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Students can Download Chapter 2 Electric Potential and Capacitance Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 2 Electric Potential and Capacitance

Plus Two Physics Electric Potential and Capacitance NCERT Text Book Questions and Answers

Question 1.
Two charges 5 × 10^-8C and -3 × 10^-8C are located 16 cm apart. At what point on the line joining the two charges is the electric potential zero?
Take the potential at infinity to be zero.
Answer:
Given q₁ = 5 × 10^-8C, r=16cm = 0.16m q₂= -3 × 10^-8C Let potential be zero at a distance × metre from positive charge q₁.
∴ r₁ = x meter
r₂ = (0.16 – x) metre
S0 V = \(\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q_{1}}{r_{1}}+\frac{q_{2}}{r_{2}}\right]\)

or 0.8 – 5x = 3x
or x = 0.1m = 10cm.

Question 2.
A regular hexagonal of side 10cm has a charge 5mC at each of its vertices. Calculate the potential at the centre of the hexagon.

Answer:
From the figure, we have
OP = OQ = OR = OS = OT = OU
= r = 10cm = 0.1m
And given q = 5µC = 5 × 10^-6C
∴ Potential at O due to all the charges

= 2.7 × 10⁶volt.

Question 3.
Two charges 2mC and -2mC are placed at points A and B 6cm apart.

1. Identify an equipotential surface of the system.
2. What is the direction of the electric field at every point on this surface?

Answer:

1. The plane normal to AB and passing through its mid-point has zero potential everywhere hence the plane is equipotential.
2. Normal to the plane is the direction AB.

Question 4.
A spherical conductor of radius 12cm has a charge of 1.6 × 10^-7C distributed uniformly on its surface. What is the electric field

1. Inside the sphere.
2. Just outside the sphere.
3. At point 18cm from the centre of the sphere?

Answer:
1. Zero

2.

3.

Question 5.
A parallel plate capacitor with air between the plates has a capacitance of 8pF (1 pF=10^-12F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant s?
Answer:
The capacitance of capacitor with air as dielectric is given by
C = \(\frac{\varepsilon_{0} A}{d}\)
Given C = 8pF = 8 × 10^-12F …………(1)
If C₁ is new capacitance when d₁ = \(\frac{d}{2}\) and space is filled with a substance of dielectric constant k=6.
Then

Using Eq.(1)
C₁ = 12 × 8 × 10^-12F
or C₁ = 96pF.

Question 6.
Three capacitors each of capacitance 9pF are connected in series.

1. What is the total capacitance of the combination?
2. What is the potential difference across each capacitor, if the combination is connected to a 120 volt supply?

Answer:
Given C₁ = C₂ = C₃ = 9pF = 9 × 10^-12F; V= 120 volt.
1. Total capacitance of the series combination is given by

∴ C = 3 × 10^-12F = 3pF.

2. Let q be the charge on each capacitor. Then, sum of the potential difference across their plates must be equal to 120 V.
ie. V₁ + V₂ + V₁ = 120

or q = 360 × 10^-12C
Since, all the capacitors are of same capacitance,

Plus Two Physics Electric Potential and Capacitance One Mark Questions and Answers

Question 1.
A hollowing metallic sphere of radius 10cm is charged such that potential on its surface is 80V. the potential at the center of the sphere would be.
(a) 80 V
(b) 800 V
(c) zero
(d) 8 V
Answer:
(a) 80 V

Question 2.
When air is replaced by a dielectric medium of constant K, the maximum force of attraction between two charges separated by a distance.
(a) increases k times
(b) remains unchanged
(c) decreases k times
(d) increases k^-1 times
Answer:
(c) decreases k times
Explanation : F_m = \(\frac{\mathrm{F}_{0}}{\mathrm{K}}\) i.e., Decrease K times.

Question 3.
Charge q₂ is at the center of a circular path with radius r. work done in carrying charged q₁ once a round this equipotential path, would be.

Answer:
(c) zero

Question 4.
Pick out a vector quantity from the following
(i) Electric potential
(ii) Electric potential gradient
(iii) Electric potential energy
(iv) Electric flux
Answer:
(ii) Electric potential gradient

Question 5.
“The surface of a charged conductor is an equipotential surface”. Comment on this statement.
Answer:
In a charged conductor, at any point inside or at the surface, the potential is same. So the surface is equipotential.

Question 6.
Pick the odd one out of the following
(i) eV
(ii) Volt
(iii) Mega Volt
(iv) Millivolt
Answer:
(i) eV (eV is the unit of energy).

Plus Two Physics Electric Potential and Capacitance Two Mark Questions and Answers

Question 1
Figure shows a parallel plate air capacitor of plate area of 100cm² and separation 5mm. A potential difference of 300v is established between its plates by a battery.

1. Calculate the capacitance and charge on the capacitor.
2. After disconnecting the battery, the space between the plate Js filled by ebonite (k=2.6). Then calculate the capacitance and charge on capacitor.

Answer:
1.

= 17.6 pF
Q = CV= 17.6 × 10^-12 × 300 = 5.2 × 10^-9C

2. C = \(\frac{A K \varepsilon_{0}}{d}=\frac{K A \varepsilon_{0}}{d}\)
= 2.6 × 17.6 × 10^-12 = 45.76pF.

Question 2.
Fill in the blanks

1. 6.25 × 10¹⁸ electrons =………..C
2. 6.25 × 10¹⁸ eV =…………J

Answer:

1. 1 coulomb
2. 1 J

Question 3.
Potential is a scalar quantity. Potential gradient is a vector quantity Negative of potential gradient is another vector quantity. It is
1.

• electric force
• electric flux
• electric field intensity
• none

2. What is magnitude and direction of above quantity in the case of a point charge
Answer:

1. Electric field
2. Direction of electric field is outward for positive charge and inward for negative change

Question 4.

1. Dielectric strength of air is 3 × 10⁶ V/m. What does it mean?
2. High power lines cannot be insulated. Why?

Answer:

1. When electric field on air exceeds the 3 × 10⁶ v/m, air becomes conductor and conduct electricity.
2. Insulator shows conducting property at high voltage.

Question 5.

Write whether true or false

1. In a charged conductor charge reside inside and outside of the conductor
2. In a charged conductor net field is zero inside the conductor
3. Potential is Zero inside the conductor
4. The entire part of the conductor (including surface of the conductor) is at constant potential.

Answer:

1. False
2. True
3. False
4. True

Plus Two Physics Electric Potential and Capacitance Three Mark Questions and Answers

Question 1.
Match the following.

Answer:

1. Electric field – force per unit (+ve) charge -NC^-1
2. Electric potential – Work done per (+ve) charge – JC^-1
3. Capacitance – charge per unit potential difference -Farad.

Question 2.
The data given below shows the variation of potential with charge during charging of a condenser.

1. Draw a graph with charge along the Y-axis and potential along the X-axis. Estimate capacity of the condenser from the graph.
2. Estimate the energy stored in the condenser from the graph.

Answer:
1.

2. Area enclosed by the graph represents total energy = 1/2 QV = 1/2 × 5 × 10^-6 × 25
= 62.5 × 10^-6 J

Question 3.
Many of the applications of capacitors depend on their ability to store energy.
1. In a charged capacitor energy is stored in the

• positively charged plate
• negatively charged plate
• electric field between the plates
• none of these

2. Draw a graph showing the variation of charge stored in a capacitor with its potential. How will you calculate the energy stored in the capacitor using the above graph?
Answer:
1. Electric field between the plates

2.

Area under the straight line graph gives the energy stored in the capacitor.

Question 4.
A and B are two points in an electric field produced by q. To bring a unit +ve charge from a to A, 10 J work is needed. To bring the same charge from. A to B, 2J work is needed.
1. What is the p.d, between A and B?

2. What are the potential at A and B?

Answer:

1. 2v
2. potential at A, V_A = 10v potential at B, V_B = 12v.

Question 5.

1. 1 coulomb = 6.25 × 10¹⁸ electrons, 1 Joule =……..eV
2. What is meant by 1 eV
3. Which is bigger unit J or eV? Justify for your answer.

Answer:
1. lev=1.6 × 10^-19 J
∴ 1J = \(\frac{1}{1.6 \times 10^{-19}}\) ev 1 J = 6.25 × 10¹⁸ ev.

2. 1 ev is the energy acquired by an electron, when it is accelerated through a potential difference of one volt.

3. Joule is bigger unit, 1J = 6.25 × 10¹⁸ eV.

Question 6.

1. Name the physical quantity which has its unit joule, coulomb^-1. Is it a vector or a scalar?
2. Two plane sheets of charge densities +σ and -σ are kept in air as shown in figure. What are the electric field intensities at point A and B?

Answer:
1. Electric potetial. Scalar.

2. Electric field at A E_A = 0
Electric field atB E_B = \(\frac{\sigma}{\varepsilon_{0}}\).

Question 7.
You are given 3 capacitors of capacities 3 µf, 2 µf, 1 µf You have to make a capacitors of capacitance less than 1 µf.

1. How do you arrange the capacitors, show by a diagram?
2. Write an expression for effective capacitance of the above combination and verify the result.

Answer:
1. Connect in series

2.

Question 8.
The arrangement of a capacitor is given below. The plate A is charged and the plate B is earthed, ‘d’ is the distance between two plates.
1. Write any one use of this capacitor.

2. Obtain an expression for capacitance of this arrangement.

3. A capacitor is made of a flat plate of area A and second plate having a stair like structure as shown in figure. The width of each stair is d and 2d. Find the capacitance of this arrangement.

Answer:
1. To store electric charges

2. C = \(\frac{A \varepsilon_{0}}{d}\)

3. It is equivalent to two capacitors connected in parallel.
C = C₁ + C₂

Question 9.
The dotted line indicate the surface such that they lie equi-distance from the charge ‘q’.

1. What is the name of this surface?
2. Write the properties of the surface.
3. Give a mathematical proof to any one of the properties.

Answer:
1. Equipotential surface.

2. properties of the surface:

• Direction of electric field is perpendicular to the equi potential surface,
• No work is done to move a charge from one point to another along the equi potential surface.

3. Work done = Potential difference × charge
= 0 × Charge = 0.

Question 10.

1. What is meant by potential at a point?
2. Obtain an expression for potential at a point
3. A wire is bent in a circle of radius 10cm. It is given a charge of 250µC which spreads on it uniformly. What is the electric potential at the centre?

Answer:
1. Potential at a point is the work done required bring a unit charge from infinity to that point without acceleration.

2. Potential due To A point charge:

Let P be a point at a distance Y from a charge +q. Let A be a point at a distance ‘x’ from q, and E is directed along PA. Consider a positive charge at A. Then the electric field intensity at ‘A’ is given by
\(E=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{x^{2}}\)
If this unit charge is moved (opposite to E} through a distance dx, the work done dw = – Edx
[-ve sign indicates that dx is opposite to E ]
So the potential at ‘P’ is given by

V = \(\frac{+q 1}{4 \pi \varepsilon_{0} r}\) (since \(\frac{1}{\infty}\) = 0).

3.

= 2.25 × 10⁷ v.

Plus Two Physics Electric Potential and Capacitance Four Mark Questions and Answers

Question 1.

1. Write any one use of capacitor
2. Obtain an expression for capacitance of a parallel plate capacitor
3. The Capacity of a parallel plate capacitor becomes 10µF when air between the plates is replaced by a dielectric slab (k =2). What is the capacity of the capacitor with air in between the plates?

Answer:
1. one use of capacitor:

• Capacitor is used to store electric charges
• It is used to prevent dc current.

2. Expression for capacitance of a capacitor:
Potential difference between two plates V= Ed

Capacitance C of the parallel plate capacitor,

C = \(\frac{\mathrm{A} \varepsilon_{0}}{\mathrm{d}}\)

3. C = 10µF
When dielectric slab is placed, New capacitance
C¹ = KC
10 × 10^-6 = 2 × C, C=5 × 10^-6F.

Question 2.
A charge of +5µC is placed in free space. P and Q are two points at 3mm and 5mm respectively from 5µC.

1. The work done to bring a + IC charge from infinity to a point ‘P’ is called

• capacitance
• dielectric constant
• potential energy
• electric potential

2. Calculate the workdone in above process.

3. Calculate the workdone to move a +IC charge from ‘P’to ‘Q’.

Answer:
1. Electric potential.

2. Work done w = VQ

3. The potential energy at p,

PE₂ = 9V
Work done to move IC from P to Q, W = PE₂ – PE₁
= 9 – 15
w = -6 J

Question 3.

1. “Electric field lines are always parallel to Equipotential surfaces.” Correct the statement if there is any mistake.
2. Draw the equipotential for a single positive point charge.

Answer:
1. Electric field lines are always perpendicular to Equipotential surfaces.

2.

Question 4.
E is the electric field intensity at any point in a uniform electric field.

1. What is meant by uniform electric field?
2. Represent a uniform electric field using lines of force.
3. Calculate energy stored per unit volume of the space if E=2V/m.

Answer:
1. Region where magnitude and direction of electric field remain same.

2. lines of force are parallel and equidistant.

3. Energy density = \(\frac{1}{2}\) ε_o E² = 1/2 × 8.85 × 1012 × 22
= 17.7 J/m3.

Question 5.
In a charged capacitors energy is stored
1.

• in+ve plate
• In dielectric
• In electric field

2. Derive an expression for energy stored in a capacitor.

3. Draw a graph between energy and charge, in electric field

Answer:
1. In electric field

2. If we supply a charge ‘dq’ to the capacitor, then work done can be written as,
dw = Vdq
dw = \(\frac{q}{c}\) dq (since v = \(\frac{q}{c}\))
∴ Total work done to charge the capacitor (from ‘0’ to ‘Q’) is

3.

Plus Two Physics Electric Potential and Capacitance Five Mark Questions and Answers

Question 1.

1. Identify the device

• Moving coil Galvanometer
• Cyclotron
• Photovoltaic cell
• Van-de-Graft electrostatic generator

2. Explain the construction and working of the above device

3. What happens if the upper metal sphere is replaced by a cubical shaped metal? Explain.

Answer:
1. Vandegraff generator

2. Van de graff generator:
Van de Graff generator is used to produce very high voltage.
Principle:
If two charged concentric hollow spheres are brought in to contact, charge will always flow from inner sphere to the outer sphere.
Construction and working:

The vande Graff generator consists of a large spherical metal shell, placed on an insulating stand. Let p₁ and p₂ be two pulleys. Pulley p₁ is at the center of the spherical shell S. A belt is wound around two pulleys p₁ and p₂.

This belt is rotated by a motor. Positive charges are sprayed by belt. Brush B₂transfer these charges to the spherical shell. This process is continued. Hence a very high voltage is produced on the sphere.

3. The charge density on the pointed edge is higher than flat surface. Hence charge will leak from the cubical shaped metal.

Question 2.
The figure below shows a non polar dielectric slab placed in between the plates of an uncharged parallel plate capacitor.
Area of each plate = A, Distance, of separation be¬tween the plates = d

1. If the dielectric slab is absent and the capacitor is charged to a surface charge density a, the electric field in between the plates is

• E = σε_o
• E = \(\frac{\sigma^{2}}{\varepsilon_{0}}\)
• E = \(\frac{\sigma}{\varepsilon_{0}}\)
• E = σε_o²

2. Redraw the given figure, which shows the align-ment of the nonpolar molecules, when the capacitor is charged.

3. Derive an expression for capacitance of the above capacitor with the dielectric slab in between the plates.
Answer:
1. E = \(\frac{\sigma}{\varepsilon_{0}}\)

2.

3. Effect of dielectric on capacitance:

Consider a capacitor of area A and charge densities +σ and -σ. Let d be the distance between the plates. If a dielectric slab is placed inside this capacitor, it undergoes polarization.

Let +σ_p and -σ_p be polarized charge densities due to polarization. Due to polarization electric field in between the plate becomes
E = \(\frac{\sigma}{\mathrm{K} \varepsilon_{0}}\) …….(1)
The potential difference between the plates,
V = Ed ………..(2)
Sub (1) in (2)
V = \(\frac{\sigma}{\mathrm{K} \varepsilon_{0}}\) d
Then the capacitance of capacitor

C = \(\frac{A \varepsilon_{0} K}{d}\)
The product ε₀K is the permittivity of the medium.

Question 3.
Three capacitors C₁, C₂ and C₃ are connected to a cell of emf V as shown in the figure.

1. The arrangement of these three capacitors are called………….

• parallel combination
• series combination
• LCR combination
• c-c combination

2. Find the effective capacitance of the above combination.

3.

The above graph shows the variation of potential in going from a to g. From the graph the relation among C₁, C₂ and C₃ is

• C₁ = C₂ = C₃
• 2C₁ = 2C₂ = C₃
• C₁ = C₂ = 2C₃
• C₁ = C₂ = C₃

Answer:
1. Series connection.

2. Capacitors in series:
Let three capacitors C₁,C₂ and C₃ be connected in series to p.d of V. Let V₁, V₂ and V₃ be the voltage across C₁, C₂ and C₃.

The applied voltage can be written as,
V=V₁ + V₂ + V₃ ………..(1)
Charge ‘q’ is same as in all the capacitor. So,

Substituting these values in (1),

If these capacitors are replaced by a equivalent capacitance ‘C’, then
V = \(\frac{q}{C}\)
Hence eq(2) can be written as
\(\frac{q}{C}=\frac{q}{C_{1}}+\frac{q}{C_{2}}+\frac{q}{C_{3}}\)
\(\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}\)
Effective capacitance is decreased by series combination.

3. From the graph, we get v₁ = 1 v, v₂ = 1 v, v₃ = 2v.
This arrangement is series. Hence charge stored in each capacitor is same.

C₁: C₂: C₃
1: 1: 1/2
2: 2: 1
2C₁: 2C₂: C₃

Question 4.
Two metal plates X and Y of the area ‘A’ are separated by a distance ‘d’, charged + and – respectively.

1. This arrangement is called………….
2. The arrangement store energy in the………..(Magnetic field, Electric field, Electromagnetic field, Gravitational field)
3. Derive an expression for the energy stored in the arrangement.
4. When we increase separation between two plates by keeping V constant, what happens to total energy stored in the system.

Answer:
1. Capacitor.

2. Electric field

3. If we supply a charge ‘dq’ to the capacitor, then work done can be written as,
dw = Vdq
dw = \(\frac{q}{c}\) dq (since v = \(\frac{q}{c}\))
∴ Total work done to charge the capacitor (from ‘0’ to ‘Q’) is

4. When we increase the separation between two plates, capacitance (c) decreases. The energy in the capacitor U = 1/2 CV² When c decreases, the energy decreases (because V is constant).

Question 5.
A combination of two conducting bodies separated by insulator can store electric charge
1. The above device is called………….

2. Derive on expression for capacitance of the device with air as medium.

3. If mica of dielectric constant Ks placed between the plates. What are its advantages? Explain action of mica in the case.
Answer:
1. capacitor.

2. Potential difference between two plates V= Ed

Capacitance C of the parallel plate capacitor,

3.

• When mica sheet is introduced, capacitance of capacitor increases
• Mica sheet will prevent electric breakdown.

Question 6.
When a charged particle moves in an electric field, work is done on the particle.
1. Pick out a vector quantity from the following

• electric potential
• electric potential gradient
• electric potential energy
• electric flux (1)

2. 5 J of work is done in moving a positive charge of 0.5C between two points. What is the potential difference between the points?

• 2.5V
• 10V
• 0.1V
• 5.5V (1)

3. Three-point electric charges q₁ = 6µC, q₂ = 4µC and q₃ = -8µC are placed on the circumference of a circle of radius 1 m as shown in the figure. What is the value of the charge q₄ placed on the circle if the potential at the centre of the circle is zero? (3)
Answer:
1. Electric potential gradient

2. 10V

3. Radius of the circle r = 1m
By the given condition
\(\frac{1}{4 \pi \varepsilon_{0}} \times \frac{1}{r}\) (q₁ + q₂ + q₃ + q₄) = 0
(q₁ + q₂ + q₃ + q₄) = 0
(6 + 4 – 8 + q₄) µC = 0
(2 + q₄) µC = 0
q₄ = -2µC.

Question 7.
Equipotential surface is a surface on which the electrical potential is the same at every point.
1. “Electric field lines are always parallel to Equipotential surfaces.” Correct the statement if there is any mistake. (1)

2. Draw the equipotential for a single positive point charge. (2)

3. A point charge +q is placed at the centre of a sphere of radius R. Another point charge +q is taken from a point A to another diametrically opposite point B on the surface of the sphere. Calculate the work done for this. (2)
Answer:
1. Electric field lines are always perpendicular to Equipotential surfaces.

2. Surface of the sphere is Equipotential in nature. So the potential difference between any points is zero. Since work done is the product of charge moving and the potential difference between the points, work done is equal to zero.

Question 8.
Capacitance of a capacitor depends on the size and shape of the conductors and on the dielectric material between them.
1. Draw the symbol of a variable capacitor. (1)

2. The plates of a parallel plate capacitor are connected to an ideal voltmeter. What will happen to the reading of the voltmeter if the plates of the capacitor are brought closer to each other in an insulating medium? (2)

3. The plates of a parellel plate capacitor in vacuum are 5mm apart and 1.5m² in area. A potential difference of 10kV is applied across the capacitor. Calculate
a. The capacitance
b. The charge on each plate.
c. magnitude of the intensity of electric field between the paltes.

Answer:
1.

2. Capacitance C = \(\frac{\varepsilon_{0} A}{d}\)
Potential difference V = \(\frac{q}{c}=\frac{q d}{\varepsilon_{0} A}\)
In an insulating medium charge remains the same,voltage is directly proportional to plate separation. Therefore, voltmeter reading decreases.

3.
a.

b. Q = CV = 2.66 × 10^-9 × 10 × 10³ = 2.66 × 10^-5 C.
c.

Question 9.
Capacitors can be combined to obtain any desired capacitance in an application.
1. You are given two capacitors of capacitance 20µF each. Draw a diagram to show how you will connect these capacitors to get 40µF capacitance. (1)

2. Two capacitors C₁ and C₂ are connected in series.
a. Draw a diagram of the above combination. (1)
b. If C₁ = 4µF, C₂ = 6µF and C₃ = 2.4µF and a potential difference of 100V is applied across the combination. Calculate the change stored in each capacitor.

Answer:
1.

2. Two capacitors C₁ and C₂ are connected in series:
a.

b. Effective capacitance C₁ anc C₂
C₁₂ = \(\frac{C_{1} C_{2}}{C_{1}+C_{2}}=\frac{4 \times 6}{4+6}\) = 2.4µF
Potential difference across C₁₂ = 100V
Charge stored on C₁₂, q₁₂ = C₁₂ × 100
q₁₂ = 2.4 × 10^-6 × 100
q₁₂ = 2.4 × 10^-4 C
Charge on C₁
q₁ = 2.4 × 10^-4C
Charge on C₂
q₂ = 2.4 × 10 ^-4C
Potential difference across C₃ = 100V
Charge on C₃
q₃ = C₃ × 100
q₃ = 2.4 × 10^-6 × 100
q₃ = 2.4 × 10^-4C.

Question 10.
Many of the applications of capacitors depend on their ability to store energy.
1. In a charged capacitor energy is stored in the

• positively charged plate
• negatively charged plate
• electric field between the plates
• none of these (1)

2. Draw a graph showing the variation of charge stored in a capacitor with its potential. How will you calculate the energy stored in the capacitor using the above graph?

3. An electric flash lamp has 20 capacitors each of capacitance 5µF connected in parallel. The lamp is operated at 100V. If the energy stored in the combination is completely radiated out in a single flash, how much energy will be radiated in a flash? (2)

Answer:
1. Electric field between the plates

2.

Area under the straight line graph gives the energy stored in the capacitor.

3. Effective capacitance C = 20 × 5µF = 100µF, V=100v,
Energy stored E = \(\frac{1}{2}\) CV
E = \(\frac{1}{2}\) × 100 × 10^-6 × 100² = 0.5 J

Kerala Plus Two Computer Application Model Question Paper 4

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Scores

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• Read the instructions carefully.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provied.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Part A

Answer all questions from 1 to 5

Question 1.
Name any two data type modifiers in C+ +.

Question 2.
Define the term “array traversal”.

Question 3.
Which tag is used for hyperlinking in HTML ?

Question 4.
What is SCM ?

Question 5.
Write the full form of OFDMA.

Part B

Answer any nine questions from 6 to 16

Question 6.
Write C++ statement to declare and initialise an array in the following situations.
a. An array of 5 unit prices 18.50, 27.5, 19.00, 12.50, 10.75.
b. An array with letters ‘WELCOME’.

Question 7.
Write the output of the following C+ + code.

1. char S1 [10], S2 [10] = “welcome”;
   strcpy(S1, S2);
   cout<<S1;
2. char S1 [20] = “welcome”;
   char S2 [20] = “to C+ + ”;
   strcat(S1, S2):
   cout<<S1;

Question 8.
Differentiate ‘static’ and ‘dynamic’ webpages.

Question 9.
Name any two attribute of <html> tag with use.

Question 10.
What is the use of FTP client Software ?

Question 11.
Define the following terms in DBMS.
a. Cardinality
b. Domain

Question 12.
Categorise the following SQL commands into DDL and DML.

(CREATE, INSERT, ALTER, UPDATE)

Question 13.
What is BPR ?

Question 14.
Explain the role of “System testing” in the implementation of ERP.

Question 15.
What is the use of RFID ?

Question 16.
Compare GPRS and EDGE technologies.

Part C

Answer any nine questions from 17 to 27

Question 17.
Write the output of the following code segment in C+ + .

int sum = 0;
int a[ ] = {1, 2, 3, 4, 5};
for (int i = 0; i<5; i = i + 2)
sum = sum+a [ i ];
cout<<sum;

Question 18.
Explain the use of following built in functions in C+ + .

i. abs( )
ii. sqrt( )
iii. pow( )

Question 19.
Compare call by value and call by reference method in C+ + ..

Question 20.
Write a short note on “Java Server Pages’’.

Question 21.
Name the three attribute of <MARQUEE> with its use.

Question 22.
Consider the following JavaScript Code.

<SCRIPT language = “JavaScript”>
var i, s;
for (i=1; i<= 10; i+ + )
{
s = i * i;
document.write(s);
document.write(“<BR>”);
}
</SCRIPT>
a. What will be the output of the program?
b. What is the use of <SCRIPT> tag ?
c. Write the output statement used in the above code.

Question 23.
Explain three type of “web hosting”,

Question 24.
Distinguish alternate key and foreign key.

Question 25.
Explain the different levels of data abstraction in DBMS.

Question 26.
Explain any three column constraints in SQL.

Question 27.
Consider the table ITEM with the fields Item code, Item name, Price. Write sql queries.
a. To Display the maximum and minimum price of items.
b. To remove all records where price < 1000.
c. To display the number of recods in the table.

Part D

Answer any two questions from 28 to 30

Question 28.
Consider the C++ program to find the sum of first 10 natural numbers and answer the following questions.

# include<iostream>
using namespace std;
int main( )
{
int n, sum = 0;
for (int i=1; i<= 10; i+ + )
sum = sum +i;
cout<<“sum is = ”<<sum;
}
a. Rewrite the above program using do- while loop.
b. What will be output of the above program if we replace i+ + with i = i + 2 in the for loop ?
c. What is the purpose of # include <iostream> ?

Question 29.
a. Name the HTML tag to create checkbox, textbox and radio box.
b. Write HTML statement to create the texbox with maximum length 25 and size 30.
c. Write HTML statement to create a submit button with caption “login”.

Question 30.
Consider the JavaScript code to display the square of first 10 numbers using the code answer the following.

<SCRIPT language = “JavaScript”>
var i, s;
for (i=1; i<=10; i+ + )
{
s=i*i; .
document.write(s);
document.write(“<BR>”);
}
</SCRIPT>
a. Rewrite the above program using while loop.
b. Modify the above program to print the first 10 even numbers.

ANSWERS

Answer 1.
Signed, short

Answer 2.
Accessing each element of the array atleast once is called array traversal.

Answer 3.
<A> tag

Answer 4.
Supply Chain Management (SCM) consists of all activities associated with moving goods from the supplier to the customer. It begins with collecting raw material and ends with delivering goods to customer.

Answer 5.
Orthogonal Frequency Division Multiple Access

Answer 6.
a. float price [ 5 ] = {18.50, 27.5, 19.00, 12.50, 10.75};
b. char ar [ 7 ] = {‘W, ‘E, L’, ‘C’, ‘O’, ‘M’, ‘E’};

Answer 7.

1. welcome
2. welcometo C+ +

Answer 8.
Static websites:

1. The content and layout of a web page is fixed.
2. Static web pages never use databases.
3. Static web pages directly run on the browser and do not require any server side application program.
4. Static web pages are easy to develop.

Dynamic websites:

1. The content and layout may change during run time.
2. Database is used to generate dynamic content through queries.
3. Dynamic web page runs on the server side application program and displays the results.
4. Dynamic web page development re-quires programming skills.

Answer 9.
Dir, Lang.

Answer 10.
FTP is used to transfer files from one computer to another on the Internet. FTP client software establishes a connection with a remote server and is used to transfer files from our computer to the server computer.

Answer 11.
a. Cardinality is the number of rows or tuples in a relation.
b. The set of possible values for a column (attribute) is called domain.

Answer 12.
DDL : CREATE, ALTER
DML : INSERT, UPDATE

Answer 13.
BPR is the analysis and redesign of work flow in an enterprise. It deals with changes in the structure and process of business. Re-engineering helps to reduce cost and effective use of resources. Business process re-engineering is a key component of ERP success.

Answer 14.
The software is tested to ensure that it performs properly from both the technical and functional areas of an enterprise. The validity of output can be determined with the help of sample data. If any mistakes are found out at this stage, it should be corrected before the operation of ERP system.

Answer 15.
RFID gives transportation and logistics operations increased visibility into product movement and business process. It increases efficiency by providing real-time data that gives up-to-date information about the products. RFID based business logistic softwares help to lower the operating costs, increase productivity in the distribution centers, maximize on-time deliveries and improve customer service and satisfaction.

Answer 16.
GPRS (General Packet Radio Services) is a packet oriented mobile data service on the 2G on GSM. EDGE (Enchanced Data rates for GSM Evolution) is three times faster than GPRS. It is used for voice communication as well as an internet connection.

Answer 17.
9

Answer 18.

1. abs( ). It is used to find the absolute value of an integer. It takes an integer as the argument (+ve or -ve) and returns the absolute value.
2. sqrt( ). It is used to find the square root of a number. The argument to this function can be of type int, float or double. The function returns the non-negative square root of the argument.
3. pow( ). This function is used to find the power of a number. It takes two arguments x and y. The argument of type int, float or double.

Answer 19.
Call by value method:

1. Ordinary variables are used as formal parameters.
2. Actual parameters may be constants, variables or expressions.
3. The changes made in the formal arguments are not reflected in actual arguments.
4. Exclusive memory allocation is required for the formal arguments.

Call by reference method:

1. Reference variables are used as formal parameters.
2. Actual parameters will be variables only.
3. The changes made in the formal arguments are reflected in actual arguments.
4. Memory of actual arguments is shared by formal arguments.

Answer 20.
It is a technology provides a simple and fast way to create dynamic web content. It uses Java as programming language. JSP files have the extension .jsp. To run JSP, Apache Tomcat web server is required. The JSP code consisting of HTML and Java is executed on the web server and the resulting HTML code is sent to the browser. JSP is an integral part of Java 2 Platform Enterprise Edition (J2EE).

Answer 21.
Height: Sets the height of the marquee in pixels or in percentage of browser window height.
Width: This specifies the width of the marquee in pixels or in percentage of browser window’s width value.
Direction: This specifies ttie direction in which marquee should scroN. This can have a value like up, down, left or right.
Direction: This specifies ttie direction in which marquee should scroN. This can have a value like up, down, left or right.

Answer 22.
a. 1
4
9
b. <SCRIPT> tag is used to specify the name of the scripting language used.
c. document.write(s); document.write(“<BR>”);

Answer 23.
1. Shard web hosting:
It is the most common type of web hosting. It is referred to as shared because many different wesites are stored on one single web server and they share resources like RAM and CPU. The features available on shared web servers are generally basic and are not flexible to suit a website that require specific features.
2. Dedicated web hosting:
It is the hosting where the client leases the entire web server and all its resources. The web server is not shared with any other website. Dedicated servers provide guaranteed performance, but they are very expensive. Since the bandwidth is not shared with other websites, it speeds up the access of the website. If the client is allowed to place their own purchased web server in the service providers facility, then it is called co-location.
3. Virtual Private Server (VPS):
It is a physical server that is virtually partitioned into several servers using the virtualization technology. Each VPS works similar to a dedicatd server and has its own separate server operating system, web server soft-ware and packages. VPS provides dedicated amount of RAM for each virtual web server. Each of these VPS works as a fully independent web server. VPS are provided with the rights to install and configure any software on their VPS.

Answer 24.
Alternate key: A candidate key that is not primary key is called alternate key. In the case of two or more candidate keys, only one of them can serve as the primary key. The rest of them are alternate key.
Foreign key: A key in a table can be called j foreign key if it is a primary key in another table. Since a foreign key can be used to link two or more tables it is also called a reference key.

Answer 25.
1. Physical level:
It is the lowest level of abstraction. The physical level describes complex low level data structures in detail. We must decide what file organisations are to be used to store the relations and create auxiliary data structures, called indexes, to speed up data retrieved operations.
2. Logical level:
It is the next higher level of abstraction describes what data is stored in the database, and what relation¬ship exist among those data. It describes : the entire database in terms of a small i number of relatively simple structures. The user of the logical level does not need to ‘ be aware of the complexity. Logical level ; is also referred as conceptual level.
3. View level:
It is the highest level of data-base abstraction and is closest to the users. It is concerned with the way in which 5 individual users view the data. It describes J only a part of the entire database. Most of i the users are not concerned with all the information that is contained in the data-base.

Answer 26.

1. NOT NULL: It specifies that a column can never have NULL values. NULL is a keyword that represents an empty value. Two NULL values cannot be added, subtracted and compared.
2. AUTO_INCREMENT: It performs an autoincrement feature. By default, the starting value for this constraint is 1 and will be incremented by 1 for each new record. Only one AUTO_INCREMENT column per table is allowed.
3. UNIQUE: It ensures that no two rows have the same value in the column specified with the constraint.

Answer 27.
a. SELECT MAX (Price), MIN (Price) FROM ITEM;
b. DELETE FROM ITEM WHERE Price <1000;
c. SELECT COUNT(*) FROM ITEM;

Answer 28.
a. # include<iostream>
using namespace std;
int main ( )
{
int n, sum = 0, i = 0;
do
{
sum = sum + i;
i+ + ;
} while (i<= 10)
cout<<“sum is = ”<<sum;
}
b. 25
c. # include statement, the pre-processor directive to attach a header file to provide information about predefined identifiers and functions used in the program.

Answer 29.
a. <INPUT>
b. <INPUT Type = “text” Name = “age” Size = “30” Maxlength = “25” >
c. <INPUT Type = “Submit” Value= “login”>

Answer 30.
a. <SCRIPT language = “JavaScript”>
var i = 1, s;
while (i<= 10)
{
s=i * i;
document.write(i);
document.write(“<BR>”);
i+ + ;
}
</SCRIPT>
b. <SCRIPT language = “JavaScript”>
var i;
for (i=2; i< = 10; i+2)
{
document.write(i);
document.write(“<BR>”);
}
</SCRIPT>

Plus Two Computer Application Previous Year Question Papers and Answers

Students can Download Chapter 3 Current Electricity Questions and Answers, Plus Two Physics Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

Kerala Plus Two Physics Chapter Wise Questions and Answers Chapter 3 Current Electricity

Plus Two Physics Current Electricity NCERT Text Book Questions and Answers

Question 1.
A storage battery of a car has an e.m.f. of 12V. If the internal resistance of the battery is 0.4W, what is the maximum current that can be drawn from the battery?
Answer:
E = 12 V, r = 0.4Ω, I = ?
Since I = \(\frac{E}{r}=\frac{12}{0.4}\) = 30A.

Question 2.
A battery of e.m.f. 10V and internal resistance 3W is connected to a resistor. If the current in the circuit is 0.5A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?
Answer:
Given E = 10V, r=3?, I = 0.5, R = ?
Since V = E – Ir
or V= 10 – 0.5 × 3
or V = 8.5 volt
And R = \(\frac{V}{I}=\frac{8.5}{0.5}\) = 17A.

Question 3.

1. Three resistors 1W, 2W and 3W are combined in series. What is the total resistance of the combination?
2. If the combination is connected to a battery of e.m.f. 12V and negligible internal resistance, obtain the potential drop across each resistor:

Answer:
Given

R₁ = 1Ω, R₂ = 2Ω , R₃= 3Ω
1. Total resistance of series combination
R = R₁ + R₂ + R₃
or R – 1 + 2 + 3 = 6Ω.

2. Since V=IR
I = \(\frac{V}{R}=\frac{12}{6}\)
∴ V₁ = IR₁ = 2 × 1 = 2V
V₂ = IR₂ = 2 × 2 = 4V
V₃ = IR₃ = 2 × 3 = 6V.

Question 4.
A silver wire has a resistance of 2.1W at 27.5°c, and a resistance of 2.7W at 100°C. Determine the temperature coefficient of resistivity of silver.
Answer:
R_t1 = R_27.5 = 2.1Ω and
R_t2 = R₁₀₀ = 2.7Ω
Applying the relation

a = 0.0039°C^-1

Question 5.
The number density of free electrons in a copper conductor estimated is 8.5 × 10²⁸m^-3. How long does an electron take in drifting from one end of a wire 3.0m long to its other end? The area of cross-section of the wire is 2.0 × 10^-6m² and it is carrying a current of 3.0A.
Answer:
n = 8.5 × 10²⁸m^-3; A = 2 × 10^-6m²
I – 3.0A; I = 3m
I = v_dneA
\(\frac{3}{t}=\frac{3}{8.5 \times 10 \times 1.6 \times 10 \times 2 \times 10}\)
t = 8.5 × 11²⁸ × 1 .6 × 10^-19 × 2 × 10^-6
= 27.2 × 10^-3sec.

Question 6.
Choose the correct alternative:

1. Alloys of metals usually have (greater/ less) resistivity than that of their constituent metals.
2. Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.
3. The resistivity of alloy manganin (is nearly indepent of/ increases rapidly) with increase of temperature.
4. The resistivity of a typical insulator (e.g. amber) is greater than that of a metal by a factor of the order of (10²²/10²³).

Answer:

1. Greater
2. lower
3. nearly independent of
4. 10²².

Plus Two Physics Current Electricity One Mark Questions and Answers

Question 1.
n equal resistors are first connected in series and then connected in parallel. What is the ratio of the maximum to the minimum resistance?
(a) n
(b) 1/n²
(c) n²
(d) 1/ n
Answer:
(c) n²
Explanation: In series R_s = nR
In parallel \(\frac{1}{Rp}\) = \(\frac{1}{R}\) + \(\frac{1}{R}\) ….n terms, Rp = \(\frac{R}{n}\)
∴ R_s/R_p = n²/1.

Question 2.
A car battery of emf 12 V and internal resistance 5 × 10^-2Ω, receives a current of 60 amp, from external source, then potential difference of battery is.
(a) 12V
(b) 9V
(c) 15V
(d) 20 V
Answer:
(c) 15V
When cell is changed by an external source, terminal voltage, V = E + Ir
V= 12 + 60 × 5 × 10^-2
=15.

Question 3.
A flowing of 10⁷ electron persecond in a conducting wire constitutes a current of……….A
(a) 1.6 × 10^-12
(b) 1.6 × 10²⁶
(c) 1.6 × 10^-26
(d) 1.6 × 10¹²
Answer:
(a) 1.6 × 10^-12
Flow of electrons \(\frac{n}{t}\) = 10⁷/sec.
There fore, Current (I) = \(\frac{q}{t}\) = \(\frac{ne}{t}\) = \(\frac{n}{t}\) × e
= 10⁷ × (1.6 × 10^-19)
= 1.6 × 10^-12 A.

Question 4.
Copper and silicon is cooled from 300K to 60K, the specific resistance.
(a) decrease in copper but increase in silicon
(b) increase in copper but decrease in silicon
(c) increase in both
(d) decrease in both
Answer:
(a) decrease in copper but increase in silicon.

Question 5.
State the potentiometer principle.
Answer:
Potential difference between two points of a current carrying conductor is directly proportional to the length of the wire between two points.

Question 6.
Find the current following through the network shown in the figure

Answer:
Since given circuit is in the form of Wheatstone bridge,

Question 7.
Kirchhoff’s first and second laws of electrical circuits are consequences of……(1)….and…..(2).. respectively
Answer:

• Conservation of electric charge
• Energy respectively

Question 8.
Pick the odd one out the following.
(a) Ohm’s law
(b) Lenz’slaw
(c) Coulomb’s law
(d) Gauss’s law
(e) Energy conservation law
Answer:
(a) Ohm’s law (It is not a universal law).

Question 9.
“Ohms law is not a fundamental law” Comment on this.
Answer:
Ohms law is not a universal law because metals do not obey this law at high temperature. Moreover, certain materials (diode and transistors, etc.) does not obey ohms law.

Plus Two Physics Current Electricity Two Mark Questions and Answers

Question 1.
Figure below shows a diagram of a water circuit. In many ways it behaves like an electric circuit. Draw an equivalent electric circuit. (Hint – water-wheel can be replaced by motor)

Answer:

Question 2.
A, B, C, and D are four rings on a carbon resistor. A = yellow, B = violet, C = Yellow, D, Silver

1. What is the value of resistance of above resistor?
2. The combined resistance of the above two resistor is

• 120Ω
• 45Ω
• 165Ω
• 35Ω

Answer:
1. 47 × 10⁴ ± 10%.

2. 12 × 10¹ + 45 × 10°
120 + 45 = 165Ω.

Question 3.
A = Brown B = Black C= Red D= Gold

1. What is the value of resistance without considering variation?
2. If the fourth ring is silver coloured what will be the change in accuracy?

Answer:

1. 1000Ω ± 5%
2. 10%

Question 4.
A junction of a electrical circuit is given below. Analyze the figure and answer the following

1. What is the value of I₁ and I₂?
2. State the law that can be applied to find I₁ and I₂?

Answer:
1. According Kirchoff first rule
3 + 2 = I + I₁
I₁ = 4A
I₁ = 3 +I₂
4 = 3 + I₂
I₂ = 1A

2. Total current meeting at any junction is zero.

Question 5.
The following question consists of two statements each, printed as assertion and reason. While answering these questions you should choose any one of the following responses.
Assertion: In a simple battery circuit, the point at the lowest potential is positive terminal of the battery. Reasons: The electrons flows from higher potential to lower potential.
(a) Both assertion and reason are true and the reason is a correct explanation of the assertion.
(b) Both assertion and reason are true but the reason is not a correct explanation of the assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(d) Both assertion and reason are false.

Plus Two Physics Current Electricity Three Mark Questions and Answers

Question 1.
Match the following.

Answer:

Question 2.
Match the following.

Answer:

Question 3.
“When electric current is passed through a resistance wire, it get heated up”.

1. Name the law associated with this phenomenon.
2. What happens to the heat energy developed, if the current through the wire is doubled.

Answer:

1. Joules law of heating
2. Heat developed in the conductor H = 1²R If current is doubled, Heat developed becomes 4 times.

Question 4.

1. Name the pd between terminals of the cell when

• key K is open
• K is closed

2. What is the reason for the difference in potential in the above two cases?
Answer:
1. key K is open

• e.m.f
• voltage.

2. Emf in a circuit
E = Ir + V
When key is open, I = 0
∴ E = V
When key is closed I ? 0
∴ V = E – Ir.

Question 5.
The variation of resistivity (ρ) with temperature (T) of a conductor, semiconductor and super conductor are given in the figure.

1. Identify them from the graph. (1)
2. Identify the figure in which the temperature coefficient of resistance of the material is positive. (1)
3. Write the equation connecting resistivity of the material with relaxation time. (1)

Answer:
1. From the graph:

• Fig (i) – Conductor
• Fig (ii) – Superconductor
• Fig (iii) – Semiconductor

2. Fig(i)

3. We have resistivity (ρ) from equation.
ρ = \(\frac{m}{n e^{2} \tau}\), where t is the relaxation time, m mass of electron, n number density of electron and e charge of electron.

Question 6.
When an animal touches an ‘electric-fence’ the animal gets a shock by completing an electrical circuit.

1. Draw an equivalent circuit using symbols, which shows the completed electrical circuit. (Assume that the animal has a resistance)
2. How much charge passes through the animal, if it receives a current pulse of 20mA for 0.1 seconds.
3. When a bird perches on the electric fence, the bird will not get a shock. Why?

Answer:
1.

2. 1 = Q/t
Q = 1 × t = 20 × 10^-3 × 0.1 = 20 × 10^-4C

3. Since the bird does not come in contact with the earth, its body is at same potential. So there is no potential difference and current. Hence the bird will not get a shock.

Question 7.
1. Pd between terminals when k is open is called

• emf
• lost voltage
• terminal voltage
• induced voltage

2. Give an equation connecting R, r, I and E What is the value of E for open circuit and ideal cell.

3. What happens to terminal voltage if current increases

• for an ideal cell
• for an ordinary cell.

Answer:
1. e.m.f

2. E = I(r+R).

3. For ideal cell, internal resistance is zero. Hence E = v. ie. terminal voltage does not change with current.
For ordinary cell, when current increases, V decreases.

Question 8.

1. What is the current through this circuit?
2. What is p.d. across 2Ω and 3Ω?
3. What is potential at A?
4. What is potential at B?

Answer:
1. Current I = \(\frac{\text { total voltage }}{\text { total resistance }}\)
I = \(\frac{5}{2+3}\) = 1A

2. Voltage across 2Ω
V = 1 × 2
V₂ = 2 V
Voltage across 3Ω
V = 1 × 3 = 3V

3. 5V

4. 3V

Question 9.
The relation between Voltage V (across the conductor) and current I through the conductor is given in the graph.

1. Which law establishes the relation between voltage and current?
2. A metal wire of resistivity 6.4 × 10^-5 ohm-cm and length 1.98 m has a resistance of 7Ω. Find radius of wire

Answer:
1. Ohms law

2. Resistance of wire R = \(\frac{ρl}{A}\)

= 2 × 10^-3m.

Question 10.
A circuit diagram of an instrument is given below.

1. Identify the instrument and state the principle of this instrument
2. Modify the circuit diagram to compare the emf of two cells.
3. How can increase the sensitivity of this instrument?

Answer:
1. Potentio meter

2. Comparison of e.m.f of two cells using potentiometer

a. Principle:
Potential difference between two points of a current carrying conductor (having uniform thickness) is directly proportional to the length of the wire between two points.

b. Circuit details :
A battery (B₁), Rheostat and key are connected in between A and B. This circuit is called primary circuit. Positive end of E₁ and E₂ are connected to A and other ends are connected to a two way key. Jockey is connected to a two key through galvanometer. This circuit is called secondary circuit.

c. Working and theory :
Key in primary circuit is closed and then E₁ is put into the circuit and balancing length l₁ is found out.
Then E_1∝ l₁ …………(1)
Similarly E₂ is put into the circuit and balancing length (l₂) is found out.
Then, E_{2 ∝} l₂ (2)
Dividing Eq(1)byEq(2),
\(\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}\)………..(3)

3. Increase the length of potentio meter wire.

Question 11.
A long resistance wire is stretched between two iron nails. A battery of 2V is applied across the wire. One end of a torch bulb is connected to nail and other end is made in contact as shown in figure.
1. If this wire slides over the resistance wire from nail 1 to nail 2, what happens to the brightness of the bulb.

• Increases
• Decreases
• Remains constant
• First increases then decreases.

2. How this principle is used to determine internal resistance of cell.

3. What happens to the reading, if we change 2V with 3V during the time of reading.

Answer:
1. Increases.

2. Measurement of internal resistance using potentiometer Principle:
[Same as before]
a. Circuit details :
Battery B₁ Rheostat and key K₁ are connected in between A and B. This circuit is called primary.

In the secondary circuit a battery E having internal resistance ‘r’ is connected. A resistance box (R) is connected across the battery through a key (K₂). Jockey is connected to battery through galvanometer.

b. Working and theory :
The key (K₁) in the primary circuit is closed and the key is the secondary (K₂) is open. Jockey is moved to get zero deflection in galvanometer. The balancing length l₁, (from A) is found out.
Then we can write.
E_1∝ l₁ ____(1)
Key K₂ is put in the circuit, corresponding balancing length (l₂) is found out. Let V be the applied voltage, then
V_{1 ∝} l₁ ____(2)
‘V’ is the voltage across resistance box. Current through resistance box ie, voltage across resistance,
V = \(\frac{E R}{(R+r)}\) _____(3)
Substituting eq (3) in eq (2),
\(\frac{E R}{(R+r)}\) _∝ l₂ ______(4)
Dividing eq (1) by eq (4),

r = \(\frac{\mathrm{R}\left(l_{1}-l_{2}\right)}{l_{2}}\)

3. Primary Voltage should not change while doing experiment. When we use 3v instead of 2v, potential gradient will change. Hence balancing length will change.

Question 12.
“Electric current has both direction and magnitude”

1. What is meant electric current
2. What is the conventional direction of electric current
3. Even though current has both magnitude and direction it is not a vector quantity. Why?
4. Thermal Motion of electrons in a conductor cannot constitute an electric current. Why?

Answer:

1. Rate of flow of charge is called current I = dq/dt
2. Direction of motion of positive charges
3. It does not obey vector law of addition
4. Average velocity of thermal motion is zero. Hence thermal motion does not produce current.

Question 13.
Match the following

A	B
Metals	Cooper pairs
Semiconductors	electrons and holes
superconductors	positrons
	electrons

Answer:

1. Metals – electrons
2. Semiconductors – electrons and holes
3. Superconductors – cooper pairs

Plus Two Physics Current Electricity Four Mark Questions and Answers

Question 1.
A cell arid two resistors R₁ and R₂ are provided to you.

1. Draw different combinations of resistors using R₁, R_2, and the cell.
2. Derive an expression for the effective resistance of the circuit in which current is the same in both resistors
3. lf R₁= 4Ω and R₂ = 6Ω, in which combination effective resistance is minimum? Find its value?

Answer:
1.

2. Derive an expression for effective resistance in series:
Consider three resistors R₁, R₂ and R₃ connected in series and a pd of V is applied across it.

In the circuit shown above the rate of flow of charge through each resistor will be same i.e. in series combination current through each resistor will be the same. However, the pd across each resistor are different and can be obtained using ohms law.
pd across the first resistor V₁ = I R₁
pd across the second resistor V₂ = I R₂
pd across the third resistor V₃ = I R₃
If V is the effective potential drop and R is the effective resistance then effective pd across the combination is
V = IR
Total pd across the combination = the sum pd across each resistor, V = V₁ + V₂ + V₃
Substituting the values of pds we get IR = IR₁ + IR₂ + IR₃
Eliminating I from all the terms on both sides we get
R = R₁ + R₂ + R₃ ………(1)
Thus the effective resistance of series combination of a number of resistors is equal to the sum of resistances of individual resistors.

3. The effective resistance becomes minimum in parallel connection.

R = 2.4Ω.

Question 2.
1. State whether the following statement is true or false “The value of resistance of a metal increase with the rise of temperature”.

2. Explain the reason.

3. With the help of the graph, match the following

4. Alloys like manganin, eureka, constantan, etc. are used in making standard resistance coils. Why?
Answer:
1. True.

2. When temperature increases, the amplitude of vibration of atom increases. Hence relaxation time decreases. Hence resistivity of metal increases according to the equation

3.
A_____Carbon
B_____Manganin
C_____Iron.

4. The temperature coefficient of resistance of manganin, eureka and constantan, etc are zero. Hence they are used in making standard resistance coils.

Question 3.

1. Identify the above device and give the principle behind it.

2. Obtain the mathematical condition for the galvanometer current to be zero.

3. If the balancing length T obtained fora resistance wire in the arrangement is 40cm. Find the new balancing length if the same resistance wire is folded to half its length and connected to the same gap.
Answer:
1. Meter bridge.

2. We get galvanometer current as zero, when P/Q = R/S. For derivation of P/Q=R/S
Wheatstone’s Bridge:
Four resistances P, Q, R, and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I₁, I₂, I₃ and I₄ be the four currents passing through P, R, Q, and S respectively.

Working:
The voltage across R
When key is closed, current flows in different branches as shown in figure. Under this situation
The voltage across P, V_AB = I₁P
The voltage across Q, V_BC = I₃Q ……(1)
The voltage across R, V_AD = I₂R
The voltage across S, V_DC = I₁S
The value of R is adjusted to get zero deflection in galvanometer. Under this condition,
I₁ = I₃ and I₂ = I₄……(2)
Using Kirchoff’s second law in loopABDA and BCDB, weget
V_AB = V_AD ……….(3)
and V_BC = V_DC …….. (4)
Substituting the values from eq(1) into (3) and (4), we get
I₁P = I₂R……….(5)
and I₃Q = I₄S……..(6)
Dividing Eq(5) by Eq(6)
\(\frac{I_{1} P}{I_{3} Q}=\frac{I_{2} R}{I_{4} S}\)
\(\frac{P}{Q}=\frac{R}{S}\) [since I₁ = I₃ and I₂ = I₄]
This is called Wheatstone condition.

3. When we apply this condition in meter bridge, we get

If wire is folded, New resistance x₁ = \(\frac{x}{2}\)
Substituting this in P/Q = R/S we get

Question 4.
1. Potentiometer is better than voltmeter for measuring emf because

• It is cheap
• Easy to handle
• Its measurement uses null method

2. Give the basic principle of potentiometer,

3.

If K₂ is open balancing length is 600cm, if K₂ is closed 350cm is balancing length. Calculate the internal resistance.

Answer:
1. Its measurement uses null method.

2. Principle:
Potential difference between two points of a current carrying conductor (having uniform thickness) is directly proportional to the length of the wire between two points.

3. I = 600cm, I = 350 cm

Question 5.
Under an external electric field electrons drift slowly inside the conductor.
1. The velocity of drift is

• 1 mm/s
• 10⁵ m/s
• 3 × 10⁸ m/s
• 3 × 10⁹ m/s

2. What is meant by relaxation time?

3. Write an expression for drift velocity in terms of relaxation time.

4. When temperature increases what happens to drift velocity?
Answer:
1. 1mm/s.

2. The average time between two successive collision is called relaxation time.

3. V_d = \(\frac{\mathrm{eE}}{\mathrm{m}}\) t

4. We know drift velocity
V_d = \(\frac{\mathrm{eE}}{\mathrm{m}}\) t
When temperature increases, relaxation time decreases. As a result drift velocity decreases.

Question 6.
1. State whether the following statement is correct. “If the current through the cell is from its positve to negative, against the direction of emf, then the potential drop across the internal resistance aids the emf of the cell. (1)

2. You are given two cells. Group them so that they give more voltage. Arrive at the expression for effective emf and internal resistance. (3)
Answer:
1. True. Internal resistance and lost volt always opposes current flow through the cell”.

2.

Considertwo cells in series. Let ε₁, r₁ be the emf and internal resistance of first cell. Similarly ε₂, r₂ be the emf and internal resistance of second cell. Let I be the current in this circuit.

From the figure, the P.d between A and B
V_A – V_B = ε₁ – 1 ………(1)
Similarly P.d between B and C
V_B – V_C = ε₂ – 1 ……..(2)
Hence, P.d between the terminals Aand C
V_AC = V_A – V_C = V_A – V_B + V_B – V_C
V_AC = [V_A – V_B] + [V_B – V_C]
when we substitute eqn. (1) and (2) in the above equation.
V_AC = ε₁ – Ir₁ + ε₂ – Ir₂
V_AC = (ε₁ – ε₂) – I(r₁ + r₂)
V_AC = ε_eq – Ir_eq
where ε_eq = ε₁ + ε₂, and r_eq = r₁ + r₂

Question 7.
The resistance value of a conductor depends on its physical dimensions.
1. Give the expression for resistance of a conductor in terms of its physical dimension.

2.

(l length and d diameter) A potential V is applied between the ends of two conductor of same material shown in the figure.

• Express the resistance of the second conductor in terms of the resistance of the first conductor. (1)
• Find the ratio of electric field across the two conductors.

Answer:
1. R = ρ \(\frac{1}{A}\)
l – length of the conductor and
A – Area of cross section

2.
a.

b.

Question 8.
Three resistors R₁, R₂, R₃ are to be combined as shown in the figure.

1. Identify the series and parallel combinations. (1)
2. Which combination has lowest resistance. Arrive at the expression for the effective resistance of this combination. (3)

Answer:
1. Fig(i) – parallel Fig. (ii) – series

2. Fig. (i)
Let V be the potential between A and B.

I = I₁ + I₂ + I₃
and applying Ohm’s law to R₁, R₂ and R₃ we get,
V= I₁R₁,V=I₂R₂,V=I₃R₃
So that
I = I₁ + I₂ + I₃ = V\(\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\right)\)
An equivalent resistance R_eq that replace the combination, and hence

Question 9.
An electric circuit is given in the Figure.

The potential difference between A and D is 40V.

1. Find the effective resistance between A and D in terms of R. (1)
2. Calculate the potential difference between A and O.

Answer:
1.

2. V_AD = V_AO + V_OD
But V_OD = 3V_AO
∴ V_AD = V_AO + 3V_AO
40 = 4V_AO
V_AO = \(\frac{40}{4}\) = 10V.

Plus Two Physics Current Electricity Five Mark Questions and Answers

Question 1.
Resistance are used to reduce the current flow in a circuit.
1. A carbon resistor has coloured strip and shown in the figure. What is its resistance?

2. Resistance can be connected in series and parallel to obtain the required value of resistance. Derive an expression for the effective resistance when three resistors are connected in parallel.

3. Kirchhoff’s rules are used to analyses the electric circuit. Use it to analyze the Wheatstone Bridge and arrive at Wheatstone’s condition for balancing the bridge.
Answer:
1. blue, Gray, yellow, gold
1st – Blue – 6
IInd – Gray – 8
IIIrd – Yellow – 10⁴
Iv – gold – 5%
68 × 10⁴ ± 5%.

2. Derive an expression for effective resistance in series:
Consider three resistors R₁, R₂ and R₃ connected in series and a pd of V is applied across it.

In the circuit shown above the rate of flow of charge through each resistor will be same i.e. in series combination current through each resistor will be the same. However the pd across each resistor are different and can be obtained using ohms law.
pd across the first resistor V₁ = I R₁
pd across the second resistor V₂ = I R₂
pd across the third resistor V₃ = I R₃
If V is the effective potential drop and R is the effective resistance then effective pd across the combination is
V = IR
Total pd across the combination = the sum pd across each resistor, V = V₁ + V₂ + V₃
Substituting the values of pds we get IR = IR₁ + IR₂ + IR₃
Eliminating I from all the terms on both sides we get
R = R₁ + R₂ + R₃ ………(1)
Thus the effective resistance of series combination of a number of resistors is equal to the sum of resistances of individual resistors.

3. Wheatstone’s Bridge:
Four resistances P,Q,R and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I₁, I₂, I₃ and I₄ be the four currents passing through P,R,Q and S respectively.

Working:
The voltage across R
When key is closed, current flows in different branches as shown in figure. Under this situation
The voltage across P, V_AB = I₁P
The voltage across Q, V_BC = I₃Q ……(1)
The voltage across R, V_AD = I₂R
The voltage across S, V_DC = I₁S
The value of R is adjusted to get zero deflection in galvanometer. Under this condition,
I₁ = I₃ and I₂ = I₄……(2)
Using Kirchoff’s second law in loopABDA and BCDB, weget
V_AB = V_AD ……….(3)
and V_BC = V_DC …….. (4)
Substituting the values from eq(1) into (3) and (4), we get
I₁P = I₂R……….(5)
and I₃Q = I₄S……..(6)
Dividing Eq(5) by Eq(6)
\(\frac{I_{1} P}{I_{3} Q}=\frac{I_{2} R}{I_{4} S}\)
\(\frac{P}{Q}=\frac{R}{S}\) [since I₁ = I₃ and I₂ = I₄]
This is called Wheatstone condition.

Question 2.
A rectangular conductor of length I and area of cross section A and electron density n; is shown below.

1. When the face Y is given positive potential and X negative potential what will happen to the electrons inside the block
2. What is meant by drift velocity? How is it related to the field inside the metal?
3. Deduce an expression connecting intensity of electric field and drift velocity.
4. Under the application of an electric field do all the electrons move in a same direction? Explain

Answer:
1. Electrons will be accelerated towards the side Y.

2. Drift Velocity (v_d):
The average velocity acquired by an electron under the applied electric field is called drift velocity.
Explanation :
When a voltage is applied across a conductor, an electric filed is developed. Due to this electric field electrons are accelerated. But while moving they collide with atoms, lose their energy and are slowed down. This acceleration and collision are repeated through the motion. Hence electrons move with a constant average velocity. This constant average velocity is called drift velocity.

3. Expression for drift velocity :
Let ‘V’ be the potential difference across the ends of a conductor. This potential difference makes an electric field E. Under the influence of electric field E, each free electron experiences a Coulomb force.
F = -eE
or ma = -eE
a = \(\frac{-e E}{m}\) ……….(1)
Due to this acceleration, the free electron acquires an additional velocity. A metal contains a large number of electrons.
For first electron, additional velocity acquired in a time τ,
v₁ = u₁ + aτ₁
where u₁ is the thermal velocity and τ is the relaxation time.
Similarly the net velocity of second, third……electron
v₂ = u₂ + aτ₂
v₃ = u₃ + aτ₃
v_n = u_n + aτ_n
∴ Average velocity of all the ‘n’ electrons will be

V_av = 0+ aτ (∴ average thermal velocity of electron is zero)
where

where V_av is the average velocity of electron under an external field. This average velocity is called drift velocity.
ie. drift velocity V_d = aτ……(2)
Substitute eq (1) in eq (2)
V_d = \(\frac{-e E}{m}\) τ

4. Electrons will continues its random thermal motion even in the presence of electric field.

Question 3.
To study the relation between potential difference and current in an electrical circuit, a student is provided with a resistance wire, a cell, and a key.
1. Draw a circuit which allows current flow through the resistance wire.

2. Modify the circuit by introducing an ammeter, Voltmeter and a rheostat for varying the potential difference across the resistance and to measure that potential difference and the corresponding current.

3. Let in the above experiment the student obtained the following data.

Draw a graph connecting V and I using above data. Then establish the relation between V and I as a law.

4. Instead of the resistance wire if the student uses a p-n junction diode in the forward biased condition how the relation between V and I changes? Justify.
Answer:
1.

2.

3.

The above graph shows that, current flowing through a conductor is directly proportional to potential difference across it ends.

4. The relation between V and I becomes nonlinear. Because, Pn diode does not obey ohms law.

Question 4.
In all metallic conductors, electric conduction is due to drifting of free electrons. But the resistivity of different metals are different.

1. Write the expression for resistivity of a conductor in terms of its dimensions.
2. Name the factors on which resistivity of a metal depends.
3. Arrive at an expression for electrical resistivity of a metal in terms of relaxation time.
4. Using the above expression explain the variation of resistivity with temperature.

Answer:
1. ρ = \(\frac{RA}{l}\).

2. Temperature and Nature of metal

3. We know current density
J = nv_de
But V_d = \(\frac{e E}{m}\) τ

4. When temperature increases, the amplitude of oscillation of atom increases. This will decrease . the relaxation time and hence resistivity increases.

Question 5.
You are supplied with a 1m long uniform resistance wire of resistance 3Ω and a cell of emf 1.5v.

1. Can you construct a potentiometer using 1m wire? If no, give reason; if yes, what is the least count of the arrangement?
2. Draw the connection diagram to compare the emf of Leclanche cell and Daniel cell using this arrangement.
3. How can you modify above arrangement to measure a p.d. in the range 0-1 mV with a least count of 0.15m V.

Answer:

1. Yes, LC is 1.5mV

2.

3. To get least count of 0.15mV, 10m wire has to be used.

Question 6.
A circuit diagram is given below. Analyze the figure and answer the following questions.

1. The above circuit is a modification of……..
2. What is the value of balancing length?
3. If 1.5 V cell is replaced by a 3V cell what will be the balancing length?
4. Calculate the value of X.

Answer:
1. Wheatstone bridge.

2. 100-20=80cm

3. No change in balancing length

4. P/Q = R/S
ie X/5 = 80/20
x = 80/20 × 5 = 20Ω.

Question 7.

1. Define electric power. What is its SI unit? (2)
2. Two bulbs of 50W, 220V and 100W, 220V are given. How will you connect the bulbs so that 50W, 220V bulb will glow brighter than 100W, 220V bulb. (1)
3. When the bulb 50W, 220V is connected to an 110V supply calculate the power generated. (2)

Answer:
1. The energy dissipated per unit time is the power.
P = \(\frac{\Delta \mathrm{W}}{\Delta \mathrm{t}}\)
P = IV =I²R = \(\frac{V^{2}}{R}\)
Unit is watt.

2. In series

3.

Question 8.
The figure shows the diagram of a potentiometer.

1. Give the principle of a potentiometer. (1)

2. The length of AB is 3m and resistance per unit length of the potentiometer wire is 4Ω/m. If E₁ = 4V, R = 20Ω and E₂ = 1V find the length of the potentiometer wire that balance E₂. (3)

3. If E₂>E₁ can we get the null deflection in galvanometer. Give reason. (1)
Answer:
1. When a constant current is flowing through a wire having uniform area of cross section and uniform composition the potential difference across any length of the wire is directly proportional to its length.
V ∝ l.

2. The resistance of potentiometer
R_p = 4 × 3 = 12Ω
Current through the potentiometer

Potential across potentiometer wire
V_AB = IR_p = 0.125 × 12 = 1.5V
Potential gradient k = \(\frac{V_{A B}}{I}=\frac{1.5}{3}\) = 0.5 V/m
Balancing length for cell E₂ is given from equation
E₂ = kI₂ = 0.5I₂

3. If E₂ > E₁, we will not get null deflection. The potential difference across the potentiometer wire AB should be higher than the emf of E₂.

Question 9.

1. When the switch is closed will all the bulbs glow? Give reason. (2)
2. Identify the underlying principle. Deduce the principle for a resistance network. (3)

Answer:
1. No. Since all the bulbs are identical the bridge is balanced. So potential B and C is same and no current flows through the bulb connected between B and C. So all other bulbs except R₅ will glow.

2. Four resistances P, Q, R, and S are connected as shown in figure. Voltage ‘V’ is applied in between A and C. Let I₁, I₂, I₃, and I₄ be the four currents passing through P, R, Q, and S respectively.

Kerala Plus Two Computer Application Model Question Paper 3

Time: 2 Hours
Cool off time: 15 Minutes
Maximum: 60 Scores

• There is a ‘cool off time’ of 15 minutes in addition to the writing time of 2 hrs.
• Use the ‘cool off time’ to get familiar with the questions and to plan your answers.
• Read questions carefully before you answering.
• Read the instructions carefully.
• Calculations, figures and graphs should be shown in the answer sheet itself.
• Malayalam version of the questions is also provied.
• Give equations wherever necessary.
• Electronic devices except non programmable calculators are not allowed in the Examination Hall.

Part A

Answer all questions from 1 to 5

Question 1.
Character array is also called ……………..

Question 2.
What is a Form handler?

Question 3.
What is the basic protocol followed in internet communication?

Question 4.
SAP stands for …………..

Question 5.
Define enterprise resource planning.

Part B

Answer any nine questions from 6 to 16

Question 6.
Write the output of the following web page and justify your answer.

<HTML>
<BODY>
<SCRIPT language= “JavaScript’’>
var x, y, z;
x = “10”;
y = “20”;
z = x + y;
document.write(z);
</SCRIPT>
</BODY>
</HTML>

Question 7.
Categorize the following tags into container tags and empty tags.

Question 8.
Mention any two values of Type attribute and explain its use in the form.

Question 9.
How is responsive web design implement-ed?

Question 10.
a. What is responsive web design?
b. Why is it gaining importance recently?

Question 11.
Explain about sophisticated user.

Question 12.
What is the importance of primary key in a table?

Question 13.
Write a short note about any two ERP solution providers.

Question 14.
“Infomania has become a psychological problem”. Write your opinion.

Question 15.
What is a smart card? How it is useful?

Question 16.
What do you mean by big data in business? Explain big data analytics.

Part C

Answer any nine questions from 17 to 27

Question 17.
Consider the following program:

# include<iostream>
Using namespace std;
int main ( )
{
for (int i=1; i<10; i+ + )
{
if (i % 2 = = 0)
continue;
cout<<i;
}
}
Predict the output of the program. Justify your answer?

Question 18.
Rewrite the following C++ code using switch statement.

if(choice== ‘S’)
cout<<“Science”;
else if (choice== ‘C’)
cout< <“ Commerce”;
else if (choice== ‘H’)
cout<<‘‘Humanities”;
else ‘
cout«“Invalid Option”;

Question 19.
The prototype of p function is: W fun(int, int); The following function calls are invalid. Give reason for each.

Question 20.
Discuss the scope of global and local variables.

Question 21.
In HTML how can we divide a browser window?

Question 22.
Aliya wants to display three web pages (A.htm, B.htm, C.htm) on the same scre¬en horizontally at the ratio 20%, 40%, 40%. Write the HTML code for the same.

Question 23.
Classify the following values in JavaScript into suitable data types.

Question 24.
Write a note on procedures.

Question 25.
What are the advantages of views in MySQL

Question 26.
What is copyright? How does it differ from patent?

Question 27.
Write a function which accepts an integer value as an argument and calculate its factorial. Call this function from main ( ) and print the result in main ( ).

Part D

Answer any two questions from 28 to 30

Question 28.
Read the following C++ code segment and replace it with a looping statement:

Question 29.
What are scripts? Explain the different scripting languages.

Question 30.
Explain operators in JavaScript.

ANSWERS

Answer 1.
string

Answer 2.
A form handler is a program on the web server that manage the data sent through the form.

Answer 3.
TCP/IP

Answer 4.
Systems, Applications and Products

Answer 5.
Enterprise Resource Planning (ERP) is business process management software that allows an organization to use a system of integrated applications to manage the business and automate many back office functions related to technology, services and human resources.

Answer 6.
It will display 1020. Operator + is used to add two strings. Adding two string means concatenating two strings. If the operands are numbers, it will add the numbers. If the operands are strings it will concatenate the strings. If + operator sees any one operand as string, it will treat both the operands as string type and concatenate the strings.

Answer 7.
Empty tag: <FRAME>, <INPUT>
Container tag <FRAMESET>, <A>

Answer 8.
Values of type attribute is radio, check box. Radio button: Used to select a single value from a group of values. Check box: It is useful for accepting Yes/ No inputs from the viewers.

Answer 9.
It can be implemented using flexible grid layout, flexible images and media queries. Flexible grid layouts set the size of the en-tire web page to fit the
display size of the device. Flexible images and videos set the image/ video dimensions to the percentage of display size of the device. Media queries provide the ability to specify defferent styles for individual devices.

Answer 10.
a. Responsive web design is the custom of building a website suitable to work on every device.
b. Today people visit websites using tablets and mobile phones/proper display of websites in devices.

Answer 11.
Sophisticated users include engineers, scientists, business analysts and others who are thoroughly familiar with the facilities of the DBMS. They interact with the systems through their own queries (a re-quest to a database) to meet their complex requirements.

Answer 12.
Primary key is a set of one or more attributes that can uniquely identify tuples within the relation. It cannot contain null value and duplicate value.

Answer 13.
Oracle: Head quarters at Redwood shores, California, USA. Oracle was working for its database rather than ERP. Oracle is famous for its finance and accounting module. It also provide Customer Relationship Management (CRM) software and Supply Chain management software (SCM). Tally ERP: Indian company with head quarters at Banglore. Provide ERP solution for accounting, inventory and payroll.

Answer 14.
Infomania may result in neglecting the .more important things like duties, family, etc. It is now treated as a psychological problem. Constantly checking e-mails, social networking sites, online news, etc. are the symptoms of infomania. Studies prove that people addicted to infomania lose concentration and sleep.

Answer 15.
It is a plastic card with a computer chip or memory that stores and transacts data. A smart card (may be like an ATM card) reader used to store and transmit data. The advantages of smart card is that it is secure, intelligent and is easy to carry.

Answer 16.
Big data analytics is the process of examining large data sets containing a variety of data types to uncover hidden patterns, market trends, customer preferences and useful business information. Big data analytics provide an opportunity for business firms to answer questions that were previously considered beyond their reach. This opened the doors to a new world of possibilities to interact effectively with the customer, with the knowledge of his preferences.

Answer 17.
The output is 113579. continue statement is used for skipping over depart of the code within the loop-body anchforcing the next iteration.

Answer 18.
switch(choice)
{
case ‘S’: cout<<“Science”; break;
case ‘C’: cout<<“Commerce”; break;
case ‘H’: cout<<“Humanities”; break;
default : cout<<“lnvalid Option”;
}

Answer 19.
a. The variable is not receive the value returned by fun( );
Correct is val = fun(2,4);
b. No arguments are used with function call.
cout<<fun(a,b);
c. Arguments in function call are not int type.
val = fun(2,3);

Answer 20.
Scope of local variables:

1. Declared within a function or a block of statements.
2. Available only within that function or block.
3. Memory is allocated when the function or block is active and freed when the execution of the function or block is completed.

Scope of global variables:

1. Declared outside all the functions.
2. Available to all the functions of the program.
3. Memory is allocated just before the ex-ecution of the program and freed when the program stops execution.

Answer 21.
The browser window can be divided into two or more panes to accommodate different pages simultaneously. HTML provides a facility called frameset to partition the browser window into different sections. Each section accommodates different HTML pages. Each individual section created by a frameset is called a frame. To create a frameset, we can use <FRAMESET> and <FRAME> tags.

Answer 22.
<FRAMESET ROWS = “20%, 40%, 40%”>
<FRAME Src= “A.htm”>
<FRAMESrc= “B.htm”>
<FRAME Src= “C.htm”>
</FRAMESET >

Answer 23.
Number: 67.4, .98
String: “Welcome’’, “123”, “true”, “hello”
Boolean: false

Answer 24.
Procedures refer to the instructions and rules that govern the design and use of database. The users of the system and the person that manages the database require documented procedures on how to use or run the system. These consist of instruction on how to:

• log onto the DBMS.
• use a particular DBMS facility or application program.
• start and stop the DBMS.
• make backup copies of the database or handle hardware or software failures.
• reorganise the database across multiple disks, improve performance or archive data to secondary storage.

Answer 25.

1. Without sparing extra storage space, we can use the same table as different tables (but virtual).
2. The views implement sharing along with privacy.
3. It also helps to reduce the complexity of conditions with WHERE clause while retrieving, updating or deleting records from tables.

Answer 26.
Copyright: It is a legal right given to the creators for an original work, usually for a limited period of time. The general rule is that the copyright lasts for 60 years after the death of the last surviving author. Registration of copyright gives a legal status to a creative work.
Patents: It is the exclusive rights granted for an invention. The term of every patent India is 20 years from the date of patent application. Once a patent expires, the protection ends and an invention can be used by the public freely.

Answer 27.
# include<iostream> using namespace std; int fact(int);
int main( )
{
int n, r;
cout«“Enter the value:”;
cin>>n;
r = fact(n);
cout<<r;
return 0;
}
int fact(int N)
{
int f;
for(f = 1; N>0; N- -) f = f*N; return f;
}

Answer 28.
# include <iostream> using namespace std; int main( )
{
int n, r, s = 0; cout<<“Enter the number: cin>>n; while(n!= 0)
{
r = n % 10; s = (s * 10)+ r; n = n/10;
}
cout«“The reverse of the number is”
<<s;
return 0;
}

Answer 29.
Scripts are program codes written inside HTML pages. They are written using a text editor like notepad. Scripting languages are:
JavaScript: It is a client side scripting language used to make web pages inter-active which was developed by Brendan Eich. On the client side, JavaScript is implemented as an interpreted language. JavaScript can be inserted inside HTML code or can be written as an external file and then included inside HTML file.
VBScript: It is a scripting language developed by Microsoft Corporation based on the popular programming language Visual Basic.
PHP: It is an open source general-purpose scripting language that is suited for web development and can be embedded into HTML code. It is a server side scripting tool. The main objective of PHP is to develop dynamic web pages at ease.
ASP (Active Server Pages): It can be used to create and run interactive web applications. ASP contains HTML and a scripting language code. The scripting language can be VBScript or JavaScript. ASP files have the extension .asp.
Java Server Pages (JSP): It is a technology provides a simple and fast way to create dynamic web content. It uses Java as programming language. JSP files have the extension .jsp. To run JSP,. Apache Tomcat web server is required

Answer 30.
The operators in JavaScript are:

1. Arithmetic operators: It contains +, -, *, /, %, + +, – -operators.
2. Assignment operators: =, + = , -=, *=,/=,%= are assignment operators.
3. Relational operators (Comparison operators): The relational operators are = = , !=, <, <=, >, >=. The result of a relational operation is either true or false. These operators compare the values on the two side of the operator and give the result accordingly.
4. Logical operators: &&, ∥ ! are logi-cal operators.
5. String addition operator (+): Opera-tor + is used to add two strings. Adding two string means concatenating two strings.

Plus Two Computer Application Previous Year Question Papers and Answers