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NCERT Exemplar Class 12 Physics Solutions Chapter 2 Electrostatic Potential and Capacitance

NCERT Exemplar Class 12 Physics Solutions Chapter 2 Electrostatic Potential and Capacitance

Edited By Safeer PP | Updated on Sep 14, 2022 09:53 AM IST | #CBSE Class 12th

NCERT Exemplar Class 12 Physics solutions chapter 2 revolves around the charges, their electrostatic field along with potential and capacitance. NCERT Exemplar Class 12 Physics chapter 2 solutions deal with a variety of topics relating to the electrostatic field of a static charge, and the work is done in moving such charge from one point to another in the presence of the electrostatic field of one or other charge and the used electrostatic potential energy in doing so. Class 12 Physics NCERT Exemplar solutions chapter 2 also provides different questions for finding and understanding the potential due to a number of charges as well as electric dipole and also establishes the relationship between the electrostatic field and potential. Students preparing for Class 12 examinations must make use of NCERT Exemplar Class 12 Physics solutions chapter 2 PDF download that have been prepared by experts. These solutions help the students in better understanding of concepts of NCERT and the best way to solve and answer the questions.

Also check - NCERT Solutions for Class 12 Physics

Question:1

A capacitor of 4\mu F is connected as shown in the circuit. The internal resistance of the battery is 0.5\Omega. The amount of charge on the capacitor plates will be

a) 0
b) 4\mu C
c) 16\; \mu C
d) 8\; \mu C


Answer:

The answer is the option (d) 8\mu C
Current through the 2\Omega resistance considering internal resistance of the battery 1\Omega,
I=\frac{2.5V}{\left ( 2\Omega +0.5\Omega \right )}=1A
The voltage across the internal resistance of the battery
=\left ( 0.5\Omega \right )\left ( 1A \right )=0.5V
The voltage across the 4\mu F capacitor
2.5V-0.5V=2V
Therefore, charge on the capacitor plates
, Q=CV=\left ( 4\mu F \right )\left ( 2V \right )=8\mu C

Question:2

A positively charged particle is released from rest in a uniform electric field. The electric potential energy of the charge
a) remains a constant because the electric field is uniform
b) increases because the charge moves along the electric field
c) decreases because the charge moves along the electric field
d) decreases because the charge moves opposite to the electric field

Answer:

The answer is the option (c) decreases because the charge moves along the electric field

Question:3

Figure shows some equipotential lines distributed in space. A charged object is moved from point A to point B.

a) the work is done in fig (ii) is the least
b) the work is done in fig (i) is the greatest
c) the work is done in fig (iii) is greater than fig (ii) but equal to that in fig (i)
d) the work done is the same in fig (i), fig (ii), and fig (iii)

Answer:

The answer is the option (d) the work done is the same in fig (i), fig (ii), and fig (iii).

Question:4

The electrostatic potential on the surface of a charged conducting sphere is 100V. Two statements are made in this regard:
S1: At any point inside the sphere, the electric intensity is zero
S2: At any point inside the sphere, the electrostatic potential is 100V
Which of the following is a correct statement?
a) S1 and S2, both are false
b) S1 is true, S2 is false
c) S1 and S2, both are true but each of the statements is independent
d) S1 and S2, both are true and S1 is the cause of S2

Answer:

The answer is the option (d) S1 and S2, both are true, and S1 is the cause of S2

Question:5

Equipotential at a great distance from a collection of charges whose total sum is not zero are approximately
a) Ellipsoids
b) Spheres
c) Paraboloids
d) planes

Answer:

The correct answer is (b) spheres

Question:7

Consider a uniform electric field in the \hat{z} direction. The potential is a constant
a) in all space
b) on the x-y plane for a given z
c) for any y for a given z
d) for any x for a given z

Answer:

The electric field in z-direction implies that equipotential surfaces are in x-y plane.
The answer is the option (b) on the x-y plane for a given z; (c) for any y for a given z; and (d) for any x for a given z.

Question:8

Equipotential surfaces
a) will always be equally spaced
b) are closer in regions of large electric fields compared to regions of lower electric fields
c) will be more crowded near regions of large charge densities
d) will be more crowded near sharp edges of a conductor

Answer:

The answer is the options,
b) are closer in regions of large electric fields compared to regions of lower electric fields
c) will be more crowded near sharp edges of a conductor
d) will be more crowded near regions of large charge densities

Question:9

The work is done to move a charge along an equipotential from A to B
a) is zero
b) cannot be defined as -\int_{A}^{B}E.dl
c) must be defined as -\int_{A}^{B}E.dl
d) can have a non-zero value

Answer:

The answer is the options
c) must be defined as
-\int_{A}^{B}E.dl if q=1 C
a) is zero

Question:10

In a region of constant potential
a) the electric field shall necessarily change if a charge is placed outside the region
b) The uniform electric field will be there
c) in the inside region, there can be no charge
d) zero electric fields will be there

Answer:

The correct answers are the options,
c) in the inside region there can be no charge
d) zero electric field will be there

Question:11

In the circuit shown in the figure, initially key K_{1} is closed and K_{2} is open. Then K_{1} is opened and K_{2} is closed. Then

a) charge on C_{1} gets redistributed i.e, V_{1}=V_{2}
b) charge on C_{1} gets redistributed i.e,\left ( Q_{1}+Q_{2} \right )=Q
c) charge on C_{1} gets redistributed such that Q_{1}=Q_{2}
d) charge on C_{1} gets redistributed such that C_{1}V_{1}+C_{2}V_{2}=C_{1}E

Answer:

The answer is the options,
a) charge on C_{1} gets redistributed such that V_{1}=V_{2}
d) charge on C_{1} gets redistributed such that \left ( Q_{1}+Q_{2} \right )=Q

Question:12

If a conductor has a potential V\neq 0 and there are no charges anywhere else outside, then
a) the charges must on the surface or inside itself
b) no charge will be in the body of the conductor
c) the charges must be only on the surface
d) the charges must be inside the surface

Answer:

The answer is the options,
a) the charges must on the surface or inside itself
b) no charge will be in the body of the conductor

Question:13

A parallel plate capacitor is connected to a battery as shown in the figure. Consider two situations:
A: Key K is kept closed and plates of capacitors are moved apart using the insulating handle
B: Key K is opened, and plates of capacitors are moved apart using the insulating handle
Choose the correct options


a) In A: V remains the same and hence Q changes
b) In A: Q remains the same but C changes
c) In B: Q remains the same and hence V changes
d) In B: V remains the same but C changes

Answer:

The answer is the options,
a) In A: V remains the same and hence Q changes
c) In B: Q remains the same and hence V changes

Question:14

Consider two conducting spheres of radius R_{1} and R_{2} with R_{1} > R_{2}. If the two are at the same potential, the larger sphere has more charge than the smaller sphere. State whether the charge density of the smaller sphere is more or less than that of the larger one.

Answer:

Here, \sigma _{1}R_{1}=\sigma _{2}R_{2}
Hence, \frac{\sigma _{1}}{\sigma _{2}}=\frac{R_{2}}{R_{1}}
If we consider R_{2}>R_{1}, therefore, \sigma _{1}>\sigma _{2}
Therefore, it is clear from the above statement that the charge density of the smaller sphere is greater than the charge density of the larger sphere.

Question:15

Do free electrons travel to a region of higher potential or lower potential?

Answer:

We know that the charged particle in the electric field has a force, that can be expressed as F=qE
Here, the direction of the electrostatic force experienced by the free electrons is exactly in the opposite direction of the electric field. The electrons travel from a lower potential region to a higher potential because the direction of the electric field is higher than the potential.

Question:16

Can there be a potential difference between two adjacent conductors carrying the same charge?

Answer:

Yes, it is possible and there may be a potential difference between two same charges carrying adjacent conductors. This is because the sizes of the conductors might be different

Question:17

Can the potential function have a maximum or minimum in free space?

Answer:

Because the atmosphere around the conductor that prevents the electric discharge is absent, therefore, the potential function cannot be maximum or minimum in free space.

Question:19

Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.

Answer:

In a closed equipotential surface, the potential changes from position to position.
The potential inside the surface is different from the potential gradient caused in the surface that is \frac{dV}{dr}
This also means that the electric field is not equal to zero and it is given as E=\frac{-dV}{dr}
Therefore, it could be said that the field lines are either pointing inwards or outwards the surface.
So, it can be said that the field lines originate from the charges inside which contradicts the original assumption. Therefore, the volume inside the surface must be equipotential.

Question:21

Prove that, if an insulated, uncharged conductor is placed near a charged conductor and no other conductors are present, the uncharged body must be intermediate in potential between that of the charged body and that of infinity.

Answer:

The electric potential will decrease in the direction of the electric field. The potential
E=\frac{dV}{dr}
The electric potential will also decrease again when the path from the charged conductor is taken to the uncharged conductor in the direction of the electric field.
This process will continue when another uncharged conductor is considered to the infinity lowering the potential even further.
As a result, the uncharged body is at intermediate potential and the charged body is at infinity potential.

Question:22

Calculate the potential energy of a point charge -q placed along the axis due to charge +Q uniformly distributed along a ring of radius R. Sketch PE as a function of axial distance z from the center of the ring. Looking at the graph, can you see what would happen if -q is displaced slightly from the center of the ring?

Answer:


The potential energy of a point charge q is U and this point is placed at potential V, U=qV
A negatively charged particle is placed at the axis of the ring with charge Q
Let a be the radius of the ring
The electric potential at the axial distance is given as
V=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{\sqrt{z^{2}+a^{2}}}
The potential energy, U is given as
U=\frac{1}{4\pi \epsilon _{0}}\frac{Qq}{\sqrt{1+\left ( \frac{z}{a} \right )^{2}}}
When z=infinity, U=0.
When z=0, U is given as
U=\frac{1}{4\pi \epsilon _{0}}\frac{Qq}{a}

Question:23

Calculate potential on the axis of a ring due to charge Q uniformly distributed along the ring of radius R.

Answer:


Here, in the above image, point P is perpendicular to O and is at a distance of z from point O, which is the centre of the ring.
The charge dq is at a distance z from the point P.
Therefore, V can be written as:
V=\frac{1}{4\pi \epsilon _{0}}\int \frac{dq}{r}=\frac{1}{4\pi \epsilon _{0}}\int \frac{dq}{\sqrt{z^{2}+a^{2}}}
Therefore, the net potential will be :
V=\frac{1}{4\pi \epsilon _{0}}\frac{Q}{\sqrt{z^{2}+a^{2}}}

Question:24

Find the equation of the equipotential for an infinite cylinder of radius r_{0}, carrying charge of linear density \lambda.

Answer:


We know, the equation of the equipotential for an infinite cylinder of radius r_{0} with linear charge density \lambda is:

r=r_{0}e^{-2\pi \varepsilon _{0} \left [ V(r)-V(r_{0}) \right ]/\lambda}

Question:25

Two-point charges of magnitude \text {+q and -q} are placed at \left ( \frac{-d}{2},0,0 \right ) and \left ( \frac{d}{2},0,0 \right ) respectively. Find the equation of the equipotential surface where the potential is zero.

Answer:


Here, the potential due to charges at the point P is
V_{p}=\frac{1}{4\pi\varepsilon _{0}}\frac{q}{r_{1}}+\frac{1}{4\pi\varepsilon _{0}}\frac{(-q)}{r_{2}}
The net electric potential at this point is zero,
Therefore, r_{1}=r_{2}
We know that
r_{1}=\sqrt{\left ( \frac{x+d}{2} \right )^{2}+h^{2}}
r_{2}=\sqrt{\left ( \frac{x-d}{2} \right )^{2}+h^{2}}
Solving the two equations, we the required equation in-plane x = 0 which is a y-z plane.

Question:26

A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as \varepsilon =\alpha U where \alpha =2V-1. A similar capacitor with no dielectric is charged to U_{0}=78 V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

Answer:

Here, the capacitors are connected in parallel. Therefore, the voltage across the capacitors is the same. Let us assume the final voltage is U. Here, C is the capacitance of the capacitor without the dielectric. At this point, the charge is Q_{1}=CU
If the initial charge is Q_{0} given by Q_{0}=CU_{0}
The conversion of charges is
Q_{1}=Q_{1}+Q_{2}
CU_{0}=CU+\alpha CU_{2}
\alpha U_{2}+U-U_{o}=0
Solving the equation, we get U=6V

Question:27

A capacitor is made of two circular plates of radius R each, separated by a distance d < < R. The capacitor is connected to a constant voltage. A thin conducting disc of radius r < < R and thickness t < < r is placed at the center of the bottom plate. Find the minimum voltage required to lift the disc if the mass of the disc is m.

Answer:



The thin conducting disc of radius r and thickness t is placed at the center of the bottom plate. Therefore, the potential of that thin disc is equal to the potential of that plate.
Hence, the electric field E on the disc will be
E=\frac{V}{d}
Therefore, the charge transferred to the disc
q'=-\varepsilon _{0}\frac{V}{d\; \pi r^{2}}
Therefore, the force acting on the disc,
F=\varepsilon _{0}\frac{V^2}{d^2}\pi r^{2}
Therefore,
V=\sqrt{\frac{md^2g}{\pi \; \varepsilon _{0}r^{2}}}

Question:29

Two metal spheres, one of radius R and the other of radius 2R, both have the same surface charge density \sigma. They are brought in contact and separated. What will be the new surface charge densities on them?

Answer:

Let us say, the charge stored on the first and second metal spheres before contact is Q_{1} and Q_{2} respectively.
Therefore,
Q_{1}=\sigma .4\pi R^2
Q_{2}=\sigma .4\pi (2R)^2=4\; Q_{1}
Again assume, the charge stored on the first and second metal spheres are Q{_{1}}^{'} and Q{_{2}}^{'} respectively.
Therefore,
\left ( Q{_{1}}^{'}+Q{_{2}}^{'} \right )=\left ( Q{_{1}}+Q{_{2}} \right )=5\; Q_{1}
When the metal spheres are in contact, the following is the potentials acquired by them
Q{_{1}}^{'} = \frac{Q{_{2}}^{'}}{2}
By solving the above equations, we get,
\sigma 1=5\frac{\sigma }{3}
\sigma 2=5\frac{\sigma }{6}

Question:30

In the circuit shown in the figure, initially, K_{1} is closed and K_{2} is open. What are the charges on each capacitor? Then K_{1} was opened and K_{2} was closed. What will be the charge on each capacitor now?

Answer:


When key K_{1} is closed and key K_{2} is open, then capacitors C_{1} and C_{2} are connected in series with the battery.
Therefore, the charge stored in the capacitors C_{1} and C_{2} will be same as Q_{1}=Q_{2}
Therefore, Q_{1}=Q_{2}=q=\left ( \frac{C_{1}}{\left ( C_{1}+C_{2} \right )} \right )E=18\mu C

Assuming only capacitors C_{2} and C_{3} are placed in parallel,
C_{2}V'+ C_{3}V'=Q_{2}
V'=\frac{Q_{2}}{C_{2}+C_{3}}=3V
Therefore,
Q{_{2}}^{'}=3CV'=9\; \mu C
Q{_{3}}=3CV'= \mu C
Q{_{1}}^{'}=18\; \mu C

Question:31

Calculate potential on the axis of a disc of radius R due to a charge Q uniformly distributed on its surface.

Answer:


In the above figure, we can see that the disc is divided into several charged rings. Let P be the point on the axis of the disc at a distance x from the centre of the disc.
The radius of the ring is r and the width is dr. dq is the charge on the ring which is given as
dq=\sigma dA=\sigma 2\pi rdr
The potential is given as
dV=\frac{1}{4\pi\varepsilon _{0}}\frac{dq}{\sqrt{r^{2}+x^{2}}}=\frac{1}{4\pi \varepsilon _{0}}\frac{\left ( \sigma 2\pi rdr \right )}{\sqrt{r^{2}+x^{2}}}
The total potential at P is given as
\frac{Q}{2\pi \varepsilon _{0}R^2}\left ( \sqrt{R_{2}+x_{2}-x} \right )

Question:32

Two charges q_{1} and q_{2} are placed at (0, 0, d) and (0, 0, -d) respectively. Find the locus of points where the potential is zero.

Answer:


We know that the potential at point P is V=\sum Vi
Where Vi=\frac{qi}{4\pi \varepsilon _{0}}, ri is the magnitude of the position vector P
V=\frac{1}{4 \pi \varepsilon _{0}}\sum \frac{qi}{rpi}
When the (x,y,z) plane is considered, the two charges lie on the z-axis and are separated by 2d. The potential is given as
\frac{q_{1}}{\sqrt{x^{2}+y^{2}+\left ( z-d \right )^{2}}}+\frac{q_{2}}{\sqrt{x^{2}+y^{2}+\left ( z+d \right )^{2}}}=0
Squaring the equation, we get
x^{2}+y^{2}+z^{2}+\left [ \left ( \frac{q_{1}^2+q_2^2}{q_{1}^2-q_2^2} \right )^2 \right ]\left ( 2zd \right )+d^2=0
The Centre of the sphere is
\left ( 0,0,-d\left [ \frac{q_{1}^{2}+q_{2}^{2}}{q_{1}^{2}-q_{2}^{2}} \right ] \right )
And radius is
r=\frac{2q_{1}q_{2}d}{q_{1}^{2}-q_{2}^{2}}

Question:33

Two charges -q each are separated by distance 2d. A third charge +q is kept at midpoint O. Find potential energy of +q as a function of small distance x from O due to -q charges. Sketch PE versus x and convince yourself that the charge at O is in an unstable equilibrium.

Answer:


In the above figure, +q charge is at a point away from O towards \left ( -d,0 \right ).
This can be written as
U=q\left ( V_{1}+V_{2} \right )=q\frac{1}{4\pi\; \varepsilon _{0}}[\frac{-q}{\left ( d-x \right )}+\frac{-q}{d+x}]
U=\frac{1}{2\pi\varepsilon _{0}}\frac{-q^{2}d}{d^{2}-x^{2}}
At x=0;
U=\frac{1}{2\pi\varepsilon _{0}}\frac{q^{2}}{d}
If we differentiate both sides of the equation with respect to x, we get
\frac{dU}{dx}>0, when x<0
and \frac{dU}{dx}<0, when x>0
With the help of this two, we can assume that the charge on the particle to be
F=\frac{-dU}{dx}
Hence, F=\frac{-dU}{dx}=0
When
a) \frac{d^2U}{dx^{2}} = positive, equilibrium is stable
b) \frac{d^2U}{dx^{2}} = negative, equilibrium is unstable
c) \frac{d^2U}{dx^{2}}=0, equilibrium is neutral
Therefore, when
x=0,\frac{d^2U}{dx^{2}}=\left ( \frac{-2dq2}{4\pi\varepsilon _{0}} \right )\left ( \frac{1}{d^{6}} \right )\left ( 2d^2 \right )<0
Which shows that the system is an unstable equilibrium.

NCERT Exemplar Class 12 Physics Solutions Chapter 2 Includes The Following Topics

  • Introduction
  • Electrostatic Potential
  • Potential due to a point charge
  • Potential Due To An Electric Dipole
  • Potential Due To A System Of Charges
  • Equipotential Surfaces
  • Relation Between Field And Potential
  • Potential Energy Of A System Of Charges
  • Potential Energy In An External Field
  • Electrostatics Of Conductors
  • Dielectrics And Polarization
  • Capacitors And Capacitance
  • The Parallel Plate Capacitor
  • Effect Of Dielectric On Capacitance
  • Combination Of Capacitors
  • Energy Stored In A Capacitor
  • Van De Graff Generator

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What will students learn in NCERT Exemplar Class 12 Physics Solutions Chapter 2 Electrostatic Potential and Capacitance?

The students will be able to understand the concept of electric potential and how it is produced by a single charge, different charges, and also by electric dipole and also establishes its relationship between such potential and electric field. They will also learn about the charge storing device known as a capacitor, its working, how they are used in various devices present around us, its capacitance, and also about different types of dielectrics and their polarization. Class 12 Physics NCERT Exemplar solutions chapter 2 also gives students an insight towards the working of Van De Graff Generators which are used in laboratories for experimentation purposes.

NCERT Exemplar Class 12 Physics Chapter Wise Links

Important Topics To Cover For Exams From NCERT Exemplar Class 12 Physics Solutions Chapter 2 Electrostatic Potential and Capacitance

Here are the important topics that the students must specifically learn in Class 12 Physics chapter 2 Electrostatic Potential and Capacitance:

· The students will get to learn about equipotential surfaces, and the changes in potential energy of charges in different kinds of the external field with other charges present. The chapter also provides a complete representation of electrostatic fields around any conductor along with the concept of electrostatic shielding.

· NCERT Exemplar solutions for Class 12 Physics chapter 2 tells about capacitors, its making and how their capacitance can be affected due to change in the geographical configuration of capacitors along with the explanation for the parallel plate capacitor.

· NCERT exemplar Class 12 Physics chapter 2 solutions also includes questions related to the effect of dielectric on the capacitor and the total capacitance when various different capacitors are arranged in series and parallel and the energy stored in it and Van De Graff Generator

NCERT Exemplar Class 12 Solutions

Also, check the NCERT solutions of questions given in book

Also read NCERT Solution subject wise

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Frequently Asked Question (FAQs)

1. How are the questions solved in NCERT exemplar Class 12 Physics solutions chapter 2?

All these questions given in the chapter are solved exhaustively, with proper explanation, steps, diagrams and conclusion.

2. How many questions are solved?

  In all 33 questions from main exercise is solved in detail in the NCERT exemplar Class 12 Physics solutions chapter 2 Electrostatic Potential and Capacitance

3. Who has solved these questions?

Our very own highly experienced teachers of physics have solved these questions as per the CBSE pattern

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hello,

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I hope this was helpful!

Good Luck

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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