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Measurement of density - class-VIII

Description: measurement of density
Number of Questions: 92
Created by:
Tags: physics substances in the surroundings - their states and properties chemistry measurements and units measurements and uncertainties measurement
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The density of a cuboid of mass 200 g with dimensions 2 cm $\times$ 4 cm $\times$ 5 cm is

  1. 1000 kg m$^{-3}$

  2. 3000 kg m$^{-3}$

  3. 5000 kg m$^{-3}$

  4. 2000 kg m$^{-3}$


Correct Option: C
Explanation:
The density of a body is defined as the amount of substance contained per unit volume.
Density $=\dfrac{mass}{volume}$

$\displaystyle d = \frac{0.2 kg}{40  cm^3} = \frac{0.2 \times 10^6}{40} kg  m^{-3}$

$=5000  kg  m^{-3}$

A wire of length 50 cm has a mass of 20 gm. It its radius is halved by stretching, its new mass per unit length will be

  1. 0.4 g cm$^{-1}$

  2. 0.2 kg m$^{-1}$

  3. 0.1 g cm$^{-1}$

  4. 0.2 g cm$^{-1}$


Correct Option: C
Explanation:

On stretching, volume remains constant. Hence on decreasing the radius by half, the length of the wire becomes quadrupled. 
Mass per unit length $\displaystyle = \frac{20}{4 \times 50 cm}$
$=0.1  g  cm^{-1}$

Density of water is :

  1. 1 g cm$^{-3}$

  2. 100 kg m$^{-3}$

  3. 1000 kg m$^{-3}$

  4. None


Correct Option: A,C
Explanation:

 Density of water = 1 g $ cm^{-3 }$ = 1000 kg $ m^{-3}$

so option (A) and option (C) are correct

If 5 litres of kerosene has a mass of 5 kg, then what is the density of kerosene?

  1. 500 kg/$m^3$

  2. 1000 kg/$m^3$

  3. 100 kg/$m^3$

  4. 50 kg/$m^3$


Correct Option: B
Explanation:

Density of kerosene $= \displaystyle \frac{mass}{volume}$
$\displaystyle = \frac{5  kg}{5  litres}= \frac{5kg}{5 \times 10^{-3}m^3} = 1000 kg  m^{-3}$

Two spheres are having the ratio of their densities 1:3 and masses in 3:4 respectively. Find the ratio of their volumes. i.e $\dfrac{V _1}{V _2}$

  1. 1:3

  2. 3:1

  3. 4:9

  4. 9:4


Correct Option: D
Explanation:

$\displaystyle \frac{d _1}{d _2} = \frac{1}{3} ; \frac{m _1}{m _2} = \frac{3}{4}$
$d = m/v$
$\therefore \displaystyle \frac{V _1}{V _2}  = \frac{m _1}{m _2} \times \frac{d _2}{d _1} = \frac{3}{4} \times \frac{3}{1} = 9:4$

Density of air at room temperature is

  1. $120 \ {g\ cm}^{-3}$

  2. $120 \ {g\ mm}^{-3}$

  3. $12 \ {g\ m}^{-3}$

  4. $1.3 \ {kg\ m}^{-3}$


Correct Option: D
Explanation:

The density of air at NTP is $1.293 kg m^{-3}$.

So after rounding, it is $1.3 kg m^{-3}$.

If the mass of a body is 12.1 g and its density is 2.2 g/cc, its volume is.

  1. $5.5 cm^3$

  2. $8$ cc

  3. $11$ cc

  4. $55$ cc


Correct Option: A
Explanation:

Given:

mass $(m)=12.1g$
density $(d)=2.2g/cc$
Let the volume be $v$
As $density=\dfrac{mass}{volume}$
$d=\dfrac{m}{v}$
$v=\dfrac{m}{d}$
$v=\dfrac{12.1}{2.2}=5.5cc$

The density of water is equal to :

  1. $10^{-3}$ $kgm^{-3}$

  2. $10^{-2}$ $kgm^{-3}$

  3. $10^{2}$ $kgm^{-3}$

  4. $10^{3}$ $kgm^{-3}$


Correct Option: D
Explanation:

density of water in S.I. unit= ${10}^3$ kg/${m}^3$

Brine has a density of 1.2 g/cc. 40 cc of it is mixed with 30 cc of water. The density of the resulting solution will be

  1. $2.11$ g/cc

  2. $1.11$ g/cc

  3. $12.2$ g/cc

  4. $20.4$ g/cc


Correct Option: B
Explanation:

Density of Brine$=\rho _{b}=1.2g/cc$

Volume of Brine$=v _{b}=40cc$
Mass of Brine$=m _{b}=\rho _{b}\times v _{b}=1.2\times 40=48g$
Density of Water$=\rho _{w}=1g/cc$
Volume of Water$=v _{w}=30cc$

Mass of Water$=m _{w}=\rho _{w}\times v _{w}=1\times 30=30g$
Density of mixture$=\dfrac{\text{Mass of mixture}}{\text{ Volume of mixture}}=\dfrac{m _{b}+m _{w}}{v _{b}+v _{w}}=\dfrac{48+30}{40+30}=\dfrac{78}{70}=1.11g/cc$

Calculate the mass of air enclosed in a room of length, breadth and height equal to $5\ m, 3 \ m$ and $4 \ m$ respectively. Density of air is $1.3 \ kg/m^3$.

  1. $78 \ kg$

  2. $38 \ kg$

  3. $42 \  kg$

  4. $87 \ kg$


Correct Option: A
Explanation:

The Volume of the room is given as $V = L\times B\times H= 5\times 3\times 4 = 60  { m }^{ 3 }.$
We know, Mass $=$ Density$\times$Volume.
Given that the density $=1.3  kg/m^3$.
So, Mass of the air $=1.3\times60  Kg$.
Hence, mass of the air enclosed in the room is 78 kg.

If the mass of a body is 12.1 g and the density is 2.2g/cc, its volume is :

  1. $\displaystyle 5.5{ cm }^{ 3 }$

  2. $8 cc$

  3. $11 cc$

  4. $55 cc$


Correct Option: A
Explanation:
density $\rho=\dfrac{mass}{ volume}$

$2.2=\dfrac{12.1}{volume}$

volume $=5.5{cm}^3$

The SI unit of density is $\displaystyle kg\ { m }^{ -3 }$. State whether true or false.

  1. True

  2. False


Correct Option: A
Explanation:
S.I unit of density is kg/${m}^3$
since, density=mass/volume

If two masses A and B have their masses in the ratio 1 : 4 and their volumes are equal, then their densities have the ratio

  1. 1:4

  2. 8:1

  3. 2: 4

  4. 3:1


Correct Option: A
Explanation:

Let mass of $A$ be $m$

Mass of $B$ be $4m$
Let density of $A$ be $d _a$
Density of $B$ be $d _b$
And Volume of A$=$Volume of B$=V$
As Density$=\dfrac{\text {Mass}}{\text {Volume}}$

Therefore $d _a=\dfrac{m}{V}$
$d _b=\dfrac{4m}{V}$

$\dfrac{d _a}{d _b}=\dfrac{\dfrac{m}{V}}{\dfrac{4m}{V}}$
$\dfrac{d _a}{d _b}=\dfrac{m}{4m}=\dfrac{1}{4}$
Hence the correct answer is option (A).

What is the density of an iron block which has a volume of ${12 cm^3}$ and a mass of 96g?

  1. ${ 8 kg/m ^3}$

  2. ${80 kg /m ^3}$

  3. ${8000 kg /m ^3}$

  4. ${800 kg/ m ^3}$


Correct Option: A
Explanation:

Answer is A.

As we know, Density $d = \dfrac{Mass, M}{ Volume, V}$


In this case, an iron block which has a volume of ${12 cm^3}$ = ${0.12 m^3}$ and a mass of $96 g = 0.096 kg$.

Therefore, Density $d =\dfrac{ 0.096}{0.12} = 8 kg/m^{ 3 }$.

Hence, the density of the iron block is 8 $kg/m^{ 3 }$.

The ratio of the values in SI units to values in CGS units of density is

  1. $10^3:1$

  2. $10^2:1$

  3. $10^{-2}:1$

  4. $10^{-3}:1$


Correct Option: A
Explanation:

the density of water in S.I. unit $\rho=1000 kg/{m}^3$

the density of water in CGS unit $\rho'=1 gm/cc$

the ratio of the densities of water in the different system 
$\dfrac{\rho}{\rho'}=\dfrac{{10}^3}{1}$

The density of water is $\displaystyle { 10 }^{ 3 }kg\ { m }^{ -3 }$. State whether true or false.

  1. True

  2. False


Correct Option: A
Explanation:

It is a fact that the density of water is 1000 kg / ${m}^3$

When air is cooled, its density increases. State whether true or false.

  1. True

  2. False


Correct Option: A
Explanation:

 As on cooling volume of air decreases and its mass remains constant so its density increases as  Density = $\dfrac{mass}{volume}$

hence given statement is correct so option (A) is correct

The hot air balloon rises because it is

  1. denser

  2. less dense

  3. equally dense

  4. the given statement is wrong


Correct Option: B
Explanation:

The hot air balloon has the hot air filled in it and the density of air is more than the hot air. So, the normal air or atmospheric air has the tendency to remain below than the hot air as the objects having the more density remains below than the objects having less density.

Brine has a density of $1.2  {g}/{cc}$. $40  cc$ of it are mixed with $30  cc$ of water. The density of solution is

  1. $2.11 {g}/{cc}$

  2. $1.11 {g}/{cc}$

  3. $12.2 {g}/{cc}$

  4. $20.4 {g}/{cc}$


Correct Option: B
Explanation:
given,  ${ \rho } _{ a }=1.2\quad 9/cc\quad \quad \quad { V } _{ a }=40cc$
                 ${ \rho } _{ b }=1g/cc\quad \quad \quad { V } _{ b }=30cc$

$ \rho _{mixture} = \dfrac { { \rho } _{ a }{ V } _{ a }+{ \rho } _{ b }{ V } _{ b } }{ { V } _{ a }+{ V } _{ b } } $

So  $ \rho _{mixture} = \dfrac { 1.2\times 40+1\times 30 }{ 40+30 } =\dfrac { 78 }{ 70 } $

        $\boxed { \rho _{mixture}=1.11\quad g/cc } $

With the increase in temperature, the density of a substance, in general, ____________.

  1. increases

  2. decreases

  3. first increase then decreases

  4. first decrease then increases


Correct Option: B
Explanation:

Due to increase in temperature, objects vibrational energy increases resulting in decrease in density of the substance.

If the mass of a body is $12.1  g$ and the density is $2.2  {g}/{cc}$, its volume is

  1. $5.5 {cm}^{3}$

  2. $8 cc$

  3. $11 cc$

  4. $55 cc$


Correct Option: A
Explanation:

 Density = $\dfrac {mass}{volume}$ 


 so volume =$\dfrac{mass}{density}$

hence volume =$\dfrac{12.1}{2.2} =5.5cc$

hence option (A) is correct

The density of a body is represented by the expression :

Where V is the volume and M is the mass of a body

  1. $D=\dfrac{V}{M}$

  2. D=M $ \displaystyle \times $ V

  3. $ D = \dfrac{1}{M \times V}$

  4. $ D = \dfrac{M}{V}$


Correct Option: D
Explanation:
The density of a body is defined as the amount of mass contained per unit volume of the body.
Density = $\dfrac{mass}{volume}$ = $\dfrac{M}{V}$

so option (D) is correct

In order to find the density of a solid we have to first know its :

  1. mass and area

  2. weight and area

  3. shape and volume

  4. mass and volume


Correct Option: D
Explanation:
The density of a body is defined as the amount of mass contained per unit volume of the body.

$density=mass/volume$
So, we need mass and volume to find the density.

Equal volumes of iron and aluminum have different masses because they have different _________ .

  1. shapes

  2. densities

  3. lengths

  4. times


Correct Option: B
Explanation:

If volume of iron = Volume of aluminium

then mass of iron > mass of aluminium
$\rho _{iron}\times V _{iron} > \rho _{aluminium}\times V _{aluminium}$
$\rho _{iron} > \rho _{aluminium}$

 The C.G.S. unit of density is g/$\displaystyle cm^{3}$ 

  1. True

  2. False


Correct Option: A
Explanation:

density of a substance is the characterstic property of material defined as mass per unit volume of the substance hence in CGS system its unit is gm/cc.

_______is the characteristic property of a substance.

  1. Density

  2. Velocity

  3. Temperature

  4. Displacement


Correct Option: A
Explanation:

Density of a substance is a characterstic property of a substance defined as mass of the substance per unit volume.

density = $\dfrac { mass\quad of\quad the\quad substance }{ volume\quad of\quad the\quad substance } $

Equal volumes of water and kerosene have equal masses. True or false.

  1. True

  2. False


Correct Option: B
Explanation:

The mass of the water and kerosene depends on the volume and the density of the each. Since the water and kerosene have different densities, they will have different mass for the same volumes.


Therefore, the given statement is false.

Density of a body depends on its shape and size. True or false.

  1. True

  2. False


Correct Option: B
Explanation:

Density of a liquid is a characterstic property of a substance & it is independent of any other external factors like volume etc.

The density of milk is 1.04 $ \displaystyle g/cm^{3} $ Therefore $ \displaystyle 1cm^{3} $ of milk has a mass of ______ g.

  1. $1.04$

  2. $2.08$

  3. $3.12$

  4. $4.16$


Correct Option: A
Explanation:

Given,    density = $1.04gm/{ cm }^{ 3 }$

              Volume = $1{ cm }^{ 3 }.$

 Mass of milk $= density \times  Volume$

                            = $1.04\dfrac { gm }{ { cm }^{ 3 } } \times { 1cm }^{ 3 }$

So, the mass of milk $= 1.04gm$

One kilogram of sugar has more volume than one kilogram of iron. True or false.

  1. True

  2. False


Correct Option: A
Explanation:

Since we know that density of iron is more than density of sugar hence 

$\rho _{iron} > \rho _{sugar}$

$\dfrac { mass\quad of\quad iron }{ volume\quad of\quad iron } >\dfrac { mass\quad of\quad sugar }{ volume\quad of\quad sugar } $

if mass of iron = mass of sugar

then,  $\dfrac { 1 }{ Volume\quad of\quad iron } >\dfrac { 1 }{ Volume\quad of\quad sugar } $

So, Volume of sugar > Volume of iron.

Weight per unit volume of a substance is called its density. State whether true or false.

  1. True

  2. False


Correct Option: B
Explanation:

Density = $\dfrac{mass}{volume}$  so $\rho$ = $\dfrac{m}{v}$  not $\dfrac{mg}{v}$ as weight = mg


so given statement is false 


hence option (B) is correct

The SI unit of density is __________.

  1. $ \displaystyle kg/m^{3} $

  2. $ \displaystyle kg/m^{2} $

  3. $ \displaystyle kg^{3} $

  4. $ \displaystyle m^{3} $


Correct Option: A
Explanation:

Density of substance is mass per unit volume hence its unit are $kg/{ m }^{ 3 }$ in S/units & $gm/{ cm }^{ 3 }$ in CGS.

Mass per unit volume of a substance is called its density. True or false.

  1. True

  2. False


Correct Option: A
Explanation:

Density of substance is a characterstic property of a substance defined as mass of substance per unit volume.

density = $\dfrac { mass\quad of\quad substance }{ volume\quad of\quad substance } $.

Which of the following statements are true for density?

  1. It is a physical property of matter

  2. It is a scalar quantity

  3. It is the mass per unit volume of a substance

  4. All of the above


Correct Option: D
Explanation:

Density is a physical property of a substance and is defined as the mass per unit volume of a substance. The density of a body is the amount of mass contained per unit volume of the body.

Since it does not have any direction it is a scalar quantity.

Multiplying relative density by 100 we get density of liquid in   $ \displaystyle kg/m^{3} $

  1. True

  2. False


Correct Option: B
Explanation:

Relative density of a substance is ratio of density of substance to that of water. Hence density of substance can be found by multiplying relative density with 1000$kg/{ m }^{ 3 }$ (density of water).

Give reasons as to why equal masses of different substances have different volumes.

  1. because they have different shapes

  2. because they have different densities

  3. because they have the same densities

  4. none of these


Correct Option: B
Explanation:

The mass of the substance is dependent on the density and volume of the substance. The substance having the same volume might have different mass because they have different densities.


Option $B$ is correct. 

Which of the following is a fundamental property of a substance?

  1. Density

  2. Mass

  3. Volume

  4. Current


Correct Option: B
Explanation:

The fundamental property of a substance is the property which depends on the body and does not change due to any physical factor. Mass is a fundamental property of a substance. Another such property is the charge.

Define density.

  1. Density of a substance is defined as the weight per unit volume of that substance.

  2. Density of a substance is defined as the mass per unit volume of that substance.

  3. Density of a substance is defined as the volume per unit mass of that substance.

  4. Density of a substance is defined as mass times volume of the substance.


Correct Option: B
Explanation:

Density of a substance is defined as the mass per unit volume. Its SI unit is $kgm^{-3}$. For example, the density of water at room temperature and atmospheric pressure is $1\ kgm^{-3}$.

Which one of the following conditions will cause the density of a liquid to decrease?

  1. Adding more of the liquid to the container

  2. Cooling the liquid

  3. Heating the liquid

  4. Dividing the liquid into two equal volumes


Correct Option: C
Explanation:

$Answer: C$

The density of a liquid or any substance is the material property. The density of the liquid cannot be changed by decreasing the amount of liquid. The density can be effected only by heating the liquid which leads to the change in the molecular distance and leads to a change in density.

Find the odd one out

  1. density

  2. pressure

  3. mass

  4. volume


Correct Option: B
Explanation:

Density, mass, volume are the characterstic property of a substance whereas Pressure is an extensive property which depends on area of application of force.

Density of a material varies with temperature.
  1. True

  2. False


Correct Option: A
Explanation:

Density of the material is defined as mass per unit volume. Since, volume is temperature dependent, increasing the temperature of the material (with few exceptions) decreases the density of material by increasing its volume. 

Hence, given statement is true.

Density of a substance is defined as

  1. Density = Mass x Volume

  2. Density = Mass /Volume

  3. Density = Volume Mass

  4. None


Correct Option: B
Explanation:

Density of the substance is defined as the ratio of mass of substance to its volume.

$\therefore$  Density of substance     $\rho = \dfrac{M}{V}$.

The density of silver is 13.6 times greater than that of water.
  1. True

  2. False


Correct Option: B
Explanation:

We know that, density of silver is $10.5 \times 10^3 \dfrac{kg}{m^3}$ and density of water is $1.00 \times 10^3 \dfrac{kg}{m^3}$.


Therefore, the density of silver is $10.5$ times greater than that of water.

Hence, the given statement is false.

From below substances which is more dense?

  1. Water

  2. Clay

  3. Wax

  4. none


Correct Option: B
Explanation:
The density of the following substances is:
Wax approximately has a density around $900kg/m^3$
The density of water is $1000kg/m^3$
Density of clay is $1330kg/m^3$

The density of the clay is the maximum of all the three.

Which of the following is true about the density of a substance?

  1. It is a derived quantity.

  2. The unit of density in SI system is kilogram per metre$^3$.

  3. The density of a substance can be measured if the mass of the substance for a certain volume is known.

  4. Density is a vector quantity


Correct Option: A,B,C
Explanation:

Let the mass of the substance be $m$ and its volume be $V$. Then the density of the substance is defined as the ratio of mass to the volume of the body.

Thus the density of a substance can be derived as  $\rho = \dfrac{m}{V}$

SI unit of mass is $kg$ and that of the volume is $m^3$. Thus SI unit of density is  $kg/m^3$.
The density of a substance is a scalar quantity, not a vector quantity as it has magnitude only.
Hence options A, B and C are correct.

SI unit of density is ________ times its CGS unit.

  1. 1000

  2. 10

  3. 0.1

  4. 0.001


Correct Option: D
Explanation:

SI unit of density $=\dfrac { kg }{ { m }^{ 3 } } =\dfrac { 1000 }{ 10^6 } \dfrac { g }{ cc } $

                             $\dfrac { 1kg }{ { m }^{ 3 } } ={ 10 }^{ -3 }9/cc$

Which of the following is true about the density of a substance?

  1. It is a derived quantity.

  2. The unit of density in SI system is kilogram per metre$^3$.

  3. The density of a substance can be measured if the mass of the substance for a certain volume is known.

  4. Density is a vector quantity.


Correct Option: A,B,C
Explanation:

The density of a substance is defined as the mass of the substance per unit volume of the substance.
$\rho=\dfrac mV$

Its unit is $kg/m^3$

It is a derived quantity as it uses the unit of mass and the unit of length.
The unit of density in the SI system is kilogram per meter $^3$.
The density of a substance can be measured if the mass of the substance for a certain volume is known.

A vessel contains oil (density $= 0.8\ gm/cm^{3})$ over mercury (density $= 13.6\ gm/ cm^{3})$. A uniform sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of sphere in $gm/ cm^{3}$ is

  1. $3.3$

  2. $6.4$

  3. $7.2$

  4. $12.8$


Correct Option: C
Explanation:

A vessel contains oil density $=0.8gm/cm^3$

Mercury density $=13.6gm/cm^3$
Total weight is balanced by the displaced volumes of the two different fluids. Let it $s$ and volume id $v$ and te density is $n$.
Then the equilibrium condition should be
$v _{mg}=v/2(0.8)g+\cfrac{v}{2\times136g}\n=0.4+6.8=7.2gm/cm^3$

The heart of man pumps 5 litre of blood through the arteries per minute at a pressure of 150 mm of mercury. If the density of mercury. If the density of mercury be $13.6 \times 10^3 \ kg/m^3$ and g = 10 m/$s^2$ then the power of heart in watt is: 

  1. $2.35$

  2. $3.0$

  3. $1.50$

  4. $1.70$


Correct Option: D

State whether true or false.
An iron ball and a copper ball of same mass have the same volume.

  1. True

  2. False


Correct Option: B
Explanation:

Since copper and iron have different density, hence both balls will have different volumes for same mass.

A boat floating in a tank is carrying passenger. If the passangers drink water, how will its affect the water level of the tank?

  1. It will go down

  2. It will rise

  3. It will remain unchanged

  4. It will depend on atmospheric pressure


Correct Option: C
Explanation:

If passenger drinks water then the weight of boat get increase and the boat will displace the same amount of water in the tank resulting no net change in the water level.

Mass per unit volume is called ..... of a substance.

  1. Specific Gravity

  2. Density

  3. Gravity

  4. None of these


Correct Option: B
Explanation:

It is very well known that mass per unit volume is defined as density. 

Density of water is ___________?

  1. $1000kg/m^3$

  2. $1kg/m^3$

  3. $1000g/cm^2$

  4. $100kg/m^3$


Correct Option: A
Explanation:

Density of water is $1000kg/m^3$.

ρ (rho) = density, m = mass, V = volume. The SI unit of density is kg/m3. Water of 4 °C is the reference ρ = 1000 kg/m3 = 1 kg/dm3 = 1 kg/l or 1 g/cm3 = 1 g/ml.

A tank $2\ m$ high is half filled with water and then filled to the top with oil of density $0.80\ g/cc$. What is the pressure at the bottom of the tank due to these liquids (Take $g=10\ ms^{-2}$) is

  1. $1.80\times 10^{3}\ Nm^{-2}$

  2. $0.9\times 10^{3}\ Nm^{-2}$

  3. $1.8\times 10^{4}\ Nm^{-2}$

  4. $0.9\times 10^{4}\ Nm^{-2}$


Correct Option: C
Explanation:

Density of oil=0.80 g/cc=800 kg/m$^{3}$

Density of water=1 g/cc=1000 kg/m$^{3}$
Pressure of water=Height will be 1 m half filled
$\Longrightarrow hdg=1 m \times 1000 kg/m^{3} \times 10m/s^{2}=10000 Pa$
Pressure by oil$=hdg$
Total pressure$=(1 \times 8000 \times 10)=8000 Pa+10000 Pa = 18000 Pa=1.8 \times 10^{4} N/m^{2}$

Density of ice is $\sigma$ and that of water is $\rho$. What will be the decrease in volume when a mass $M$ of ice melts 

  1. $\dfrac {M}{\sigma - \rho}$

  2. $\dfrac {\sigma - \rho}{M}$

  3. $M\left [\dfrac {1}{\sigma} - \dfrac {1}{\rho}\right ]$

  4. $M\left [\dfrac {1}{\rho} - \dfrac {1}{\sigma}\right ]$


Correct Option: D
Explanation:

Volume of ice= Mass/Density= $M/P$

Volume of water= Mass/Density= $M/\sigma$
Decrease in volume= $V _i-V _w$
$=\cfrac {M}{P}-\cfrac {M}{\sigma}= M\left[\cfrac {1}{P}-\cfrac {1}{\sigma}\right]$

By sucking through a straw, a boy can reduce the pressure in his lungs to $750\ mm$ of $Hg$ (density$ = 13.6\,\,g/c{m^3}$). Using a straw, he can drink water from a maximum depth of

  1. $13.6\ cm$

  2. $1.36\ cm$

  3. $0.136\ cm$

  4. $10\ cm$


Correct Option: A
Explanation:

Given Pressure in lungs${P} _{lungs}=750$ mm of Hg

and atmospheric pressure${P} _{atm}=760$ mm of Hg
Now pressure difference $\Delta{P}=760-750=10$ mm of Hg=1 cm of Hg
Now
1 cm of Hg$={\rho} _{water}gh$
$1{\rho} _{Hg}g={\rho} _{water}gh$
$13.6g=gh$
$h=13.6 cm$

Maximum density of $H _2O$is at the temperature

  1. $32^oF$

  2. $39.2^oF$

  3. $42^oF$

  4. $4^oF$


Correct Option: B
Explanation:

Maximum density of water is at $4^0C$


Here, the value in degree celsius is converting to degree Fahrenheit.

$T(°F) = (T(°C) × \dfrac 95 )+ 32$

or

$T(°F) = (T(°C) × 1.8) + 32$

We have,

$T(^0C)=4^0C$

Then,

$T(°F) = (4 × 1.8) + 32=39.2^oF$

A body of density ${d _1}$ is counterpoised by $Mg$ of weights of density ${d _2}$ in air of density $d.$ Then the true mass of the body is

  1. $M$

  2. $M\left( {1-\dfrac{d}{{{d _2}}}} \right)$

  3. $M\left( {1 - \dfrac{d}{{{d _1}}}} \right)$

  4. $\dfrac{{M\left( {1 - d/{d _1}} \right)}}{{\left( {1 - d/{d _2}} \right)}}$


Correct Option: D

A $10L$ container is filled with a gas to a pressure of $2atm$ at $0^0C.$ At what temperature will the pressure inside the container be $2.50atm?$   

  1. $68^0C$

  2. $50^0C$

  3. $40^0C$

  4. $45^0C$


Correct Option: A
Explanation:

Since at both times, container is same . which means that volume remains constant. By applying ideal gas equation.

$PV=nRT$

or $\dfrac{P _{1}}{T _{1}}=\dfrac{P _{2}}{T _{2}}$

Substituting as per question,
$P _{1}=2 atm$ 
$p _{2}=2.5atm$
$T _{1}=0^\circ or \ 273k \ use \ S.I.unit $
$T _{2}=?$

$\dfrac{2}{273}=\dfrac{2.5}{T _{2}}$ or $T _{2}=\dfrac{2.5\times 273}{2}$

$=341.25K$           $(341.25-273=68.25)$
$=68.25^\circ C$

A copper ball of density $8.6 g cm^{-3}$, 1 cm is diameter is immersed in oil of density $0.8cm^{-3}$ . if the ball remains suspended in oil in a uniform electric field of intensity $36000 NC^{-1} $acting in upward direction, what is the charge on the ball ? 

  1. $1.1 \mu C$

  2. $4.2 \mu C$

  3. $2.4 \mu C$

  4. $3.7 \mu C$


Correct Option: A

A block of mass 20 kg and volume $ { 10 }^{ 3 }{ cm }^{ 3 }$ is suspended vertically from a ceiling by  a wire . The liner mass density of the its length  is 50 cm. The wire is vibrating in its  fundamental mode and producing beats with a tuning fork f frequency block is just completely immersed in a liquid and vibrated in its fundamental mode , it produces the same number  of beats with  earlier . Density of the liquid is $\left( g={ 10 }{ m/s }^{ 2 } \right) $

  1. $3.8 gm/{ cm }^{ 3 }$

  2. $7.6 gm/{ cm }^{ 3 }$

  3. $1.9 gm/{ cm }^{ 3 }$

  4. $5.0 gm/{ cm }^{ 3 }$


Correct Option: A

 A body of mass 20.00 g has volume $5.0 cm^3$. The maximum possible error in the measurement of mass and volume respectively are 0.01 g and$0.1 cm^3$. The maximum percentage error in the density will be nearest to 

  1. 1%

  2. 2%

  3. 11%

  4. 25%


Correct Option: B
Explanation:

$m = 20.00g$

$v = 5.0\,c{m^2}$
$d = \dfrac{m}{v}$

$ \Rightarrow \dfrac{{\Delta d}}{d} \times 100 = \dfrac{{\Delta m}}{m} \times 100 + \dfrac{{\Delta v}}{v} \times 100$

$ \Rightarrow \dfrac{{\Delta d}}{d} \times 100 = \dfrac{{0.01}}{{20}} \times 100 + \dfrac{{0.1}}{5} \times 100 = \dfrac{1}{{20}} + 2$

$ \Rightarrow \dfrac{{\Delta d}}{d} \times 100 = 2.005$

$\therefore \% $ error in density $ \approx 2\% $
Hence,
option $(B)$ is correct answer.

A metallic sphere with an internal cavity weighs 40 g weight in air and 20 g weight in water. If the density of the material with  cavity be 8 g per $c{m^3}$ then the volume of cavity is:

  1. zero

  2. 15 $c{m^3}$

  3. 5 $c{m^3}$

  4. 20 $c{m^3}$


Correct Option: B
Explanation:

$V _s=5cc$ As given

When it is placed in water net force$=$ new downward force
$\Rightarrow Hog-(V+V _s)g=20g\Rightarrow V+V _s=20\ \therefore V=20-V _s cc\V=20-5cm^2=15cm^2$

The density of mercury is $13600 kg m^{-3} $.Its  value in CGS system will be:

  1. $13.6 g cm^{-3}$

  2. $1360 g cm^{-3}$

  3. $136 g cm^{-3}$

  4. $1.36 g cm^{-3}$


Correct Option: A
Explanation:

The $C.G.S.$ unit of kg is Gm$.$ and that of meter is cm$.$ When we convert $13600\,kg{m^{ - 3}}$ then $13600 \times 1000\left( {100 \times 100 \times 100} \right) = 13.6\,gc{m^{ - 3}}$

Hence,
option $(A)$ is correct answer.

A solid ball of radius R has a charge density p given by $p=p _0(1 -r/R)$ for $ 0 \leq r \leq R.$ The electric field outside the ball is:

  1. $\dfrac{p _0R^3}{\epsilon _0r^2}$

  2. $\dfrac{p _0R^3}{12\epsilon _0r^2}$

  3. $\dfrac{4p _0R^3}{3\epsilon _0r^2}$

  4. $\dfrac{3p _0R^3}{4\epsilon _0r^2}$


Correct Option: B
Explanation:

$According\, to\, question..................... \ q=\int  _{ 0 }^{ R }{ pdv }=\int  _{ 0 }^{ R }{ po }\left( { 1-\dfrac { r }{ R }  } \right) \, .\, 4\pi { r^{ 2 } }drr \ \Rightarrow q=po.4\pi \, \left[ { \int  _{ 0 }^{ R }{ { r^{ 2 } }dr-\int  _{ 0 }^{ R }{ \dfrac { { { r^{ 3 } } } }{ R } dr } } } \right]  \ \Rightarrow q=po.4\pi \left( { \dfrac { { { R^{ 3 } } } }{ 3 } -\dfrac { { { R^{ 3 } } } }{ 4 }  } \right) =po\left( { 4\pi  } \right) \dfrac { { { R^{ 3 } } } }{ { 12 } }  \ \, \, \, \, \, \, \therefore \, \, \, q=\dfrac { { po.\pi { R^{ 3 } } } }{ 3 }  \ E=\dfrac { 1 }{ { 4\pi { \in _{ 0 } } } } \times \, \dfrac { q }{ { { r^{ 2 } } } } =\dfrac { 1 }{ { 4\pi { \in _{ 0 } } } } \, \times \dfrac { { po\, .\pi { R^{ 3 } } } }{ { 3{ r^{ 2 } } } } =\dfrac { { po\, .{ R^{ 3 } } } }{ { 12{ \in _{ 0 } }{ r^{ 2 } } } }  \ \, \, \, \therefore \, \, \, \, E=\dfrac { { po\, .{ R^{ 3 } } } }{ { 12{ \in _{ 0 } }{ r^{ 2 } } } }  \ there\, for\, the\, correct\, option\, is\, B.$

A body of uniform cross-sectional area floats in a liquid of dentisty thrice its value. The portion of exposed height will be:

  1. 2/3

  2. 5/6

  3. 1/6

  4. 1/3


Correct Option: A

The surface density of charge on a large flat metal plate is $2.25 x { 10 }^{ 9 } coul/{m}^{2}$. The intensity of electric field at a point at a distance $20 cm$ from the plate is

  1. $90\pi N/C$

  2. $180\pi N/C$

  3. $45\pi N/C$

  4. $81\pi N/C$


Correct Option: A

A cylindrical tree has a breaking stress of $10^6\ N/m^2$. The maximum possible height of the tree is $5\ m$. the density of material of the tree is (take $g = 10\ m/s^2$)

  1. $10^3\ kg/m^3$

  2. $10^4\ kg/m^3$

  3. $2 \times 10^4\ kg/m^3$

  4. $1\ kg/m^3$


Correct Option: C

A block of ice floats on a liquid of density $1.2$ in a beaker then level of liquid when ice completely melt.

  1. Remains same

  2. Rises

  3. Lowers

  4. Either (b) or (c)


Correct Option: B
Explanation:

Rises.

The level of the liquid in the beaker will rise. This is because the density of the water formed by the melting of ice is less than that of the density of liquid in the beaker 

An orifice of radius $r$ is present at the bottom of a container $r$. The maximum height to which the container can be filled with a liquid of density $\rho$ for which liquid will not come out of the orifice is?

  1. $\dfrac {T}{r\rho g}$

  2. $\dfrac {3T}{r\rho g}$

  3. $\dfrac {2T}{r\rho g}$

  4. $\dfrac {4T}{r\rho g}$


Correct Option: A

A rectangular block 5m x 4m x 2m lies on a table with its largest surface in contact with the table. The work done to keep it so that the block rests on the smallest surface is, if its density is 600 k $m ^ { - 3 }$

  1. $352800 \mathrm { J }$

  2. zero

  3. $376000 \mathrm { J }$

  4. $24,0000 \mathrm { J }$


Correct Option: A
Explanation:

The volume of Rectangular block $ = 5m \times 4m \times 2m$

$ = 40\,\,{m^3}$
Mass = Volume $ \times$ Density
$\begin{array}{l} =40\, \, { m^{ 3 } }\times 600\, \, k{ m^{ -3 } } \ =24,000=2.4\times { 10^{ 4 } }\, \, hg \end{array}$
Potential energy of block when it is lying on its largest surface $= mgh$
$ = 2.4 \times {10^4} \times 9.8 \times 1\,\,joule$
Potential energy of the block when it is lying on 
$ = mgh = 2.4 \times {10^4} \times 9.8 \times 2.5$
Work done = difference in potential energy
$\begin{array}{l} =\left( { 2.4\times { { 10 }^{ 4 } }\times 9.8\times 2.5-2.4\times { { 10 }^{ 4 } }\times 9.8\times 1 } \right) joule \ =2.4\times { 10^{ 4 } }\times 9.8\times 1.5\, \, joule \ =352800\, \, joule \end{array}$
Option A

When two liquid of same volume but different densities $\rho _{1}$ and $\rho _{2}$ are mixed together, then the density of the mixture is

  1. $\dfrac {p _{1}+p _{2}}{2}$

  2. $p _{1}+p _{2}$

  3. $\dfrac {2p _{1}p _{2}}{p _{1}+p _{2}}$

  4. $2p _{1}+2p _{2}$


Correct Option: A
Explanation:

$Density \,\,of \,\,mixture = \dfrac{Total\,\, mass }{ Total \,\,volume}$

Let $1^{st}$ liquid have mass $M _1$, density $p _1$ and Volume $V$

and

$2^{st}$ liquid have mass $M _2$, density $p _2$ and Volume $V$

So,Density of mixture  $= \dfrac{M _1 +M _2}{2V}$$=$$\dfrac {p _1V+p _2V}{2V}$

                                 $\rho _m=\dfrac{p _1+p _2}{2}$

If the length of the cylinder is measured as 25 mm, the diameter is 3.09cm and mass of the cylinder is measured as 50.0 gm find the density of the cylinder in proper significant figures.(in $gm/cm^3$)

  1. 2.700 $gm/cm^3$

  2. 2.7 $gm/cm^3$

  3. 0.27 $gm/cm^3$

  4. 2.70$gm/cm^3$


Correct Option: B
Explanation:

$p\, =\, \displaystyle \frac{m}{\pi (d^2\, /\, 4)h}$
$p\, =\, \displaystyle \frac{(50.0) gm}{3.14\, \times\, (3.09/2)^2\, \times\, (25\, \times\, 10^{-1}) cm^3}\, =\, 2.7\, gm/cm^3$
(in two S.F.)

The greatest length of the wire made of material of breaking stress $ 8 \times 10^8 N/m^2 $ and density $ 8 \times 10^3kg m^3 $ that can be suspended from a rigid support without breaking is $ ( g= 10m/s^2) $ 

  1. $1 km$

  2. $0.1 km$

  3. $10 km$

  4. $20$


Correct Option: C
Explanation:

$ stress= length \times desity \times g $
$ length = \dfrac {stress}{g \times density} $
$= \dfrac { 8 \times 10^8}{10 \times 8 \times 10^3 } $


$=10km$

A tank  $5 { m }$  high is half filled with water and then is filled to the top with oil of density  $0.85 g / c m ^ { 3 } .$  The pressure at the bottom of the tank, due to these liquids is

  1. $1.85{ g }/{ cm }^{ { 2 } }$

  2. $89.25{ g }/{ cm }^{ { 2 } }$

  3. $462.5{ g }/{ cm }^{ { 2 } }$

  4. $500{ g }/{ cm }^{ { 2 } }$


Correct Option: C
Explanation:
Given,

Total height of tank $=5\,m$

Density of water, $\rho _1=1\,g/cm^3$

Density of oil, $\rho _2=0.85\,g/cm^3$

We know that,

Half tank is filled with water and half tank is filled with oil.

So,  $h _1=h _2=\dfrac 52=2.5\,m=250\,cm$

Pressure at the bottom of the tank is

$P=h _1\rho _1g+h _2\rho _2g$

$P=g(250\times 1+250\times 0.85)$

$P=462.5\,g/cm^2$

Suppose that a student finds that $24 \ mL$ of a certain liquid weighs $36 \ g$. What is the density of this liquid in SI unit?

  1. $1500  \ kg m^{-3}$

  2. $1200  \ kg m^{-3}$

  3. $1550 \  kg m^{-3}$

  4. $1300  \ kg m^{-3}$


Correct Option: A
Explanation:

Given, mass of liquid, $m=36g=36\times 10^{-3}kg$;
density of liquid, $\rho=?$
Volume of liquid, $V=24mL=24\times 10^{-3}L$
$=24\times 10^{-3}\times 10^{-3}m^3$
$=24\times 10^{-6}m^3$
Always remember, $1L=10^{-3}m^3=10^3 cm^3$,
Also, $1mL=1cm^3$(also called $1 \ cc$)
Density, $\rho=\displaystyle \frac{Mass}{Volume}=\frac{m}{V}=\frac{36\times 10^{-3}}{24\times 10^{-6}}$
$=1500kg m^{-3}$

Brine has a density of 1.2 g/cc. 40 cc of it is mixed with 30 cc of water. The density of the solution is:

  1. 2.11 g/cc

  2. 1.11 g/cc

  3. 12.2 g/cc

  4. 20.4 g/cc


Correct Option: B
Explanation:

${ \rho  } _{ mixture }=\dfrac { { \rho  } _{ 1 }{ V } _{ 1 }+{ \rho  } _{ 2 }{ V } _{ 2 } }{ { V } _{ 1 }+{ V } _{ 2 } } $


               $=\dfrac { 1.2\times 40+1\times 30 }{ 70 } =\dfrac { 78 }{ 70 } $


${ \rho  } _{ mixture }=1.11gm/cc$

The densities of three liquids are D, 2D and 3D. What will be the density of the resulting mixture if equal volumes of the three liquids are mixed? 

  1. 6D

  2. 1.4D

  3. 2D

  4. 3D


Correct Option: C
Explanation:

let $V$ be the volume of each liquid then the total volume of the mixture becomes $3V$. 

$\rho = \dfrac{total\ mass}{total\ volume}$

Therefore, the mass of the liquids can be written as:
$m _1=D\times V=DV$
$m _1=2D\times V=2DV$
$m _1=3D\times V=3DV$

the total mass of the liquids is
$M= DV+2DV+3DV=6DV$

Therefore, the density of the mixture is:
$\rho=\dfrac{6DV}{3V}$

$\rho = 2D$

One litre of cool air weighs heavier than one litre of hot air because of the : 

  1. Increased number of collisions between . the molecules

  2. Increased number of molecules at. low temperature

  3. Greater energy of molecules at high temperature

  4. Lower energy of molecules at high temperature


Correct Option: B
Explanation:

Their exists less no. of molecules at the high temperature and high no. of molecules at the low temperature i.e. in the cold conditions.

$60  cc$ of a liquid of relative density $1.4$ are mixed with $40  cc$ of another liquid of relative density $0.8$. The density of the mixture is

  1. $1.16 {g}/{cc}$

  2. $2.26 {g}/{cc}$

  3. $11.6 {g}/{cc}$

  4. $116 {g}/{cc}$


Correct Option: A
Explanation:
given : - ${ S } _{ 1 }=1.4\quad \quad \quad \quad { V } _{ 1 }=60cc$
               ${ S } _{ 2 }=0.89/cc\quad \quad { V } _{ 2 }=40cc$


The density of the mixture is given by:
$S _{mixture }= \dfrac { { S } _{ 1 }{ V } _{ 1 }+{ S } _{ 2 }{ V } _{ 2 } }{ { V } _{ 1 }+{ V } _{ 2 } } $

$S _{mixture} = \dfrac { 1.4\times 60+0.8\times 40 }{ 100 } $

$S _{mixture} = 1.16 g/cc$

If two masses A and B have their masses in the ratio 1 : 4 and their volumes are equal, then their densities have the ratio :

  1. 1 : 4

  2. 4 : 1

  3. 2 : 1

  4. 3 : 1


Correct Option: A
Explanation:
We know that density is defined as:
$\rho=\dfrac{mass}{volume}$

It means that density is directly proportional to the mass.
$\dfrac{d}{d'}=\dfrac{m}{m'}=\dfrac14$

Since, $m:m'=1:4$

$\therefore, d:d'=1:4$

A wire of length 50 cm has a mass of 20 g. If its radius is halved by stretching, its new mass per unit length will be :

  1. 0.4 g $cm^{-1}$

  2. 0.2 kg $m^{-1}$

  3. 0.1 g $cm^{-1}$

  4. 0.2 g $cm^{-1}$


Correct Option: C
Explanation:

Mass, $m=\rho V=\rho (\pi r^2 l)$
where $\rho=$ density of the material of the wire, $r=$ radius of wire and $l=$ length of wire. 

When its radius is halved by stretching, the new mass will be ,
 $m'=\rho \pi (r/2)^2l'$

Thus, $\dfrac{m'}{m}=\dfrac{l'}{4l}$ or $\dfrac{m'}{l'}=\dfrac{m}{4l}$

Hence, the new mass per unit length $=m'/l'=m/4l=\dfrac{20}{4\times 50}=0.1 $ $g$ $cm^{-1}$

Ventilators are provided at the top of room

  1. To bring oxygen for breathing

  2. So that sunlight may enter the room

  3. To maintain convectional currents to keep the air fresh in the room

  4. To provide an outlet for carbon dioxide


Correct Option: C
Explanation:

Ventilators are provided in the rooms at the top of the roofs because if the air inside the room gets hot, the hot air rises up and flows through these ventilators and thus cool air remains at bottom. It brings cool and fresh air in the room. Thus ventilators maintain conventional currents to keep the air fresh in the room.

A vessel contains a mixture consisting of ${m} _{1}=7kg$ of nitrogen $\left( { M } _{ 1 }=28 \right) $ and ${m} _{2}=11g$ of carbon dioixide $\left( { M } _{ 2 }=44 \right) $ at temeprature $T=300K$ and pressure ${ P } _{ 0 }=1\quad atm$. The density of the mixture is:

  1. $1.446g$ per litres

  2. $2.567g$ per litre

  3. $3.752g$ per litre

  4. $4.572g$ per litre


Correct Option: A
Explanation:

Let V is the volume of the vessel.

Now, let $p _{1}$ and $p _{2}$ be the partial pressure, then using gas law: 

$p _{1}V = \dfrac{m _1}{M _1}RT\\$

$p _{2}V = \dfrac{m _2}{M _2}RT,\ p _{0}  = p _{1} +  p _{2}\\$

$p _{0} = \left(\dfrac{m _1}{M _1} + \dfrac{m _2}{M _2}\right)\dfrac{RT}{V}\\$

$V = \left(\dfrac{m _1}{M _1} + \dfrac{m _2}{M _2}\right)\dfrac{RT}{p _{0}}\\$

$\because \rho _{mix}=\dfrac{(m _{1} + m _{2})}{V}\\$

$rho _{mix}=\dfrac {(m _1 + m _2)M _1 M _2} {(m _1M _2 + m _2M _1)} \times \dfrac{p _0}{RT}\\$

Substituting values,

$\rho _{mix}=\dfrac {(7 + 11) \times 28 \times 44\times 10^{-3}} {(7 \times 44 + 11\times 28))} \times \dfrac{10^{5}}{8.3 \times 300}\\$

$= 1.446 \ per \ litre$

Option A is correct.

Two non-mixing liquids of densities $\rho$ and $n \rho ( n >$ 1) are put in a container. The height of each liquids $h$ . A solid cylinder of length $L$ and density $d$ is put in this container. The cylinder floats with its its axis vertical and length $p L ( p < 1 )$ in the denser liquid. The density $d$ is equal to

  1. ${ 1 + ( n - 1 ) p } p$

  2. ${ 1 + ( n + 1 ) p } p$

  3. ${ 2 + ( n + 1 ) p } p$

  4. ${ 2 - ( n + 1 ) p } p$


Correct Option: A
Explanation:

$\begin{array}{l} d=density\, \, of\, \, cylin{ { de } }r \ A=area\, \, of\, \, cross-sectional\, \, of\, \, cylinder \ U\sin  g\, \, law\, \, of\, \, floation, \ weight\, \, of\, \, cylinder=up\, thrust\, \, by\, \, two\, \, liquids \ L\times A\times d\times g \ =n\rho \times \left( { pL\times A } \right) g+\rho \left( { L-pL } \right) Ag \ d=np\rho +\rho \left( { 1-p } \right) =\left( { np+1-p } \right) \rho  \ d=\left{ { 1+\left( { n-1 } \right) p } \right} \rho  \end{array}$

Hence,
option $(A)$ is correct answer.

Calculate the mass of air enclosed in a room of length, breadth and height equal to $5m,3m$ and $4m$ respectively.
Density of air $=1.3kg/{m}^{3}$.
  1. 60 kg

  2. 78 kg

  3. 18 kg

  4. 10 kg


Correct Option: B
Explanation:

The density of a body is given by:

$density=\dfrac{mass}{volume}$

$1.3=\dfrac{mass}{(5 \times 3 \times 4)}$


$mass=1.3\times 60=78kg$

Density of solid sphere is varied by $\rho = \rho _0 \lgroup 1 + \frac{r}{R} \rgroup$ where $0 \leq r \leq R$, R is the radius of the sphere. Moment of inertia of sphere w.r.t. axis passing through its centre will be : ($\rho _0$ is constant)

  1. $\dfrac{44}{45} \pi \rho _0 R^5$

  2. $\dfrac{44}{45} \pi \rho _0 R^4$

  3. $\dfrac{44}{35} \pi \rho _0 R^5$

  4. $\dfrac{48}{45} \pi \rho _0 R^5$


Correct Option: A
Explanation:

$\begin{array}{l} dM=s\left( { 4\pi { x^{ 2 } }dx } \right)  \ ={ \rho _{ 0 } }\left( { 1+\dfrac { x }{ R }  } \right) \left( { 4\pi { x^{ 2 } }dx } \right)  \ dI=\dfrac { 2 }{ 3 } dM{ x^{ 2 } } \ \dfrac { 2 }{ 3 } { \rho _{ o } }\int _{ 0 }^{ R }{ \left( { 4\pi { x^{ 4 } }dx+\dfrac { { 4\pi  } }{ R } { x^{ 5 } }dx } \right)  }  \ =\dfrac { { 8\pi { \rho _{ 0 } } } }{ 3 } \left[ { \dfrac { { { x^{ 5 } } } }{ 5 } +\dfrac { { { R^{ 5 } } } }{ 6 }  } \right]  \ =\dfrac { 8 }{ 3 } \pi { \rho _{ o } }\times \dfrac { { 11{ R^{ 5 } } } }{ { 30 } }  \ =\dfrac { { 44 } }{ { 45 } } \pi { \rho _{ 0 } }{ R^{ 5 } } \end{array}$

Hence,
option $(A)$ is correct answer.

The blades of a windmill sweep out a circle of area $A$. If the wind flows at a velocity $v$ perpendicular to the circle, then the mass of the air of density $\rho$ passing through it in time $t$ is:

  1. $Av\rho t$

  2. $2Av\rho t$

  3. $Av^{2}\rho t$

  4. $\dfrac {1}{2}Av\rho t$


Correct Option: A
Explanation:

Volume of wind flowing per second $= Av$
Mass of wind flowing per second $= Av\rho$
Mass of air passing in time $t\ s = Av\rho t$.

The clouds float in the atmosphere because of their low.

  1. Pressure

  2. Velocity

  3. Temperature

  4. Density


Correct Option: D
Explanation:
As the rule of thumb, you can assume that the things with less density float on things with higher density. And water do have less density than air that's the reason clouds float in the atmosphere.

Sea water at frequency $\nu \  =\  4\  x\  { 10 }^{ 8 }$ Hz has permittivity $\varepsilon  \  \approx \  80\  { \varepsilon   } _{ 0 }$, permeability $\mu \  \approx \  { \mu  } _{ 0 }$ and resistivity $\rho \  =\  0.25\  \Omega m$. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source V(t) = ${ V } _{ 0 }\  \sin { \  (2\pi \nu t) }$. The of amplitude of the displacement current density to the conduction current density is

  1. $\dfrac { 2 }{ 3 }$

  2. $\dfrac { 4 }{ 9 }$

  3. $\dfrac { 9 }{ 4 }$

  4. 2


Correct Option: B
Explanation:

Suppose distance between the parallel plates is $D$ and applied voltage $V _{(t)} = V _02\pi vt$.

thus electric field
$E = \dfrac{V _0}{d} \sin (2\pi vt)$
Now using Ohm's law 
$J _c = \dfrac{1}{\phi} \dfrac{V _0}{d}\sin (2\pi vt)$

$\dfrac{V _0}{\phi d}\sin  (2 \pi vt) = J _0^c \sin  2 \pi vt$

Here $J _0^c = \dfrac{V _0}{pd}$
Now the displacement current density is given as
$Jd = \in \dfrac{\delta E}{dt} =\dfrac{\in \delta}{dt}$    $\left[\dfrac{V _0}{dt} \sin (2\pi vt)\right]$

$= \dfrac{\in 2\pi v V _0}{d} \cos (2\pi vt)$

$\Rightarrow = J^d _0 \cos (2\pi vt)$

Where $J _0^d = \dfrac{2\pi V\in V _0}{d}$

$\Rightarrow \dfrac{J^d _0}{J^c _0} = \dfrac{2\pi v \in V _0}{d}. \dfrac{pd}{V _0} = 2\pi v \in \rho$

$= 2\pi \times 80\in _0v\times 0.25 = 4\pi \in _0v \times 10$ 

$= \dfrac{10v}{9\times 10^9} = \dfrac{4}{9}$

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