Class 12 English Worksheet-13 · current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section.

ST. XAVIER’S SR. SEC. SCHOOL, CHANDIGARH

Class 12 English Worksheet-13

Proposal

A Proposal, literally means a plan or a suggestion. It refers to formal or written

document, put forward for consideration by others. Based on content, a proposal can

be of different types such as:

Research Proposal

Grants Proposal

Business Proposal

Project based Proposal

In class 12th , students are required to write a proposal for implementing a project /

event and seeking its approval from the head of the institute, usually the Principal of

the school.

Requirements

The proposal requires the following headings:

Heading

Statement of objective

List of measures

Closing line

Signing off your full name

Designation

Date

SPECIMEN PAPER

English Paper 1 (Language) Specimen for Proposal Writing Question 2(b): [10] As a

member of the Student Council of your school, you have been given the responsibility of

setting up a Science Club. Write a proposal in about 150 words, stating the steps you

would take to successfully establish this particular club.

Answer:

PROPOSAL FOR SETTING UP A SCIENCE CLUB

Heading/Introduction: To foster an interest in Science outside the classroom and

introduce students to the wonders and relevance of Science in our lives, we propose to set

up a Science Club in school.

(maximum 2 sentences – 2 marks)

Objectives: A Science Club will help students overcome their phobias regarding Science.

It will be instrumental in developing the scientific curiosity of students through its

activities and programmes.

(minimum 2 points - 2 marks)

List of Measures: • The middle- school activity room will be used as the room for all

Science Club meetings and activities. • The meetings will take place once a week after

school from 2.00 pm till 3.30 pm. Any activities such as talks by scientists or competitions

will take place on Saturdays. • Membership of the Science Club will be open to all

students from Classes VI to XII. The Club President will be Mr. Sinha, our Senior Physics

Teacher. Eight other office bearers will be elected from the members of the Club. • Club

membership has been fixed at Rs. 250/- per member per year. • The Club will have a range

of activities ranging from Science Fairs, Robot making, creating slogans and posters,

documentaries and so on.

(minimum 4 points - 4 marks)

We hope that the proposal will be accepted so that the Science Club becomes a reality in

the life of the school.

XYZ (full name)

Member of student’s council (designation)

(linguistic ability - 2 marks)

[Total - 10 marks]

Assignments

As the Head Boy/Head Girl of your school , you have been given the responsibility

of organising the Literary Fest in your school. Write a proposal in about 150

words, stating the steps you would take to successfully organise this fest.

As a member of the Arts and Crafts Society of your school, you have been

assigned the task of organising the sale of Greeting Cards and other Handicrafts

made by the students of your school. The proceeds of this sale would go to an

NGO working for helping the children of the slums. Write a proposal in about 150

words, stating the steps you would take to successfully implement this project.

Class 12 Physics Worksheet-13

NUMERICAL CURRENT ELECTRICITY

Question 3.1:

The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is

0.4Ω, what is the maximum current that can be drawn from the battery?

Answer

Emf of the battery, E = 12 V

Internal resistance of the battery, r = 0.4 Ω

Maximum current drawn from the battery = I

According to Ohm’s law,

The maximum current drawn from the given battery is 30 A.

Question 3.2:

A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current

in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage

of the battery when the circuit is closed?

Answer


Internal resistance of the battery, r = 3 Ω

Current in the circuit, I = 0.5 A

Resistance of the resistor = R

The relation for current using Ohm’s law is,

Terminal voltage of the resistor = V

According to Ohm’s law,

V = IR

= 0.5 × 17

= 8.5 V

Therefore, the resistance of the resistor is 17 Ω and the terminal voltage is

8.5 V.

Question 3.3:

Three resistors 1 Ω, 2 Ω, and 3 Ω are combined in series. What is the total resistance of

the combination?

If the combination is connected to a battery of emf 12 V and negligible internal

resistance, obtain the potential drop across each resistor.

Answer

Three resistors of resistances 1 Ω, 2 Ω, and 3 Ω are combined in series. Total resistance of

the combination is given by the algebraic sum of individual resistances.

Total resistance = 1 + 2 + 3 = 6 Ω

Current flowing through the circuit = I


Total resistance of the circuit, R = 6 Ω

The relation for current using Ohm’s law is,

Potential drop across 1 Ω resistor = V1

From Ohm’s law, the value of V1 can be obtained as

V1 = 2 × 1= 2 V … (i)


Again, from Ohm’s law, the value of V2 can be obtained as

V2 = 2 × 2= 4 V … (ii)


Again, from Ohm’s law, the value of V3 can be obtained as

V3 = 2 × 3= 6 V … (iii)

Therefore, the potential drop across 1 Ω, 2 Ω, and 3 Ω resistors are 2 V, 4 V, and 6 V

respectively.

Question 3.4:

Three resistors 2 Ω, 4 Ω and 5 Ω are combined in parallel. What is the total resistance of

the combination?

If the combination is connected to a battery of emf 20 V and negligible internal

resistance, determine the current through each resistor, and the total current drawn from

the battery.

Answer

There are three resistors of resistances,

R1 = 2 Ω, R2 = 4 Ω, and R3 = 5 Ω

They are connected in parallel. Hence, total resistance (R) of the combination is given by,

Therefore, total resistance of the combination is

Emf of the battery, V = 20 V

Current (I1) flowing through resistor R1 is given by,



Total current, I = I1 + I2 + I3 = 10 + 5 + 4 = 19 A

Question 3.5:

At room temperature (27.0 °C) the resistance of a heating element is 100 Ω. What is the

temperature of the element if the resistance is found to be 117 Ω, given that the

temperature coefficient of the material of the resistor is

Answer

Room temperature, T = 27°C

Resistance of the heating element at T, R = 100 Ω

Let T1 is the increased temperature of the filament.

Resistance of the heating element at T1, R1 = 117 Ω

Temperature co-efficient of the material of the filament,

Therefore, at 1027°C, the resistance of the element is 117Ω.

Question 3.6:

A negligibly small current is passed through a wire of length 15 m and uniform cross-

section 6.0 × 10−7 m2, and its resistance is measured to be 5.0 Ω. What is the resistivity of

the material at the temperature of the experiment?

Answer

Length of the wire, l =15 m

Area of cross-section of the wire, a = 6.0 × 10−7 m2

Resistance of the material of the wire, R = 5.0 Ω

Resistivity of the material of the wire = ρ

Resistance is related with the resistivity as

Therefore, the resistivity of the material is 2 × 10−7 Ω m.

Question 3.7:

A silver wire has a resistance of 2.1 Ω at 27.5 °C, and a resistance of 2.7 Ω at 100 °C.

Determine the temperature coefficient of resistivity of silver.

Answer

Temperature, T1 = 27.5°C

Resistance of the silver wire at T1, R1 = 2.1 Ω

Temperature, T2 = 100°C

Resistance of the silver wire at T2, R2 = 2.7 Ω

Temperature coefficient of silver = α

It is related with temperature and resistance as

Therefore, the temperature coefficient of silver is 0.0039°C−1.

Question 3.8:

A heating element using nichrome connected to a 230 V supply draws an initial current of

3.2 A which settles after a few seconds toa steady value of 2.8 A. What is the steady

temperature of the heating element if the room temperature is 27.0 °C? Temperature

coefficient of resistance of nichrome averaged over the temperature range involved is

1.70 × 10−4 °C −1.

Answer

Supply voltage, V = 230 V

Initial current drawn, I1 = 3.2 A

Initial resistance = R1, which is given by the relation,

Steady state value of the current, I2 = 2.8 A

Resistance at the steady state = R2, which is given as

Temperature co-efficient of nichrome, α = 1.70 × 10−4 °C −1

Initial temperature of nichrome, T1= 27.0°C

Study state temperature reached by nichrome = T2

T2 can be obtained by the relation for α,

Therefore, the steady temperature of the heating element is 867.5°C

Question 3.9:

Determine the current in each branch of the network shown in figure.

Answer

Current flowing through various branches of the circuit is represented in the given figure.

I1 = Current flowing through the outer circuit

I2 = Current flowing through branch AB

I3 = Current flowing through branch AD

I2 − I4 = Current flowing through branch BC I3

+ I4 = Current flowing through branch CD I4 =

Current flowing through branch BD

For the closed circuit ABDA, potential is zero i.e.,

10I2 + 5I4 − 5I3 = 0

2I2 + I4 −I3 = 0

I3 = 2I2 + I4 … (1)

For the closed circuit BCDB, potential is zero i.e.,

5(I2 − I4) − 10(I3 + I4) − 5I4 = 0

5I2 + 5I4 − 10I3 − 10I4 − 5I4 = 0

5I2 − 10I3 − 20I4 = 0

I2 = 2I3 + 4I4 … (2)

For the closed circuit ABCFEA, potential is zero i.e.,

−10 + 10 (I1) + 10(I2) + 5(I2 − I4) = 0

10 = 15I2 + 10I1 − 5I4

3I2 + 2I1 − I4 = 2 … (3)

From equations (1) and (2), we obtain

I3 = 2(2I3 + 4I4) + I4

I3 = 4I3 + 8I4 + I4

3I3 = 9I4

3I4 = + I3 … (4)

Putting equation (4) in equation (1), we obtain

I3 = 2I2 + I4

4I4 = 2I2

I2 = − 2I4 … (5)

It is evident from the given figure that,

I1 = I3 + I2 … (6)

Putting equation (6) in equation (1), we obtain

3I2 +2(I3 + I2) − I4 = 2

5I2 + 2I3 − I4 = 2 … (7)

Putting equations (4) and (5) in equation (7), we obtain

5(−2 I4) + 2(− 3 I4) − I4 = 2

10I4 − 6I4 − I4 = 2

17I4 = − 2

Equation (4) reduces to

I3 = − 3(I4)

Therefore, current in branch

In branch BC =

In branch CD =

In branch AD

In branch BD =

Total current =

Question 3.10:

resistors in a Wheatstone or meter bridge made of thick copper strips?

Determine the balance point of the bridge above if X and Y are interchanged.

What happens if the galvanometer and cell are interchanged at the balance point of the

bridge? Would the galvanometer show any current?

In a metre bridge Fig the balance point is found to be at 39.5 cm from the end A, when

the resistor Y is of 12.5 Ω. Determine the resistance of X. Why are the connections

between

Answer

A metre bridge with resistors X and Y is represented in the given figure.

Balance point from end A, l1 = 39.5 cm

Resistance of the resistor Y = 12.5 Ω

Condition for the balance is given as,

Therefore, the resistance of resistor X is 8.2 Ω.

The connection between resistors in a Wheatstone or metre bridge is made of thick

copper strips to minimize the resistance, which is not taken into consideration in the

bridge formula.

If X and Y are interchanged, then l1 and 100−l1 get interchanged.

The balance point of the bridge will be 100−l1 from A.

100−l1 = 100 − 39.5 = 60.5 cm

Therefore, the balance point is 60.5 cm from A.

Question 3.11:

A storage battery of emf 8.0 V and internal resistance 0.5 Ω is being charged by a 120 V

dc supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery

during charging? What is the purpose of having a series resistor in the charging circuit?

Answer

Emf of the storage battery, E = 8.0 V

Internal resistance of the battery, r = 0.5 Ω

DC supply voltage, V = 120 V

Resistance of the resistor, R = 15.5 Ω

Effective voltage in the circuit = V1

R is connected to the storage battery in series. Hence, it can be written as

V1 = V − E

V1 = 120 − 8 = 112 V

Current flowing in the circuit = I, which is given by the relation,

Voltage across resistor R given by the product, IR = 7 × 15.5 = 108.5 V

DC supply voltage = Terminal voltage of battery + Voltage drop across R

Terminal voltage of battery = 120 − 108.5 = 11.5 V

A series resistor in a charging circuit limits the current drawn from the external source.

The current will be extremely high in its absence. This is very dangerous.

Question 3.12:

In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm

length of the wire. If the cell is replaced by another cell and the balance point shifts to

63.0 cm, what is the emf of the second cell?

Answer

Emf of the cell, E1 = 1.25 V

Balance point of the potentiometer, l1= 35 cm

The cell is replaced by another cell of emf E2.

New balance point of the potentiometer, l2 = 63 cm

Therefore, emf of the second cell is 2.25V.

Question 3.13:

The number density of free electrons in a copper conductor estimated in Example 3.1 is

8.5 × 1028 m−3. How long does an electron take to drift from one end of a wire 3.0 m long

to its other end? The area of cross-section of the wire is 2.0 × 10−6 m2 and it is carrying a

current of 3.0 A.

Answer

Number density of free electrons in a copper conductor, n = 8.5 × 1028 m−3 Length of the

copper wire, l = 3.0 m

Area of cross-section of the wire, A = 2.0 × 10−6 m2

Current carried by the wire, I = 3.0 A, which is given by the relation,

I = nAeVd

Where,

e = Electric charge = 1.6 × 10−19 C

Vd = Drift velocity

Therefore, the time taken by an electron to drift from one end of the wire to the other is

2.7 × 104 s.

Question 3.14

Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Ω

are joined in series to provide a supply to a resistance of 8.5 Ω. What are the current

drawn from the supply and its terminal voltage?

A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Ω.

What maximum current can be drawn from the cell? Could the cell drive the starting motor of

a car?

Answer

Number of secondary cells, n = 6

Emf of each secondary cell, E = 2.0 V

Internal resistance of each cell, r = 0.015 Ω

series resistor is connected to the combination of cells.

Resistance of the resistor, R = 8.5 Ω

Current drawn from the supply = I, which is given by the relation,

Terminal voltage, V = IR = 1.39 × 8.5 = 11.87 A

Therefore, the current drawn from the supply is 1.39 A and terminal voltage is

11.87 A.

After a long use, emf of the secondary cell, E = 1.9 V

Internal resistance of the cell, r = 380 Ω

Hence, maximum current

Therefore, the maximum current drawn from the cell is 0.005 A. Since a large current is

required to start the motor of a car, the cell cannot be used to start a motor.

Question 3.15:

Two wires of equal length, one of aluminium and the other of copper have the same

resistance. Which of the two wires is lighter? Hence explain why aluminium wires are

preferred for overhead power cables. (ρAl = 2.63 × 10−8 Ω m, ρCu = 1.72 × 10−8 Ω m,

Relative density of Al = 2.7, of Cu = 8.9.)

Answer

Resistivity of aluminium, ρAl = 2.63 × 10−8 Ω m

Relative density of aluminium, d1 = 2.7

Let l1 be the length of aluminium wire and m1 be its mass.

Resistance of the aluminium wire = R1

Area of cross-section of the aluminium wire = A1

Resistivity of copper, ρCu = 1.72 × 10−8 Ω m

Relative density of copper, d2 = 8.9

Let l2 be the length of copper wire and m2 be its mass.

Resistance of the copper wire = R2

Area of cross-section of the copper wire = A2

The two relations can be written as

R1=R2 and

Mass of the aluminium wire,

m1 = Volume × Density

= A1l1 × d1 = A1 l1d1 … (3)

Mass of the copper wire,

m2 = Volume × Density

= A2l2 × d2 = A2 l2d2 … (4)

Dividing equation (3) by equation (4), we obtain

It can be inferred from this ratio that m1 is less than m2. Hence, aluminium is lighter than

copper.

Since aluminium is lighter, it is preferred for overhead power cables over copper.

Question 3.16:

What conclusion can you draw from the following observations on a resistor made of

alloy manganin?

Current

A

Voltage

V

Current

A

Voltage

V

0.2 3.94 3.0 59.2

0.4 7.87 4.0 78.8

0.6 11.8 5.0 98.6

0.8 15.7 6.0 118.5

1.0 19.7 7.0 138.2

2.0 39.4 8.0 158.0

Answer

It can be inferred from the given table that the ratio of voltage with current is a constant,

which is equal to 19.7. Hence, manganin is an ohmic conductor i.e., the alloy obeys

Ohm’s law. According to Ohm’s law, the ratio of voltage with current is the resistance of

the conductor. Hence, the resistance of manganin is 19.7 Ω.

Question 3.17:

Answer the following questions:

A steady current flows in a metallic conductor of non-uniform cross- section. Which of

these quantities is constant along the conductor: current, current density, electric field,

drift speed?

Is Ohm’s law universally applicable for all conducting elements?

If not, give examples of elements which do not obey Ohm’s law.

A low voltage supply from which one needs high currents must have very low internal

resistance. Why?

A high tension (HT) supply of, say, 6 kV must have a very large internal resistance.

Why?

Answer

When a steady current flows in a metallic conductor of non-uniform cross-section, the

current flowing through the conductor is constant. Current density, electric field, and drift

speed are inversely proportional to the area of cross-section. Therefore, they are not

constant.

No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode

semi-conductor is a non-ohmic conductor. Ohm’s law is not valid for it.

According to Ohm’s law, the relation for the potential is V = IR

Voltage (V) is directly proportional to current (I).

R is the internal resistance of the source.

If V is low, then R must be very low, so that high current can be drawn from the source.

In order to prohibit the current from exceeding the safety limit, a high tension supply

must have a very large internal resistance. If the internal resistance is not large, then the

current drawn can exceed the safety limits in case of a short circuit.

Question 3.18:

Choose the correct alternative:

Alloys of metals usually have (greater/less) resistivity than that of their constituent

metals.

Alloys usually have much (lower/higher) temperature coefficients of resistance than pure

metals.

The resistivity of the alloy manganin is nearly independent of/increases rapidly with

increase of temperature.

The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor

of the order of (1022/103).

Answer

Alloys of metals usually have greater resistivity than that of their constituent metals.

Alloys usually have lower temperature coefficients of resistance than pure metals.

The resistivity of the alloy, manganin, is nearly independent of increase of temperature.

The resistivity of a typical insulator is greater than that of a metal by a factor of the order

of 1022.

Question 3.19:

Given n resistors each of resistance R, how will you combine them to get the (i)

maximum (ii) minimum effective resistance? What is the ratio of the maximum to

minimum resistance?

Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent

resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?

Determine the equivalent resistance of networks shown in Fig.

Answer

Total number of resistors = n

Resistance of each resistor = R

When n resistors are connected in series, effective resistance R1is the maximum, given by

the product nR.

Hence, maximum resistance of the combination, R1 = nR

When n resistors are connected in parallel, the effective resistance (R2) is the minimum,

given by the ratio .

Hence, minimum resistance of the combination, R2 =

The ratio of the maximum to the minimum resistance is,

The resistance of the given resistors is,

R1 = 1 Ω, R2 = 2 Ω, R3 = 3 Ω2

Equivalent resistance,

Consider the following combination of the resistors.

Equivalent resistance of the circuit is given by,


Consider the following combination of the resistors.


Equivalent resistance, R’ = 6 Ω

Consider the series combination of the resistors, as shown in the given circuit.

Equivalent resistance of the circuit is given by the sum,

R’ = 1 + 2 + 3 = 6 Ω


Consider the series combination of the resistors, as shown in the given circuit.


(a) It can be observed from the given circuit that in the first small loop, two resistors of

resistance 1 Ω each are connected in series. Hence,

their equivalent resistance = (1+1) = 2 Ω

It can also be observed that two resistors of resistance 2 Ω each are connected in series.

Hence, their equivalent resistance = (2 + 2) = 4 Ω.

Therefore, the circuit can be redrawn as

It can be observed that 2 Ω and 4 Ω resistors are connected in parallel in all the four

loops. Hence, equivalent resistance (R’) of each loop is given by,

The circuit reduces to,

All the four resistors are connected in series.

Hence, equivalent resistance of the given circuit is

It can be observed from the given circuit that five resistors of resistance R each are

connected in series.

Hence, equivalent resistance of the circuit = R + R + R + R + R = 5 R

Question 3.20:

Determine the current drawn from a 12 V supply with internal resistance 0.5 Ω by the

infinite network shown in Fig. 3.32. Each resistor has 1 Ω resistance.

Answer

The resistance of each resistor connected in the given circuit, R = 1 Ω

Equivalent resistance of the given circuit = R’

The network is infinite. Hence, equivalent resistance is given by the relation,Negative value

of R’ cannot be accepted. Hence, equivalent resistance,

Internal resistance of the circuit, r = 0.5 Ω

Hence, total resistance of the given circuit = 2.73 + 0.5 = 3.23 Ω

Supply voltage, V = 12 V

According to Ohm’s Law, current drawn from the source is given by the ratio, =3.72 A

Question 3.21:

Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a

potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02

V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire.

To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in

series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell

of unknown emf ε and the balance point found similarly, turns out to be at 82.3 cm length of the

wire.What is the value ε ?

What purpose does the high resistance of 600 kΩ have?

Is the balance point affected by this high resistance?

Is the balance point affected by the internal resistance of the driver cell?

Would the method work in the above situation if the driver cell of the potentiometer

had an emf of 1.0 V instead of 2.0 V?

Answer

Constant emf of the given standard cell, E1 = 1.02 V

Balance point on the wire, l1 = 67.3 cm

A cell of unknown emf, ε,replaced the standard cell. Therefore, new balance point on

the wire, l = 82.3 cm

The relation connecting emf and balance point is,

The value of unknown emfis 1.247 V.

The purpose of using the high resistance of 600 kΩ is to reduce the current through

the galvanometer when the movable contact is far from the balance point.

The balance point is not affected by the presence of high

resistance. The point is not affected by the internal resistance of

the driver cell.

The method would not work if the driver cell of the potentiometer had an emf of 1.0 V instead

of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf

of the other cell, then there would be no balance point on the wire.

Question 3.22:

Figure shows a potentiometer circuit for comparison of two resistances. The balance point

with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown

resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find

a balance point with the given cell of emf ε?

Answer

Resistance of the standard resistor, R = 10.0 Ω

Balance point for this resistance, l1 = 58.3 cm

Current in the potentiometer wire = i

Hence, potential drop across R, E1 = iR

Resistance of the unknown resistor = X

Balance point for this resistor, l2 = 68.5 cm Hence,

potential drop across X, E2 = iX

The relation connecting emf and balance point is,

Therefore, the value of the unknown resistance, X, is 11.75 Ω.

If we fail to find a balance point with the given cell of emf, ε, then the potential drop across R

and X must be reduced by putting a resistance in series with it. Only if the potential drop

across R or X is smaller than the potential drop across the potentiometer wire AB, a balance

point is obtained.

Question 3.23:

Figure shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V

cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in

the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiomete wire.

Determine the internal resistance of the cell.

Answer

Internal resistance of the cell = r

Balance point of the cell in open circuit, l1 = 76.3 cm

An external resistance (R) is connected to the circuit with R = 9.5

Ω New balance point of the circuit, l2 = 64.8 cm

Current flowing through the circuit = I

The relation connecting resistance and emf is,

Therefore, the internal resistance of the cell is 1.68Ω.

Class 12 Chemistry Worksheet-13

SURFACE CHEMISTRY CONTINUED

PURIFICATION OF COLLOIDAL SOLUTIONS:

Colloids have a number of electrolytic impurities. The process used for reducing the

amount of impurities to a required minimum is called purification of colloids.

Colloids are purified by the following methods:

1.DIALAYSIS: The process of separating the particles of a colloid by diffusion through

a suitable membrane. An apparatus called dialyser is used. A bag with a suitable

membrane containing the colloid is suspended in a vessel through which fresh water

flows continuously. The impurities diffuse through the membrane into the water leaving

behind the colloid

2.ELECTRO- DIALYSIS: Dialysis is a slow method. It is made faster by

electrodialysis. In this method, an electric filed is applied using metal electrodes. These

ions present in the colloidal solution migrate to the oppositely charged electrodes. It is

possible only if the dissolved substance in the impure colloid is an electrolyte.

3.ULTRAFILTRATION: This is the process of separating colloidal particles from the

soluble solutes(impurities) using specially prepared filters, which are permeable to all

substances except the colloid.

Colloidal particles can usually pass through filter papers as the pores are too large. An

ultrafilter paper can be made by soaking the filter paper in a colloidal solution, hardening

by formaldehyde and then finally drying it. As this is slow process, pressure or suction is

applied to speed it up. The colloidal particles left on the ultra-filter paper are stirred with

fresh dispersion medium(solvent) to form a pure colloid.

4.ULTRACENTRIFUGATION: Ultracentrifugation is a specialized technique used to

spin samples at exceptionally high speeds. The basis of ultracentrifugation is to separate

the components of a solution based on their size. The denser particles sediment in a

centrifugal field.

PROPERTIES OF COLLOIDS

1.Tyndall effect: Tyndall effect is the scattering of the light by the particles present in

the colloidal solution when viewed at right angles to the passage of light. The bright cone

of light is called Tyndall cone.

It is observed only when-

o The diameter of the dispersed particles is not much smaller than the wavelength of

light used

o The refractive indices of the dispersed phase and dispersed medium have a large

difference

o This effect was used to make an ultramicroscope and differentiate between true

solution and colloids.

2.Electrophoresis: It is the movement of colloidal particles under the influence of an

electric field

Negatively charged particles move towards the cathode and Positively charged particles

moves towards anode.When the movement of particles is prevented, it is observed that

the dispersion medium starts to move in the electric field. This is called electro-osmosis.

3.Coagulation: It is process of settling of colloidal particles. It is also called

precipitation of sol.

COAGULATION OF LYOPHOBIC SOL: can be done by the following methods

By electrophoresis - The colloidal particles move towards oppositely charged

electrodes get discharged and precipitate.

o By mixing two oppositely charged sols - Oppositely charged sols when mixed

together in almost equal proportion, neutralise their charges and get partially or

completely precipitated.

o By Boiling- When a sol is boiled the adsorbed layer is disturbed due to increased

number of collisions with the molecules of the dispersion medium. This reduces the

charge on the particles and they ultimately settle down in the form of a precipitate.

o By Persistent dialysis - On prolonged dialysis, traces of the electrolyte present in the

sol are removed almost completely. Colloids become unstable and coagulate.

o By addition of electrolyte - When excess of electrolyte is added, colloidal particles

precipitate as colloids interact with ions carrying charge opposite to that present on

themselves. This causes neutralisation leading to their coagulation. Example- A

negatively charged ion when added to a positively charged sol causes coagulation and

vice- versa. The negatively charged ion is called coagulating ion/flocculating ion as it

neutralises the colloid to cause coagulation.

It has been observed that, the greater the valency of the flocculating ion added, the

greater is its precipitation. This is known as HARDY-SCHULZE RULE.

For negative sols, when positive ions are added

Al3+> Ba2+>Na+ is the order in terms of flocculating power

For positive sols, when negative ions are added

[Fe(CN)6] 4-> PO43->SO4

2->Cl- is the order in terms of flocculating power

COAGULATION OF LYOPHILLIC SOLS:

Lyophilic sols are stable because of charge and solvation of colloidal particles. So we

remove these two factors to coagulate them. This is done by

o Addition of an electrolyte

o Addition of a suitable solvent

4.PROTECTION OF COLLOIDS -

o Lyophilic sols are more stable than lyophobic sols because lyophilic are extensively

solvated.

o Lyophilic colloids have a unique ability to protect lyophobic colloids from

electrolytes

o When a lyophilic sol is added to lyophobic sol, the lyophilic particles (colloids) form

a layer around the particles of lyophobic sol

o Lyophilic colloids used for this purpose are called protective colloids

5. BROWNIAN MOVEMENT: is the continuous zigzag movement of colloidal

particles in a colloidal solution. It depends on the size of the particles, smaller the size of

particles faster is its motion. This movement is responsible for the stability of sols.

Emulsions

o Emulsions are colloids where both the dispersed phase and dispersion medium are

liquids. These two liquids are immiscible or partially miscible. Generally, one of the

liquids is water.

o There are two types of emulsions

1. Oil dispersed in water (o/w type)- Water acts as dispersion medium

Example –Milk, vanishing cream

2. Water dispersed in oil (w/o type)- Oil acts as dispersion medium

Example- Butter, Cream

o Emulsions like water and oil separate into two layers and make the emulsion unstable.

So, emulsions are stabilised by stabilising agents.

Emulsification: It is a process of stabilizing an emulsion by means of an emulsifier.

Emulsifying agent or emulsifier: These are the substances which are added to stabilize

the emulsions. Examples- soaps, gums.

o/w emulsions are stabilised by proteins, gum, natural and synthetic soaps

o w/o emulsions are stabilised by metal salts of fatty acids, long chain alcohols

Demulsification: It is the process of breaking an emulsion into its constituent liquids by

freezing, boiling, centrifugation or some chemical methods.

SURFACTANTS: These are compounds that lower the surface tension between two

liquids or a gas and a liquid or a liquid and a solid.

Examples: Detergents, wetting agents, emulsifiers, foaming agents etc.

.

Applications of Colloids

A. NATURAL APPLICATION

1. Blue colour of the sky: Colloidal particles scatter blue light which reaches our

eyes and the sky looks blue to us.

2. Fog, mist and rain: When air having dust particles is cooled below its dew point,

the moisture from the air condenses on the surface of these particles forming fine

droplets .These droplets being colloidal in nature continue to float in the air as

mist or fog.

3. Food articles: Milk, butter, ice creams are all colloids.

4. Blood: is a colloidal solution of an albuminoid substance.

5. Delta formation: River water is a colloidal solution of clay. Sea water has a no. of

electrolytes. When river water meets sea water the electrolytes present in sea water

coagulate the colloidal solution of clay which gets deposited with the formation of

DELTA.

6. Soils: Fertile soils are colloidal in nature in which humus acts as a protective

colloid.

B. TECHNICAL APPLICATION

1.Electro precipitation of smoke – The smoke is led through a chamber containing

plates having a charged opposite to that carried by smoke particles. The particles on

coming in contact with these plates lose their charge and get precipitated. The particles

settle down on the floor of the chamber. The precipitator is called Cottrell precipitator.

2.Purification drinking water – Alum is added to impure water to coagulate the

suspended impurities and make water fit for drinking.

3.Medicines – Most of the medicines are colloidal in nature. Colloidal medicines are

more effective because they have a larger surface area and are more easily absorbed by

the body. E.g. – Argyrol is a silver sol used as an eye lotion, milk of magnesia is used to

cure stomach disorders

4.Tanning – Animal hides are colloidal in nature. When a hide that has

positively charged particles is soaked in tannin/chromium salts, which contains

negatively charged particles, mutual coagulation takes place. This results in the hardening

of leather. This process is termed as tanning.

5.Cleansing action of soaps- Soap is a kind of molecule in which both the ends have

different properties .

Hydrophilic end (head)

Hydrophobic end (tail)

Soap molecules are sodium or potassium salts of long chain carboxylic acids.The

hydrophilic end (short polar end) is the carboxylate ion which is water soluble but insoluble

in oil. The hydrophobic end (long non-polar end) is a hydrocarbon chain which is insoluble

in water but soluble in oil

Soap on mixing with water forms a concentrated solution and causes foaming. The

hydrophobic end(tail) surrounds the dirt since it is insoluble in water. The hydrophilic

(short polar end) is on the surface in contact with water. This results in clusters of molecules

known as micelles. Thus, the soap molecule dissolves the dirt and our clothes get clean.

6.Photographic plates and films – Photographic plates and films are prepared by coating

an emulsion of the light sensitive silver bromide in gelatin over glass plates or celluloid

films.

7.Rubber industry- Latex is a colloidal solution of rubber particles which are negatively

charged. Rubber is obtained by coagulation of latex.

8.Industrial products- Paints inks, synthetic plastics, rubber, cement, graphite

lubricants are all colloids

Class 12 Maths Worksheet-13

INVERSE TRIGNOMETRIC FUNCTIONS:

Consider the sine function f(x) = y= sin x, Df = R and Rf = [-1,1]

Therefore y = sin-1x called inverse sine function or arc sine function has Domain [-1,1] and

Range [ −𝜋

2 ,

𝜋

2 ] . The angles lying in [ −

𝜋

2 ,

𝜋

2 ] is called its Principal Value Branch.

Lets learn all the Principal Values

Inverse Function Principal Value Branch

or Principal Values

Remarks ( Just for understanding)

sin−1 𝑥 [−𝜋

2 ,

𝜋

2 ] Because sin

𝜋

2 =1 and sin (-

𝜋

2 ) = -1

cos−1 𝑥 [ 0, 𝜋 ] Because cos 0 = 1 and cos 𝜋 = -1

tan−1 𝑥 (−𝜋

2 ,

𝜋

2 ) Because tan x is not defined at −

𝜋

2 and

𝜋

2

cosec−1 𝑥 [−𝜋

2 , 0 ) ∪ ( 0,

𝜋

2 ] Because cosec x is not defined at 0

sec−1 𝑥 [ 0, 𝜋

2 ) ∪ (

𝜋

2 , 𝜋 ] Because sec x is not defined at

𝜋

2

cot−1 𝑥 (0, 𝜋 ) Because cot x is not defined at 0 and 𝜋 both

Finding the Principal Values

The following simple steps should be followed in order to find the Principal value of a given

inverse function.

STEP 1: Find out the quadrants in which Principal value lies.

STEP 2: Find out the quadrant according to sign convention of II I

Trigonometric function.

STEP 3: Principal value will lie in the common quadrant from

STEP 1 and STEP 2

STEP 4: The quadrants can be accessed with this rule III IV

shown in the image.

If f is a real function f defined by y = f(x) is invertible ( one –one and onto) then f-1 is given

by x = f -1 (y) for all x ∈ Df and y ∈Rf

In other words, Domain of f becomes Range of f -1 and vice versa.

0 + 𝝅 -

−𝝅 + 0 -

Illustration 1: Find the principal value of sin−1 (−√3

2 )

Solution: As we can see that Principal Value of The negative sign of sin function

sin-1x ∈ [−𝜋

2 ,

𝜋

2 ] I, IV Quadrants III, IV Quadrants

Therefore required Principal value will lie in common quadrant IV which can be accessed by

x = 0- 𝜋

3 , therefore the principal value of sin−1 (

−√3

2 ) is -

𝜋

3

Illustration 2: Find the principal value of sec−1 ( −2 )

Solution: As we can see that Principal Value of The negative sign of sin function

sec-1x ∈ [ 0, 𝜋

2 ) ∪ (

𝜋

2 , 𝜋 ] I, II Quadrants II, III Quadrants

Therefore required Principal value will lie in common quadrant II which can be accessed by

x = 𝜋 - 𝜋

3 =

2𝜋

3 , therefore the principal value of sec−1 ( −2 ) is

2𝜋

3

Illustration 3:Find the Principal value of sin−1(sin3𝜋

5 )

Solution: Let sin−1(sin3𝜋

5 ) = x , x ∈ [−

𝜋

2 ,

𝜋

2 ] and

3𝜋

5 ∈ [−

𝜋

2 ,

𝜋

2 ]

⇒ sin x = sin 3𝜋

5 = sin ( 𝜋 -

𝜋

5 ) = sin (

2𝜋

5 )

Now 2𝜋

5 ∈ [−

𝜋

2 ,

𝜋

2 ] , Therefore x =

2𝜋

5

Illustration 4: Evaluate sin ( cot−1(cot17𝜋

3 )

Solution: Let cot−1(cot17𝜋

3 ) = x , x ∈ (0, 𝜋 ) [ Principal value branch of cot-1x]

⇒ cot x = cot 17𝜋

3 = cot ( 5𝜋 +

𝜋

3 ) = cot (

2𝜋

3 )

And now 2𝜋

3 ∈ (0, 𝜋 ),

Therefore cot−1(cot17𝜋

3 ) =

2𝜋

3

So sin ( 2𝜋

3 ) = sin ( 𝜋 -

𝜋

3 ) = sin

𝜋

3 =

√3

2

Hence sin ( cot−1(cot17𝜋

3 ) =

√3

2

Illustration 5: Evaluate sin−1(sin 2)

Solution: In such cases when angles are given in radians [ 1 rad ≅ (57.27)0 ]

Therefore 2 radians ≅ (114.54) 0 and (114.54) 0 ∈ [−𝜋

2 ,

𝜋

2 ]

Therefore sin−1(sin 2) = sin−1{ 𝑠𝑖𝑛 (𝜋 -2) } = π -2 [ ∵ π -2 = 1800 – 114.540 ]

Which lies in the first quadrant or Principal value branch of sin-1 x .

Hence sin−1(sin 2) = π -2

Illustration 6: Evaluate cos { cos−1(−√3

2 ) +

𝜋

6 }

Solution: First of all we will find cos−1(−√3

2 ) by the same method done earlier.

Principal value of cos−1(𝑥) lies in Negative sign of cos comes in

I, II Quadrant II, III Quadrant

Therefore required Principal value will lie in common quadrant II

Hence cos−1(−√3

2 ) = 𝜋 -

𝜋

6 =

5𝜋

6

Therefore cos { cos−1(−√3

2 ) +

𝜋

6 } = cos {

5𝜋

6 +

𝜋

6 } = cos π = -1

SOLVE YOURSELVES:

1. Evaluate tan−1(tan 4) Ans: 4 – π

2. Find Principal value of cot−1( −1

√3 ) Ans:

2𝜋

3

3. Find Principal value of sec−1( −2

√3 ) Ans:

5𝜋

6

4. Evaluate tan−1(tan9𝜋

8 ) Ans:

𝜋

8

5. Evaluate cos−1(cos5𝜋

3 ) Ans:

𝜋

3

6. Evaluate sin { 𝜋

6 - sin−1(

−√3

2 ) } Ans: 1

7. Prove that tan−1(−1) + cos−1( −1

√2 ) =

𝜋

2

8. Using Principal values , solve cosec−1(−1) + cot−1( −1

√3 ) Ans:

𝜋

6

9. Find the value of tan−1(𝑡𝑎𝑛 5𝜋

6 ) + cos−1(𝑐𝑜𝑠

13𝜋

6 ) Ans: 0

[ Check: −𝝅

𝟔

𝝅

𝟔 ]

Illustration 7: Evaluate sin { 2 cos-1 −3

5 }

Solution: Let cos-1 −3

5 = x or cos x =

−3

5

Since we know that the Principal value of cos-1 x 𝜖 [0, 𝜋] where sin x will be positive

Now sin x = √1 − 𝑐𝑜𝑠2𝑥 = √1 − (−3

5 )2 =

4

5

sin { 2 cos-1 −3

5 }= sin 2x = 2 sin x cos x = 2 (

4

5 ) (

−3

5 ) =

−24

25

Illustration 8: Evaluate the following: tan ( 1

2 sin−1 3

5 )

Solution: Let 1

2 sin−1 3

5 = x , It means sin−1 3

5 = 2x or sin 2x =

3

5

⇒ 2 𝑡𝑎𝑛𝑥

1+ 𝑡𝑎𝑛2𝑥 =

3

5 or 3 + 3 tan2x = 10 tan x or 3 tan2x - 10 tan x + 3 = 0

( tan x -3 ) ( 3 tan x -1 ) = 0 Which gives tan x = 3 or 1

3

Now since x 𝜖 [−𝜋

4 ,

𝜋

4 ] therefore tan x 𝜖 [ -1, 1] Hence tan x =

1

3 or tan (

1

2 sin−1 3

5 ) =

1

3

SOLVE YOURSELVES:

10. Evaluate cosec { cos−1(−12

13 ) } Ans:

13

5

11. Evaluate sin { 2 cot−1( −5

12 ) } Ans:

−120

169

12. Find the value of tan−1{ 2 sin( 2 cos−1 √3

2 ) } Ans:

𝜋

3

INVERSE TRIGONOMETRIC FUNCTIONS CONTD….. IN WORKSHEET-14

REMEMBER:

𝐬𝐢𝐧−𝟏(−𝒙 ) = - 𝐬𝐢𝐧−𝟏(𝒙 )

𝐜𝐨𝐬−𝟏(−𝒙) = 𝝅 - 𝐜𝐨𝐬−𝟏 𝒙

𝐭𝐚𝐧−𝟏(−𝒙) = - 𝐭𝐚𝐧−𝟏 𝒙

𝐜𝐨𝐭−𝟏(−𝒙)= - 𝐜𝐨𝐭−𝟏 𝒙

𝐬𝐞𝐜−𝟏(−𝒙) = 𝝅- 𝐬𝐞𝐜−𝟏 𝒙

𝐜𝐨𝐬𝐞𝐜−𝟏(−𝒙)= - 𝐜𝐨𝐬𝐞𝐜−𝟏 𝒙

ALSO REMEMBER: Inverse functions can be interconverted by our

evergreen triangle method.

Supposing sin-1 3

5 = x then sin x =

3

5 3 5

4

Therefore sin-1 3

5 can be written as cos−1 4

5 or tan−1 3

4

[As it is clear from the right angled triangle]

IMPORTANT OBSERVATION: Since Principal value of inverse sine function 𝜖 [−𝜋

2 ,

𝜋

2 ]

Therefore 2x 𝜖 [−𝜋

2 ,

𝜋

2 ] or x 𝜖 [

−𝜋

4 ,

𝜋

4 ]

Class 12 Biology Worksheet-13

CHAPTER-Sexual Reproduction in Flowering Plants (CONTINUED)

-----------------------------------------------------------------------------------------------------

.

Class 12 Physical Education Worksheet-12

CHAPTER 2: TRAINING METHODS: TOPIC 5 COOLING DOWN

Meaning of cooling down/limbering down : The main aim of the cool down is to promote

recovery and return the body to a pre exercise, or pre work out level during a strenuous

work out your body goes through a number of stressful processes e.g. muscle fibers,

tendons and ligaments get damaged, and waste products build up within your body.

For appropriate cooling down, we should perform we should perform jogging as well as

walking for 5 to 10 minutes. This will help in decreasing the body temperature and

removing the waste products from the working muscles. After that static stretching

exercises should be performed for 5 to 10 minutes. The stretches should be held for 10 to

20 seconds. The repetition of stretch should be done at least 2 to 3 times.

Major stretching exercises of muscles for cooling down:

Quadriceps (one of the more powerful muscles of the body)

Lying on your right side

Pull left heel into left glute

Feeling the stretch in the front of the thigh.

Repeat with the right leg.

Hamstrings (group of muscles rear of the upper leg)

Lying on your back

Lift and straighten left leg directly above hips

Holding the calf or thigh

Press heel toward ceiling as you pull leg back towards chest

Repeat with the right leg

Glutes

Lying on your back

Cross right over bent left knee

Then bring left knee to chest

Holding on to back of your thigh

Gently pressing right knee wide

Repeat with right leg

Chest

Standing straight

Interlace fingers behind your back as you straighten out your arms, lift your chin

towards ceiling.

Triceps/shoulders

Take left arm overhead

Bend at elbow joint

Extend palm down centre of your back

Gently pulling elbow with opposite hand

Take same arm across the chest

Gently pulling at the elbow joint, to extend through the shoulder

Repeat with the right arm.

Core/ back

Round out your back and then invert it, making a C- shape with your spine

Repeat three times

Then sit back between your heels

Forehead on the mat

Arms extended in front of you, as you lengthen your back

Now, hit yourself on the back.

Advantages of cooling down:

There are two part5s of a cool-down. First a person does al light cardiovascular activity

such as walking to lower the heart rate, body temperature, etc. then static stretching

should be followed. Exercisers should stretch all the major muscle groups for at least 10

to 30 seconds. Some of the benefits of a cool down are listed below:

1. Normal blood circulation: during the cool-down stage, the legs help return the

blood to your heart. Muscle in the legs acts as a pump to bring blood back to the

heart. Hence, cooling down helps in bringing the blood flow in a normal mode.

2. Helps the systems to work efficiently respiration, cardiovascular system,

circulatory system, body temperature and heart rate are gradually returned to

normal by doing cooling down which preventing an irregular beat that may be

life threatening.

3. Preventing injury: the cool down is as instrumental to the prevention of injury as

the warm up, stopping an activity without cooling down will contribute to a

buildup of toxic substances and lactic acid which will cause muscular pain and

stiffness the day after, this can restrict movement and be very painful.

4. Body temperature become normal: during high intensity, exhaustive exercises or

competition the body temperature increases more than 160 degree Fahrenheit. A

good session of cooling down helps in reducing the body temperature to normal

condition.

5. Removal of waste products: when a sports person performs training or takes part

in competition the waste products such as lactic acid, uric acid, phosphate,

sulphate chloride etc. accumulated in the body. An affecting cooling down

reduces the accumulation of such products very rapidly forms the muscle

suitably.

6. Supply of oxygen: appropriate cooling down helps in supplying the blood and

oxygen to muscles, restoring them to the position they were in before performing

training. Along this, recovery becomes fast.

7. Muscle does not remain stiff: by performing cooling down properly muscles do

not remain stiff but get relaxed speedily.

8. Reduces the chances of dizziness or fainting: the most significant function

appropriate cooling down is to reduce the chances of dizziness or fainting. When

exercises stopped spontaneously without taking time to cool down or limber

down, the heart rate slows abruptly (all of sudden) and that blood can pool (blood

to collect) in the lower body (legs and feet), causing dizziness or fainting.

Topic 6 : Isometric and isotonic exercises

Meaning and advantages of isometric exercises: isometric exercises are those exercises,

which are not visible. In fact there are no direct movements, hence they cannot be

observed. It these exercises, work are performed but it is not seen directly. In these

exercises, a group of muscles carry out tension against the other group of muscles. When

these exercises are done, muscles do not change their length. They remain fixed or

constant.

For example, if we push a concrete wall, we will be unable to move it from its place. So

we should not consider it as work. our muscles exert force, while pushing a wall, but we3

see that wor4k is not done. Because works is said to be done when the point of

application of a force moves, or we can say that ------

Work done = force x distance moved in the direction of force.

Advantages of isometric exercises:

It needs less time

It can be done any where

Every muscle can be exercised within a short period of time

It is good to maintain strength by dynamic exercises.

Increased blood flow

Decrease blood pressure

Meaning of isotonic exercises: (done against resistance) *resistance training is a form of

exercises that improves muscular strength and endurance. During a resistance training

workout, you move your limbs against resistance provided by your body weight, gravity,

bands, weighted bars or dumbbells.

The exercise in simple terms relates to muscles contractions. Any exercise in which your

muscle contracts or there is a strain or tension on the muscle is called isotonic exercises.

For example weight lifting. Ninety percent of Gym workout s is isotonic exercises. The

word isotonic is derived from the Greek words ‘iso’ which means equal and ‘tonus’

which means tone; i.e., the word isotonic implies maintaining equal muscle tone. When

you flex your biceps, it is isotonic contraction.

In isotonic exercises the weight that can be lifted 10 consecutive times but not more than

10. This is called 10 repetitions maximum (10 RM’s). The exercises are done in 3 sets of

10 repetitions each. (*RMs = repetitions maximum).

Advantages of isotonic exercise:

1. Isotonic exercise is very useful, not only in helping participants bulk up, but in

providing specific muscle responses that will be useful in a range of athletic and

recreational activities.

2. Isotonic exercise promotes the development of muscle endurance, muscle tone and

muscle strength.

3. These movements have also been shown to improve ligament and tendon strength.

4. Helps in preventing injuries.

ISOTONIC EXERCISES

5. Improve posture and develop joint stability.

6. Isotonic training helps you strengthen a muscle throughout a range of movement.

7. It is also easier to choose sports-specific exercises that mimic movements in your

sport of choice.

8. One of the main benefits of isotonic exercise is that it doesn’t require extensive

equipment. Portable items like dumb ells, kettle bells, medicine balls and other

similar tools are all ways to fit isotonic exercise into any space or environment.

------------------------------------------------------------------------------------------------------------

Questions regarding above topic 5 and 6 are given below:

1. What do you mean by cooling/limbering down? Enumerate the advantages of

cooling down in detail.

2. What is isometric and isotonic exercises and explain its advantages?

3. Write short notes on the following:

Isotonic exercises

Isometric exercises

Cooling down

Stretching exercises of cooling down

Class 12 Computer Science Worksheet-12

Question 1.

Design a class BinSearch to search for a particular value in an array.

Some of the members of the class are given below:

Class name: BinSearch

Data members/instance variables:

arr[]: to store integer elements.

n: integer to store the size of the array.

Member functions/methods:

BinSearch(int num): parameterized constructor to initialize n = num.

void fillArray(): to enter elements in the array.

void sort(): sorts the array elements in ascending order using any

standard sorting technique.

int binSearch(int l, int u, int v):

searches for the value ‘v’ using binary search and returns its location if found otherwise

returns -1.

Question 2

Write a Program in Java to input a word and print its anagrams..

Note: Anagrams are words made up of all the characters present in the original word by

re-arranging the characters.

Example: Anagrams of the word TOP are: TOP, TPO, OPT, OTP, PTO and POT

Question 3:

A class Admission contain the admission numbers of 100 students. Some of the data

members/ member functions are given below:

Class name: Admission

Data member/instance variable:

Adno[ ]:

Integer array to store admission numbers

Member functions/methods:

Admission():

constructur to initialize the array elements

void fillArray():

to accept the element of the array in ascending order

int binSearch(int l, int u, int v):

to search for a particular admission number(v) using binary search and recursive technique

and return 1 if found otherwise returns -1

Specify the class Admission giving details of the constructor, void fillArrray() and int

binSearch(int, int, int).

Define the main() function to create an object and call the functions accordingly to enable

task.

Define the class BinSearch giving details of the constructor, void fillArray(), void sort()

and int binSearch(int, int, int). Define the main() function to create an object and call the

functions accordingly to enable the task.

Class 12 English Worksheet-13 · current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section.

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