standing on one leg.
(ii) When the person is standing on both legs.
Answer(i) Given : Area of the sole A $=0.01 m^2$
$\begin{aligned} \text { Force applied on the floor } & =\text { Weight of the person } \\ & = mg =30 kg \times 10 m / s ^2 \\ & =300 \text { Newton }\end{aligned}$
Pressure exerted on the floor $P =\frac{\text { Force }( F )}{\operatorname{Area}( A )}$
$P=\frac{300}{0.01}=3 \times 10^4 N / m ^2$
(ii) Area of both soles of a man standing on both feet
$A=2 \times 0.01=0.02 m^2$
$\begin{aligned} \text { Force applied on the floor } & =\text { Weight of the floor } \\ & =300 \text { Newton }\end{aligned}$
Pressure exerted on the floor $P =\frac{\text { Force }( F )}{\text { Area }( A )}$
$P=\frac{300 \text { Newton }}{0.02 m^2}=1.5 \times 10^4 N / m ^2$
View full question & answer→Question 212 Marks
Explain the reason for change of path of a ball in spinning motion.
Question 222 Marks
If a capillary tube is immersed in water in the condition of weightlessness, then how would the rise of water in it be different from the rise of water in normal condition?
AnswerIn normal condition, the force of surface tension (due to which water rises in the capillary) becomes equal to the weight of water column in the capillary tube and the water stops rising. In the condition of weightlessness the effective weight of the water column rising in the tube would be zero. So the water would reach the other end of the capillary tube, whatever be the length of the capillary tube.
View full question & answer→Question 232 Marks
To calm the sea waves, oil is put on them. Why?
AnswerWhen oil is poured, the air spreads the oil on the surface of water in the direction of air flow to a large distance. The surface tension of water is more than the surface tension of oil covered water. So the water without oil pulls the water with oil in the opposite direction of air, due to which the sea waves calm down.
View full question & answer→Question 242 Marks
Explain the reason why the liquid rise to a higher height in the capillary tube when its diameter decreases.
Answer From Zurian law of capillary tube, the height of the liquid column in capillary tube of radius $r$ is $h \propto \frac{1}{r}$ hence, if the diameter of the tube i.e. the valve of radius is less than the value of $h$ is more i.e. the liquid rise to a higher height.
View full question & answer→Question 252 Marks
When small drops of water are sprayed, why is cooling produced?
AnswerSpraying in small drops increases the surface area. Therefore, there is an increase in the surface energy but in this work there is a decrease in the internal energy (because the total energy is conserved). Hence, the fall in temperature produces coolness, which we experience in daily life. Feeling more cold while bathing under a water fountain is an example of this.
View full question & answer→Question 262 Marks
Water mixed with soap is better for clothes than pure water cleans well, explain.
AnswerCleaning clothes with detergent and soap solutions : When there are grease and oil stains on clothes, they cannot be cleaned with water. The reason for this is that water does not spread on a smooth cloth and wet it, rather if forms round droplets like mercury on it. When soap solution or detergent powder is added to water, its surface tension reduces. This causes the solution to wet the cloth and separates the grease from the cloth by creating a layer between the grease and the cloth.
View full question & answer→Question 272 Marks
Write the factors affecting surface tension?
AnswerThe surface tension of a liquid is affected by many factors. Some examples are given below :
1. Effect of temperature : On increasing the temperature the surface tension of liquid decreases and at critical temperature the value of surface tension becomes zero.
2. Effect of solute : If the solute is highly soluble then the surface tension of the liquid increases (like salt or sugar in water). If the solute is less soluble (like kerosene oil in water) then the surface tension of the liquid decreases.
3. Effect of contamination : The presence of dust particles and oily substance on the surface of water decreases its surface tension.
View full question & answer→Question 282 Marks
From a capillary tube of insufficient length water does not spread outside, why?
AnswerIf one end of a capillary tube of insufficient length is immersed in a liquid, then the liquid rises up to its upper end in the liquid tube and at the upper end it turns into a drop of radius whose internal pressure and the resultant pressure of the liquid column in the tube. Due to which the liquid does not come out from the upper end of the tube despite the hike being of insufficient length.
View full question & answer→Question 292 Marks
When a glass rod is heated at high temperature, why does its ends becomes rounded?
AnswerWhen an end of a glass tube is heated it melts and becomes liquid. Due to surface tension liquid tries to occupy minimum surface area. For any given volume the surface area of a sphere is minimum so the molten glass at the ends becomes spherical. So the ends of a glass tube become round on heating.
View full question & answer→Question 302 Marks
Prove that the energy required to break a large drop of radius $R$ into n smaller drops of radius $r$ is $4 \pi T\left(r^2 n- R ^2\right)$. Where $T$ is the surface tension.
AnswerIncrease in surface area $\Delta A=\eta \times 4 \pi r^2-4 \pi R^2$
$=T\left(4 \pi r^2 n-4 \pi R^2\right)$
$\begin{aligned} \therefore \text { Required energy } W & = T \times \Delta A \\ & = T \left(4 \pi r^2 n -4 \pi R ^2\right) \\ & =4 \pi T\left(r^2 n- R ^2\right)\end{aligned}$
View full question & answer→Question 312 Marks
What is stokes' law? Write its application of this formula.
AnswerStokes' law : According to this law the radius of the sphere $r$, which is the viscosity $\eta$ in a fluid, moving with velocity $v$, experiences a gravitational force F due to the viscosity of the fluid which can be expressed by $\vec{F}=$ $-6 \pi \eta r \vec{v}$
Application of Stokes' law :
(i) When water vapour condenses on dust particles then their terminal velocity is proportional to the square of the radius. So smaller drops fall with low velocity and bigger drops with high velocity.
(ii) When a soldier jumps down with a parachute, then initially his velocity increases rapidly but after the parachute opens the air exerts more viscous force upwards due to which the soldier falls down with a constant terminal velocity and lands safely.
(iii) To find the charge of electrons using Millikan's oil drop method, to find the terminal velocity of oil drops Stokes formula is used.
(iv) When water vapour present in air gets condensed on dust particles, small drops are formed. The drops acquire terminal velocity easily because of light weight and appear to be floating in space because of low velocity.
View full question & answer→Question 322 Marks
Write the application of viscosity?
AnswerFollowing application of viscosity are as follow :
(1) It is necessary to provide lubricant oil in machines. While choosing any lubricant oil, its coefficient of viscosity has to be kept in mind as well as it has to be seen that there is minimum change in it due to increase in temperature. In different machine, different lubricant have to be supplied according to the season. The lubricant oils provided in bicycles and motors are different.
(2) The coefficient of viscosity is used to equalize the velocity of a moving object in order to slow it down.
(3) Molecular weight is also determined from the value of coefficient of viscosity and it is also used in determining the complex compound structure.
View full question & answer→Question 332 Marks
What is the dimension of coefficient of viscosity?
AnswerWe know that $F =-\eta A \frac{d v}{d x}$
where $\eta$ is the constant of proportion. This is called coefficient of viscosity. Its value is depend on the nature of the liquid. For its dimension \[ \begin{array}{l} \eta=\frac{F}{A \cdot \frac{d v}{d z}}=\frac{\text { Force }}{\text { Area } \times \frac{\text { Velocity }}{\text { Distance }}} \\ \eta=\frac{MLT^{-2}}{L^2 \times LT^{-1}} \times L=ML^{-1} T^{-1} \end{array} \] In CGS system unit of coefficient of viscosity is gm per cm per second. It can also he called poise.
View full question & answer→Question 342 Marks
AnswerThere is a velocity gradient in a flowing fluid due to which the upper layer flows at a higher velocity than the lower layer. Thus there is relative lag in successive layers.
Just as the force of friction acts between two surface moving with relative velocity, in the same way a type of friction occurs due to relative velocity in successive layers of liquid. For this reason, the slower moving layer creates obstructions on the faster moving layer.
Energy is required to overcome this obstruction and maintain the flow.
The force creating obstruction are called viscous forces and this property is called viscosity.
View full question & answer→Question 352 Marks
How does Bernoulli's principle help in understanding the flow of blood in vein?
AnswerDue to accumulation of plague on the inner wall of the artery, the artery becomes narrow from within. To get blood flowing through these narrow arteries, there is more burden on the heart's activity. In this area the speed of blood flow increases and the internal pressure decreases and due to external pressure the artery gets compressed. The heart pushes blood to open this artery. As soon as the blood flows out at high speed through the opening, the internal pressure drops again and the artery get compressed again, which can lead to a heart atttack.
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Define critical velocity.
AnswerCritical velocity : The critical velocity is that maximum velocity of the flow of a fluid, at a velocity below which the flow of fluid is streamline and at a velocity above which the flow of fluid is trubulent.
Critical velocity is directly proportion to coefficient of viscosity $\eta$ of fluid and it is inversely proportional to the density of liquid ( $\rho$ ) and inversely proportional to the diameter of tube (D).
Therefore, $v _{ C } \propto \frac{\eta}{\rho D }$
$\therefore\quad v _{ C }=\frac{ R \eta}{\rho D }$
where $R \approx 2000$ is Reynolds number.
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Write the definition of turbulent flow.
AnswerWhen the velocity of flow of a liquid is more than the critical velocity, then the movement of particles in the liquid flow ceases to be systematic and becomes irregular zig-zag and eddy current are generated inside the liquid. This type of flow of liquid is called turbulent flow.
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What is called stream line flow?
AnswerWhen the flow of a fluid is such that the direction of every particle of the fluid passing through a point is the same as that of first particle that passed through that point, then the flow of the fluid is on streamline represents the direction of velocity of the liquid at that point.
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The density of water is greater than the density of air, yet the cloud containing small drops of water does not fall but keeps floating in the sky. Why?
AnswerA cloud consisting of small water drop lets acquires a marginal velocity downwards due to the viscosity of the air. This speed is directly proprotional to the square of the droplets, which is very small, hence the speed of the cloud containing the droplets is so low that they appear to the floating in the air.
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Does he velocity of falling rain drops continuously increases? Do all the drops of rains (big and small) reach the earth with the same velocity?
Answer Rain drops fall below the terminal velocity due to the viscosity of the air, hence their velocity does not increase continuously since the terminal velocity $V _t \propto r^2$, Hence, the big drop with reach will a more velocity than the small drops.
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Near the bank of a river the velocity of water is less but is more in the centre, why?
AnswerOn moving away from steady surface the velocity of water layers increases.
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There is glycerine in one container and water in the other container. Both are Shaken vigorously and placed on the table. The liquid in which container will come to rest first and why?
Answer Glycerine will come to rest first because the viscosity of glycerine is higher than that of water.
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Why does an iceberg float on water?
AnswerThe density of the iceberg is less than the density of water due to which the buoyancy force of water equal to the volume of the iceberg exceeds the weight of the iceberg and the iceberg remains floating on water. While floating, only that much volume of the iceberg sinks in water, which is equal to the weight of the water displaced by the volume.
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If air is blow speedily below a pan of a balance of physical balance what would be the effect on its balance?
AnswerDue to increase in air velocity below the pan pressure would decrease. So the pan would go down.
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When blown between two light balls hanging near each other, they are attracted towards each other why?
AnswerOn blowing the velocity of air between them increase so the pressure decreases.
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Why is there a hole in the lid of a tea kettle?
Answer If there is no hole in the lid of the tea kettle, the pressure inside the kettle filled with tea will be less than the atmospheric pressure. Obviously if the kettle is tilted tea will not come out easily from its spout because the pressure outside will be more than the pressure inside the kettle
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Why are the tires of the wheels of heavy vehicles made wider?
AnswerDue to wider tires of the wheels of heavy vehicles (the value of area A is more) the pressure (P = F/A) applied on the road or ground reduces because the weight of the vehicle is applied on a larger area. Therefore the wheels of the vehicle are saved from sinking on the road.
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When you stand on the sand, your feet sink deeper into it, but when you lie down, your body sinks less into the sand, why?
AnswerIn standing position the area is less but when you lie down the area is more. Whereas in both the cases the weight force is the same. Therefore, in the lying position the body sinks less due to less pressure $P = F/A$ as compared to the standing positon.
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