Yearly Archives: 2022

In This Post we are  providing Chapter- 5 MAGNETISM AND MATTER NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON MAGNETISM AND MATTER

1. A particle of mass m and charge q moving with a uniform speed  normal to a uniform magnetic field B describes a circular path of radius & Derive expressions for (1) Radius of the circular path (2) time period of revolution (3) Kinetic energy of the particle?
Ans. A particle of mass (m) and change (q) moving with velocity normal to  describes a circular path if
 


Since Time period of Revolution
During circular path = 
=> ()
=> T = 

Kinetic energy K.E = 
=> KE = 

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2. Write an expression for the force experienced by the charged particle moving in a uniform magnetic field B With the help of labeled diagram explain the working of cyclotron? Show that cyclotron frequency does not depend upon the speed of the particle?
Ans. Force experienced by the charged particle moving at right angles to uniform magnetic field  with velocity  is given by  = q () Initially Dee is negatively charged and Dee is positively charged so, the positive ion will get accelerated towards Dee since the magnetic field is uniform and acting at right angles to the plane of the Dees so the ion completes a circular path in when ions comes out into the gap, polarity of the Dee’s gets reversed used the ion is further accelerated towards Dee with greater speed and cover a bigger semicircular path. This process is separated time and again and the speed of the ion becomes faster till it reaches the periphery of the dees where it is brought out by means of a deflecting plate and is made to bombard the target.

Since F = qVBsin90⁰ provides the necessary centripetal force to the ion to cover a circular path so we can say 
=> r = 
Time period = 
V = 
 frequency is independent of velocity

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3. (a) Obtain an expression for the torque acting on a current carrying circular loop.
(b) What is the maximum torque on a galvanometer coil 5 cm 12 cm of 600 turns when carrying a current of 10^-5 A. in a field where flux density is?
Ans. ABCD is a rectangular loop of length (L), breadth (b) and area (A). Let I be the Current flowing in the anti clockwise direction. Let  be the angle between the normal to the loop and magnetic field 

Force acting on arm AB of the loop

Force on arm CD

Force on arm BC

Force on arm DA

Since are equal and opposite and also acts along the same line, hence they cancel each other.
are also equal and opposite but their line of action is different, so they form a couple and makes the rectangular loop rotate anti clockwise.
Thus = either force  distance





For loop of N turns



Where M is magnet
ic moment of the loop.

Torque will be maximum when  = 90^o
 

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4. The current sensitivity of a moving coil galvanometer increases by 20% when its resistance is increased by a factor of two. Calculate by what factor, the voltage sensitivity changes?
Ans. Current sensitivity 
Voltage sensitivity 
Resistance of a galvanometer increases when n and A are changed
Given  = 2R
Then n =  and A = 
New current sensitivity

New voltage sensitivity


From (i) and (iii)


n’A’=
Using equation (iv)



Thus voltage sensitivity decreases by a factor of .

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5. (a) Show how a moving coil galvanometer can be converted into an ammeter?
(b) A galvanometer has a resistance 30 and gives a full scale deflection for a current of 2mA. How much resistance in what way must be connected to convert into?
(1) An ammeter of range 0.3A
(2) A voltammeter of range 0.2V.
Ans. (a) A galvanometer can be converted into an ammeter by connecting a low resistance called shunt parallel to the galvanometer.
Since G and R_S are in parallel voltage across then is same 

 
(b) (1) I = 0.3A G = 30 Ig = 2mA = 
Sheent (S) = 

S = 0.2
(2) G = 30, Ig = 2mA =, V = 0.2V
Shunt Resistance (R) 

R = 70 

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6. A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.04 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me=). [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]
Ans. Energy of an electron beam, E= 18 keV =
Charge on an electron, e= 
E=
Magnetic field, B = 0.04 G
Mass of an electron, me= 
Distance up to which the electron beam travels, d = 30 cm = 0.3 m
We can write the kinetic energy of the electron beam as:




The electron beam deflects along a circular path of radius, r.
The force due to the magnetic field balances the centripetal force of the path.



Let the up and down deflection of the electron beam be 
Where,
θ= Angle of declination





Therefore, the up and down deflection of the beam is 3.9 mm.

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7. A sample of paramagnetic salt contains  atomic dipoles each of dipole moment . The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)
Ans. Number of atomic dipoles, n=
Dipole moment of each atomic dipole, M= 
When the magnetic field,  = 0.64 T
The sample is cooled to a temperature,  = 4.2°K
Total dipole moment of the atomic dipole, 
= 
= 
Magnetic saturation is achieved at 15%.
Hence, effective dipole moment, 
When the magnetic field,  = 0.98 T
Temperature,  = 2.8°K
Its total dipole moment =
According to Curie’s law, we have the ratio of two magnetic dipoles as:



Therefore, is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K.

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8. A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm below the cable?
Ans. Number of horizontal wires in the telephone cable, n= 4
Current in each wire,  = 1.0 A
Earth’s magnetic field at a location, H= 0.39 G =
Angle of dip at the location, 
Angle of declination, 
For a point 4 cm below the cable:
Distance, r = 4 cm = 0.04 m
The horizontal component of earth’s magnetic field can be written as:
 Where,
B= Magnetic field at 4 cm due to current I in the four wires

 = Permeability of free space = 
= 0 = 0.2 G
∴ 
= 
The vertical component of earth’s magnetic field is given as:
Hv= Hsin

The angle made by the field with its horizontal component is given as:


The resultant field at the point is given as:

s
For a point 4 cm above the cable:
Horizontal component of earth’s magnetic field:

= 0.39 cos 35° + 0.2 = 0.52 G
Vertical component of earth’s magnetic field:
 = 0.39

Angle, 
And resultant field:

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9. Answer the following questions:
(a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.
(b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy?
(c) ‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?’ Explain the meaning of this statement.
(d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer?
(e) A certain region of space is to be shielded from magnetic fields.
Suggest a method.
Ans. The hysteresis curve (B–H curve) of a ferromagnetic material is shown in the following figure.

(a) It can be observed from the given curve that magnetisation persists even when the external field is removed. This reflects the irreversibility of a ferromagnet.
(b) The dissipated heat energy is directly proportional to the area of a hysteresis loop. A carbon steel piece has a greater hysteresis curve area. Hence, it dissipates greater heat energy.
(c) The value of magnetisation is memory or record of hysteresis loop cycles of magnetisation. These bits of information correspond to the cycle of magnetisation. Hysteresis loops can be used for storing information.
(d) Ceramic is used for coating magnetic tapes in cassette players and for building memory stores in modern computers.
(e) A certain region of space can be shielded from magnetic fields if it is surrounded by soft iron rings. In such arrangements, the magnetic lines are drawn out of the region.

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10. Answer the following questions:
(a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?
(b) Why is diamagnetism, in contrast, almost independent of temperature?
(c) If a to roid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?
(d) Is the permeability of a ferromagnetic material
independent of the magnetic field? If not, is it more for lower or higher fields?
(e) Magnetic field lines are always nearly normal to the surface of a ferromagnetic at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point.) Why?
(f) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetization of a ferromagnet?
Ans. (a)Owing to therandom thermal motion of molecules, the alignments of dipoles get disrupted at high temperatures. On cooling, this disruption is reduced. Hence, a paramagnetic sample displays greater magnetisation when cooled.
(b) The induced dipole moment in a diamagnetic substance is always opposite to the magnetising field. Hence, the internal motion of the atoms (which is related to the temperature) does not affect the diamagnetism of a material.
(c) Bismuth is a diamagnetic substance. Hence, a toroid with a bismuth core has a magnetic field slightly greater than a toroid whose core is empty.
(d) The permeability of ferromagnetic materials is not independent of the applied magnetic field. It is greater for a lower field and vice versa.
(e) The permeability of a ferromagnetic material is not less than one. It is always greater than one. Hence, magnetic field lines are always nearly normal to the surface of such materials at every point.
(f) The maximum possible magnetisation of a paramagnetic sample can be of the same order of magnitude as the magnetisation of a ferromagnet. This requires high magnetising fields for saturation.

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In This Post we are  providing Chapter- 6 ELECTROMAGNETIC INDUCTION NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ELECTROMAGNETIC INDUCTION

1. IF the rate of change of current of 2A/s induces an emf of 1OmV in a solenoid. What is the self-inductance of the solenoid?
Ans.

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2. A circular copper disc. 10 cm in radius rotates at a speed of 2 rad/s about an axis through its centre and perpendicular to the disc. A uniform magnetic field of 0.2T acts perpendicular to the disc.
1) Calculate the potential difference developed between the axis of the disc and the rim.
2) What is the induced current if the resistant of the disc is 2?
Ans. (1) Radius = 10cm, B = 0.2T w = 2 rad/s




I = 0.0314 A

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3. An ideal inductor consumes no electric power in a.c. circuit. Explain?
Ans. P = E rms I rms cos 
But for an ideal inductor 

P=0

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4. Capacitor blocks d.c. why?
Ans. The capacitive reactance

For d.c.  = 0

Since capacitor offers infinite resistance to the flow of d.c. so d.c. cannot pass through the capacitor.

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5. Why is the emf zero, when maximum number of magnetic lines of force pass through the coil?
Ans. The magnetic flux will be maximum in the vertical position of the coil. But as the coil rotates 
Hence produced emf 

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6. An inductor L of reactance  is connected in series with a bulb B to an a.c. source as shown in the figure.

Briefly explain how does the brightness of the bulb change when
(a) Number of turns of the inductor is reduced.
(b) A capacitor of reactance is included in series in the same circuit.
Ans. (a) Since Z = 
When number of turns of the inductor gets reduced and Z decreases and in turn current increases
Hence the bulb will grow more brightly
(b) When capacitor is included in the circuit

But (given)
Z = R (minimum)
Hence brightness of the bulb will become maximum.

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7. A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location has a magnitude of  and the dip angle is 
Ans. Speed of the jet plane, v = 1800 km/h = 500 m/s
Wing span of jet plane, l = 25 m
Earth’s magnetic field strength, B = 
Angle of dip, s
Vertical component of Earth’s magnetic field,
BV = B sin
= 
= 
Voltage difference between the ends of the wing can be calculated as:

=
= 3.125 V
Hence, the voltage difference developed between the ends of the wings is
3.125 V.

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8. A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Ans. Mutual inductance of a pair of coils, µ = 1.5 H
Initial current,  = 0 A
Final current  = 20 A
Change in current, 
Time taken for the change, t = 0.5 s
Induced emf, 
Where is the change in the flux linkages with the coil.
Emf is related with mutual inductance as:
Equating equations (1) and (2), we get



Hence, the change in the flux linkage is 30 Wb.

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9. A horizontal straight wire 10 m long extending from east to west is falling with a speed of , at right angles to the horizontal component of the earth’s magnetic field, Wb .
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
Ans. Length of the wire, l = 10 m
Falling speed of the wire, v = 5.0 m/s
Magnetic field strength, B = 
(a) Emf induced in the wire,
e = Blv


(b) Using Fleming’s right hand rule, its can be inferred that the direction of the induced emf is from West to East.
(c) The eastern end of the wire is at a higher potential.

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10. A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.
Ans. Length of the rod, l = 1 m
Angular frequency, = 400 rad/s
Magnetic field strength, B = 0.5 T
One end of the rod has zero linear velocity, while the other end has a linear velocity of.
Average linear velocity of the rod,
Emf developed between the centre and the ring,


Hence, the emf developed between the centre and the ring is 100 V.

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In This Post we are  providing Chapter-1 ELECTRIC CHARGES AND FIELDS NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ELECTRIC CHARGES AND FIELDS

Question 1.
Using Gauss’ law deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius R at a point
(i) outside and
(ii) inside the shell.
Plot a graph showing variation of electric field as a function of r > R and r < R (r being the distance from the centre of the shell)
Answer:
(i) Field Outside Shell :
Consider a thin spherical shell of radius R with centre O. Let charge +q be distributed uniformly over the surface of shell. To calculate electric field intensity at P where OP = r, imagine a sphere S, with centre at O and radius r. The surface of sphere is Gaussian surface over at every point. Electric field is same and directed radially outwards.

Applying Gauss’ theorem

r→ is distance of point P from centre where E is calculated]
(ii) Inside Shell: As we know charge is located on its surface,

(iii) at r < R E→ is zero and r  = R, E is maximum at r > R, E is decreasing at E ∝ 1r2

Question 2.
Using Gauss’s law, derive the expression for the electric field at a point
(i) outside and
(ii) inside a uniformly charged thin spherical shell. Draw a graph showing electric field E as a function of distance from the centre.
Answer:
Electric field due to a uniformly charged spherical shell:
Suppose a thin spherical shell of radius R and centre O
Let the charge + q be distributed over the surface of sphere
Electric field intensity E→ is same at every point on the surface of sphere directed directly outwards
Let a point P be outside the shell with radius vector

Question 3.
(a) An electric dipole of dipole moment p→ consists of point charges + q and – q separated by a distance 2a apart. Deduce the expression for the electric field E→ due to the dipole at a distance x from the centre of the dipole on its axial line in terms of the dipole moment p→?. Hence show that in the limit x >> a, E→ —>2p→ (4πε₀x³).
(Delhi 2015)
Answer:
(a) Expression for magnetic field due to dipole on its axial lane:

(b)Only the faces perpendicular to the direction of x-axis, contribute to the Electric flux. The remaining faces of the cube given zero

Question 4.
(a) Define electric flux. Write its S.I. unit. “Gauss’s law in electrostatics is true for any closed surface, no matter what its shape or size is”. Justify this statement with the help of a suitable example.
(b) Use Gauss’s law to prove that the electric field inside a uniformly charged spherical shell is zero. (All India)
(a) Electric flux. The electric lines of force passing through that area, when held normally to the lines of force.
Answer:
Electric flux. The electric lines of force passing through that area, when held normally to the lines of force.

S.I. units: Vm, Nm²C^-1
Gauss’s Law states that the electric flux through a closed surface is given by

The law implies that the total electric flux through a closed surface depends on the quantity of total charge enclosed by the surface, and does not depend on its shape and size.

For example, net charge enclosed by the electric dipole (q, -q) is zero, hence the total elec¬tric flux enclosed by a surface containing electric dipole is zero.
(b) Electrical field inside a uniformly charged spherical shell. Let us consider a point ‘P’ inside the shell. The Gaussian surface is a sphere through P centred at O.

The flux through the Gaussian surface is E × 4πr².

However, in this case, the Gaussian surface encloses no charge. Gauss’s law then gives
E × 4πr² = 0
or E = 0
(r < R)
that is, the field due to a uniformly charged thin shell is zero at all points inside the shell.

Question 5.
(a) Derive the expression for the energy stored in a parallel plate capacitor. Hence obtain the expression for the energy density of the electric field.
(b) A fully charged parallel plate capacitor is connected across an uncharged identical capacitor. Show that the energy stored in the combination is less than that stored initially in the single capacitor. (All India 2015)
Answer:
(a) (i) Energy of a parallel plate capacitor. Consider a capacitor of capacitance C. Initial charge on plates is zero. Initial potential difference between the capacitor plates is zero. Let a charge Q be given to it in small steps. When charge is given to capacitor, the potential difference between its plates increases. Let at any instant when charge on capacitor be q, the potential difference between its plates be,

Now work done in giving an additional infinitesimal charge dq to capacitor

The total work done in giving charge from 0 to Q will be equal to the sum of all such infinitesimal works, which may be obtained by integration. Therefore total work


If V is the final potential difference between capacitor plates, then Q = CV

This work is stored as electrostatic potential energy of capacitor i.e., Electriostatic potential energy,

(ii) Expression for Energy Density.
Consider a parallel plate capacitor consisting of plates, each of area A, separated by a distance d. If space between the plates is filled with a medium of dielectric constant K, then
Capacitence of Capacitor,

If σ is the surface charge density of plates, then electric field strength between the plates.

This is the expression for electrostatic energy density in medium of dielectric constan K. In air of free space  (K = 1), therefore energy density,

(b) The energy of the capacitor when fully charged is

When this charged capacitor is connected to an identical capacitor C, then the charge will be distributed equally, q2 on each of the capacitors, then


Hence, the total energy stored is half of that stored initially in one capacitor which means the energy stored in combination is less than that stored initially in the single capacitor.

Question 6.
(i) Use Gauss’s law to find the electric field due to a uniformly charged infinite plane sheet. What is the direction of field for positive and negative charge densities?
(ii) Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors Cj and C2 with their capacitances in the ratio 1 : 2 so that the energy stored in the two cases becomes the same.
Answer:
(i)
(a) Electric flux: The electric flux through a given area held inside an electric field is the measure of the total number of electric lines of force passing normally through that area.

(b) Consider a thin, infinite plane sheet of charge with uniform surface charge density 0. We wish to calculate its electric field at a point P at distance r from it.

By symmetry, electric field E points outwards normal to the sheet. Also, it must have same magnitude and opposite direction at two points P anel F equidistant from the sheet and on opposite sides. We choose cylindrical Gaussian surface of cross¬sectional area A and length 2r with its axis perpendicular to the sheet.

As the lines of force are parallel to the curved surface of the cylinder, the flux through the curved surface is zero. The flux through the plane-end faces of the cylinder is :

(i) For positively charged sheet ➝ away from the sheet
(ii)For negatively charged sheet ➝towards the sheet

(b) and

Question 7.
(a) Derive an expression for the electric field E due to a dipole of length ‘2a’ at a point distant r from the centre of the dipole on the axial line. (b) Draw a graph of E versus r for r >> a.
(c) If this dipole were kept in a uniform external electric field diagrammatically represent the position of the dipole in stable and unstable equilibrium and write the expressions for the torque acting on the dipole in both the cases.
Answer:
(a) Expression for magnetic field due to dipole on its axial lane:

(b)Only the faces perpendicular to the direction of x-axis, contribute to the Electric flux. The remaining faces of the cube given zero

(b) Graph between E Vs r

(i) Diagrammatic representation
(ii) Torque acting on these cases
(i) In stable equilibrium, torque is zero (θ = 0)

(ii) In unstable equilibrium also, torque is zero (θ = 180°)

Question 8.
(a) Use Gauss’s theorem to find the electronic field due to a uniformly charged infinitely large plane thin sheet with surface charge density a.
(b) An infinitely large thin plane sheet has a uniform surface charge density +a. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point, distant r, in front of the charged plane sheet.

Answer:
(a) Electric flux: The electric flux through a given area held inside an electric field is the measure of the total number of electric lines of force passing normally through that area.

(b) Consider a thin, infinite plane sheet of charge with uniform surface charge density 0. We wish to calculate its electric field at a point P at distance r from it.

By symmetry, electric field E points outwards normal to the sheet. Also, it must have same magnitude and opposite direction at two points P anel F equidistant from the sheet and on opposite sides. We choose cylindrical Gaussian surface of cross¬sectional area A and length 2r with its axis perpendicular to the sheet.

As the lines of force are parallel to the curved surface of the cylinder, the flux through the curved surface is zero. The flux through the plane-end faces of the cylinder is :

(i) For positively charged sheet ➝ away from the sheet
(ii)For negatively charged sheet ➝towards the sheet

Question 9.
(a) State Gauss’ law. Using this law, obtain the expression for the electric field due to an infinitely long straight conductor of linear charge density X.
(b) A wire AB of length L has linear charge density λ = kx, where x is measured from the end A of the wire. This wire is enclosed by a Gaussian hollow surface. Find the expression for the electric flux through this surface.

Answer:
(a)
Gauss’s law in electrostatics : It states that “the total electric flux over the surface S in vaccum is 1ε0 times the total charge (q).”

Electric field due to an infinitely long straight wire : Consider an infinitely long straight line charge having linear charge density X to determine its electric field at distance r. Consider a cylindrical Gaussian surface of radius r and length l coaxial with the charge. By symmetry, the electric field E has same magnitude at each point of the curved surface Sj and is directed radially outward.
Total flux through the cylindrical surface,

(b) Given : Length of wire = L, Charge density (λ) = kx, ϕ = ?
We know

Question 10.
Two point charges 4 (J.C and +1 pC are separated by a distance of 2 m in air. Find the point on the line-joining charges at which the net electric field of the system is zero. (Comptt. Outside Delhi 2017)
Answer:
q₁ = 4 µC, q₂ = 1 µC, r = 2 m
At this point, the net electric field of the system is zero.

In This Post we are  providing Chapter-2 ELECTROSTATIC POTENTIAL AND CAPACITANCE NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ELECTROSTATIC POTENTIAL AND CAPACITANCE

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1. Explain what would happen if in the capacitor , a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates,
1. While the voltage supply remained connected.
2. After the supply was disconnected.
Ans. Dielectric constant of the mica sheet, k = 6
1. Initial capacitance, 
New capacitance, 
Supply voltage, V = 100 V
New charge s
Potential across the plates remains 100 V.
2. Dielectric constant, k = 6
Initial capacitance
New capacitance 
If supply voltage is removed, then there will be no effect on the amount of charge in the plates.
Charge = 
Potential across the plates is given by,

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2. A 12 pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?
Ans. Capacitor of the capacitance, 
Potential difference, V = 50 V
Electrostatic energy stored in the capacitor is given by the relation,















Therefore, the el
ectrostatic energy stored in the capacitor is 

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3. A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Ans. Capacitance of the capacitor, C = 600 pF Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,



If supply is disconnected from the capacitor and another capacitor of capacitance C = 600 pF is connected to it, then equivalent capacitance  of the combination is given by,



New electrostatic energy can be calculated as



Loss in electrostatic energy =



Therefore, the electrostatic energy lost in the process is.

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4. A spherical conducting shell of inner radius and outer radius has a charge
1. A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
2. Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.
Ans. (a) Charge placed at the centre of a shell is +q. Hence, a charge of magnitude –q will be induced to the inner surface of the shell. Therefore, total charge on the inner surface of the shell is –q.
Surface charge density at the inner surface of the shell is given by the relation,


A charge of +q is induced on the outer surface of the shell. A charge of magnitude Q is placed on the outer surface of the shell. Therefore, total charge on the outer surface of the shell is Q + q. Surface charge density at the outer surface of the shell,
1. Yes
The electric field intensity inside a cavity is zero, even if the shell is not spherical and has any irregular shape. Take a closed loop such that a part of it is inside the cavity
along a field line while the rest is inside the conductor. Net work done by the field in carrying a test charge over a closed loop is zero because the field inside the conductor is zero. Hence, electric field is zero, whatever is the shape.

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5. If one of the two electrons of a molecule is removed, we get a hydrogen molecular ion. In the ground state of an, the two protons are separated by roughly 1.5 , and the electron is roughly 1  from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
Ans.

The system of two protons and one electron is represented in the given figure.
Charge on proton 1, 
Charge on proton 2, 
Charge on electron, 
Distance between protons 1 and 2, 
Distance between proton 1 and electron, 
Distance between proton 2 and electron, 
The potential energy at infinity is zero.
Potential energy of the system,

Substituting 



Therefore, the potential energy of the system is .

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6. Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Ans .Let a be the radius of a sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal.
Let EAbe the electric field of sphere A and EB be the electric field of sphere B. Therefore, their ratio,


However 
And 

Putting the value of (2) in (1), we obtain

Therefore, the ratio of electric fields at the surface is.
 

7. What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realize from your answer why ordinary capacitors are in the range of  or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]
Ans. Capacitance of a parallel capacitor, V = 2 F
Distance between the two plates, d = 0.5 cm = Capacitance of a parallel plate capacitor is given by the relation,
Where,
 = Permittivity of free space = 

Hence, the area of the plates is too large. To avoid this situation, the capacitance is taken in the range of .

---

8.A spherical capacitor consists of two concentric spherical conductors, held in position by suitable insulating supports

Show that the capacitance of a spherical capacitor is given by
where  and  are the radii of outer and inner spheres, respectively.
Ans.Radius of the outer shell = 
Radius of the inner shell = 
The inner surface of the outer shell has charge +Q.

The outer surface of the inner shell has induced charge  Potential difference between the two shells is given by,
Where,
= Permittivity of free space


Capacitance of the given system is given by,

=
Hence, proved.

---

9. A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).
Ans.Length of a co-axial cylinder, l = 15 cm = 0.15 m
Radius of outer cylinder,  = 1.5 cm = 0.015 m
Radius of inner cylinder,  = 1.4 cm = 0.014 m
Charge on the inner cylinder, q = 3.5 
Capacitance of a co-axil cylinder of radii and is given by the relation,

Where,
 = Permittivity of free space = 


Potential difference of the inner cylinder is given by,

---

10. A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
Ans.Potential rating of a parallel plate capacitor, V = 1 kV = 1000 V
Dielectric constant of a material,  =3 Dielectric strength = 
For safety, the field intensity never exceeds 10% of the dielectric strength. Hence, electric field intensity, E = 10% of = 
Capacitance of the parallel plate capacitor, C = 50 pF =
Distance between the plates is given by,


Capacitance is given by the relation,

Where,
A = Area of each plate
= Permittivity of free space =


Hence, the area of each plate is about 19.Search

In This Post we are  providing Chapter- 3 CURRENT ELECTRICITY NCERT MOST IMPORTANT QUESTIONS for Class 12 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON CURRENT ELECTRICITY

Question 1.
A cell of emf ‘E’ and internal resistance ‘r’ is connected across a variable load resistor R. Draw the plots of the terminal voltage V versus (i) R and (ii) the current l.
It is found that when R = 4 Ω the current is 1 A and when R is increased to 9 Ω, the current reduces to 0.5 A. Find the values of the emf E and internal resistance r.
Answer:
The plots are as shown

Here l₁ = 1.0 A, R₁ =4 ohm, l₂ = 0.5 A, R₂ =9 ohm
Using the equation l = E(R+r) Or E = l(R + r)

we have
1.0 × (4 + r) = 0.5 × (9 + r)

Solving the above equation for r we have r = 1 ohm
Also E = 0.5 (9 + 1) = 5 V

Question 2.
A wire of resistance R, length l and area of cross-section A is cut into two parts, having their lengths in the ratio 1:2. The shorter wire is now stretched till its length becomes equal to that of the longer wire. If they are now connected in parallel, find the net resistance of the combination.
Answer:
Since the wires are cut in the ratio of 1:2 therefore,
Resistance of the shorter wire R₁ = R3 and

Resistance of the longer wire R₂ = 2R3

Since the shorter wire is stretched to make it equal to the longer wire therefore, it is stretched by n = 2 times its length. Hence New resistance of the shorter wire

Question 3.
In the figure, a long uniform potentiometer wire AB is having a constant potential gradient along its length. The null points for the two primary cells of EMFs ε₁ and ε₂ connected in the manner shown are obtained at a distance of 120 cm and 300 cm from the end A. Find (i) ε₁/ε₂ and (ii) position of null point for the cell ε₁. How is the sensitivity of a potentiometer increased?

Answer:
From the diagram we have
ε1−ε2ε1+ε2=120300=25
Or
5ε₁ – 5ε₂ = 2ε₁ + 2ε₂

Solving we have
ε1ε2=73 ….(1)

Also let L be the balancing length for cell of emf ε1, then
ε1ε1+ε2=L300

Using equation (1) we have
ε1ε1+37ε1=L300

Solving for L we have L = 210 cm
The sensitivity of a potentiometer can be increased by increasing the length of the potentiometer wire.

Question 4.
The network PQRS, shown in the circuit diagram, has batteries of 4 V and 5 V and negligible internal resistance. A milli- ammeter of 20 Ω resistance is connected between P and R. Calculate the reading in the milliammeter.

Answer:
Using Kirchhoff’s junction rule to distribute current we have

Consider the loop SRPS, by Kirchhoff’s loop rule we have
200 l₂ + 20 (l₁ + l₂ ) – 5 = 0 …(1)
Or
220 l₂ + 20 l₁ = 5 …(2)

Consider the loop PRQP, by Kirchhoff’s loop rule we have
– 60 l₁ + 4 – 20 (l₁ + l₂) = 0 …(3)
80 l₁ + 20 l₂ =4 …(4)

Multiplying equation (2) by (4) we have
880 l₂ + 80 l₁ = 20 …(5)

Subtracting equation (4) from equation (5)
we have
860 l₂ = 16 or l₂ = 4/215 A

Substituting in equation (4) we have
l₁ = 39860A

Therefore reading of the milliammeter is
l₁+ l₂ = 4215+39860 = 0.063 A = 63 mA

Question 5.
A set of ‘n’ identical resistors, each of resistance ‘R’ when connected in series have an effective resistance ‘X’. When they are connected in parallel, their effective resistance becomes ‘Y’. Find out the product of X and Y.

Answer:
In series
R_s = R₁ + R₂ + R₃ + ……
R_s = X = R + R + R + …. upto n
X=nR

In Parallel

Question 6.
In the following circuit, a metre bridge is shown in its balanced state. The metre bridge wire has a resistance of 1 ohm per centimetre. Calculate the value of the unknown resistance X and the current drawn from the battery of negligible internal resistance.

Answer:
Using the Wheatstone bridge principle we have
4060=X3
or X = 2 Ω

Now total resistance of the combination is
R = 5×1005+100=500105 = 4.76 Ω

Current drawn is
l = V/R = 6/4.76 = 1.26 A

Question 7.
Calculate the electrical conductivity of the material of a conductor of length 3 m, area of cross-section 0.02 mm² having a resistance of 2 ohms.
Answer:
Given L = 3 m,
A = 0.02 mm2 = 0.02 × 10-6 m2
R = 2 ohm.

Using the equation
R = ρLA
Or
ρ = RAL

σ = LAR = 30.02×10−6×2 = 7.5 × 107 Sm-1

Question 8.
A potential difference of 2 volts is applied between points A and B has shown in the network drawn in the figure. Calculate (i) equivalent resistance of the network across the points A and B and (ii) the magnitudes of currents in the arms AFCEB and AFDEB.

Answer:
The circuit can be redrawn as shown below.

As seen the circuit is a balanced Wheatstone bridge; therefore the resistance in the arm CD is superfluous.
(i) Resistance of arm FCE = 2 + 2 = 4 Ω
Resistance of arm FDE = 2 + 2 = 4 Ω
Hence net resistance of the circuit between A and B is
R = 4×44+4=168 = 2 Ω

(ii) current in the arm AFCEB
l = V/R = 2/4 = 0.5 A

Current in the arm AFDEB
l = V/R = 2/4 = 0.5 A

Question 9.
A cell of emf E and internal resistance ‘r’ gives a current of 0.8 A with an external resistor of 24 ohms and a current of 0.5 A with an external resistor of 40 ohms.
Calculate
(i) emf E and
(ii) internal resistance ‘r’ of the cell.
Answer:
Given l₁ = 0.8 A, R₁ = 24 ohm l₂ = 0.5 A, R₂ = 40 ohm
Using the equation
E = l(R + r) we have
0.8 × (24 + r) = 0.5 × (40 + r)

Solving for r we have r = 2.67 ohm
Also E = 0.5( 40 + 2.67) = 21.3 V

Question 10.
In the circuit diagram of the metre bridge given below, the balance point is found to be at 40 cm from A. The resistance of X is unknown and Y is 10 ohms.
(i) Calculate the value of X;
(ii) if the positions of X and Y are interchanged in the bridge, find the position of the new balance point from A; and
(iii) if the galvanometer and the cell are interchanged at the balance point, would the galvanometer show any current.

Answer:
(i) Using the Wheatstone bridge principle we have
4060=X10
Or
X = 6.67 Ω

(ii) If X and Y are interchanged then
L(100−L)=106.67 solving for L we have
L = 59.9cm

(iii) The galvanometer will not show any current.

In This Post we are  providing Chapter-16 CHEMISTRY IN EVERYDAY LIFE NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON CHEMISTRY IN EVERYDAY LIFE

Question 1.
Differentiate between disinfectants and antiseptics. (Delhi 2012)
Answer:

Antiseptics	Disinfectants
1. They are chemical substances which prevent the growth of micro-organisms and may even kill them.	1. They are chemical substances which kill micro-organisms.
2. They are safe to be applied to the living tissues.	2. They are not safe to be applied to the living tissues.
3. They are generally applied on wounds, cuts, ulcers and diseased skin surfaces. Example : Furacin, soframycin, dettol and savlon, 0.2% solution of phenol.	3. They are used to kill micro-organisms present in the drains, toilets, floors etc. Example: Phenol (> 1% solution) and chlorine (0.2 to 0.4 ppm).

Question 2.
What are limited spectrum antibiotics? Give one example. (Comptt. Delhi 2012)
Answer:
Those antibiotics which are specific for certain diseases are called limited spectrum antibiotics. Example: Streptomycin for tuberculosis.

Question 3.
Name the important by-products of soap industry. (Comptt. Delhi 2012)
Answer:
Glycerol is the important by-product of soap industry.

Question 4.
Why do we require artificial sweetening agents? (Comptt. All India 2012)
Answer:
To reduce calorie intake and to protect teeth from decaying, we need artificial sweetners.

Question 5.
What are food preservatives? Name two such substances. (All India 2012)
Answer:
Food preservatives : Food preservatives are the compounds which prevent spoilage of food due to microbial growth.
Two substances : Example : Sodium benzoate, vinegar.

Question 6.
Explain the cleaning action of soap. Why do soaps not work in hard water? (All India 2012)
Answer:
Cleaning action of soap : The cleansing action of soap is due to the fact that soap molecules form micelles around the oil droplets in such a way that hydrophobic part of stearate ions is in the oil droplet and hydrophilic part projects out of the grease droplet like the bristles. Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed from the dirty surface. Thus soap helps in emulsification and washing away of oils and fats.

Reason : Hard water contains calcium and magnesium ions. These ions form insoluble Ca and Mg salts. These salts act as scum. The insoluble scum sticks on the clothes, so the cleaning capacity of soap is reduced when Na or K soaps are dissolved in hard water.

Question 7.
Explain the following terms with suitable examples :
(a) Cationic detergents
(b) Anionic detergents (Comptt. Delhi 2013)
Answer:
(a) Cationic detergents :
(i) Cationic detergents : They are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions and the cationic part possess a long hydrocarbon chain, and a positive charge on nitrogen atom. Therefore they are called cationic detergents.

(b) Anionic detergents : Those detergents in which large part of their molecules are anions and used in cleansing action, are called anionic detergents.
Example : Sodium alkyl sulphates These are obtained from long straight chain alcohols containing 12-18 carbon atoms by treatment with cone. H2S04 followed by neutralization with NaOH.
Example : Sodium lauryl sulphate.

Question 8.
Explain the following types of substances with one suitable example, for each case :
(i) Cationic detergents.
(ii) Food preservatives.
(iii) Analgesics. (Delhi 2009)
Answer:
(i) Cationic detergents : They are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions and the cationic part possess a long hydrocarbon chain, and a positive charge on nitrogen atom. Therefore they are called cationic detergents.

(ii) Food preservatives : They are used to prevent spoilage of food due to microbial growth.
Example : Table salt, vegetable oils, sodium benzoate etc.

(iii) Analgesics : Analgesics reduce or abolish pain without causing impairment of consciousness, mental confusion, in coordination or paralysis or some other disturbance of nervous system.
They are of two types :
(a) Non-narcotic analgesics Example : Aspirin
(b) Narcotic analgesics Example : Morphine

Question 9.
How do antiseptics differ from disinfectants? Give one example of each type. (Delhi 2008)
Antiseptics Disinfectants
Answer:

Antiseptics	Disinfectants
1. They are chemical substances which prevent the growth of micro-organisms and may even kill them.	1. They are chemical substances which kill micro-organisms.
2. They are safe to be applied to the living tissues.	2. They are not safe to be applied to the living tissues.
3. They are generally applied on wounds, cuts, ulcers and diseased skin surfaces. Example : Furacin, soframycin, dettol and savlon, 0.2% solution of phenol.	3. They are used to kill micro-organisms present in the drains, toilets, floors etc. Example: Phenol (> 1% solution) and chlorine (0.2 to 0.4 ppm).

Question 10.
What are the following substances? Give one example of each type.
(i) Antacid
(ii) Non-ionic detergents
(iii) Antiseptics (All India 2008)
Answer:
(i) Antacid : Those substances which neutralize the excess acid and raise the pH to an appropriate level in stomach are called antacids.
Example : Sodium bicarbonate, Mg(OH)₂

In This Post we are  providing Chapter-15 POLYMERS NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON POLYMERS

Question 1.
Mention two important uses of each of the following :
(i) Bakelite
(ii) Nylon 6
Answer:
(i) Bakelite :
(i) It is used in making handles of utensils.
(ii) Also used in production of billiard balls, dominoes and pieces for games like chess.
(ii) Nylon 6 :
(i) Nylon is used in making stockings.
(ii) It is also used for making parachutes.

Question 2.
Name the sub-groups into which polymers are classified on the basis of magnitude of intermolecular forces.
Answer:

1. Elastomers
2. Fibres
3. Thermoplastic polymers
4. Thermosetting polymers.

Question 3.
Draw the structure of the monomer for each of the following polymers : (Delhi 2012)
(i) Nylon-6
(ii) Polypropene
Answer:
(i) Nylon-6 :

Question 4.
Define thermoplastic and thermosetting polymers. Give one example of each.
Answer:
Thermoplastic polymers : Linear polymers in which the intermolecular forces of attraction are in between those of elastomers and fibres and can be melted again and again on heating followed by moulding to give desired shape.
Example : Polyethene, Polyvinyl chloride (PVC) etc.

Thermosetting polymers : These are semifluid substances with low molecular masses which when heated in a mould, undergo change in chemical composition to give a hard, infusible and insoluble mass. These cannot be re-melted.
Example : Bakelite, Melamine etc.

Question 5.
Write the name and structure of the monomer of each of the following polymers :
(i) Neoprene
(ii) Buna-S
(iii) Teflon
Answer:

Question 6.
Differentiate between thermoplastic and thermosetting polymers. Give one example of each. (Delhi 2013)
Answer:
Thermoplastic polymers : Linear polymers in which the intermolecular forces of attraction are in between those of elastomers and fibres and can be melted again and again on heating followed by moulding to give desired shape.
Example : Polyethene, Polyvinyl chloride (PVC) etc.

Thermosetting polymers : These are semifluid substances with low molecular masses which when heated in a mould, undergo change in chemical composition to give a hard, infusible and insoluble mass. These cannot be re-melted.
Example : Bakelite, Melamine etc.

Question 7.
Draw the structures of the monomers of the following polymers :
(i) Polythene
(ii) PVC
(iii) Teflon
Answer:

(ii)

(iii)

Question 8.
Write the names and structures of the monomers of the following polymers :
(i) Buna-S
(ii) Dacron
(iii) Neoprene
Answer:

Question 9.
How are polymers classified on the basis of mode of polymerization? Explain with suitable examples. (Comptt. Delhi 2012)
Answer:
Polymers can be classified on the basis of mode of polymerisation into two groups :
(i) Addition polymers : The addition polymers are formed by the repeated addition of monomer molecules possessing double or triple bonds.

(ii) Condensation polymers : They are formed by repeated condensation reaction between two different bi-functional or tri-functional monomeric units.

Question 10.
Explain the term co-polymerization and give two examples of copolymers and the reactions for their preparations.
Answer:
Co-polymerization : Co-polymerization is a polymerization reaction in which a mixture of more than one monomeric species is allowed to polymerise and form a copolymer.
Example : Buna- S, Buna-N.
Equations :

In This Post we are  providing Chapter-13 AMINES NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON AMINES

Question 1.
Write the main products of the following reactions:

Answer:

Answer:

Answer:

Question 2.
Write the main products of the following reactions:

Answer:

Answer:

Answer:

Question 3.
Account for the following:
(i) Primary amines (R-NH₂) have a higher boiling point than tertiary amines (R₃N).
(ii) Aniline does not undergo Friedel — Crafts reaction.
(iii) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution.
OR
Give the structures of A, B, and C in the following reactions:

Answer:
(i) Primary amines (RNH₂) have two hydrogen atoms on the N atom and therefore, form intermolecular hydrogen bonding.

Tertiary amines (R₃N) do not have hydrogen atoms on the N atom and therefore, these do not form hydrogen bonds. As a result of hydrogen bonding in primary amines, they have higher boiling points than tertiary amines of comparable molecular mass. For example, b.p. of n-butylamine is 351 K while that of tert-butylamine is 319 K.

(ii) Aniline being a Lewis base reacts with Lewis acid such as AlCl₃ to form a salt.

As a result, N of aniline acquires +ve charge and hence it acts as a strong deactivating group for electrophilic substitution reactions. Hence aniline does not undergo Friedel Crafts reaction.

(iii) Due to the presence of lone pair of electrons on the N atom, amines are basic in nature. The methyl group is the electron releasing group (+I inductive effect) and therefore, it increases the electron density on the N atom, and therefore, basic character increases, so that (CH₃)₃N should be more basic than (CH₃)₂NH. But tertiary ammonium ion formed from tertiary amines is less hydrated than secondary ammonium ion formed from secondary amine. Therefore, (CH₃)₃N has less tendency to form ammonium ion, and consequently, it is less basic than (CH₃)₂NH. Thus, (CH₃)₂NH is more basic than (CH₃)₃N due to the combined effect of inductive effect and hydration effect.
OR

Question 4.
Give the structures of A, B, and C in the following reactions:

OR
How will you convert the following:
(i) Nitrobenzene into aniline
(ii) Ethanoic acid into methanamine
(iii) Aniline into N-phenylethylamine
(Write the chemical equations involved.) (CBSE Delhi 2014)
Answer:

OR

Question 5.
Write chemical equations for the following conversions:
(i) Nitrobenzene to benzoic acid.
Answer:

(ii) Benzyl chloride to 2-phenytethanamine.
Answer:

(iii) Aniline to benzyl alcohol. (CBSE Delhi 2012)
Answer:

Question 12.
(a) Identify ‘A’ and ‘B’ in the following reaction:

Answer:

(b) Why is ethylamine soluble in water whereas aniline is not?
Answer:
Ethylamine dissolves in water due to intermolecular hydrogen bonding as shown below:

However, because of the large hydrophobic part (i.e. hydrocarbon part) of aniline, the extent of hydrogen bonding is less and therefore, aniline is insoluble in water.

Question 6.
(a) Identify X and Y in the following:

Answer:
X =
benzene diazonium ch(orlde,
Y =
 Cyanobenzene

(b) Amino group is o, p-directing for aromatic electrophilic substitution reactions. Why does aniline on nitration give m-nitroaniline? (CBSE 2019C)
Answer:
Under strongly acidic conditions of nitration, most of the aniline is converted into anilinium ion having an NH₃⁺ group. This group is an m-directing group, therefore, m-nitro aniline is also obtained along with o- and p-products.

Question 14.
Give the structures of A, B, and C in the following reactions: (CBSE Delhi 2013)

Answer:

Answer:

Question 7.
Do as directed:
(i) Arrange the following compounds in the increasing order of their basic strength in an aqueous solution:
CH₃NH₃, (CH₃)₃N, (CH₃)₂NH.
Answer:
(CH₃)₃N < CH₃NH₂ < (CH₃)₂ NH

(ii) Identify ‘A’ and ‘B’:

Answer:
A: C6H5N2+ Cl- B: C6H5OH

(iii) Write the equation of carbylamine reaction. (CBSE 2018C)
Answer:

Question 8.
(i) Illustrate the following reactions giving a suitable example in each case:
(a) Hoffmann bromamide degradation reaction
(b) Diazotisation
(c) Gabriel phthalimide synthesis
(ii) Distinguish between the following pairs of compounds:
(a) Aniline and N-methylaniline
(b) (CH₃)₂NH and (CH₃)₃N
OR
(i) Write the structures of main products when benzene diazonium chloride (C₆H₅N₂⁺Cl^–) reacts with the following reagents:
(a) CuCN/KCN
(b) H₂0
(c) CH₃CH₂OH
(ii) Arrange the following:
(a) C₂H₅NH₂, C₂H₅OH, (CH₃)₃N – in the increasing order of their boiling point.
(b) Aniline, p-nitroaniline, p-methyl aniline – in the increasing order of their basic strength. (CBSE Delhi 2015)
Answer:
(i) (a) Hoffmann bromamide degradation reaction: Primary amines can be prepared from amides by treatment with Br₂ and KOH solution. The amine formed contains one carbon atom less than the parent amide.

(b) Diazotisation: The reaction of aniline or other aromatic amines, with nitrous acid at 0-5 °C to form diazonium salts is called diazotization. Nitrous acid needed for this reaction is prepared in situ by the action of dil. HCl on NaNO₂.

(c) Gabriel’s phthalimide synthesis. This method is used for preparing only primary amines. In this method, phthalimide is treated with alcoholic KOH to give potassium phthalimide, which is treated with an alkyl halide or benzyl halide to form N-alkyl or aryl phthalimide. The hydrolysis of N-alkyl phthalimide with 20% HCl under pressure or refluxing with NaOH gives primary amine.

Phthalic acid can again be converted into phthalimide and is used again and again. This method is very useful because it gives pure amines. Aryl halides cannot be converted to arylamines by Gabriel synthesis because they do not undergo nucleophilic substitution with potassium phthalimide.

(ii) (a) Add an alcoholic solution of KOH and CHCl3 to the compounds. Aniline gives the foul smell of isocyanide whereas N-methyl aniline does not give a foul smell.

(b) When treated with Hinsberg’s reagent (benzene sulphonyl chloride, C₆H₅SO₂CI), dimethylamine, (CH₃)₂NH gives precipitate which is insoluble in aqueous KOH.

(CH₃)₃N does not react with Hinsberg’s reagent.
Or

(ii) (a) (CH₃)₂ N < C₂H₅NH₂ < C₂H₅OH
(b) p-nitroaniline < aniline < p-methylaniline

Question 9.
Give the IUPAC names of the following compounds:

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

(CBSE Delhi 2017)
Answer:

(CBSE Delhi 2017)
Answer:

Answer:

Question 10.
Complete the following reactions:

Answer:

Answer:

Answer:

Answer:

Answer:

In This Post we are  providing Chapter-14 BIOMOLECULES NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON BIOMOLECULES

Question 1.
Define the following terms:
(i) Glycosidic linkage
Answer:
Glycosidic linkage: The condensation of hydroxyl groups of two monosaccharides to form a link between them is called glycosidic linkage.

(ii) Invert sugar
Answer:
Invert sugar: The sugar which on hydrolysis with dilute adds or enzymes gives mixture having specific rotation opposite to the original is called invert sugar. For example, sucrose is inverted sugar.

(iii) Oligosaccharides
Answer:
Oligosaccharides: These are the carbohydrates that give two to ten monosaccharide molecules on hydrolysis. These are further classified as disaccharides, trisaccharides, tetrasaccharides, etc. depending upon the number of monosaccharide units present in their molecules. For example, Disaccharides: Sucrose, lactose, maltose. All these have the molecular

Formula C₁₂H₂₂O₁₁.
Trisaccharide: Raffinose (C₁₈H₃₂O₁₆).
Tetrasaccharides: Stachyose (C₂₄H₄₂O₂₁ ).

Question 2.
Define the following terms:
(i) Nucleotide
Answer:
Nucleotide: A unit formed by the combination of a nitrogen-containing heterocyclic base, a pentose sugar and a phosphoric acid group.

(ii) Anomers
Answer:
Anomers: The anomers are the isomers formed due to the change in the configuration of the -OH group at C-1 of glucose. For example, α-and β-forms of glucose are anomers.

(iii) Essential amino acids.
Answer:
Essential amino acids: The amino acids which cannot be made by our bodies and must be supplied in our diet for the growth of the body are called essential amino acids.

Question 3.
Differentiate between the following:
(a) Fibrous protein and Globular protein
Answer:
Difference between fibrous protein and globútar protein:

Fibrous	Globular
(i) The polypeptide chains run parallel and are held together by hydrogen and disulphide bonds.	1. The polypeptide chains colt around to give a spherical shape.
(ii) These are insoluble in water.	2. These are soluble in water.
(iii) For example, keratin in hair	3. For example, albumin in egg.

(b) Essential amino acids and Non-essential amino acids
Answer:
Difference between essential amino acids and non-essential amino acids:

Essential amino acids	Non-essential amino acids
These are not synthesìsed in our body and must be supplied in the diet.	These are synthesised in our body and not required in our diet.
For example, valine.	For example, alanine

(C) Amylose and Amylopectin (CBSE AI 2019)
Answer:
Difference between amylose and amylopectin:

Amylose	Amylopectin
It consists of branched polymeric chains of α – D – glucose.	It consists of a long straight chain of α – D – glucose.
It is water-insoluble.	It is water-soluble.

Question 4.
What is essentially the difference between the α-form of D-glucose and β-form of D-glucose? Explain.
Answer:
α-form and β-form of glucose differ in the orientation of —H and —OH groups around the C₁ atom. The isomer having the -OH group on the right is called α-D-glucose while the one having -OH group on the left is called β-D-glucose. Such pairs of optical isomers which differ in the configuration only around C₁ are called anomers. The structures of these two may be shown below:

These two forms are crystalline and have different melting points and optical rotations. For example, α-form of glucose has m.p. 419 K and |α|_D = +111° and the β-form of glucose has m.p. 423 K and |α|_D = +19.2°.

Question 5.
Explain the following:
(i) Amino acids behave like salts rather than simple amines or carboxylic acids. (CBSE 2018C)
Answer:
Due to the formation of zwitterion.

(ii) The two strands of DNA are complementary to each other.
Answer:
The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases.

(iii) Reaction of glucose indicates that the carbonyl group is present as an aldehydic group in the open structure of glucose.
Answer:

Glucose gets oxidised to gluconic acid on reaction with a mild oxidising agent like Bromine water.

Question 6.
What is essentially the difference between α-glucose and β-glucose? What is meant by the pyranose structure of glucose?
Answer:
α-form and β-form of glucose differ in the orientation of -H and -OH groups around C, atom. The isomer having the -OH group on the right is called α-D-glucose while the one having -OH group on the left is called β-D-glucose. Such pairs of optical isomers which differ in the configuration only around C₁ are called anomers. The structures of these two may be shown below:

These two forms are crystalline and have different melting points and optical rotations. For example, α-form of glucose has m.p. 419 K and |α|_D = + 111° and the β-form of glucose has m.p. 423 K and |α|_D = + 19.2°.

The α-D-glucose and β-D-glucose can be drawn in a simple six-membered ring form called pyranose structures. These resemble pyran which is a six-membered heterocyclic ring containing five carbon atoms and one oxygen atom.

These are known as pyranose structures and are shown below:

Question 7.
Which sugar is called invert sugar? Why is it called so?
Answer:
Sucrose is called invert sugar. The sugar obtained from sugar beet is a colourless, crystalline and sweet substance. It is very soluble in water and its aqueous solution is dextrorotatory having [α]_D = + 66.5°.

On hydrolysis with dilute acids or enzyme invertase, cane sugar gives an equimolar mixture of D-(+)-glucose and D-(-)-fructose.

So, sucrose is dextrorotatory but after hydrolysis, gives dextrorotatory glucose and laevorotatory fructose. D-(-)-fructose has a greater specific rotation than D-(+)- glucose. Therefore, the resultant solution upon hydrolysis is laevorotatory in nature with a specific rotation of (-39.9°). Since there is a change in the sign of rotation from Dextro before hydrolysis to Laevo after hydrolysis, the reaction is called Inversion reaction and the mixture (glucose and fructose) is called invert sugar.

Question 8.
How do you explain the presence of an aldehydic group in a glucose molecule?
Answer:
Glucose reacts with hydroxylamine to form a monoxime and adds one molecule of hydrogen cyanide to give cyanohydrin.

Therefore, it contains a carbonyl group which can be an aldehyde or a ketone. On mild oxidation with bromine water, glucose gives gluconic acid which is a carboxylic acid-containing six carbon atoms.

This indicates that the carbonyl group present in glucose is an aldehydic group.

Question 9.
Define the following with an example of each: (CBSE 2018)
(i) Polysaccharides
(ii) Denatured protein
(iii) Essential amino acids
OR
(i) Write the product when D-glucose reacts with the cone. HNO₃.
(ii) Amino acids show amphoteric behaviour. Why?
(iii) Write one difference between a-helix and p-pleated structures of proteins.
Answer:
(i) Carbohydrates that give a large number of monosaccharide units on hydrolysis or a large number of monosaccharide units joined together by glycosidic linkage, e.g. starch, glycogen, cellulose.
(ii) Proteins that lose their biological activity or proteins in which secondary and tertiary structures are destroyed, e.g. curdling of milk.
(iii) Amino acids cannot be synthesised in the body. e.g. Valine / Leucine

OR
(i) Saccharic acid / COOH-(CHOH)₄-COOH
(ii) Due to the presence of carboxyl and amino group in the same molecule or due to formation of zwitterion or dipolar ion.
(iii) a-helix has intramolecular hydrogen bonding while p-pleated has intermolecular hydrogen bonding / a-helix results due to regular coiling of polypeptide chains while in p-pleated all polypeptide chains are stretched and arranged side by side.

Question 10.
Describe the term D- and L-configuration used for sugars with examples.
Answer:
The sugars are divided into two families: the D-family and L-family which have definite configurations. These configurations are represented with respect to glyceraldehyde as the standard. The glyceraldehyde may be presented in two forms:

The D-configuration has —OH attached to the carbon adjacent to —CH₂OH on the right while L-configuration has —OH attached to the carbon adjacent to —CH₂OH on left.

The sugars are calLed D- or L- depending upon whether the configuration of the molecule is related to D-glyceraldehyde or L-glyceraldehyde. It has been found that all naturally occurring sugars beLong to D-series, e.g. D-glucose, D-ribose and D-fructose.

In This Post we are  providing Chapter-12 ALDEHYDES, KETONES AND CARBOXYLIC ACIDS NCERT MOST IMPORTANT QUESTIONS for Class 12 CHEMISTRY which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter

NCERT MOST IMPORTANT QUESTIONS ON ALDEHYDES, KETONES AND CARBOXYLIC ACIDS

Question 1.
(a) Write the structures of A and B in the following reactions:

(b) Distinguish between:
(i) C₆H₅— COCH₃ and C₆H₅— CHO
(ii) CH₃COOH and HCOOH
(c) Arrange the following in the increasing order of their boiling points:
CH₃CHO, CH₃COOH, CH₃CH₂OH
Answer:

(b) (i) Benzaldehyde and acetophenone :
By Iodoform test: Acetophenone being a methyl ketone on treatment with I₂ and NaOH (NaOI) undergoes iodoform test to give yellow ppt. of iodoform on heating whereas benzaldehyde does not.

(ii) CH₃COOH (Acetic acid) and HCOOH (Formic acid). Formic acid is the only acid which contains aldehydic group and thus shows reactions with Tollen’s reagent (silver nitrate) and Fehling’s solution which Acetic acid does not show.
Tollen’s Test:
Add ammonical solution of silver nitrate to both the compounds, HCOOH gives silver mirror but CH₃COOH does not

(c) CH₃CHO < CH₃CH₂OH < CH₃COOH

Question 2.
(a) Write the chemical reaction involved in Wolff-Kishner reduction.
(b) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction:
C₆H₅COCH₃, CH₃— CHO, CH₃COCH₃
(c) Why carboxylic acid does not give reactions of carbonyl group?
(d) Write the product in the following reaction

(e) A and B are two functional isomers of compound C₃H₆O. On heating with NaOH and I₂, isomer B forms yellow precipitate of iodoform whereas isomer A does not form any precipitate. Write the formulae of A and B.
Answer:
(a) Wolff-Kishner reduction reaction : The reduction of aldehydes and ketones to the corresponding hydrocarbons by heating them with hydrazine and KOH or potassium tert-butoxide in a high boiling solvent like ethylene glycol is called Wolff-Kishner reduction.

(b) C₆H₅COCH₃ < CH₃COCH₃ < CH₂CHO
(c) The carboxylic carbon is less electrophilic than carbonyl carbon because of the possible resonance structure.

(e) The given compound has molecular formula C3H60. One of its functional isomer i.e., B shows iodoform test which can be only shown by compounds having methyl ketone so the compound B will be Acetone or 2-propanone. Its functional isomer A will be propanal.

Question 3.
(a) Write the product(s) in the following :

(b) Give simple tests to distinguish the following pairs of compounds :
(i) Ethanal and Propanal
(ii) Benzaldehyde and Acetophenone
(iii) Benzoic acid and Ethyl benzoate
Answer:

(b) (i) On heating with NaOH and I₂, ethanal forms yellow ppt of CHI₃ whereas propanal can not.
CH₃CHO + 3I₂ → NaOH -4 CHI₃ + 3NaI + HCOONa + 3H₂O
(ii) On heating with NaOH and I2, aceptophenone forms yellow ppt of CHI3 whereas benzaldehyde does not.
C₆H₅COCH₃ + 3NaOI → C₆H₅COONa + CHI₃4 + 2NaOH
(iii) On adding NaHCO,, benzoic acid produces brisk effervescence of C02 gas whereas ethylbenzoate does not.

Question 4.
(a) Give reasons :
(i) CH₃—CHO is more reactive than CH₃COCH₃ towards HCN.
(ii) 4-nitrobenzoic acid is more acidic than benzoic acid.
(b) Describe the following :
(i) Acetylation (ii) Cannizzaro reaction (iii) Cross aldol condensation
Answer:
(a) (i) Because carbonyl carbon of CH₃—CHO is more electrophilic than CH₃COCH₃ due to only one electron donating CH₃– group.
(ii) Because of electron withdrawing nature of -NO₂ group.
(b) (i) Acetylation : Introduction of an acetyl group/CH3CO- by heating an organic compound with acetyl chloride/acetic anhydride.

(ii) Cannizzaro reaction : Aldehydes having no a-hydrogen atom when treated with cone. NaOH, undergoes self-oxidation and self-reduction simultaneously

(iii) Cross Aldol Condensation : When aldol condensation is carried out between two different aldehydes or ketones, it is called cross aldol condenstation.

Question 5.
(a) Complete the following equations :

Answer:

(b) (i) On adding NaHCO₃, CH₃COOH produces brisk effervescence of CO₂ gas whereas phenol does not.
(ii) On heating with Tollen’s reagent, CH₃CHO forms silver mirror whereas CH₃COCH₃ does not.

Question 6.
(a) What is meant by the following terms? Give an example of the reaction in each case.
(i) Aldol (ii) Semicarbazone
(b) Complete the following :

Answer:
(a) (i) Two molecules of aldehyde and ketones containing a-hydrogen atom react in the presence of
aqueous alkali giving product known as Aldol.
Example :

Question 7.
Write the product(s) in the following reactions.

Answer:

Question 8.
(a) Write the product(s) in the following reactions:

(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Butanal and Butan-2-one (ii) Benzoic acid and Phenol
Answer:

(b) (i) Pollen’s reagent test. Add ammoniacal solution of sliver nitrate (Tollen’s Reagent) in both the solutions. Butanal gives silver mirror whereas Butan-2-one does not. Therefore Butanal gives Tollen’s test.
(ii) Ferric chloride test. Add neutral FeCl₃ in both the solutions, phenol reacts with neutral FeCl₃ to form an iron-phenol complex giving violet colour but benzoic acid does not.

Question 9.
(a) Write the reactions involved in the following:
(i) Etard reaction (ii) Stephen reduction
(b) How will you convert the following in not more than two steps:
(i) Benzoic acid to Benzaldehyde (ii) Acetophenone to Benzoic acid
(iii) Ethanoic acid to 2-Hydroxyethanoic acid (All India 2017)
Answer:
(a) (i) Etard reaction

(ii) Stephen reduction:

(b) (i) Benzoic acid to Benzaldehyde

(ii) Acetophenone to Benzoic acid

Question 10.
(a) How will you convert:
(i) Benzene to acetophenone (ii) Propanone to 2-Methylpropan-2-ol
(b) Give reasons :
(i) Electrophilic substitution in benzoic acid takes place at meta position.
(ii) Carboxylic acids are higher boiling liquids than aldehydes, ketones and alcohols of comparable molecular masses.
(iii) Propanal is more reactive than propanone in nucleophilic addition reactions. (Comptt. Delhi 2017)
Answer:
(i) Benzene to acetophenone

(b) (i) Because -COOH group is electron withdrawing group and deactivates the benzene ring. As a result of this ortho and para position acquires positive charge but only meta does not, so electrophile can attack on rneta position.
(ii) Because -COOH group of carboxylic acids is capable to do intermolecular hydrogen bonding forming a dimer while alcohols, aldehydes and ketones can not.

(iii) Because of smaller +1 effect of one alkyl group in propanal as compared to larger + I effect ol 2 alkyl groups of propanone, the magnitude of positive charge on the carbonyl carbon is more in propanal than propanone.