Sastry CBSE

Get Free NCERT Solutions for Class 10 Maths Chapter 11 Ex 11.1 PDF. Constructions Class 10 Maths NCERT Solutions are extremely helpful while doing your homework. Exercise 11.1 Class 10 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 11 Maths Class 10 Constructions Exercise 11.1 provided in NCERT TextBook.

Topics and Sub Topics in Class 10 Maths Chapter 11 Constructions:

Section Name	Topic Name
11	Constructions
11.1	Introduction
11.2	Division Of A Line Segment
11.3	Construction Of Tangents To A Circle
11.4	Summary

• Constructions Class 10 Ex 11.1
• प्रश्नावली 11.1 का हल हिंदी में
• Constructions Class 10 Ex 11.2
• प्रश्नावली 11.2 का हल हिंदी में
• Extra Questions for Class 10 Maths Constructions

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 11
Chapter Name	Constructions
Exercise	Ex 11.1
Number of Questions Solved	5
Category	NCERT Solutions

Ex 11.1 Class 10 Maths Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.
Solution:

Ex 11.1 Class 10 Maths Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac { 2 }{ 3 }\) of the corresponding sides of the first triangle.
Solution:

You can also download the free PDF of  Ex 11.1 Class 10 Constructions NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

Download NCERT Solutions For Class 10 Maths Chapter 11 Constructions PDF

Ex 11.1 Class 10 Maths Question 3.
Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides are \(\frac { 7 }{ 5 }\) of the corresponding sides of the first triangle.
Solution:

Ex 11.1 Class 10 Maths Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1\(\frac { 1 }{ 2 }\) times the corresponding sides of the isosceles triangle.
Solution:

Ex 11.1 Class 10 Maths Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac { 3 }{ 4 }\) of the corresponding sides of the triangle ABC.
Solution:

Ex 11.1 Class 10 Maths Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are \(\frac { 4 }{ 3 }\) times the corresponding sides of ∆ABC.
Solution:

Ex 11.1 Class 10 Maths Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac { 5 }{ 3F }\) times the corresponding sides of the given triangle.
Solution:
Steps of Construction:
1. Construct a ∆ABC, such that BC = 4 cm, CA = 3 cm and ∠BCA = 90°
2. Draw a ray BX making an acute angle with BC.
3. Mark five points B₁, B₂, B₃, B₄ and B₅ on BX, such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B_5.
4. Join B₃C.
5. Through B₅, draw B₅C’ parallel to B₃C intersecting BC produced at C’.
6. Through C’, draw C’A’ parallel to CA intersecting AB produced at A’.
Thus, ∆A’BC’ is the required right triangle.

Justification:

Class 10 Maths Constructions Mind Maps

Construction

Construction implies drawing geometrical figures accurately such that triangles, quadrilateral and circles with the help of ruler and compass.

Division of a Line Segment

A line segment can be divided in a given ratio (both internally and externally)
Example:
Divide a line segment of length 12 cm internally in the ratio 3:2.
Solution :
Steps of construction :
(i) Draw a line segment AB = 12 cm. by using a ruler.
(ii) Draw a ray making a suitable acute angle ∠BAX with AB.

(iii) Along AX, draw 5 ( = 3 + 2) arcs intersecting the ray AX at A₁? A₂, A₃, A₄ and A₅ such that
AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅
(iv) Join BA₅.
(v) Through A₃ draw a line A₃P parallel to A₅B making ∠AA₃P = ∠AA₅B, intersecting AB at point P.
The point P so obtained is the required point, which divides AB internally in the ratio 3 : 2.

Similar Triangles

(i) This Construction involves two different situation.
(a) Construction of a similar triangle smaller than the given triangle.
(b) Construction of a similar triangle greater than the given triangle.
(ii) The ratio of sides of the triangle to be constructed with the corresponding sides of the given triangle is called scale factor.
Example:
Draw a triangle ABC with side BC = 7 cm. ∠B = 45°, ∠A = 105°. Construct a triangle whose sides are (4/3) times the corresponding side of ∆ABC.

Solution :
Steps of construction :
(i) Draw BC = 7 cm.
(ii) Draw a ray BX and CY such that ∠CBX= 45° and
∠BCY = 180° – (45° + 105°) = 30°
Suppose BX and CY intersect each other at A.
∆ABC so obtained is the given triangle.
(iii) Draw a ray BZ making a suitable acute angle with BC on opposite side of vertex A with respect to BC.
(iv) Draw four (greater of 4 and 3 in 4/3) arcs intersecting the ray BZ at B₁, B₂, B₃, B₄ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
(v) Join B₃ to C and draw a line through B₄ parallel to B₃C, intersecting the extended line segment BC at C’.
(vi) Draw a line through C’ parallel to CA intersecting the extended line segment BA at A’. Triangle A’BC’ so obtained is the required triangle.

Tangents to a Circle

Two tangents can be drawn to a given circle from a point outside it.
Example:
Draw a circle of radius 4 cm. Take a point P outside the circle. Without using the centre of the circle, draw two tangents to the circle from point P.
Solution :

Steps of construction :
(i) Draw a circle of radius 4 cm.
(ii) Take a point P outside the circle and draw a secant PAB, intersecting the circle at A and B.
(iii) Produce AP to C such that AP = CP.
(iv) Draw a semi-circle with CB as diameter.
(v) Draw PD ⊥ CB, intersecting the semi-circle at D.
(vi) With P as centre and PD as radius draw arcs to intersect the given circle at T and T’
(vii) Join PT and PT’. Then, PT and PT’ are the required tangents.
Note:
If centre of a circle is not given, then it can be located by finding point of intersection of perpendicular bisector, of any two nonparallel chords of a circle.

NCERT Solutions for Class 10 Maths Chapter 11 Constructions (Hindi Medium) Ex 11.1

We hope the NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 11 Constructions Exercise 11.1, drop a comment below and we will get back to you at the earliest.

NCERT Solutions For Class 10 Science Chapter 12 Electricity: In this article, we will provide you all the necessary information regarding NCERT solutions for class 10 science physics chapter 12 electricity. Working on CBSE class 10 physics electricity questions and answers will help candidates to score good marks in-class tests as well as in the CBSE Class 10 board exam.

Electricity class 10 NCERT solutions will not only help in board exam preparation but also helps in clearing the competitive exams like Engineering. Also, candidates can find electricity class 10 numericals with solutions which helps candidates solving their assignments. Read on to find out everything NCERT Solutions For Class 10 Science Chapter 12 Electricity.

NCERT Solutions for Class 10 Science Chapter 12 Electricity

Before getting into the details of NCERT Solutions For Class 10 Science Chapter 12 Electricity, let’s have an overview of the list of topics and subtopics under Electricity class 10 NCERT solutions:

1. Electricity
2. Electric Current And Circuit
3. Electric Potential And Potential Difference
4. Circuit Diagram
5. Ohm’S Law
6. Factors On Which The Resistance Of A Conductor Depends
7. Resistance Of A System Of Resistors
8. Heating Effect Of Electric Current
9. Electric Power

Free download NCERT Solutions for Class 10 Science Chapter 12 Electricity PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

• विद्युत कक्षा 10 विज्ञान हिंदी में
• Class 10 Electricity Important Questions
• Electricity Class 10 Notes
• Electricity NCERT Exemplar Solutions
• Class 10 Science Electricity Mind Map

NCERT Solutions for Class 10 Science Chapter 12 Intext Questions

Page Number: 200

Question 1
What does an electric circuit mean ?
Answer:
A continuous and closed path along which an electric current flows is called an electric circuit.

Question 2
Define the unit of current.
Answer:
Unit of current is ampere. If one coulomb of charge flows through any section of a conductor in one second then the current through it is said to be one ampere.
I = \(\frac { Q }{ t }\) or 1 A = I C s^-1

Question 3
Calculate the number of electrons constituting one coulomb of charge.
Answer:
Charge on one electron, e = 1.6 x 10^-19 C
Total charge, Q = 1 C
Number of electrons, n = \(\frac { Q }{ e }\) = \(\frac { 1C }{ 1.6x{ 10 }^{ -19 } }\) = 6.25 x 10¹⁸

Page Number: 202

Question 1
Name a device that helps to maintain a potential difference across a conductor.
Answer:
A battery.

Question 2
What is meant by saying that the potential difference between two points is IV?
Answer:
The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to the other.

Question 3
How much energy is given to each coulomb of charge passing through a 6 V battery ?
Answer:
Energy given by battery = charge x potential difference
or W = QV = 1C X 6V = 6J.

Page Number: 209

Question 1
On what factors does the resistance of a conductor depend ?
OR
List the factors on which the resistance of a conductor in the shape of a wire depends. [CBSE2018]
Answer:
The resistance of a conductor depends (i) on its length (ii) on its area of cross-section and (iii) on the nature of its material.

Question 2
Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source ? Why ?
Answer:
The current will flow more easily through a thick wire than a thin wire of the same material. Larger the area of cross-section of a conductor, more is the ease with which the electrons can move through the conductor. Therefore, smaller is the resistance of the conductor.

Question 3
Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it ?
Answer:
When potential difference is halved, the current through the component also decreases to half of its initial value. This is according to ohm’s law i.e., V ∝ I.

Question 4
Why are coils of electric toasters and electric irons are made of an-alloy rather than a pure metal ?
OR
Why are alloys commonly used in electric heating devices? Given reason. [CBSE 2018]
Answer:
The coils of electric toasters, electric irons and other heating devices are made of an alloy rather than a pure metal because (i) the resistivity of an alloy is much higher than that of a pure metal, and (ii) an alloy does not undergo oxidation (or burn) easily even at high temperature, when it is red hot.

Question 5
Use the data in Table 12.2 (in NCERT Book on Page No. 207) to answer the following :
(i) Which among iron and mercury is a better conductor ?
(ii) Which material is the best conductor ?
Answer:
(i) Resistivity of iron = 10.0 x 10^-8 Ω m
Resistivity of mercury = 94.0 x 10^-8 Ω m.
Thus iron is a better conductor because it has lower resistivity than mercury.
(ii) Because silver has the lowest resistivity (= 1.60 x 10^-8 Ω m), therefore silver is the best conductor.

Page Number: 213

Question 1
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Answer:
The required circuit diagram is shown below :

Question 2
Redraw the circuit of Questions 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter ?
Solution:
The required circuit diagram is shown on the right.
Total voltage, V = 3 x 2 = 6V
Total resistance, R = 5Ω + 8Ω + 12Ω = 25Ω

Page Number: 216

Question 1
Judge the equivalent resistance when the following are connected in parallel :
(i) 1 Ω and 106 Ω,
(if) 1 Ω and 103 Ω and 106 Ω.
Answer:
When the resistances are connected in parallel, the equivalent resistance is smaller than the smallest individual resistance.
(i) Equivalent resistance < 1 Ω.
(ii) Equivalent resistance < 1 Ω.

Question 2
An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it ?
Solution:
Resistance of electric lamp, R₁ = 100 Ω
Resistance of toaster, R₂ = 50 Ω
Resistance of water filter, R₃ = 500 Ω
Equivalent resistance R_p of the three appliances connected in parallel, is

Resistance of electric iron = Equivalent resistance of the three appliances connected in parallel = 31.25 Ω
Applied voltage, V = 220 V
Current, I = \(\frac { V }{ R }\) = \(\frac { 220V }{ 31.25\Omega }\)

Question 3
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series ?
Answer:
Advantages of connecting electrical devices in parallel with the battery are :

1. In parallel circuits, if an electrical appliance stops working due to some defect, then all other appliances keep working normally.
2. In parallel circuits, each electrical appliance has its own switch due to which it can be turned on turned off independently, without affecting other appliances.
3. In parallel circuits, each electrical appliance gets the same voltage (220 V) as that of the power supply line.
4. In the parallel connection of electrical appliances, the overall resistance of the household circuit is reduced due to which the current from the power supply is high.

Question 4
How can three resistors of resistances 2Ω, 3 Ω, and 6Ω be connected to give a total resistance of (i) 4 Ω, (ii) 1 Ω ?
Solution:
(i) We can get a total resistance of 4Ω by connecting the 2Ω resistance in series with the parallel combination of 3Ω and 6Ω.

(ii) We can obtain a total resistance of 1Ω by connecting resistors of 2 Ω, 3 Ω and 6 Ω in parallel.

Question 5
What is (i) the highest, (ii) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Solution:
(i) Highest resistance can be obtained by connecting the four coils in series.
Then, R = 4Ω + 8Ω + 12Ω + 24Ω = 48Ω
(ii) Lowest resistance can be obtained by connecting the four coils in parallel.

Page Number: 218

Question 1
Why does the cord of an electric heater not glow while the heating element does ?
Solution:
Heat generated in a circuit is given by I²R t. The heating element of an electric heater made of nichrome glows because it becomes red-hot due to the large amount of heat produced on passing current because of its high resistance, but the cord of the electric heater made of copper does not glow because negligible heat is produced in it by passing current because of its extremely low resistance.

Question 2
Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Solution:
Here, Q = 96,000 C, t =1 hour = 1 x 60 x 60 sec = 3,600 s, V = 50 V
Heat generated, H = VQ = 50Vx 96,000 C = 48,00,000 J = 4.8 x 10⁶ J

Question 3
An electric iron of resistance 20Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Solution:
Here, R = 20 Ω, i = 5 A, t = 3s
Heat developed, H = I² R t = 25 x 20 x 30 = 15,000 J = 1.5 x 10⁴ J

Page Number: 220

Question 1
What determines the rate at which energy is delivered by a current ?
Answer:
Resistance of the circuit determines the rate at which energy is delivered by a current.

Question 2
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
Here, I = 5 A, V = 220 V, t = 2h = 7,200 s
Power, P = V I = 220 x 5 = 1100 W
Energy consumed = P x t = 100 W x 7200 s = 7,20,000 J = 7.2 x 10⁵ J

NCERT Solutions for Class 10 Science Chapter 12 Textbook Chapter End Questions

Question 1
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is :
(a) \(\frac { 1 }{ 25 }\)
(b) \(\frac { 1 }{ 5 }\)
(c) 5
(d) 25
Answer:
(d) 25

Question 2
Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) \(\frac { { v }^{ 2 } }{ 2 }\)
Answer:
(fa) IR2

Question 3
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be :
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:
(d) 25 W

Question 4
Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be :
(a) 1 : 2
(b) 2 : 1
(c) 1 : 4
(d) 4 : 1
Answer:
(c) 1 : 4

Question 5
How is a voltmeter connected in the circuit to measure the potential difference between two points ?
Answer:
A voltmeter is connected in parallel to measure the potential difference between two points.

Question 6
A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10^-8 Ω m. What will be the length of this wire to make its resistance 10 Ω ? How much does the resistance change if the diameter is doubled ?
Answer:


If a wire of diameter doubled to it is taken, then area of cross-section becomes four times.
New resistance = \(\frac { 10 }{ 2 }\) = 2.5 Ω, Thus the new resistance will be \(\frac { 1 }{ 4 }\) times.
Decrease in resistance = (10 – 2.5) Ω = 7.5 Ω

Question 7
The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below :

Plot a graph between V and I and calculate the resistance of the resistor.
Solution:
The graph between V and I for the above data is given below.
The slope of the graph will give the value of resistance.
Let us consider two points P and Q on the graph.
and from P along Y-axis, which meet at point R.
Now, QR = 10.2V – 34V = 6.8V
And PR = 3 – 1 = 2 ampere


Thus, resistance, R = 3.4 Ω

Question 8
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Solution:
Here, V = 12 V and I = 2.5 mA = 2.5 x 10^-3 A
∴ Resistance, R = \(\frac { V }{ I }\) = \(\frac { 12V }{ 2.5\times { 10 }^{ 3 }A }\) = 4,800 Ω = 4.8 x 10^-3 Ω

Question 9
A battery of 9V is connected in series with resistors of 0.2 O, 0.3 O, 0.4 Q, 0.5 Q and 12 £1, respectively. How much current would flow through the 12 Q resistor?
Solution:
Total resistance, R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω – 13.4 Ω
Potential difference, V = 9 V
Current through the series circuit, I = \(\frac { V }{ R }\) = \(\frac { 12V }{ 13.4\Omega }\) = 0.67 A
∵ There is no division of current in series. Therefore current through 12 Ω resistor = 0.67 A.

Question 10
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line? [CBSE (Delhi) 2013]
Solution:
Suppose n resistors of 176 Ω are connected in parallel.

Thus 4 resistors are needed to be connect.

Question 11
Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4Ω
Solution:
Here, R₁ = R₂ = R₃ = 6 Ω.

(i) When we connect R₁ in series with the parallel combination of R₂ and R₃ as shown in Fig. (a).
The equivalent resistance is

(ii) When we connect a series combination of R₁ and R₂ in parallel with R₃, as shown in Fig. (b), the equivalent resistance is

Question 12
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A ?
Solution:
Here, current, I = 5 A, voltage, V = 220 V
∴ Maxium power, P = I x V = 5 x 220 = 1100W
Required no. of lamps \( =\frac { Max.Power }{ Power\quad of\quad 1\quad lamp } \quad =\quad \frac { 1100 }{ 10 } =110\)
∴ 110 lamps can be connected in parallel.

Question 13
A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases ?
Solution:
(i) When the two coils A and B are used separately. R = 24 Ω, V = 220 V

(ii) When the two coils are connected in series,

(iii) When the two coils are connected in parallel.

Question 14
Compare the power used in the 2 Ω resistor in each of the following circuits
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Solution:
(i) The circuit diagram is shown in figure.
Total resistance, R = 1Ω + 2Ω = 3Ω
Potential difference, V = 6 V

Power used in 2Ω resistor = I²R = (2)² x 2 = 8 W

(ii) The circuit diagram for this case is shown :
Power used in 2 resistor = \(\frac { { v }^{ 2 } }{ R }\) =\(\frac { { 4 }^{ 2 } }{ 2 }\) = 8 W.

[ ∵ Current is different for different resistors in parallel combination.]

Question 15
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V ? [CBSE 2018]
Solution:
Power of first lamp (P₁) = 100 W
Potential difference (V) = 220 V

Question 16
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes ?
Solution:
Energy used by 250 W TV set in 1 hour = 250 W x 1 h = 250 Wh
Energy used by 1200 W toaster in 10 minutes = 1200 W x 10 min
= 1200 x \(\frac { 10 }{ 60 }\) = 200 Wh 60
Thus, the TV set uses more energy than the toaster.

Question 17
An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
Solution:
Here, R = 8 Ω, 1 = 15 A, t = 2 h
The rate at which heat is developed in the heater is equal to the power.
Therefore, P = I² R = (15)² x 8 = 1800 Js^-1

Question 18
Explain the following:
(i) Why is tungsten used almost exclusively for filament of electric lamps ?
(ii) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal ?
(in) Why is the series arrangement not used for domestic circuits ?
(iv) How does the resistance of a wire vary with its area of cross-section ?
(v) Why are copper and aluminium wires usually employed for electricity transmission?
Answer:
(i) The tungsten is used almost exclusively for filament of electric lamps because it has a very high melting point (3300°C). On passing electricity through tungsten filament, its temperature reaches to 2700°C and it gives heat and light energy without being melted.
(ii) The conductors of electric heating devices such as bread-toasters and electric irons, are made of an alloy rather than a pure metal because the resistivity of an alloy is much higher than that of pure metal and an alloy does not undergo oxidation (or burn) easily even at high temperature.
(iii) The series arrangement is not used for domestic circuits because in series circuit, if one electrical appliance stops working due to some defect, than all other appliances also stop working because the whole circuit is broken.
(iv) The resistance of a wire is inversely proportional to its area of cross-section, i.e., Resistance R ∝ (1/πr²). If the area of cross section of a conductor of fixed length is increased, then resistance decreases because there are more free electrons for movement in conductor.
(v) Copper and aluminium wires usually employed for electricity transmission because they have very low resistances. So, they do not become too hot on passing electric current.

NCERT Solutions for Class 10 Science Chapter 12 Electricity

Electric current, potential difference and electric current, Ohms law, Resistance, Resistivity factors on which the resistance of a conductor depends; Series combination of resistors, parallel combination of resistors; and its application on daily life; Heating effect of Electric current, electric Power, Interrelation between P, V, and R.

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Science
Chapter	Chapter 12
Chapter Name	Electricity
Number of Questions Solved	41
Category	NCERT Solutions

Formulae Handbook for Class 10 Maths and Science

Page 200
What does an electric circuit mean?
Electric circuit is a continuous and closed path made of conducting wires, through which the electric current flows. It comprises a cell, ammeter, voltmeter, plug key, etc.
Define the unit of current.
SI unit of electric current is ampere (A).
Ampere is the flow of electric charges through an area at the rate of one coulomb per second, i.e. if 1 coulomb of electric charge flows through a cross-section of wire for 1 second, then it would be equal to 1 ampere.

Page 202

Question 1:
Name a device that helps to maintain a potential difference across a conductor.
Answer:
Cell or battery eliminator.

Question 2:
What is meant by saying that the potential difference between two points is 1 V?
Answer:
As we know that V = W / q
Thus, the potential difference between two points is one volt when one joule of work is done to carry a charge of one coulomb between the two points in the electric field.

More Resources for CBSE Class 10

• NCERT Solutions
• NCERT Solutions for Class 10 Science
• NCERT Solutions for Class 10 Maths
• NCERT Solutions for Class 10 Social
• NCERT Solutions for Class 10 English
• NCERT Solutions for Class 10 Hindi
• NCERT Solutions for Class 10 Sanskrit
• NCERT Solutions for Class 10 Foundation of IT
• RD Sharma Class 10 Solutions

Question 3:
How much energy is given to one coulomb of charge passing through a 6 V battery
Answer:

Page 209

Question 1:
On what factors does the resistance of a conductor depend
Answer:
Resistance of a conductor depends upon:
(i) Resistivity of the material.
(ii) Length of the conductor.
(iii) Cross-sectional area of the conductor.

Question 2:
Will current flow more easily through a thick wire or thin wire of the same material when connected to the same source? Why
Answer:
The current flows more easily through a thick wire than through a thin wire because the resistance of thick wire is less than that of a thin wire as R ∝ 1/A.

Download NCERT Solutions for Class 10 Science Chapter 12 Electricity PDF

Question 3:
Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Answer:

Hence, the current through an electrical component also becomes half of its previous value.

Question 4:
Why are the coils of electric toasters and electric irons made of an alloy rather than a pure metal
Answer:
The coils of electric toaster and electric iron are made of an alloy rather than a pure metal because of the following reasons;
(i) The resistivity of an alloy is higher than that of a pure metal.
(ii) It has high melting point and does not oxidise.

Question 5:
Use the data in Table 12.2 of NCERT book to answer the following:
(a) Which among iron and mercury is a better conductor?
(b) Which material is the best conductor? ‘
Answer:
(a) Iron because its resistivity is less than mercury.
(b) Silver is the best conductor as it has least resistivity.

Page 213

Question 1:
Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, a 8 Ω resistor and a 12 Ω resistor and a plug key, all connected in series.
Answer:

Question 2:
Redraw the circuit of the above question, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the voltage across the 12 resistor. What would be the reading in the ammeter and the voltmeter?
Answer:

Total resistance of the circuit = R
Since all the three resistors are connected in series, so, the equivalent resistance R is equal to the sum of all resistance.
R = 5 Ω + 8 Ω + 12 Ω = 25 Ω

Page 216

Question 1:
Judge the equivalent resistance when the following are connected in parallel.
(a) 1 Ω and 10⁶ Ω
(b) 1 Ω , 10³ Ω and 10⁶ Ω
Answer:
Equivalent resistance in parallel combination of resistors is always less than the least resistance of any resistor in the circuit.
Hence, in both the given cases, the equivalent resistance is less than 1 Ω.

Question 2:
An electric lamp of 100 Ω, a toaster of resistance 50 Ω and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances and what is the current flows through it?
Answer:

Question 3:
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series ?
Answer:
Advantages of connecting electrical devices in parallel:

1. When the appliances are connected in parallel with the battery, each appliance gets the same potential difference as that-of battery which is not possible in series connection.
2. Each appliance has different resistances and requires different currents to operate properly. This is possible only in parallel connection, as in series connection, same current flows through all devices, irrespective of their resistances.
3. If one appliance fails to work, other will continue to work properly.

Question 4:
How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω (b) 1 Ω?
Answer:
(a) In order to get 4 Ω, resistance 2 Ω should be connected in series with the parallel combination of 3 Ω and 6 Ω.

Question 5:
What is (a) the highest (b) the lowest total resistance that can be secured by combination of four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answer:
(a) The highest resistance is secured by combining all four coils of resistance in series.
R_s = 4 Ω+ 8 Ω + 12 Ω + 24 Ω = 48 Ω
(b) The lowest resistance is secured by combining all four coils of resistance in parallel.

Page 218

Question 1:
Why does the cord of an electric heater not glow while the heating element does?
Answer:
The cord of an electric heater is made up of metallic wire such as copper or aluminum which has low resistance while the heating element is made up of an alloy which has more resistance than its constituent metals. Also heat produced ‘H’ is
H = I²Rt
Thus, for the same current H oc R, so for more resistance, more heat is produced by heating element and it glows.

Question 2:
Compute the heat generated while transferring 96000 C of charge in one hour through a potential difference of 50 V.
Answer:

Question 3:
An electric iron of resistance 20 Q takes a current of 5 A. Calculate the heat developed in 30 s.
Answer:
Given R = 20 Ω, I = 5 A, t = 30 s
H = I²Rt = (5)² x 20 x 30 = 15000 J = 1.5 x 10⁴ J

Page 220

Question 1:
What determines the rate at which energy is delivered by a current?
Answer:
Electric power determines the rate at which energy is delivered by a current.

Question 2:
An electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
Answer:
Given I = 5 A, V = 220 V, t = 2 h Power,
p = VI = 220 x 5 = 1100 W
Energy consumed = Vlt = Pt
= 1100 x 2 = 2200 Wh

Textbook Questions

Question 1:
A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio R/R’ is

Answer:

Question 2:
Which of the following terms does not represent electrical power in a circuit?
(a) I²R
(b) IR²
(c) VI
(d) V^2/R
Answer:
(b) P = V^2/R = I²R = VI Option (b) does not represent electrical power.

Question 3:
An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be
(a) 100 W
(b) 75 W
(c) 50 W
(d) 25 W
Answer:

Question 4:
Two conducting wires of same material and of equal lengths and diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be
(a) 1:2
(b) 2:1
(c) 1:4
(d) 4:1
Answer:

Question 5:
How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer:
A voltmeter is connected in parallel across any two points in a circuit to measure the potential difference between them with its +ve terminal to the point at higher potential and -ve terminal to the point at lower potential of the source.

Question 6:
A copper wire has a diameter 0.5 mm and resistivity of 1.6 X 10^-8 Ωm. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Answer:

Question 7:
The values of the current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:

Plot a graph between V and I and calculate the resistance of that resistor.
Answer:

Question 8:
When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
Answer:

Question 9:
A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω, respectively. How much current would flow through the 12 Ω resistor?
Answer:
Since all the resistors are in series, the same current, 0.67 A flows through the 12 Ω resistor.

Question 10:
How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Answer:

Question 11:
Show how you would connect three resistors, each of resistance 6 Ω , so that the combination has a resistance of (i) 9 Ω , (it) 4 Ω .
Answer:
(i) When two 6 Ω resistances are in parallel and the third is in combination to this, the equivalent resistance will be 9 Ω.

(ii) When two 6 Ω resistances are in series and the third is in parallel to them, then it will be 4 Ω.

Question 12:
Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
Answer:
Since, N bulbs of power P each connected in parallel will make the total power of NP,

Question 13:
A hot plate of an electric oven connected to 220 V line has two resistance coils A and B, each of 24 Q resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
Answer:

Question 14:
Compare the power used in the 2 Ω resistor in each of the following circuits.
(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and
(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
Answer:

Question 15:
Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Answer:

Question 16:
Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
Answer:
Energy consumed by 250 W TV set in 1 h = 250 x 1 = 250 Wh.
Energy consumed by 1200 W toaster in 10 min = 1200 X 1/6 = 200 Wh.
∴ Energy consumed by TV set is more than the energy consumed by toaster in the given timings.

Question 17:
An electric heater of resistance 8 f2 draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.
Answer:

Question 18:
Explain the following.
(a) Why is the tungsten used almost exclusively for filament of electric lamps?
(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?
(c) Why is the series arrangement not used for domestic circuits?
(d) How does the resistance of a wire vary with its area of cross-section?
(e) Why are copper and aluminum wires usually employed for electricity transmission?
Answer:
(a) It has high melting point and emits light at a high temperature.
(b) It has more resistivity and less temperature coefficient of resistance.
(c) (i) All appliances do not get same potential in series arrangement.
(ii) All appliances cannot be individually operated.
(d) R ∝ =1 / Area of cross – section.
(e) They are very good conductors of electricity.

Short Answer Type Questions

Question 1:
Three 2 Ω resistors, A, B and C are connected as shown in figure. Each of them dissipates energy and can withstand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors.
Answer:

Question 2:
Should the resistance of an ammeter be low or high? Give reason.
Answer:
The resistance of an ammeter should be low so that it will not disturb the magnitude of current flowing through the circuit when connected in series in a circuit.

Question 3:
How does use of a fuse wire protect electrical appliances?
Answer:
The fuse wire is always connected in series with the live wire or electrical devices. If the flow of current exceeds the specified preset value due to some reason, the heat produced melts it and disconnects the circuit or the device from the mains. In this way, fuse wire protects the electrical appliances.

Question 4:
What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?
Answer:
The resistance offered by a metallic wire of unit length and unit cross-sectional area is called electrical resistivity.

Hence, when the length of wire is doubled, the resistance becomes double and current decreases to half.

Question 5:
A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp.
Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place? Give reason.
Answer:

Question 6:
Why is parallel arrangement used in domestic wiring?
Answer:
Parallel arrangement is used in domestic wiring because
(i) Each appliance gets the same voltage as that of the mains supply.
(ii) If one component is switched off, others can work properly.
(iii) Fault in any branch of the circuit can be easily identified.

Question 7:
B_1, B₂ and B₃ are three identical bulbs connected as shown in figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

(i) What happens to the glow of the other two bulbs when the bulb B j gets fused?
(ii) What happens to the reading of A₁ ,A₂ , A₃ and A when the bulb B₂ gets fused?
(iii) How much power is dissipated in the circuit when all the three bulbs glow together?
Answer:
(i) Since B₁ ,B₂ and B₃ are in parallel, the potential difference across each of them will remain same. So when the bulb B₁ gets fused, B₂ and B₃ have the same potential and continues with the same energy dissipated per second, i.e. they will glow continuously as they were glowing before.

Long Answer Type Questions

Question 1:
Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit, another set of three bulbs of the same wattage are connected in parallel to the same source.
(а) Will the bulb in the two circuits glow with the same brightness? Justify your answer.
(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.
Answer:

(a) The bulbs in the two circuits will not glow equally bright as the current through them is not the same.

(b) As one bulb fuses, the other bulbs in the series circuit will not glow because the circuit becomes an open circuit. While the rest of bulbs in parallel circuit will continue to glow without getting disturbed because in parallel combination, current gets additional paths to flow.

Question 2:
Find out the following in the electric circuit given in figure:

Answer:

(e) No difference, since the ammeters are connected in series and same current will pass through them, so reading of both ammeters will be same.

Multiple Choice Questions (MCQs) [1 Mark each]

Question 1.
To determine the equivalent resistance of two resistors when connected in series, a student arranged the circuit components as shown in the diagram. But he did not succeed to achieve the objective. [CCE 2010]

Which of the following mistakes has been committed by him in setting up the circuit?
(a) Position of ammeter is incorrect
(b) Position of voltmeter is incorrect
(c) Terminals of ammeter are wrongly connected
(d) Terminals of voltmeter are wrongly connected
Answer:
(c) Because positive terminal of ammeter must be connected with positive terminal of cell and negative terminal of an ammeter must be connected to negative terminal of a cell.

Question 2.
For the given circuit, name the components which are connected in parallel. [CCE 2011]

(a) R₁ and R₂
(b) R₁, R₂ and V
(c) R₂ and V
(d) R₁ and V
Answer:
(b) The components R₁, R₂ and V are connected in parallel combination. Because terminals of the resistance and voltmeter are connected together.

Question 3.
A student arranges the following circuit to get equivalent resistance of a series combination of two resistors R₁ and R₂.

Which one of the following statements will be true for this circuit [CCE 2007]
(a) It gives incorrect reading for current I as well as potential difference V
(b) It gives correct reading for current I but incorrect reading for potential difference V
(c) It gives correct reading for potential difference V but incorrect reading for current I
(d) It gives correct reading for both I and V
Answer:
(b) The voltmeter should be connected across the components of and R₂ to give correct reading for potential difference.

Question 4.
An ammeter has 20 divisions between 0 mark and 2A mark on its scale. The least count of ammeter is
(a) 0.01A
(b) 0.2A
(c) 0.1A
(d) 1A
Answer:
(c) Number of divisions = 20
Maximum reading of ammeter = 2 A
Least count of ammeter = 2/20 = 1/10 = 0.1 A

Question 5.
A student finds that there are 20 divisions between zero mark and 1V mark of a voltmeter. The least count of voltmeter is
(a) 0.1 V
(b) 0.01 V
(c) 0.05 V
(d) 1.0 V
Answer:
(c) Number of divisions = 20
Maximum reading of the voltmeter = 1 V
Least count of voltmeter = 1/20 = 0.05 V

Question 6.
The current flowing through a resistor connected in an electric circuit and the potential difference applied across its ends are shown in figure alongside.

The value of the resistance of the resistor is [CCE2013]
(a) 1 Ω
(b) 5 Ω
(c) 8 Ω
(d) 10 Ω
Answer:
(d) Reading from ammeter (7) = 180 mA= 0.18 A,
reading from voltmeter (V) = 1.8 V
Resistance of the resistor R = V/I = 1.8/0.18 = 180/18 = 10Ω

Question 7.
Which of the following is the correct method to connect the ammeter and voltmeter with resistance in the circuit to verify Ohm’s law? [CCE 2012]
(a) Ammeter and voltmeter in series
(b) Ammeter in series and voltmeter in parallel
(c) Ammeter in parallel and voltmeter in series
(d) Ammeter and voltmeter in parallel
Answer:
(b) In a circuit, ammeter should be connected in series, while voltmeter in parallel.

Question 8.
In an experiment on studying the dependence of the current I flowing through a given resistor on the potential difference V applied across it, a student has to change the value of the current. For doing this, he should change the
(a) number of cells used
(b) resistor itself
(c) ammeter used in the circuit
(d) Voltmeter used in the circuit
Answer:
(a) If we change the number of cells in electric circuit, the potential difference will change and as a result current flowing in the circuit changes.

Question 9.
A milliammeter had graduations marked 0, 100, 200, 300, 400 and 500. The space between 0 mark and 100 mark is divided into 20 divisions. If the pointer of the milliammeter is indicating the seventh graduation after 300 mark, the current flowing in the circuit is
(a) 335 mA
(b) 330 mA
(c) 331mA
(d) 340 mA
Answer:
(a) Number of divisions = 20
Least count of milliammeter = (100-0) / 20 = 5 mA
Milliammeter reading = 300 + 7 x 5 = 335 mA

Question 10.
If a student while studying the dependence of current on the potential difference keeps the circuit closed for a long time to measure the current and potential difference, then
(a) ammeter’s zero error will change
(b) ammeter will give more reading
(c) voltmeter will show constantly higher readings
(d) resistor will get heated up and its value will change
Answer:
(d) If the circuit is closed for a long time, then current flows in it for a long time which results that the resistor is heated.

Question 11.
To determine the eguivalent resistance of two resistors connected in series, a student prepared two electric circuits, correct reading of ammeter in the circuits is [CCE 2015]

(a) In circuit I, 1.0 A and in II, 0.1 A
(b) In both circuits I and II, 1.0 A
(c) In circuit I, 0.1 A and in II, 1.0 A
(d) In both circuits I and II, 0.1 A
Answer:
(b) Equivalent resistance of two resistors 3.5Ω and 1Ω in both the circuits I and II is R = 3.5 + 1 = 4.5 Ω
As, I = V/R = 4.5/4.5 = 1A
Therefore, current in both the circuits I and II is 1.0 A.

Question 12.
When parallel resistors are of three different values, the potential difference across its terminals is [CCE 2015]
(a) greatest across smallest resistance
(b) greatest across largest resistance
(c) equal across each resistance
(d) least across the smallest resistance
Answer:
(c) Potential difference across each resistor is same in parallel combination of resistors.

NCERT Solutions for Class 10 Science Chapter 12 Electricity (Hindi Medium)

Class 10 Science Electricity Mind Map

Electricity
Study of Electric Charges at Rest and in Motion

Charge
Something associated with the matter due to which it produces and experiences electric and magnetic effects. Resides on the outer surface of the conductor.
Q = ne S.I. unit coulomb (C)

Electric Current (I)
The time rate of flow of charge (Q) through any cross-section
I = \(\frac{Q}{t}\) S.I. unit ampere (A)

Types of Current
Direct Current
Current whose magnitude and direction does not vary with time.

Alternating Current
Current whose magnitude and direction periodically changes with time.

Electric Potential
Work done per unit charge
V = \(\frac{W}{Q}\)
S.I. unit volt

Ohm’s law: If the physical conditions remain same, current I ∝ V => V = IR
R-electric resistance Substances which obey ohm’s law called ohmic and that do not obey called non-ohmic substances.

Dependence of Resistance

On length (l) and area of cross-section (A)
R ∝ l
∝ \(\frac{1}{A}\)
R = \(\rho \frac{l}{A}\)
ρ = resistivity
Resistivity depends on the material of the conductor only.

On Temperature
R_t = R₀( 1 + αt)
α = temperature coefficient of resistance

Resistance (R): Obstruction offered to flow of electrons.
SI unit ohm
Resistance, R ∝ \(\frac{\ell^{2}}{m}\)
l = length and
m = mass of conducting wire

After stretching, if length increases by n times then resistance will increase by n² times i.e., R₂ = n² R₁. Similarly
if radius be reduced to \(\frac{1}{n}\) times then area of cross-section decreases \(\frac{1}{n^{2}}\) times so the resistance becomes n⁴ times i.e.. R₂ = n⁴ R₁
After stretching, if length of a conductor increases by x%, then resistance will increase by 2x% (valid only if x< 10%).

• Using n conductors of equal resistance, the number of possible combinations is 2^n-1 .
• If the resistances of n conductors are totally different, then the number of possible combinations will be 2ⁿ .
• If n identical resistances are first connected in series and then in parallel, the ratio of the equivalent resistance is given by
  \(\frac{R_{s}}{R_{p}}=\frac{n^{2}}{1}\)
• If a wire of resistance R is cut in n equal parts and then these parts are collected to form a bundle, then equivalent resistance of combination will be \(\frac{\mathrm{R}}{\mathrm{n}^{2}}\)
• If equivalent resistance of R₁ and R₂ in series and parallel be R_s and R_p respectively, then
  R₁ = \(\frac{1}{2}\left[\mathrm{R}_{\mathrm{s}}+\sqrt{\mathrm{R}_{\mathrm{s}}^{2}-4 \mathrm{R}_{\mathrm{s}} \mathrm{R}_{\mathrm{p}}}\right]\)
  and R₂ = \(=\frac{1}{2}\left[\mathrm{R}_{\mathrm{s}}-\sqrt{\mathrm{R}_{\mathrm{s}}^{2}-4 \mathrm{R}_{\mathrm{s}} \mathrm{R}_{\mathrm{p}}}\right]\)

Grouping of Resistances

Series Grouping of Resistances
Equivalent resistance, resistance, R_s = R₁ + R₂ + … + R_n
In this case same current flows through each resistance but potential difference in the ratio of resistance

Parallel Grouping of Resistances
\(\frac{1}{R_{P}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\ldots+\frac{1}{R_{n}}\)
In this case same potential across each resistance but current distributes in the reverse ratio of their resistances

Electric Circuit
The arrangement of various electrical components along which electric current flow

Heating Effect of Electric Current
As current flows through a conductor, the free electrons lose energy which is converted into heat.
Joule’s heating law
H ∝ I²
H ∝ R
H ∝ t
H = I²Rt = VIt

Practical Applications

• Electric heater, electric iron and water heater, etc. work on the principle of heating effect of current
• Electric bulb glows when electric current flows through the filament of the bulb

Electric Power
Rate at which electric energy is dissipated or consumed in a circuit,
P = VI ,
or P = I²R = \(\frac{\mathrm{V}^{2}}{\mathrm{R}}\)

Watt is a smaller unit of power, its other bigger units are kilowatt (KW),
Megawatt (MW) and Horsepower (HP)
1 KW = 10³W 1 MW = 10⁶W
1 hp =746 W
The commercial unit of electrical energy is 1 Kwh.
1 Kwh = 3.6 × 10⁶J

Elements of Circuit
Cell
Direct current source of electromotive force. Combination of two or more cells is called battery.

Rheostat
Wire of special type of alloy like manganin, Eureka, nichrome, etc. is wound on a hollow cylinder of china clay. It controls the current in the electric circuit.

Switch
It is used to close or open the electric circuit, controls the movement of electrons in a circuit.

Voltmeter
Measures the potential difference between two points in the circuit. Its resistance is high and it is used in parallel with the resistance wire.

Fuse
It is a safety device having very thin wire which is made up of either tin or alloy of tin and lead.
This wire has low melting point so it melts and breaks the circuit easily if the current in the circuit exceeds.

Ammeter
Measures the value of current flowing in the circuit.
The resistance of ammeter is small and it is used in series with the circuit.

LED
It is a device which glows even if a weak electric current is allowed to flow through it

Now that you are provided all the necessary information regarding Electricity class 10 NCERT solutions and we hope this detailed article on NCERT Solutions For Class 10 Science Chapter 12 Electricity is helpful. If you have any doubt regarding this article or  NCERT Solutions For Class 10 Science Chapter 12 Electricity, leave your comments in the comment section below and we will get back to you as soon as possible.

NCERT Solutions For Class 10 Science Chapter 6 Life Processes: In this article, you will find all the necessary information regarding NCERT Solutions For Class 10 Science Chapter 6 Life Processes.  Students who are planning to build their career stream in the field of medicine can refer to this article as biology plays a major role in the medical field. Candidates having strong command over life processes class 10 NCERT solutions will be able to easily crack the competitive exams like NEET, AIIMS, JIPMER, etc.,

Did you check the Life Process Important Questions which were given in the previous Board Papers?

In this article, we have covered all the important topics in the exercises and each answer comes with a detailed explanation to help the class 10 students to understand concepts better. Read on to find out everything about NCERT class 10 science book activities solutions chapter 6.

NCERT Solutions for Class 10 Science Chapter 6 Life Processes

Before getting into the details of NCERT Solutions For Class 10 Science Chapter 6 Life Processes, let’s have an overview of topics and subtopics under class 10 science chapter 6 activities solutions:

1. Life Processes
2. What Are Life Processes?
3. Nutrition
4. Respiration
5. Transportation
6. Excretion

Free download NCERT Solutions for Class 10 Science Chapter 6 Life Processes PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

• जैव-प्रक्रम कक्षा 10 विज्ञान हिंदी में
• Life Processes Class 10 Notes
• Life Processes NCERT Exemplar Solutions
• Life Processes Class 10 Extra Questions
• Class 10 Life Processes Mind Map

NCERT Solutions for Class 10 Science Chapter 6 Intext Questions

Page Number: 95

Question 1
Why is diffusion insufficient to meet the oxygen requirements of multicellular organisms like humans ?
Answer:
In multicellular organisms like humans, all the body cells are not in direct contact with the surrounding environment. Therefore, every cell of the body will not get oxygen as per need by the process of diffusion from the environment. Therefore diffusion is insufficient to meet the oxygen requirements of multicellular organisms.

Question 2
What criteria do we use to decide whether something is alive ?
Answer:
The main criteria used to decide whether something is alive are breathing and respiration. However, living beings also show growth and movement.

Question 3
What are outside raw materials used by an organism ?
Answer:
Any organism uses organic molecules as raw material. Heteroptrophs use food and autotrophs use carbon dioxide, minerals, water and all organisms use oxygen (for respiration) as raw materials.

Question 4
What processes would you consider essential for maintaining life ?
Answer:
Processes essential for maintaining life are :
(i) Nutrition
(ii) Respiration
(iii) Transportation
(iv) Excretion

Page Number: 101

Question 1
What are the differences between autotrophic nutrition and heterotrophic nutrition ?
Answer:

Autotrophic nutrition	Heterotrophic nutrition
(i) In this mode of nutrition an organism makes or synthesizes its own food.	(i) In this mode of nutrition an organism cannot make or synthesize its own food
Organisms use simple inorganic materials like carbon dioxide and water and synthesise their food in presence of sunlight.	(ii) Organisms cannot make their own food from simple inorganic matter and depend on other organisms for their food.
(iii) All green plants and some algae undergo this mode of nutrition.	(iii) All the animals, most bacteria and fungi undergo this mode of nutrition.

Question 2
Where do plants get each of the raw materials required for photosynthesis ?
Answer:
(i) Carbon dioxide : Plants get carbon dioxide from the environment/atmosphere through stomata.
(ii) Water : Plants absorb water from the soil through roots and transport to leaves.
(iii) Sunlight : Plants get sunlight from the sun.
(iv) Chlorophyll : It is present in chloroplast found in green leaves and green parts of plants.

Question 3
What is the role of the acid in our stomach ?
Answer:
Role of acid in our stomach is :
(i) To make acidic medium which is necessary for the activation of the enzyme pepsin.
(ii) To kill bacteria which the food may contain.

Question 4
What is the function of digestive enzymes ?
Answer:
The food we eat is complex in nature, i.e., it contains complex molecules. Digestive enzymes break down these complex molecules into smaller simpler molecules so that they can be absorbed by the walls of the intestine.

Question 5
How is the small intestine designed to absorb digested food ?
Answer:
The small intestine is designed to provide maximum area for absorption of digested food and its transfer into the blood for its circulation into the body. For this the inner lining of the small intestine has numerous finger-like projections called villi. The villi are richly supplied with blood vessels which take the absorbed food to each and every cell of the body.

Page Number: 105

Question 1
What advantage over an aquatic organism does a terrestrial organism have with regard to obtaining oxygen for respiration ?
Answer:
Aquatic organisms use oxygen dissolved in surrounding water. Since air dissolved in water has fairly low concentration of oxygen, the aquatic organisms have much faster rate of breathing.
Terrestrial organisms take oxygen from the oxygen-rich atmosphere through respiratory organs. Hence, they have much less breathing rate than aquatic organisms.

Question 2
What are the different ways in which glucose is oxidised to provide energy in various organisms ?
Answer:
First step of breakdown of glucose (6 carbon molecules) takes place in the cytoplasm of cells of all organisms. This process yields a three carbon molecule compound called pyruvate.
Further break down of pyruvate takes place in different ways in different organisms.

(i) Anaerobic respiration : The anaerobic respiration in plants (like yeast) produces ethanol and carbon dioxide as end products.
(ii) Aerobic respiration : In aerobic respiration break down of pyruvate takes place in presence of oxygen to give rise three molecules of carbon dioxide and water. The release of energy in aerobic respiration is much more than in anaerobic respiration.
(iii) Lack of oxygen : Sometimes, when there is lack of oxygen especially during physical exercise, in our muscles, pyruvate is converted into lactic acid (3 carbon molecule compound). Formation of lactic acid in muscles causes cramp.

Question 3
How is oxygen and carbon dioxide transported in human beings ?
Answer:
(i) Transport of oxygen : Haemoglobin present in the blood takes up the oxygen from the air in the lungs. It carries the oxygen to tissues which are deficient in oxygen before releasing it.
(ii) Transport of carbon dioxide : Carbon dioxide is more soluble in water. Therefore, it is mostly transported from body tissues in the dissolved form in our blood plasma to lungs. Here it diffuses from blood to air in the lungs.

Question 4
How are the lungs designed in human beings to maximise the area for exchange of gases ?
Answer:
Within the lungs, the air passage divides into smaller and smaller tubes, called bronchi which in turn form bronchioles. The bronchioles terminate in balloon-like structures, called alveoli. The alveoli present in the lungs provide maximum surface for exchange of gases. The alveoli have vary thin walls and contain an extensive network of blood vessels to facilitate exchange of gases.

Page Number: 110

Question 1
What are the components of the transport system in human beings ? What are the functions of these components ?
Answer:
The transport system (circulatory system) in human beings mainly consists of heart, blood and blood vessels.

(i) Function of heart : The heart receives deoxygenated blood from the body parts and pumps it to lungs for enriching with oxygen. It receives purified blood from lungs and pumps it around the body.
(ii) Function of blood : Blood transports oxygen, carbon dioxide, digested food, hormones and nitrogeneous waste like urea. It also protects the body from diseases and regulates the body temperature.
(iii) Function of blood vessels : The blood pushed by the heart flows through the blood vessels (arteries, veins and capillaries) and also comes back to the heart through them.

Question 2
Why is it necessary to separate oxygenated and deoxygenated blood in mammals and birds ?
Answer:
Separation of oxygenated and deoxygenated blood allows good supply of oxygen to the body. This system is useful in animals that have high energy requirement. Mammals and birds constantly need oxygen to get energy to maintain their body temperature constant.

Question 3
What are the components of the transport system in highly organised plants?
Answer:
In highly organised plants there are two conducting tissues xylem and phloem.
Xylem consists of vessels, tracheids and other xylem tissues. The interconnected vessels and tracheids form a continuous system of water conducting channels reaching all parts of the plant. Xylem carries water and minerals.
Phloem conducts soluble products of photosynthesis from leaves to different parts of the plant body.

Question 4
How are water and minerals transport in plants ? [AICBSE 2015]
Answer:
The roots of a plant have hair called root hair.
The root hair are directly in contact with the film of water in between the soil particles. Water and dissolved minerals get into the root hair by the process of diffusion. The water and minerals absorbed by the root hair from the soil pass from cell to cell by osmosis through the epidermis, root cortex, endodermis and reach the root xylem.

The xylem vessels of the root of the plant are connected to the xylem vessels of its stem.
Therefore the water containing dissolved minerals enters the root xylem vessels into stem xylem vessels. The xylem vessels of the stem branch into the leaves of the plants. So, the water and minerals carried by the xylem vessels in the stem reach the leaves through the branched xylem vessels which enter from the petiole (stalk of the leaf) into each and every part of the leaf. Thus the water and minerals from the soil reach through the root and stem to the leaves of the plants. Evaporation of water molecules from the cells of a leaf creates a suction which pulls water from the xylem cells of roots. The loss of water in the form of vapour from the aerial parts of the plant is known as transpiration.

Question 5
How is food transported in plants ?
Answer:
The movement of food in phloem (or translocation) takes place by utilizing energy. The sugar (food) made in leaves is loaded into the sieve tubes of phloem tissue by using energy from ATR Water now enters the sieve tubes containing sugar by the process of osmosis due to which the pressure in the phloem tissue rises. This high pressure produced in the phloem tissue moves the food to all parts of the plant having less pressure in their tissues. This allows the phloem to transport food according to the needs of the plant.

Page Number: 112

Question 1
Describe the structure and functions of nephrons.
Answer:
Structure of nephron : Each nephron is composed of two parts. First one is a cup-shaped bag at its upper end which is called Bowman’s capsule.
The Bowman’s capsule contains a bundle of blood capillaries which is called glomerulus. One end of the glomerulus is attached to the renal artery which brings the impure blood containing the urea waste into it. These impurities are filtered. The other part of the nephron is coiled. In this part, the substances like sugar (glucose), amino acid, ions and excess water which are required by the body, are reabsorbed. The substance remained in the nephron is mainly urine containing dissolved urea in water which is expelled from the body through urethra from time to time.

Functions of nephron : Filtration of blood takes place in Bowman’s capsule from the capillaries of glomerulus. The filtrate passes into the tubular part of the nephron. This filtrate contains glucose, amino acids, urea, uric acid, salts and water.
Reabsorption : As the filtrate flows along the tubule, useful substances such as glucose, amino acids, salts and water are selectively reabsorbed into the blood by capillaries surrounding the nephron tubule.
Urine : The filtrate which remained after reabsorption is called urine. Urine contains dissolved nitrogenous waste like urea and uric acid, excess salts and water. Urine is collected from nephrons to carry it to the ureter from where it passes into urinary bladder.

Question 2
What are the methods used by plants to get rid of excretory products ?
Answer:
(i) The plants get rid of gaseous products-through stomata in leaves and lenticels in stems.
(ii) The plants get rid of stored solid and liquid waste by the shedding off leaves, peeling off bark and felling off fruits.
(iii) The plants get rid of wastes by secreting them in the form of gums and resins.
(iv) Plants also excrete some waste substances into the soil around them.

Question 3
How is the amount of urine produced regulated ?
Answer:
The amount of urine is regulated by kidney. It depends on the quantity of excess water and wastes dissolved in water.

(i) Quantity of water : When water is abundant in the body tissues, large quantities of dilute urine is excreted out. When water is less in quantity in the body tissues, a small quantity of concentrate urine is excreted.
(ii) Quantity of dissolved wastes : Dissolved wastes, especially nitrogenous wastes, like urea and uric acid and salts are excreted from the body. When there is more quantity of dissolved wastes in the body, more quantity of water is required to excrete them. Therefore, the amount of urine produced increases.
(iii) Hormones : The amount of urine produced is also regulated by certain hormones which control the movement of water and Na+ ions in and out of the nephrons.

NCERT Solutions for Class 10 Science Chapter 6 Textbook Chapter End Questions

Question 1
The kidneys in human beings are a part of the system for
(i) nutrition
(ii) respiration
(iii) excretion
(iv) transportation
Answer:
(iii) Excretion

Question 2
The xylem in plants are responsible for
(i) transport of water
(ii) transport of food
(iii) transport of amino acids
(iv) transport of oxygen
Answer:
(i) Transport of water

Question 3
The autotrophic mode of nutrition requires
(i) carbon dioxide and water
(ii) chlorophyll
(iii) sunlight
(iv) all of the above
Answer:
(iv) All of the above

Question 4
The breakdown of pyruvate to give carbon dioxide, water and energy takes place in
(i) cytoplasm
(ii) mitochondria
(iii) chloroplast
(iv) nucleus
Answer:
(ii) Mitochondria

Question 5
How are fats digested in our bodies ? Where does this process take place ?
Answer:
Digestion of fats takes place in the small intestine.
Bile juice secreted by the liver poured in the intestine along with pancreatic juice. The bile salts present in the bile juice emulsify fhe large globules of fats. Therefore, by enulsification large globules break down into fine globules to provide larger surface area to act upon by the enzymes.
Lipase enzyme present in the pancreatic juice causes break down of emulsified fats. Glands present in the wall of small intestine secrete intestinal juice which contains lipase enzyme that converts fats into fatty acids and glycerol.

Question 6
What is the role of saliva in the digestion of food ?
Answer:
Saliva contains salivary amylase enzyme that breaks down starch into sugars like maltose.

Saliva keeps the mouth cavity clean and moistens the food that help in chewing and breaking down the big pieces of food into smaller ones.

Question 7
What are the necessary conditions (or autotrophic nutrition and what are its by-products ?
Answer:
Necessary conditions for autotrophic nutrition :
(i) Presence of chlorophyll in the living cells.
(if) Provision of supply of water to green plants or cells of the plant.
(iii) Sufficient sunlight.
(iv) Sufficient supply of carbon dioxide.
By-product of auto tropic nutrition is oxygen.

Question 8
What are the differences between aerobic and anaerobic respiration ? Name some organisms that use the anaerobic mode of respiration.
Answer:

Aerobic respiration	Anaerobic respiration
1. It takes place in the presence of oxygen.	1. It takes place in the absence of oxygen.
2. Complete breakdown of food occurs in aerobic respiration.	2. Partial breakdown of food occurs in anaerobic respiration.
3. The end products in aerobic respiration are carbon dioxide and water.	3. The end products in anaerobic respiration may be ethanol and carbon dioxide (as in yeast plants) or lactic acid (as in animal muscles).
4. Aerobic respiration produces a considerable amount of energy.	4. Much less energy is produced in anaerobic respiration.

Some organisms which use anaerobic respiration are yeast, bacteria etc.

Question 9
How are the alveoli designed to maximise the exchange of gases ?
Answer:
(i) The alveoli are thin walled and richly supplied with a network of blood vessels to facilitate exchange of gases between blood and the air filled in alveoli.
(ii) Alveoli have balloon-like structure. Hence, provide maximum surface for exchange of gases.

Question 10
What would be the consequences of a deficiency of haemoglobin in our bodies?
Answer:
Due to the deficiency of haemoglobin in blood, its oxygen carrying capacity decreases. As a result the production of energy by oxidation will become slower. Therefore, one would fall sick and would feel fatigue most of the time.

Question 11
Describe double circulation in human beings. Why is it necessary ?
Answer:
In our heart blood enters twice and also pumped out twice from the heart. The deoxygenated blood from the body is brought to the right atrium through vena cava from where it is sent to right ventricle. From right ventricle, the blood is pumped to the lungs for oxygenation through pulmonary artery. The oxygenated blood from lungs again enters the left atrium of the heart through pulmonary veins. From left atrium it is send to left ventricle, from where this oxygenated blood is pumped to different parts of body through the arteries. In this way the blood flows through the heart twice, that’s why it is called ‘double circulation’.

Necessity of double circulation: The right side and the left side of the human heart are useful to keep deoxygenated and oxygenated blood from mixing. This type of separation of oxygenated and deoxygenated blood ensures a highly efficient supply of oxygen to the body. This is useful in case of humans who constantly need energy to maintain their body temperature.

Question 12
What are the differences between the transport of materials in xylem and phloem ?
Answer:

Xylem	Phloem
1.  Xylem conducts water and dissolved minerals from roots to leaves and other parts.	1. Phloem conducts prepared food material from leaves to other parts of plant in dissolved form.
2. In xylem, the transport of material takes place through vessels and tracheids which are dead tissues.	2. In phloem, transport of material takes place through sieve tubes with the help of companion cells, which are living cells.
3. In xylem upward movement of water and dissolved minerals is mainly achieved by transpiration pull. It is caused due to suction created by evaporation of water molecules from the cells of a leaf.	3. In translocation, material is transferred into phloem tissue using energy from ATP. This increases the osmotic pressure that moves the material in the phloem to tissues which have less pressure

Question 13
Compare the functioning of alveoli in the lungs and nephrons in the kidneys with respect to their structure and functioning.
Answer:

Alveoli	Nephron
1. Alveoli are functional unit of lungs.	1. Nephrons are functional unit of kidney.
2. A mature lung has about 30 crore alveoli.	2. A kidney has about 10 lakh nephrons.
3. Alveoli provide a wide surface for gaseous exchange.	3. The surface area of a nephron is not much more.
4. The exchange of O2 and CO2 takes place through the network of capillaries in alveoli.	4. The Bowman’s capsule in nephron regulates the concentration of water and salts.

NCERT Solutions for Class 10 Science Chapter 6 Life Processes

Life processes: ‘Living being’. Basic concept of nutrition, respiration, transport and excretion in plants and animals.

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Science
Chapter	Chapter 6
Chapter Name	Life Processes
Number of Questions Solved	34
Category	NCERT Solutions

Question 1
How are fats digested in our bodies? Where does this process take place?
Solution:
The small intestine is the site of the complete digestion of carbohydrates, proteins and fats. It receives the secretions of the liver and pancreas for this purpose. The food coming from the stomach is acidic and has to be made alkaline for the pancreatic enzymes to act. Bile juice from the liver accomplishes this in addition to acting on fats. Fats are present in the intestine in the form of large globules, which make it difficult for enzymes to act on them. Bile salts break them down into smaller globules increasing the efficiency of enzyme action. The pancreas secretes pancreatic juice, which contains enzymes like trypsin for digesting proteins and lipase for breaking down emulsified fats. The walls of the small intestine contain glands, which secrete intestinal juice. The enzymes present in it finally convert the proteins to amino acids, complex carbohydrates into glucose and fats into fatty acids and glycerol.

Question 2
What is the role of saliva in the digestion of food?
Solution:
When we eat something we like, our mouth ‘waters’. This is actually not only water, but also a fluid called saliva secreted by the salivary glands. Another aspect of the food we ingest is its complex nature. If it is to be absorbed from the alimentary canal, it has to be broken into smaller molecules. This is done with the help of biological catalysts called enzymes. The saliva contains an enzyme called salivary amylase that breaks down starch, which is a complex molecule to give sugar. The food is mixed thoroughly with saliva and moved around the mouth while chewing by the muscular tongue.

More Resources for CBSE Class 10

• NCERT Solutions
• NCERT Solutions for Class 10 Science

Question 3
What are the necessary conditions for autotrophic nutrition and what are its byproducts?
Solution:
Carbon and energy requirements of the autotrophic organism are fulfilled by photosynthesis. It is the process by which autotrophs take in substances from the outside and convert them into stored forms of energy. This material is taken in the form of carbon dioxide and water, which is converted into carbohydrates in the presence of sunlight and chlorophyll. Carbohydrates are utilised for providing energy to the plant. The carbohydrates, which are not used immediately, are stored in the form of starch, which serves as the internal energy reserve to be used as and when required by the plant.

Download NCERT Solutions for Class 10 Science Chapter 6 Life Processes PDF

Question 4
What are the differences between aerobic and anaerobic respiration? Name some organisms that use the anaerobic mode of respiration.
Solution:
The food material taken in during the process of nutrition is used in cells to provide energy for various life processes. Diverse organisms do this in different ways – some use oxygen to breakdown glucose completely into carbon dioxide and water; some use other pathways that do not involve oxygen. In all cases, the first step is the breakdown of glucose, a six-carbon molecule, into a three-carbon molecule called pyruvate. This process takes place in the cytoplasm. Further, the pyruvate may be converted into ethanol and carbon dioxide. This process takes place in yeast during fermentation. Since this process takes place in the absence of air (oxygen), it is called anaerobic respiration. Breakdown of pyruvate using oxygen takes place in the mitochondria. This process breaks up the three-carbon pyruvate molecule to give three molecules of carbon dioxide. The other product is water. Since this process takes place in the presence of air (oxygen), it is called aerobic respiration. The release of energy in this aerobic process is a lot greater than in the anaerobic process.

Question 5
How are the alveoli designed to maximise the exchange of gases?
Solution:
Within the lungs, the passage divides into smaller and smaller tubes, which finally terminate in balloon-like structures, which are called alveoli. The alveoli provide a surface where the exchange of gases can take place. The walls of the alveoli contain an extensive network of blood vessels. As we have seen in earlier years, when we breathe in, we lift our ribs and flatten our diaphragm, and the chest cavity becomes larger as a result. Because of this, air is sucked into the lungs and fills the expanded alveoli. The blood brings carbon dioxide from the rest of the body for release into the alveoli, and the oxygen in the alveolar air is taken up by blood in the alveolar blood vessels to be transported to all the cells in the body. During the breathing cycle, when air is taken in and let out, the lungs always contain a residual volume of air so that there is sufficient time for oxygen to be absorbed and for the carbon dioxide to be released.

Question 6
Describe double circulation in human beings. Why is it necessary?
Solution:
The double circulatory system of blood flow refers to the separate systems of pulmonary circulation and the systemic circulation.
The adult human heart consists of two separated pumps, the right side with the right atrium and ventricle which pumps deoxygenated blood into the pulmonary circulation.
The oxygenated blood re-enters the left side of the heart through the pulmonary vein into the left atrium and passes to the left ventricle where it is pumped to the rest of the body. This part of the circulation is called as systemic circulation. This type of circulation is called double circulation. The advantage of a double circulatory system is that blood can be pumped to the rest of the body at a higher pressure.

Multiple Choice Questions (MCQs) [1 Mark each]

Question 1.
Yeast respires anaerobically using sugar as a substrate. Out of the options given below, choose the correct combination of condition and product?

	Condition	Product
(a)	Aerobic	Alcohol
(b)	Aerobic	Lactic acid
(c)	Anaerobic	Alcohol
(d)	Anaerobic	Lactic acid

Answer:
(c) Under an aerobic condition, yeast respires and converts glucose to alcohol and CO₂.

Question 2.
The table shows the percentage composition of four samples of air. Which sample could have been breathed out by a person after vigorous exercise?

Samples	Oxygen	Carbondioxide	Water Vapour
(a)	16	0.3	Saturated
(b)	16	4	Saturated
(c)	21	0.03	Trace
(d)	21	3	Trace

Answer:
(b) This is because rapid aerobic respiration occurs – during vigorous exercise in order to obtain more energy.

Question 3.
Cramps caused during sudden activities are due to the formation of
(a) lactic acid
(b) acetic acid
(c) excess of water
(d) ethanol
Answer:
(a) Lactic acid is formed by the breakdown of pyruvate when oxygen is insufficient in muscles instead of forming C02 and water. Accumulation of excess lactic acid in the muscles causes cramps.

Question 4.
Which of the following plays nose like function in plants?
(a) Flower
(b) Phloem
(c) Stomata
(d) Chlorophyll
Answer:
(c) Stomata are pores which help in the passage of air in the plants.

Question 5.
Which changes occur when a person breathe in deeply?

	Diaphragm Muscle	External Intercostal Muscles
(a)	Contracts	Contract
(b)	Contracts	No change
(c)	Relaxes	Contract
(d)	Relaxes	Relax

Answer:
(a) When a person breathes deeply the external intercostal muscles contract causing the rib cage to swing up and out. Also, the diaphragm contracts and flattens causing the thoracic cavity to increase in volume and decrease in pressure.

Question 6.

The diagram given above shows part of the lining of the human trachea. What is the function of X?
(a) Gaseous exchange
(b) Mucus removal
(c) Phagocytosis
(d) Secretion of mucus
Answer:
(b) The cilia (X) of the cells lining the air passages move in a sweeping motion to keep the air passages clean. The constant action of these cilia carry mucus and debris upward into the pharynx where they are swallowed.

Question 7.
The table given below shows the percentage composition of a gas in inspired and in expired air.

%Composition
Inspired Air	Expired Air
21.0	16.0

What is the gas?
(a) Carbon dioxide
(b) Nitrogen
(c) Oxygen
(d) Water vapour
Answer:
(c) The gas is oxygen as atmospheric air has approximately 21% of oxygen

NCERT Solutions for Class 10 Science Chapter 6 Life Processes (Hindi Medium)

Class 10 Science Life Processes Mind Map

Nutrition
Nutrition is the process by which source of energy (food) is transferred from outside the body of the organism to the inside. Most of the food sources are also carbon-based on Earth and depending on the complexity of these carbon sources different organisms use different kinds of nutritional processes.
Autotrophic Nutrition: Carbon and energy requirements of the autotrophic organism are fulfilled by photosynthesis.

• It is the process by which autotrophs convert carbon dioxide & water into carbohydrate in the presence of sunlight and chlorophyll. Oxygen is the byproduct.
• The following events occur during this process:
• Absorption of light energy by chlorophyll.
• Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen.
• Reduction of carbon dioxide to carbohydrates.

Heterotrophic Nutrition: Heterotrophs depend on other organisms for their nutrition.

• Saprophytes: They break-down the food material outside the body and then absorb it, also termed as extra-cellular digestion. E.g. fungi like bread moulds, yeast, mushrooms etc.
• Parasites: Derive nutrition from plants or animals without killing them. E.g. cuscuta (amar-bel), ticks, lice, leeches, tape-worms etc.
• Holozoic nutrition: These organisms take in whole material & break it down inside their bodies. E.g. cow, deer, lion, tiger, humans etc. What can be taken in and broken down depends on body design and functioning

Respiration
It is the process by which organism uses the food material to produce energy. Diverse organisms do this in different ways:

Energy released during cellular respiration is immediately used to synthesise ATP which is used to fuel all other activities in the cell. Aerobic organisms need to ensure that there is sufficient intake of oxygen:
• Plants: Exchange of gases takes place through stomata by simple diffusion. Large inter-cellular spaces ensure that all cells are in contact with air. Direction of diffusion depends upon the environmental conditions and the requirements of the plant. For e.g. CO₂ elimination majorly takes place at night while oxygen release is the major event of the day time.
• Aquatic animals such as fishes take in water through their mouths & force it past the gills where the dissolved oxygen is taken up by blood.
• In human beings, the passage of air can be written as nostril → trachea → bronchi → bronchioles → alveolar sac. The alveoli provide a surface where the exchange of gases can take place. Blood releases the dissolved CO₂ into the alveoli & carries O₂ from alveolar air. Haemoglobin in RBC of blood transport O₂ from lungs to various tissues of the body.

Life Process
The processes which maintain the body functions and are required for the survival of living being are called life processes. Some of the important life processes are nutrition, respiration, transportation, excretion etc.

Nutrition In Human Beings
The alimentary canal is a long tube extending from the mouth to the anus. The nutrition in human being is divided into five steps:
• Ingestion: Intake of food from outside source. Teeth & saliva crush the food to generate the particles of same size & texture. The food is then passed to stomach via oesophagus. The peristaltic movements occur all along the gut which helps in pushing the food forward.
• Digestion: In mouth, salivary amylase helps in carbohydrates digestion. In stomach, pepsin helps in protein digestion. However, small intestine is the main site of complete digestion of carbohydrates, proteins & fats. It receives pancreas and liver secretions. Bile juice emulsifies fats and pancreatic enzymes, trypsin & lipases digest proteins & emulsified fats. It finally converts proteins to amino acids, complex carbohydrates into glucose & fats into fatty acids & glycerol.
• Absorption: The digested food is taken up by the walls of the intestine. The inner lining of the small intestine has numerous finger-like projections called villi which increase the surface area for absorption. Large intestine absorbs water from the unabsorbed food.
• Assimilation: The villi are richly supplied with blood vessels which take the absorbed food to each & every cell of the body, where it is required either for energy, build up or repair.
• Excretion: The waste material is removed from the body via anus which is regulated by anal sphincter.

Transportation
Transportation in Human Beings

• Blood consists of fluid medium called plasma in which
  the cells are suspended. Plasma transports food, CO₂ & nitrogenous wastes in dissolved form. Oxygen is carried by RBC.
• Heart: Heart is the muscular organ made up of cardiac muscles and is as big as our fist. It is composed of four chambers (2 atria & 2 ventricles) to prevent the mixing of oxygenated & deoxygenated blood.
• Ventricles are thick wailed as they have to pump the blood to various organs of the body. In addition, valves are also present in heart and veins to prevent the backflow of the blood.

Circulation of blood: Oxygenated blood is carried out from lungs to the left atrium with the help of pulmonary’ veins.

• Left atrium contracts to release blood into the left ventricle which relaxes while collecting it. It then pumped out the blood to whole body via aorta.
  a Deoxygenated blood from whole body then enters the right atrium via vena cava vein.
• Right atrium contracts to pump the blood in right ventricle. It then pumps the blood towards lungs via pulmonary’ artery for oxygenation.

Oxygenation of blood: Invertebrates such as birds, mammals etc which constantly use energy to maintain their body temperature, blood goes through heart twice during each cycle which is known as double circulation.

• In contrast, animals like amphibians or many reptiles have three-chambered hearts as they can tolerate some mixing of the oxygenated & de-oxygenated blood streams. They do not use energy for thermoregulation and body temperature depends on the temperature in the environment.
• Fishes, on other hand, have only two chambered heart. Blood is pumped to the gills for oxygenation and passes directly to the rest of the body.

Transportation In Plants
There are two main pathways present in plants: xylem pathway- moves water & minerals from the soil & phloem transports products of photosynthesis from leaves (where they are synthesized) to other parts of the plant.
Transport of Water

• In xylem tissue, vessels and tracheids of roots, stems & leaves are interconnected to form a continuous system of water-conducting channels reaching all parts of the plant.
• At root site, cells actively take up ions from soil which creates concentration gradient. Water then diffuses into the root cells in order to eliminate this gradient.
• It provides steady movement of water into root xylem, creating a column of water that is steadily pushed upwards.
• However, it is not efficient enough to push water over the heights of tall plants.
• So, plants use other method which is known as transpiration to push water upwards. The loss of water in the form of vapour from aerial parts of plant is known as transpiration.
• Evaporation of water molecules from the cells of a leaf creates a suction which pulls water from the xylem cells of roots. It also aids in thermoregulation.
• Transport of food and other substances
• Transport of soluble products of photosynthesis is called translocation.
• The translocation takes place in sieve tubes with the help of adjacent companion cells both in upward & downward directions.
• It utilizes energy (ATP) in contrast to xylem transport.
• Material like sucrose is transferred into phloem tissue using energy from ATP.
• It increases osmotic pressure of tissue causing water to move into it.
• This pressure moves the material in phloem to tissues which have less pressure.
• It allows phloem to move material according to plant’s needs.

Excretion
The biological process involved in removal of harmful metabolic wastes from body is called excretion.
Many unicellular organisms remove these wastes by simple diffusion from body surface into surrounding water. However, complex multi-cellular organisms use specialised organs to perform this function.
Excretion in Human Beings: The excretory system includes pair of kidneys, pair of ureters, urinary bladder & urethra.

• Nephrons are the functional units of kidneys. They are the clusters of thin-walled capillaries. Each cluster is associated with cup-shaped end (Bowmans capsule) of a tube that collects the filtered urine.
• Substance such as glucose, amino acids, salts & a major amount of water are selectively re-absorbed as the urine flows along the tube. The amount of water depends up on amount of excess water & dissolved waste in the body.
• The urine formed in each kidney is carried to urinary bladder by ureter. Urine is stored in urinary bladder until the pressure of the expanded bladder leads to the urge to pass it out through the urethra.
• Excretion in Plants: They get rid of excess water by transpiration.
• Many plant waste products are stored in cellular vacuoles.
• Waste products may be stored in leaves that fall off.
• In addition, some waste products are stored as resins & gums, especially in old xylem.
• Lastly, plants excrete some waste substances into the soil around them.

We hope the detailed information regarding  NCERT Solutions For Class 10 Science Chapter 6 Life Processes is helpful. For any query related to  NCERT Solutions For Class 10 Science Chapter 6 Life Processes, kindly drop your questions in the comment box below and we will get back to you as soon as possible.

CBSE Class 6 English Paragraph Writing

Paragraph For Class 6 CBSE

1. A Day That I Can’t Forget
There are many days in my life that I can’t forget. Of these is the day when the result of my class V examination was declared. I had done all my papers well. I expected a good result. But I had stood first in the entire district. I felt very very happy. My parents and other family members were very happy. All friends and my teachers came to congratulate me on this great success. Sweets were distributed in the colony. My father decided to send me to a very good school for further education. I can’t forget that day.
Word-Notes : Declared घोषत किया गया| Expected—आशा करवा था| Entire—संपूर्ण। Congratulate–बधाई देने के लिए। Success—सफलता। Distributed—बाँटी। Further—आगे की|

2. How I Celebrated My Birthday
I celebrated my 15th birthday last Saturday. I invited all my friends to a birthday party at our house. My parents performed a havan in the morning. In the afternoon, the guests began to arrive. The house was decorated beautifully. In the evening, I cut the birthday cake. My friends and family wished me a happy birthday. The guests were served sweets and hot and cold drinks. My friends danced and sang. Everybody was in a joyful mood. The guests took their dinner. They went away around midnight.
Word-Notes : Celebrated—मनाया Performed-पूरा किया| Havan-हवन| Decorated—सजाया गया था। Joyful-खुशी से भरा हुआ।

Class 6 Paragraph Writing CBSE

3. An Accident That You Have Seen
I happened to see a very horrible accident. It occurred last month. It was between a truck and a motor cycle. The road was slippery due to rain. A truck was going on the road. Suddenly a motor cycle overtook it with a great speed. The motor cycle was being driven by two young students. The riders couldn’t see the mini-truck coming from the opposite direction. He applied-brakes. But the motor cycle slipped. The rider and the other student fell off it. They had head injuries. Both died there. I can’t forget this accident.

4. The Night Before The Examination
Examinations are always fearful. They are a great cause of fear, anxiety, tension and worry. The night before the examination is really painful. The students have no sleep. They stop eating food. They want to read and revise everything. They are disturbed if someone tells them about an important question. They at once read for its answer. They are irritated also if anyone disturbs them. They pass the whole night in tension. They do not take anything in the morning. They go to the examination hall on empty stomachs.

Paragraph Class 6 CBSE

5. Holi Celebration In My Colony
Holi is a festival of fun and colours. It comes at the end of the winter season. Our colony celebrated it as in previous years. On the Holi day young boys of the colony gathered at a particular spot. They coloured one another’s face and of the passers-by. They also threw water balloons on other persons. They went to the nearby colony to put colours on its residents. Soon the women collected. They started throwing coloured water on one another. They offered sweets also. Everybody seemed to be very happy.

6. Morning Assembly In My School
Morning Assembly is a very important ritual. It is also a part of the ageold culture of our country. Every new work starts with prayers to our gods. So it happens in a school too. Schools are like our temples. Students and teachers stand in rows. They say prayers before starting the day’s work. The scene is very calm and peaceful. All students stand in rows in school uniform. In many schools, some physical training is also given in the morning assembly. Important information is also given to the students by the Principal.

English Paragraph For Class 6 CBSE

7. The Scene At A Railway Platform
The scene at a Railway Platform is always enjoyable. It presents a picture of mini India. Last Sunday I went to the city Railway staan to see my uncle off. The platform was overcrowded. Passengers of all states like Rajasthan, Tamilnadu, Assam etc, could be seen. There were many stalls on the platform. Small children were running here and there. Coolies in red dress were going to and fro. Passengers were taking tea. Some were reading newspapers. They were waiting for the train. Soon the train came. My uncle got a seat and the train started. I returned home.

Class 6 English Paragraph Writing CBSE

8. My Visit To A Historical Place
A visit to a historical place is always educative. Last year I went to Agra. Agra is a historical place. It has many buildings of historical interest. The most famous is Taj Mahal. I have no words to express its beauty. The white marble looks like silver. The reflection of it in the water is simply enchanting. The four minarets add to its glory. The grassy lawns increase it. It looks like a dream in the moon-lit night. I can’t forget my visit to Taj Mahal.

9. One Day Cricket Match
One Day Cricket Matches have become the most favourite of cricket lovers. They stop work to enjoy the match. So these one-dayers have become more popular than the test matches. One day match is a one-day game. Time and overs are fixed. Each team plays a maximum of fifty overs. So there is suspense and romance in this match. It seldom ends in a draw. It gives results. So the players try to win the match. They put in their best. Every ball played is filled with great fun and suspense.

Paragraph Writing For Class 6 Pdf CBSE

10. Bad Effects Of Watching Television
Television has become an essential part of every house. It is the cheapest source of entertainment. Children find it very interesting. But it has bad effects also. Seeing it for long hours harms the eyes. The students are left with less time to study. So their studies suffer. Many programmes influence young minds in a wrong way. Foreign programmes are not in good taste. The youth are easily attracted towards them. They forget their own age-old culture. Their lives are spoiled. So television should be watched under the guidance of the elders.

NCERT Solutions for Class 6 English

On Children Question and Answers