Bhagya

CBSE Class 9 Maths Lab Manual – Quadrilateral Formed by Joining Mid-points of Sides of a Quadrilateral

Objective
To show that the quadrilateral formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram by paper folding.

Prerequisite Knowledge

1. Concept of finding mid-point of a line segment by performing paper folding activity.
2. Properties of a parallelogram.

Materials Required
Glazed papers, pencil, a pair of scissors, gluestick and tracing paper.

Procedure

1. Take any coloured glazed paper.
2. Draw a quadrilateral of any dimensions on glazed paper and name it as ABCD.
3. Cut that quadrilateral from the glazed paper.
4. Now, find the mid-point of each side AB, BC, CD, DA by paper folding and name them E, F, G, H respectively as shown infig. (i).
   
5. Now, fold the figure along EF, GF, GH and EH. Press it and then unfold it as shown in fig. (ii).
   
6. We will get creases along EF, GF, GH, HE.
7. Make a replica (true copy) of EFGH (say PQRS) by using a tracing paper [fig.(iii)].
   
8. Cut the quadrilateral PQRS along any diagonal (say RP) [fig.(iv)].
   
9. We will get two triangles ∆PSR and ∆PQR.
10. Now, overlap these two triangles. Two triangles coincide with each other [fig.(v)] such that side PS overlaps with QR and PQ with SR.
    

Observation
We observe that two triangles coincide with each other which means two triangles are congruent to each other. In a quadrilateral, two triangles cover each other completelv along any diagonal, then the quadrilateral will be a parallelogram.
∴ ∆PQR = ∆PSR
i.e., ar (∆PQR) = ar(∆PSR)
∴ PQRS is a parallelogram.

Result
As the replica of ∆PQR exactly covers the replica of ∆PSR
∴ PQ = RS, QR=SP
∴ PQRS is a parallelogram.

Learning Outcome
We have verified by paper folding that the quadrilateral formed by joining the mid-points of adjacent sides of a quadrilateral will be a parallelogram. We also learnt that a diagonal always divides the parallelogram into two triangles of equal areas.

Activity Time
What type of figures do you obtain?

• If you join mid-points of the sides of a rectangle (Do it by paper folding).
• If you join the mid-points of thesidesof a square (Do it by paper folding).

Viva Voce

Question 1.
What do you mean by a quadrilateral ?
Answer:
A quadrilateral is a plane closed figure bounded by four line segments.

Question 2.
What are two main properties of a quadrilateral ?
Answer:

• Sum of four angles is 360°.
• It has 4 sides.

Question 3.
Is a parallelogram a quadrilateral ?
Answer:
Yes

Question 4.
Write two main properties of a parallelogram.
Answer:

• Diagonals bisect each other.
• Opposite sides are equal.

Question 5.
In a parallelogram, if one angle is 90°, then what type of parallelogram you will get ?
Answer:
Rectangle.

Question 6.
Do you know any difference between a parallelogram and a trapezium ?
Answer:
In a parallelogram, two pairs of opposite sides are parallel. In a trapezium, one pair of opposite sides is parallel.

Question 7.
What is the area of a parallelogram ?
Answer:
Base x corresponding altitude.

Question 8.
If base and altitude of a parallelogram are same, then what will be area of parallelogram ?
Answer:
Base x altitude.

Question 9.
If base and altitude of a parallelogram are same, then what type of parallelogram will be obtained ?
Answer:
Square.

Question 10.
What do you mean by a parallelogram ?
Answer:
A parallelogram is a quadrilateral in which opposite sides are equal and parallel.

Question 11.
Write the name of different kinds of parallelograms.
Answer:
Rectangle, square and rhombus.

Question 12.
If you join the mid-points of consecutive sides of a quadrilateral, what shape will you obtain ?
Answer:
Parallelogram.

Question 13.
Which theorem is used in this activity ?
Answer:
Mid-point theorem.

Question 14.
If you join the mid-points of consecutive sides of a rectangle, what figure will you obtain ?
Answer:
Rhombus.

Question 15.
If you join the mid-points of consecutive sides of a rhombus, what figure will you obtain ?
Answer:
Rectangle.

Multiple Choice Questions

Question 1.
Name the quadrilateral formed by joining the mid¬points of the consecutive sides of a square:
(i) rectangle
(ii) square
(iii) rhombus
(iv) none of these

Question 2.
The four triangles formed by joining the mid-points of three sides of a triangle are:
(i) congruent
(ii) non-congruent
(iii) similar
(iv) none of these

Question 3.
In the given figure ABCD, if P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively, then:

(i) SR = \(\frac { 1 }{ 2 }\) AC
(ii) SR = AC
(iii) SR = \(\frac { 1 }{ 3 }\) AC
(iv) none of these

Question 4.
In ∆ABC, if E is the mid-point of AC, F lies on BC and EF // AB then:
(i) EF = \(\frac { 1 }{ 3 }\) AB
(ii) EF = \(\frac { 1 }{ 2 }\) AB
(iii) EF = AB
(iv) none of these

Question 5.
In a parallelogram, the figure formed by joining the mid-points of consecutive sides is :
(i) a rectangle
(ii) a rhombus
(iii) a square
(iv) none of these.

Question 6.
In a rhombus, diagonals bisect each other at an angle of:
(i) 45° and 135°
(ii) 60° and 120°
(iii) 90°
(iv) none of these

Question 7.
In a rectangle, diagonals are:
(i) equal
(ii) not equal
(iii) half of each other
(iv) none of these

Question 8.
The straight line joining the mid-points of the non¬parallel sides of a trapezium is parallel to:
(i) parallel sides
(ii) non-parallel sides
(iii) one non-parallel side
(iv) none of these

Question 9.
The triangle formed by joining the mid-points of the sides of a right triangle is :
(i) a right triangle
(ii) an obtuse-angled triangle
(iii) an isosceles triangle
(iv) none of these

Question 10.
The triangle formed by joing the mid-points of the sides of an isosceles triangle is:
(i) an equilateral triangle
(ii) an isosceles triangle
(iii) a right-angled triangle
(iv) none of these

Answers

1. (ii)
2. (i)
3. (i)
4. (ii)
5. (iv)
6. (iii)
7. (i)
8. (i)
9. (i)
10. (ii)

Math Lab ManualMath Labs with ActivityMath LabsScience LabsScience Practical Skills

Looking for an easy way to Learning of new elementary english grammar and composition for class 7 answers, Solutions. You have to learn basic English Grammar topics like Tenses Verbs, Nouns, etc… In this article, we will review the best English Grammer Topics and compare them against each other.

Possessive Exercises With Answers for Class 7 CBSE PDF

Possessives are used to indicate ownership. Possessive Nouns are usually formed by adding an apostrophe (‘) and s.
Examples:

• Pawan’s book/Kiran’s car/Grandma’s mirror

When a noun is plural and ends in s, we just add an apostrophe (‘) at the end to show possession.
The kids’ toys / My parents’ house / The teachers’ lounge

If two people own one thing, add the apostrophe and s only to the second person only.
Examples:

• Neena and Anil’s new house/Deepak and Jyoti’s wedding/Rahul and Rohit’s car

If two people own separate things, add the apostrophe and s for each person.
Examples:

• Suman’s and Babita’s books/Jiwan’s and Dilip’s pants/Bunty’s and Jiwan’s offices

Possessive Pronouns:
Examples:
in the sentence, “These books are mine, not yours”, the words “mine” and “yours” are possessive pronouns and stand for “my books” and “your books,” respectively.

Look at the following Examples:
Possessive Pronouns

• I like your drawing.
  Do you like mine.
• My flowers have withered.
  Yours are lovely
• Your photos are good.
  Mine are terrible.

The words mine, yours, his, hers, its, ours, theirs are possessive pronouns. They show who or what something belongs to.

Possessive Adjective:
Possessive adjectives – my, your, his, her, its, our, your, their – modify the noun following it in order to show possession.

Possessive Pronouns

• We have a dog his name is victor
• I like singing my mother sings with me.
• Namite is from sechi his husband is from Kolkata:
  

  Subject Pronouns		I	you	he	she	it	we	you	they
  Possessive	Adjectives	my	your	his	her	its	our	your	their
  	Pronouns	mine	yours	his	hers	its	ours	yours	theirs

Presentation:
Read the picture story.

Answer the following questions.

1. What did father bring for his family?
2. What did father bring for his sons, wife and daughter?
3. What was the most exciting thing?

A. Complete the blanks using the possessive adjectives.

1. Amit is from New Delhi and ___________ teacher is from Lucknow.
2. Maya is from Mumbai and ___________ doctor is from Chandigarh.
3. I am from Bihar and ___________ architect is from Rajasthan.
4. Maya and Kavita are Indian and ___________ friends are Brazilian.
5. Pierre and I are French and ___________ students are English.
6. Tilak and you are engineers and ___________ telephone number is 555-5678.
7. Anu and ___________ teacher are from Punjab.
8. Vijay is ___________ first name.

B. Form sentences using suitable possessive adjectives.
Examples:

• I/address/ II Baker Street
• My address is II Baker Street.

1. She/telephone number/888-0000
______________________
2. He/mechanic/Vijay Mathur
______________________
3. We/teacher/Sunita Rai.
______________________
4. They/zip code/55909
______________________
5. You/address/10 Violet Street
______________________
6. She/nickname/Mamta
______________________
7. I/number/ten
______________________
8. He/telephone/8822-2323/number
______________________
9. She/nickname/Vinita
______________________

C. Choose and circle the correct alternative.

1. Alok always types (his/hers) letters, but we never type (our/ours).
2. Kiran and Suman are waiting for (their/theirs) parents.
3. I know (my/mine) family very well. Do you know (yours/your)?
4. (Your/Yours) is an excellent car. How much did you pay for it?
5. She always shouts at (her/hers) children.
6. Children ask (theirs/their) parents difficult questions.
7. Give him (your/yours) address.

D. Write the correct possessive pronoun for each sentence.

1. I’ll do ___________ job, you do ___________
2. She can do ___________ homework alone. He can’t do ___________
3. We must wash ___________ hands before we eat.
4. The boys aren’t going to clean ___________ rooms today.
5. Dr. Mathur washes ___________ car every morning.

E. Choose the correct option to fill in the blanks.

1. Does ___________ bring ___________ books to the class?
a) he – him b) she – her c) her – she d) she – hers

2. Do you prefer listening to ___________ records or ___________?
a) your – yours b) my – him c) your – our d) your – mine

3. Lalit must talk to ___________ coordinator.
a) ours b) mine c) his d) him

4. Mrs. Khanna and ___________ children are waiting for ___________ answer.
a) his – our b) their – yours c) her – our d) her – ours

F. Read the text and choose the right word to complete the following paragraph.
Hi, I’m Ganesh and this is (1) ___________ sister Kiran. This is a photo of (2) ___________ mum and dad.

(3) ___________ teachers. My mum is 35 and (4) ___________ a maths teacher. (5) ___________. favourite place is California. My dad is 38 and (6) ___________ an English teacher.

(7) ___________ favourite place is Blackpool! (8) ___________ favourite people are Kiran and me, of course.

1. a) my – b) me
2. a) we – b) our
3. a) They’re – b) Their
4. a) she’s – b) he’s
5. a) Her – b) His
6. a) she’s – b) he’s
7. a) His – b) Her
8. a) They – b) Their

G. Fill in the blanks with the correct possessive adjective.

1. The children drink ___________ milk.
2. The woman said it was ___________ car.
3. You could see that the boy liked ___________ father very much.
4. We always complete ___________ homework.
5. I would like to go to ___________ cousin’s house next week.

H. Fill in the blanks with the correct possessive pronoun.

1. This is not her dress. Where’s ___________ ? (her dress)
2. They drank out of my glass yesterday, now I’m drinking out of ___________ (their glasses)
3. This piece of cake is ___________ (his piece)
4. These toys are ___________ (your toys)
5. These pictures are ___________ (Pawan and Maya’s picture)

I. Fill in the blanks with the pronouns from the box. One pronoun can be used more than once.
my – your – his – her – our – their

1. She’s really beautiful ___________ name is Suman.
2. He’s very intelligent ___________ name is Rohan.
3. They’re really smart. ___________ names are Mita and Gita.
4. ___________ name is Deepak.
5. ___________ names are Dipika and Gitika. We’re sisters.
6. ___________ brothers’ names are Deepak and Sandeep.
7. ___________ friend’s name is Pradip.

J. Fill in the blanks with his, her, your or their.

1. ___________ name is Monica.
2. ___________ name is Jatin.
3. What’s name? My ___________ name is Anu.
4. What’s ___________ nationality? They’re Brazilian.

K. Rewrite the following sentences using possessive pronouns/adjectives. The first one has been done for you.
1. Harshit’s book is in the classroom.
His book is in the classroom.

2. These watches belong to me.
______________________
3. Excuse me, does this mobile belong to you?
______________________
4. Those pens belong to Mayank and Chetna.
______________________
5. Meena’s shoes are in her bedroom.
______________________
6. The children’s pencils are on the table.
______________________
7. Manshu’s bag is in the living room.
______________________
8. Those trousers belong to Sahil.
______________________
9. Anita’s skirt is in the wardrobe.
______________________

CBSE Previous Year Solved  Papers  Class 12 Physics Outside Delhi 2014 

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1. 1.  All questions are compulsory. There are 26
      questions in all.
   2.  This question paper has five sections : Section A, Section B, Section C, Section D and Section E.
   3. Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each.
   4.  There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weight age. You have to attempt only one of the choices in such questions.
   5. You may use the following values of physical constants wherever necessary:

SET I

SECTION – A

Question.1. Using the concept of force between two infinitely long parallel current carrying conductors, define one ampere of current.
Answer : The amount of current flowing in two parallel wire in same direction through two infinitely long parallel wire separated by one metre which produces an attractive force of 2 x 10^-7 N/m, provided the wires, have negligible circular cross section and placed in vacuum, this flowing current can be defined as “One Ampere of Current.

Question.2. To which part of the electromagnetic spectrum does a wave of frequency 5 x 1019 Hz belong ?
Answer : The frequency 5 x 10¹⁹ Hz lies in the gamma region of the. electromagnetic spectrum.

Question.3. What is the force between two small charges of 2 x 10^-7 C and 3 x 10^-7 C placed 30 cm apart in air?
Answer : Taking all the units of physical quantities in S.I. units, we have,

Question.4. Define intensity of radiation on the basis of photon picture of light. Write its S.I. unit.
Answer : The intensity of radiation can be defined as the energy related with photons emitted from a unit surface area in unit time. Its S.I units is joule/metre² second (J/m²s).

Question.5. The electric current flowing in a wire in the direction from B to A is decreasing. Find out the direction of the induced current in the metallic loop kept above the wire as shown.

Question.6. Why is it found experimentally difficult to detect neutrinos % in nuclear β -decay?
Answer : Neutrinos are difficult to detect experimentally in β decay because they do not have any charge with almost zero mass and also they do not interact easily with any matter.

Question.7. Why is the use of A.C. voltage preferred over D.C. voltage? Give two reasons.
Answer : The use of A.C. voltage is preferred over the use of D.C. voltage because of the following reasons :
(i) The energy losses while transmission of A.C. voltage are very less as compared to D.C. voltage.
(ii) A.C. voltage can be controlled as required by using a transformer (i. e., stepped up or stepped down).

Question.8. A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason.
Answer : In this case the biconvex lens will behave as a diverging lens because the refractive index of water (1.33) is more than that of the material (1.25) of the lens.

Question.9. Using Rutherford model of the atom, derive the expression for the total energy of the electron in hydrogen atom. What is the significance of total negative energy possessed by the electron?
OR
Using Bohr’s postulates of the atomic model, derive the expression for radius of n^th electron orbit. Hence obtain the expression for Bohr’s radius.
Answer : The magnitude of force according to Rutherford’s model of the atom is,


Here, negative sign indicates that the revolving electron is bound to the positive nucleus.
OR
Suppose m be the mass of an electron and v_nbe its speed in n^th orbit of radius r_n . From Rutherford model, the centripetal force for revolution is produced by electrostatic attraction between electron and nucleus.

Question.10. A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.
Answer : Let q be the charge on the charged capacitor. Energy stored in it is,

Question.11. Considering the case of a parallel plate capacitor being charged, show how one is required to generalize Ampere’s Circuital law to include the term due to displacement current.

The total current passing through any surface, of which the closed loop is the perimeter, is the sum of the conduction and displacement current.

Question.12. A cell of emf “E’ and internal resistance V is connected across a variable resistor ‘R’. Plot a graph showing variation of terminal voltage ‘V’ of the cell versus the current ‘I’. Using the plot, show how the emf of the cell and its internal resistance can be determined.

Question.13. Explain, with the help of a circuit diagram, the working of a p-n junction diode as a half-wave rectifier.
Answer : The circuit diagram for a p-n junction diode as a half wave rectifier is shown below :

Working : During the positive half cycle of the input a.c., the p-n junction is forward biased i.e., the forward current flows from p to n. In the forward biasing, the diode provides a very low resistance and allows the current to flow. Thus we get output across-load i.e. a.c. input will be obtained as d.c. output.
During the negative half cycle of the input a.c., the p-n junction is reversed biased i.e., the reverse current flows from n to p. In the reversed biasing, the diode provides a high resistance and hence a very small amount of current will flow through the diode which is of negligible amount. Thus no output is obtained across the load. The wave form of input and output is shown below :

Question.14. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 x 10^-7 m² carrying a current of 1.5 A. Assume the density of conduction electrons to be 9 x 10²⁸ m^-3.

Question.15. Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.35 and 1.45. Trace the path of these rays after entering through the prism.

Ray ‘1’ and ‘2’ will fall on the side AC at an angle of incidence (i) if 45°. Critical angle of ray T ’ is greater than i, so it will get refracted from the prism. Critical angle of ray ‘2’ is less than that of i, so it will undergo total internal reflection.

Question.16. Write the functions of the following in communication systems:
(i) Transducer
(ii) Repeater
Answer : (i) Transducer : A transducer is used to convert one form of energy to another form.
(ii) Repeater : A repeater is used to receive a signal from a transmitter, amplifies it and retransmits it to the receiver.

Question.17. Show diagrammatically the behaviour of magnetic field lines in the presence of (i) paramagnetic and (ii) diamagnetic substances. How does one explain this distinguishing feature.
Answer : (i) The behaviour of magnetic field lines in the presence of a paramagnetic substance is shown below :

(ii) The behaviour of magnetic field lines in the presence of a diamagnetic substance is shown below :

This distinguishing feature is because the relative permeability of the diamagnetic substance is less than 1, so the magnetic lines of force do not prefer passing through the substance and the relative permeability of a paramagnetic substance is greater than 1, so the magnetic lines of force prefer passing through the substance.

Question.18. Draw a circuit diagram n-p-n transistor amplifier, CE configuration. Under what condition does the transistor act as an amplifier?
Answer : The circuit diagram of an n-p-n transistor .amplifier in CE configuration is given below :

Condition : For a transistor to act as a amplifier, it must be operated close be to the centre of its active region.
The transistor behaves as an amplifier when the input circuit (emitter —base) is forward biased having low voltage VBB and the output circuit (collector-base) is reverse biased having high voltage Vcc.

Question.19. (a) Using the phenomenon of polarization, show how transverse nature of light can be demonstrated. (b) Two polaroids P₁ and P₂ are placed with their pass axes perpendicular to each other. Unpolarised light of intensity IQ is incident on P₁. A third Polaroid P₃  is kept in between P₁  and P₂ such that its pass axis makes an angle of 30° with that of P₁. Determine the intensity of light transmitted through P₁, P₂ and P₃ .
Answer : (a) Suppose that an ordinary light is incident normally on a pair of crystals C₁  and C₂ . When the incident ray of light passes through crystal C₁ , it gets plane polarised in the direction perpendicular-to the length of crystal. Now, we see that when the axes of two crystals are parallel, the intensity of the emerging light will be maximum. When the second crystal is placed perpendicular with respect to the first crystal, the intensity of light observed is zero. This is due to the electric field of the plane polarised light obtained from C₁  can vibrate only in one direction. Hence, when the axis of the crystal is perpendicular to its direction of vibration of electric field, it gets blocked. This shows the transverse nature of light.

(b) As given in the question, the polaroids P₁  and P₂ are placed with their axes perpendicular to each other. Also, polaroid P₃ . is placed at an angle of 30° with respect to P₁ .

Question.20. Define the term ‘mutual inductance’ between the two coils. Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of length l and radii r₁  and r₂ (r₁  >> r₂ ). Total number of turns in the two solenoids are N₁  and N₂ respectively.
Answer : The ratio of magnetic flux passing through one coil to the current passing through the other is known as coefficient of mutual inductance between the two coils.

Question.21. Answer the following:
(a) Why are the connections between the resistors in a meter bridge made of thick copper strips?
(b) Why is it generally preferred to obtain the balance point in the middle of the meter bride wire?
(c) Which material is used for the meter bridge wire and why?
OR
A resistance of R Ω draws current from a potentiometer as shown in the figure. The potentiometer has a total resistance Ro Ω. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer.

Answer : (a) The connection between the resistors in a meter bridge is made of thick copper strips because the resistivity of a copper wire is very low. As, the connections are thick, so the area becomes large and the resistance of the wires becomes almost negligible.
(b) It is preferred to obtain the balance point in the middle of the meter bridge wire because it improves the sensitivity of the meter bridge.
(c) Constant an nichrome/Manganine is used for meter bridge wire because its temperature coefficient of resistance is almost negligible due to which the resistance of the wire does not get affected on increasing temperature of the wire during flow of current.
OR
While the slide is in the middle of the potentiometer only half of its resistance (Ro/2) will be between the points A and B. Hence, the total resistance between A and B, say. R1 will be given by the following expression :

Question.22. A convex lens of focal length 20 cm is placed co-axially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm apart from each other. A point object lies 60 cm in front of the convex lens. Draw a ray diagram to show the formation of the image by the combination. Determine the nature and position of the image formed.
Answer : Let us first locate the image of the point object S formed by the convex lens.
Here : u = -60 cm and f= 20 cm
From the lens formula, we have :

The image I1 is formed behind the mirror and acts as a virtual source for the mirror. The convex mirror forms the image H, whose distance from the mirror can be determined as :

Question.23. A voltage V = V₀  sin ωt is applied to a series LCR circuit. Derive the expression for the average power dissipated over a cycle. Under what condition is (i) no power dissipated even though the current flows through the circuit, (ii) maximum power dissipated in the circuit ?

The average power over a cycle is average of the two terms on the R.H.S. of the above equation. The second term is time dependent, so, its average is zero.

Question.24. Write any two distinguishing features between conductors, semiconductors and insulators on the basis of energy band diagrams.
Answer: Conductors :
(i) In case of conductors, the valence band is completely filled and the conduction band can have two cases-either it is partially filled with an extremely small energy gap between the valence and conduction bands or it is empty, with the two bands overlapping each other as shown below :

(ii) Even when a small current is applied, conductors can conduct electricity.
Insulators :
(i) In case of insulators, the energy gap between the conduction and valence bands is very large. And the conduction band is practically empty.

(ii) When an electric field is applied to such kind of material, the electrons find hard to receive such a large amount of energy to attain the conduction band. Thus, the conduction band remains to be empty. That is why no current flows through insulators.
Semiconductors :
(i) In case of semiconductor, the energy band structure of semiconductors is similar to insulators, but in this case, the size of forbidden energy gap is quite smaller than that of the insulators.

(ii) When an electric field is applied to a semiconductor, the electrons in the valence band find it relatively easier to jump to the conduction band. So, the conductivity of semiconductors lies between the conductivity of conductors and insulators.

Question.25. For the past some time, Aarti had been observing some erratic body movement, unsteadiness and lack of coordination in the activities of her sister Radha, who also used,to complain of severe headache occasionally. Aarti suggested to her parents to get a (a) What, according to you, are the values displayed by Aarti?
(b) How can radioisotopes help a doctor to diagnose brain tumour?
Answer : (a) Aarti has showed awareness and responsibility towards her sister.
(b) A little amount radio-isotope like radioiodine is inserted into the body along with organic dyes which are absorbed strongly by the tumour tissue than the normal tissues. By detecting the emitted radiation, the radiologist get information about the size and location of the tumour.

Question.26. Write two basic modes of communication. Explain the process of amplitude modulation. Draw a schematic sketch showing how amplitude modulated signal is obtained by superposing a modulating signal over a sinusoidal carrier wave.
Answer : The two basic modes of communication are :
(1) Point-to-point communication.
(2) Broadcast communication
Amplitude Modulation: Amplitude modulation is obtained by changing the amplitude of the carrier waves according to the amplitude of the modulating wave.
Amplitude modulated signal is obtained by superposing a modulating signal over a sinusoidal carrier wave.

Question.27. An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture etc. to be same, how does the resolving power of an electron microscope Compare with what of an optical microscope which uses yellow light ?
Answer : The de-Broglie wavelength of the electrons is given by:

This formula shows that to enhance resolution, we have to use minimized wavelength and media with large indices of refraction.
For an electron microscope, p is equal to 1 (vacuum).
For an electron microscope, the electrons are accelerated through a 60,000 V potential difference.

As, λ is very little (roughly 10^-5 times smaller) for electron microscope than an optical microscope which use yellow light of wavelength (5700 A to 5900 A). So, the resolving power of electron microscope is about 10⁵ greater than that of optical microscope.

Question.28. Draw a labelled diagram of Van de Graff generator. State its working principle to show how by introducing a small charged sphere into a larger sphere, a large amount of charge can be transferred to the outer sphere. State the use of this machine and also point out its limitations.
OR

(b) Consider two hollow concentric spheres, S₁  and S₂ , enclosing charges 2Q and 4Q respectively as shown in the figure, (i) Find out the ratio of the electric flux through them, (ii) How will the electric flux through the sphere Si change if a medium of dielectric constant ‘ε_r ’ ‘ε_r ’ is introduced in the space inside S₁ in place of air? Deduce the necessary expression.

Answer : Van de Graff generator is a device used for building up high potential differences of the order of a few million volts. Principle : The working of Vande Graff Graff generator is based on the following two electrostatic phenomena.
(i) Discharging action at sharp points (corona discharge i.e., electric discharge takes place in air or gases readily at the pointed ends of conductors.
(ii) If a charged conductor is brought into internal contact with a hollow conductor, all of its charge transfers to the hollow conductor, howsoever high the potential of the latter may be.
Construction : It has a big spherical conducting shell (S) kept over insulating pillars. A long narrow insulating belt is wound around 2 pulleys’ P₁  and P₂ . B₁ and B₂ are 2 metal combs with sharp points. B₁ is known as spray comb and B₂ collecting comb.
Working : The spray comb provides positive potential by high tension source. The positive charge is sprayed on belt. As belt moves and touches the sphere, a negative charge is induced on the sharp ends of collecting comb B₂ and similar positive charge is induced on the further end of B₂ . This positive charge moves immediately to the outer surface of S. because of discharging action of sharp points of B₂ , the positive charge on the belt is neutralized. The uncharged belt moves downwards and collects the positive charge from B₁ , which is then collected by B₂ . This process is repeated and the positive charge on S goes on accumulating. In this way, voltage differences of as much as 6 or 8 million volts (with respect to the ground) can be created.
Use : Van de Graff generator used to create high potential differences that are used to accelerate charged particles such as electrons, protons, ions, etc. used for nuclear breakdown.
Limitations :
1. It is a series of combination that allows only one way for moving charge.
2. It can accelerate only the charged particles and not the uncharged particles.

Question.29. (a) In Young’s double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.
(b) The ratio of the intensities at minima to the maxima in the Youngs double slit experiment is 9 : 25. Find the ratio of the widths of the two slits.
OR
(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit illuminated by a monochromatic source of light. Hence obtain the conditions for the angular width of secondary maxima and secondary minima.
(b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2 x 10^-16 m. The distance between the slit and the screen is 1.5 m. Calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases.
Answer : (a) In Youngs double slit experiment, the wave fronts from the two illuminated slits superpose on the screen. This results in formation of alternate bright and dark fringes because of constructive and destructive interference, respectively. The intensity of light is maximum at the centre C of the screen and it is called central maxima.

Let S₁  and S₂ be two slits separated by a distance d. GG’ is the screen at a distance D from the slits S₁ and S₂ . Both the slits are equidistant from point C. The intensity of light will be maximum at this point due to the path difference of the waves reaching this point will be zero.
At point P, the path difference between the rays coming from the slits S₁  and S₂ is S₂ P — S₁P.

OR
(a) The phenomenon of bending of light round the sharp corners of an obstacle and spreading into the regions of the geometrical shadow is called diffraction.

Consider a parallel light beam from a lens is incident on slit AB. As diffraction happens, the pattern is focussed on screen XY with help of lens L₂ . We will get a diffraction pattern that is a central maximum at the centre O flanked by a number of dark and bright fringes known as secondary maxima and minima.
Central Maximum: Each point on the plane wave front AB sends secondary wavelets in all directions: The waves from points equidistant from the centre C kept on the upper and lower half reach point O with zero path difference and so, reinforce each other, making maximum intensity at point O.
Positions and Widths of Secondary Maxima and Minima
Consider a point P on screen at which wavelets moving in a direction making angle θ with CO are brought to focus by the lens. The wavelets from points A and B will have a path difference similar to BN :

Question.30. (a) Deduce an expression for the frequency of revolution of a charged particle in a magnetic field and show that it is independent of velocity or energy of the particle. (b) Draw a schematic sketch of a cyclotron. Explain, giving the essential details of its construction, how it is used to accelerate the charged particles.
OR
(a) Draw a labelled diagram of a moving coil galvanometer. Describe briefly its principle and working.
(b) Answer the following :
(i) Why is it necessary to introduce a cylindrical soft iron core inside the coil of a galvanometer.
(ii) Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity. Explain, giving reason.



Therefore, the frequency of the revolution of the charged particle is independent of the velocity or the energy of the particle.
(b) The working principle of a cyclotron is that a charge particle can be accelerated to high energy by an oscillating electric field. A cyclotron uses an electric field to accelerate charge particles across the gap between the two D-shaped magnetic field regions. The magnetic field is perpendicular to the paths of the charged particles that makes them follow in circular paths within the two Ds. Each time the charged particles cross the Ds, it is accelerated by an alternating voltage. As its speed increases the radius of path of each particle also increases. So, the accelerated particles move in a spiral path to the other wall of the cyclotron.

Square wave electric fields are used to accelerate the charged particles in a cyclotron.

At the time the charge particle finishes its half circle, the accelerating electric field reverse so that it gets accelerated across the gap between the Ds.
The particle gets accelerated again and again, and its velocity increases. Therefore, a high kinetic energy is achieved.
OR
(a) Moving coil galvanometer : It is a device used for detecting and measuring small electric current.

Principle : The working is based upon the principle, when a current carrying coil suspended in a magnetic field experiences a torque.
Construction : It is consists of a coil with a large number of turns of insulated copper wire wounded on a metallic frame.
The coil is suspended by means of a phosphor bronze strip and a horse shoe magnet NS surrounds it. The lower end of the coil is attached with a hair spring. The scale of the pointer is attached to the other end of the spring.
Working : When current is passed, the couple acts on it. Since the plane remains parallel to the magnetic field in all position of the coil, the force on the vertical arms always remains perpendicular to the place of the coil.

(b) (i) When the cylindrical soft iron core placed inside the coil of a galvanometer, the magnetic field gets stronger and radial in the space between it and pole pieces, such that whatever the position of the rotation of the coil, the magnetic field is always parallel to its plane.
(ii) Current Sensitivity : It is defined as the small deflection produced in the galvanometer when a unit current is passed through it.

SET II

Note : Except for the following questions, all the remaining questions have been asked in Previous Set.

Question.1. A conducting loop is held above a current carrying wire ‘PQ’ as shown in the figure. Depict the direction of the current induced in the loop when the current in the wire PQ is constantly increasing.

Question.4. Why do the electrostatic field lines not form closed loops ?
Answer : Electrostatic field lines do not form closed loops because electric field is conservative in nature.

Question.5. A biconvex lens made of a transparent material of refractive index 1.5 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason.
Answer: The refractive index of material of lens (1.5) is greater than the refractive index of water (1.33). So, it will behave as a converging lens.

Question.7. To which part of the electromagnetic spectrum does a wave of frequency 3 x 10¹³ Hz belong ?
Answer : The frequency 3 x 10¹³ Hz belongs to infrared, region of electromagnetic spectrum.

Question.9. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 2.5 x 10^-17m² carrying a current of 1.8. A. Assume the density of conduction electrons to be 9 x 10²⁸m^-3.

Question.13. Write the functions of the following in communication systems:
(i) Transmitter
(ii) Modulator
Answer: (i) Transmitter: A transmitter converts the message signal from the source of information into a form suitable for transmission through a channel.
(ii) Modulator: Modulator : is a device which changes some characteristics like amplitude/frequency/phase angle of a high frequency carrier wave in accordance with the instantaneous value of the low frequency message signal through the appropriate superposition.

Question.21. (a) Show, with the help of a diagram, how unpolarised sunlight gets polarised due to scattering.

Explanation : The dots in Diagram show Vibrations Perpendicularte to the plane of paper and double arrow show vibrations in the plane of paper. The elctrons in the molecule define to vibrate in both of these directions. The electrons vibrating parallel to the double arrows cannot send energy toward an observer looking at 90° to the direction of sun because their acceleration has no transverse component. The light scattered by the molecules in this direction has only dots. It is polarised perpendicular to the plane of paper.
(b) Two polaroids P₁ and P₂ are placed with their pass axes perpendicular to each other. Unpolarised light of intensity Io is incident on P₁. A third Polaroid P₃ is kept in between P₁ and P₂ such that its pass axis makes an angle of 45° with that of P₁. Determine the intensity of light transmitted through P₁, P₂ and P₃.
Answer : (a) Unpolarized light scattering from air molecules shakes their electrons perpendicular to the direction of the original ray. The scattered light therefore, has a polarization perpendicular to the original direction and none parallel to the original direction.
(b) As given in the question, the polaroids P₁ and P₂ are placed with their pass axes perpendicular to each other. Also, P₃ is placed at an angle of 45° with respect of P₁.
Now, we have :
Intensity of light after falling on P₁, I’ = I₀ /2

Question.22.Define the term self-inductance of a solenoid. Obtain the expression for the magnetic energy stored in an inductor of self-inductance L to build up a current I through it.
Answer : The ratio of magnetic flux through the solenoid to the current passing through it is called self-inductance of a solenoid. It is given by

Energy stored in an inductor : When a current grows through an inductor, a back e.m.f. is set up which opposes the growth of current. So work needs to be done against back e.m.f. (e) in building up the current. This work done is stored as magnetic potential energy.
Let I be the current through the inductor L at any instant t. The current rises at the rate dl/dt.
So the induced e.m.f. is

Question.24. A convex lens of focal length 20 cm is placed co axially with a concave mirror of focal length 10 cm at a distance of 50 cm apart from each other. A beam of light coming parallel to the principal axis is incident on the convex lens. Find the position of the final image formed by this combination. Draw the ray diagram showing the formation of the image.
Answer : The beam incident on lens L is parallel to principal axis. Hence the lens forms an image I₁ at its focus, i. e., at a distance OI₁ (= 20 cm) from the lens.

The image I₁ is formed in front of the mirror and hence, acts as a real source for the mirror. The concave mirror forms the image I₂, whose distance from the mirror can be calculated as;

SET III

Note : Except for the following questions, all the remaining questions have been asked in Previous Set.

Question.1. A conducting loop is held below a current carrying wire PQ as shown. Predict the direction of the induced current in the loop when the current in the wire is constantly increasing.

Question.2. The graph shows variation of stopping potential V₀ versus frequency of incident radiation v for two photosensitive metals A and B. Which of the two metals has higher threshold frequency and why ?

Answer : Metal A has higher threshold energy because from the graph it is clear that the minimum frequency required to start photo emission is more in A than that of B.

Question.5. Why do the electric field lines never cross each other ?
Answer : At any point, if electric field lines cross each other it means at that point there are two directions of electric field, which is impossible.

Question.6. To which part of the electromagnetic spectrum does a wave of frequency 5 x 10¹¹ Hz belong ?
Answer : A wave of frequency 5 x 10¹¹ Hz will belong to the microwaves of electromagnetic spectrum.

Question.10. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 2.5 x 10^-17 m² carrying a current of 2.7 A. Assume the density of conduction electrons to be 9 x 10¹⁸m^-3.

Question.18. Write the functions of the following in communication systems:
(i) Receiver
(ii) Demodulator
Answer:
(i) Receiver : The receiver functions on the received signal collected from transmitted signal at the channel output and reproduce the original message signal for sending.
(ii) Demodulator : Demodulator is a device which recovers the original information signal from the modulated wave at the receiver end.

Question.19. A convex lens of focal length 20 cm is placed co axially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm from each other. A point object placed 40 cm in front of the convex lens. Find the position of the image formed by this combination. Draw a ray diagram to show the formation.

The positive sign describes that the image is. formed to the right of the lens.

The image I₁ is formed behind the mirror and thus acts as a virtual source for the mirror. The convex mirror forms the image I₂, whose distance from the mirror is given by :

Question.25. (a) A rod of length l is moved horizontally with a uniform velocity V in a direction perpendicular to its length through a region in which a uniform magnetic field is acting vertically downward. Derive the expression for the emf induced across the ends of the rod.
(b) How does one understand this motional emf by invoking the Lorentz force acting on the free charge carriers of the conductor? Explain.


After this, resultant force on the free electrons of the wire PQ becomes zero. The potential difference between the ends Q and P is given by, ‘
V = El = vBl
Thus, the potential difference is maintained by the magnetic force on the moving free electron and hence, produces an emf, e = Bvl
(b) Lorentz force acting on a charge q which is moving with a speed v in a (normal) uniform magnetic field B, is Bqv.
All the charges will experience the same force.

Question.26. (a) Show, giving via suitable diagram, how unpolarized light can be polarised by reflection.
(b) Two polaroids P₁ and P₂ are placed with their pass axes perpendicular to each  other. Unpolarised light of intensity Io is incident on P₁. A third Polaroid P₃ is kept in between P₁ and P₂ such that its pass axis makes an angle of 60° with that of P₁. Determine the intensity of light transmitted through P₁, P₂ and P₃.
Answer: (a) On reflection from a transparent medium a normal light beam becomes partially polarised. As the angle of incidence gets higher, the degree of polarization gets higher.

The reflected light beam becomes fully polarised at a certain  value. This angle of incidence is known as polarizing angle (p). At the interface of a refracting medium when light is incident at polarizing angle, the refractive index of the medium is similar to the tangent of the polarizing angle.

(b) As given, the Polaroid P₁ and P₂ are placed with their axes perpendicular to each other and Polaroid P₃ placed at 60° with respect to P₁.

CBSE Class 9 Maths Lab Manual – Equal Division of a Line Segment

Objective
To divide a line segment into equal parts.

Prerequisite Knowledge
Intercept Theorem: If there are three (or more) parallel lines and intercepts made by them on any transversal are equal, then intercepts made by them on any other transversal are also equal.

Materials Required
A sheet of ruled paper in which all the lines are parallel and equidistant, tracing paper or transparent sheet, geometry box.

Procedure

1. Take a piece of tracing paper sheet and draw a line segment of length 10 cm or 9 cm (let line segment is to be divided into 7 equal parts). Name it as PQ.
   
2. Now, take a ruled sheet, and mark the lines as 0, 1, 2, 3, 4…10 from top to bottom.
3. Place the paper strip (tracing paper) on the ruled sheet such that the point P coincides as with the line (ruled sheet) marked 0, and other point Q coincides with the line marked 7 as shown in fig.(i).
4. Mark the points on PQ which are the points of intersection with lines of ruled sheet as A₁, A₂, A₃,…A₆.
5. Now remove the ruled sheet.

Observation
7 equal parts will be obtained.
PA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ = A₅A₆ = A₆Q.
Here, marked axis PQ is a transversal on parallel lines of ruled sheet with equal intercept. Therefore, PQ also has equal intercepts (By intercept theorem).

Result
Line PQ is divided into 7 equal parts.

Learning Outcome
By paper activity, a line segment of any measurement can be divided into equal parts.

Activity Time

1. Draw a line segment of 7 cm, and divide it into 6 equal parts and draw perpendicular on the 4th point of line segment, by paper folding.
2. Take any rectangular sheet of 5 cm by 9 cm and divide the edges of sheet in 8 equal parts.
3. Take any rectangular sheet of 10 cm by 7 cm and divide along length in 10 equal parts and along the breadth in 7 equal parts (By paper folding).

Viva Voce

Question 1.
What do you mean by parallel lines ?
Answer:
Two distinct lines which are not intersecting are called as parallel lines.

Question 2.
What is transversal line ?
Answer:
A line intersecting two or more than two distinct lines at distinct points is called as transversal line.

Question 3.
What is intercept?
Answer:
If a transversal line intersects two parallel lines, the line segment intersected between the parallel lines on the transversal is called intercept.

Question 4.
How will you state intercept theorem ?
Answer:
If there are three (or more) parallel lines and the intercepts made by them on one transversal are equal, the intercepts on any other transversal are also equal.

Question 5.
What is line segment ?
Answer:
Line segment is a part of a line.

Question 6.
What do you mean by fractions ?
Answer:
Equal parts of a whole are called fractions.

Question 7.
What do you mean by equivalent fractions ?
Answer:
Two or more fractions representing the same part of the whole are known as equivalent fractions.

Question 8.
How will you bisect a rope without measuring it ?
Answer:
By folding rope and joining its end points, two equal parts of a rope will be obtained.

Question 9.
To divide a thread in 8 equal parts, how many successive folds will you give ?
Answer:
7 folds.

Question 10.
Explain the trisection of a strip.
Answer:
A strip is divided into three equal parts.

Question 11.
Explain the six equal divisions of a rope.
Answer:
If a rope is divided into six equal parts then each part is one-sixth of rope.

Question 12.
If the intercept made by a transversal between three parallel lines is in ratio 1:2, what intercept made by another transversal?
Answer:
Same ratio (1:2).

Multiple Choice Questions

Question 1.
If AB = 10 cm, is divided into 7 equal parts then what will be the value of each part ?
(i) \(\frac { 1 }{ 7 }\) cm
(ii) \(\frac { 10 }{ 7 }\) cm
(iii) \(\frac { 3 }{ 7 }\) cm
(iv) none of these

Question 2.
If ratio of intercepts made by a transversal is \(\frac { 2 }{ 3 }\) then what will be the ratio of intercepts made by another transversal ?
(i) \(\frac { 3 }{ 2 }\)
(ii) \(\frac { 2 }{ 3 }\)
(iii) \(\frac { 1 }{ 3 }\)
(iv) none of these

Question 3.
The length of a line segment is 12 cm, what is the length of each part, if it is divided into 10 equal parts ?
(i) 1.2 cm
(ii) 2.2 cm
(iii) \(\frac { 10 }{ 12 }\) cm
(iv) none of these

Question 4.
If PQ = 9 cm, is divided into three equal parts, what is ratio of first two parts to the remaining part ?
(i) 2:1
(ii) 1:2
(iii) 3 :2
(iv) none of these

Question 5.
Which theorem is used to get the equal division of a line segment ?
(i) Pythagoras theorem
(ii) mid-point theorem
(iii) intercept theorem
(iv) none of these.

Question 6.
What do you mean by transversal line ? A line which intersects:
(i) one line
(ii) two or more lines
(iii) angle
(iv) none of these

Question 7.
Write first two equivalent fractions of \(\frac { 5 }{ 2 }\)
(i) \(\frac { 5 }{ 2 } ,\frac { 10 }{ 2 }\)
(ii) \(\frac { 10 }{ 4 } ,\frac { 15 }{ 6 }\)
(iii) \(\frac { 2 }{ 5 } ,\frac { 4 }{ 10 }\)
(iv) none of these

Question 8.
If \(\frac { 5 }{ 7 } =\frac { x }{ 35 }\) what is the value of x?
(i) 25
(ii) 10
(iii) 1
(iv) none of these

Question 9.
Find x, if \(\frac { 21 }{ 9 } =\frac { 7 }{ x }\)
(i) 3
(ii) \(\frac { 1 }{ 3 }\)
(iii) 21
(iv) none of these

Question 10.
In the given figure, AB = BC = 4 cm, find DE, if EF = 5 cm

(i) 5 cm
(ii) 6 cm
(iii) 2 cm
(iv) none of these

Answers

1. (i)
2. (ii)
3. (i)
4. (i)
5. (iii)
6. (ii)
7. (ii)
8. (i)
9. (i)
10. (i)

Math Lab ManualMath Labs with ActivityMath LabsScience LabsScience Practical Skills

CBSE Class 9 Maths Lab Manual – Angle at Centre is Double the Angle Subtended by Same Arc at Any Point on Circumference of Circle

Objective
To verify that the angle subtended by an arc at the centre of circle is double the angle subtended at any point on the remaining part of the circle, experimentally.

Prerequisite Knowledge

1. Basic terms related to circle.
2. Concept of an angle subtended by an arc at the centre and at the circumference of the circle.

Materials Required
Glazed papers, white sheet, pencil, a pair of scissors, gluestick.

Procedure

1. Cut one circle of radius 2.5 cm with centre O from red coloured glazed paper.
2. Cut three more circles from different coloured glazed papers of same radius.
3. Keep all four circles one on the other. Now fold along any part and press them to make a crease. On unfolding, we get chords of same length on each circle.
4. Name the chords, AB in first circle with centre O.
5. Join OA and OB with pencil [fig. (i)].
6. Take two other circles of yellow and green colour and put one on the other and fold it such as to form an angle on the circumference with the same chord AB [fig. (ii) and (iii)].
7. Name these angles as ACB where AB is a chord.
8. Cut angles from fig. (ii) and (iii), ∠ACB from yellow circle and ∠ACB from green circle.
9. Cut the small portion of ∠ACB from both the circles [fig (ii) and (iii)].
   
   
   
   
10. Paste these two cut outs of fig. (ii) and (iii) on the another (blue) circle [fig. (iv)] at centre O such that their arms lie on the radius OA and OB of circle.

Observation
We observe that two cut outs of angles fully cover ∠AOB in fig.(iv).
∠AOB = ∠ECF + ∠GCH (as ∠ECF = ∠GCH = ∠ACB).
= 2 ∠ACB

Result
Hence we verified that the angle subtended by an arc at the centre of circle is double the angle subtended by the same arc at any point on the remaining part of the circle.

Learning Outcome
Verification of above theorem can be done for arc AB as major arc or semicircular arc. For semicircle, angle on the diameter is of 90°.

Activity Time

1. Prove this theorem by taking different situations on the circle such as chord, a diameter, chord in minor segment, chord on major segment.
2. What do you observe if you take chord as a diameter?

Viva Voce

Question 1.
What is the segment of a circle ?
Answer:
A chord divides the circle into two parts. Each part is known as segment.

Question 2.
What divides a circle into two equal segments ?
Answer:
Diameter.

Question 3.
Two diameters of a circle are perpendicular to each other, how many equal sectors will they form in circle ?
Answer:
4 sectors.

Question 4.
Two chords AB and CD in a circle form ∠AOB and ∠COD at the centre O of the circle. If AB and CD are not equal in length what can you say about the angles ∠AOB and ∠COD ?
Answer:
∠AOB ≠ ∠COD.

Question 5.
If the angle at the centre is 60°, what will be the angle on the remaining part of circle subtended by an arc ?
Answer:
30°.

Question 6.
If the angles subtended by the chords of a circle at the centre are equal, then what will be the length of chords ?
Answer:
Length of the chords will be equal.

Question 7.
What will be the distance of the two equal chords from the centre ?
Answer:
Equal distance.

Question 8.
If the two chords are equal, then what will be the length of their corresponding arcs ?
Answer:
Equal length.

Multiple Choice Questions

Question 1.
In a circle, arc BC subtends 40° at the centre, the value of angle subtended by the arc BC at the remaining part of the circle is:
(i) 20°
(ii) 40°
(iii) 80°
(iv) none of these

Question 2.
In a circle, chord AB subtends 32° at circle, the value of angle subtended by it at the centre is:
(i) 16°
(it) 64°
(iii) 30°
(iv) none of these.

Question 3.
Angle subtended by the diameter of the semicircle is:
(i) 90°
(ii) 45°
(iii) 180°
(iv) none of these

Question 4.
Triangle formed by the chord and two radii in a circle is:
(i) right angled triangle
(ii) isosceles triangle
(iii) equilateral triangle
(iv) none of these

Question 5.
In the given figure, O is the centre of the circle, ∠BAC = 40°. Find x:

(i) 50°
(ii) 100°
(iii) 80°
(iv) none of these

Question 6.
In the given figure, find x, if ∠ACB = 30°:

(i) 60°
(ii) 15°
(iii) 120°
(iv) none of these

Question 7.
The angle formed by a chord is 110° at the centre of the circle, the value of angle formed by it at the remaining part of circle is:
(i) 55°
(ii) 220°
(iii) 100°
(iv) none of these

Question 8.
The angle formed by a chord in minor segment is half of the angle formed by it at centre is:
(i) straight angle
(ii) reflex
(iii) acute angle
(iv) none of these

Question 9.
Find the value of x, in the given figure, if ∠BOC = 42°:

(i) 21°
(ii) 84°
(iii) 40°
(iv) none of these

Question 10.
The angle at the centre of the circle is:
(i) 260°
(ii) 180°
(iii) 360°
(iv) none of these

Answers

1. (i)
2. (ii)
3. (i)
4. (ii)
5. (i)
6. (i)
7. (i)
8. (ii)
9. (i)
10. (iii)

Math Lab ManualMath Labs with ActivityMath LabsScience LabsScience Practical Skills

CBSE Previous Year Solved  Papers  Class 12 Computer Science Delhi 2015

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions :

1.  There are a total of 26 questions and five sections in the question paper, All questions are compulsory.
2. Section A contains question number 1 to 5, Very Short Answer type questions of one mark each.
3.  Section B contains question number 6 to 10, Short Answer type I questions of two marks each.
4.  Section C contains question number 11 to 22, Short Answer type II questions of three marks each.
5.  Section D contains question number 23, Value Based Question of four marks.
6. Section E contains question number 24 to 26, Long Answer type questions of five marks each.
7. There is no overall choice in the question paper, however, an internal choice is provided in one question of two marks, one question of three marks and all three questions of five marks. An examined is to attempt any one of the questions out of two given in the question paper with the same question number.

SET I

SECTION B

Question.1. (a) Name the function/method required to
(i) check if a string contains only upper case letters
(ii) give the total length of the list.
(b) Rewrite the following Code in python after removing all syntax error(s). Underline each correction done in the code.

(c) Find and Write the output of the following python code :

(d) Find and write the output of the following python code :


(e) What are the possible outcome(s) executed from the following code ? Also specify the maximum and minimum values that can be assigned to variable NUMBER.

Question.2. (a) Illustrate the concept of inheritance with the help of a python code.
(b) What will be the output of the following python code ? Explain the try and except used in the code.

(c) Write a class PICTURE in Python with following specifications:

(d) What is operator overloading with methods ? Illustrate with the help of an example using a python code.
(e) Write a method in python to display the elements of list thrice, if it is a number and display the element terminated with ‘#’ if it is not a number.
For example, if the content of fist is as follows :


Answer: (a) Inheritance: Python supports inheritance, it even supports multiple inheritance. Classes can inherit from other classes. A class can inherit attributes and behaviour methods from another class, called the super class. A class which inherits from a super class is called .a subclass, also called heir class or child class. Super classes are sometimes called ancestors as well. There exists a hierarchy relationship between classes, we define a class Calendar Clock, which inherits both from the class Calendar and from the class Clock.





Here try and its except statement form a unit. The scope of the except clause is restricted to those statements specified by the immediately preceding try statement. A except statement cannot catch an exception thrown by another try statement.

(d) Operator overloading is a specific case of polymorphism in which’ some or all operators like +, = or == are treated as polymorphic functions and as such have different behaviours depending on the types of its arguments. here we overload the + sign, in the class Point.

Question.3. (a) What will be the status of the following list after fourth pass of bubble sort and fourth pass of selection sort used for arranging the following elements in descending order ?
14, 10, -12, 9, 15, 35
(b) Write a method in python to Search for a value in a given list (assuming that the elements in list are in ascending order) with the help of Binary Search method. The method sould return -1, if the value not present else it should return position of the value present in the list.
(c) Write PUSH (Books) and POP (Books) methods in python to add Books and remove Books considering them to act as Push and Pop operations of Stack.
(d) Write a method in python to find and display the prime numbers between 2 to N. Pass N as argument to the method.
(e) Evaluate the following post fix notation of expression.
Show status of stack after every operation.
84, 62, -, 14, 3, *, +

Question.4. (a) Differentiate between the following :
(i) f = open (‘diary.txt’, V)
(ii) f = open (‘diary, text’ , ‘w’)
(b) Write a method in python to read the content from a text file diary.txt line by fine and display the same on screen.
(c) Consider the following definition of class Member, write a method in python to write the content in a pickled file member.dat.

SECTION C

Question.5. (a) Observe the following table carefully and write the names of the most appropriate columns, which can be considered as (i) candidate keys and (ii) primary key:

(b) Consider the following DEPT and WORKER tables. Write SQL queries for (i) to (iv) and find outputs for SQL queries (v) to (viii).

Note: DOJ refers to date of joining and DOB refers to date of Birth of workers. in ascending order of Wno.
(ii) To display the Name of all the FEMALE workers from the table WORKER.
(iii) To display the Wno and Name of those workers from the table WORKER who are born between ‘1987 01, 01 and ‘1991-12-01’.
(iv) To count and display MALE workers who have joined after T986-01-01’.
(v) SELECT COUNT (*), DCODE FROM WORKER GROUP BY DCODE HAVING COUNT (*1 > 1;
(vi) SELECT DISTINCT DEPARTMENTf ROM DEPT;
(vii) SELECT NAME, DEPARTMENT, CITY FROM WORKER W, DEPT D WHERE W. DCODE=D. DCODE AND WNO< 1003;

Question.6. (a) Verily the following using Boolean Laws :
X+Y’ = X.Y+X.Y’+ X’.Y‘
(b) Draw the Logic Circuit for the following Boolean Expression :
(U+V>.W + Z
(c) Derive a Canonical SOP expression for a Boolean function F represented by the following truth table.

Question.7. (b) What is a spam mail ?
(c) Differentiate between ftp and http.
(e) What is Worm ? How is it removed ?
(f) Out of the following, which all comes under cyber crime ?
(i) Stealing away a brand new computer from a showroom.
(ii) Getting in someone’s social networking account without his consent and posting pictures on his behalf to harass him.
(iii) Secretly copying files from server of a call center and selling it to the other organization.
(iv) Vewing sites on a internet browser.
(e) Perfect Edu Services Ltd. is an educational organization. – It is planning to set up its India campus at Chennai with its head office at Delhi. The Chennai campus has 4 main buddings – ADMIN, ENGINEERING, BUSINESS and MEDIA.
you as a network expert have to suggest the best network related solutions for their problems raised in (i) to (iv), keeping in mind the distances between the buildings and other given parameters.

(i) Suggest the most appropriate location of the server inside the CHENNAI campus (out of the 4 buildings), to get the best connectivity for maximum number of computers. Justify your answer.
(ii) Suggest and draw the cable layout to efficiently connect various’ buildings within the CHENNAI campus for connecting the computers.
(iii) Which hardware device will you suggest to be procured by the company to be installed to protect and control the internet uses within the campus ?
(iv) Which of the following will you suggest to establish the online face-to-face communication between the people in the Admin Office of CHENNAI campus and DELHI-I Head Office ?
(i) Cable TV
(ii) Email
(iii) Video Conferencing
(iv) Text Chat .
Answer : (b) spam Email, also known as junk email or * unsolicited bulk email (UBE), is a subset of electronic spam involving nearly identical messages sent to numerous
recipients by email. Clicking on links in spam email may send users to phishing web sites or sites that are hosting malware,
(c) As far as file transfer speeds, FTP can be a bit faster for large files, while HTTP can be faster for multiple smaller files. HTTP is’ easier to work with, and it was developed specifically to address some of the inefficiencies of FTP.
(e) A program that when run, has the ability to spread to other computers on its own using either mass-mailing techniques to email addresses found on your computer or by using the Internet to infect a remote computer using known security holes.
There are several steps that should be taken for computer worm removal.

• Check that all antivirus signatures are up-to-date.
• Scan the computer with antivirus software.
• If the scan detects a computer worm or other malware, use the software to remove malware and clean or delete infected files. A scan that detects no malware is usually indicative that symptoms are being caused by hardware or software problems.
• Check that the computer’s operating system is up-to-date and all software and applications have current patches installed.
• If a worm is difficult to remove, check online for specific computer worm removal utilities.

(f) Out of the following, following comes under cyber crime Getting in someone’s social networking account without his consent and posting pictures on his behalf to Secretly copying files from server of a call center and selling it to the other organization.
(e) (i) The most appropriate location of the server inside the CHENNAI campus (out of the 4 buildings), is ADMIN block because this block have maximum number of computers.

(iii) Firewall device will be produced by the company to be installed to protect and control the internet uses within the campus.
(iv) Video Conferencing will be best to establish the online face-to-face communication between the people in the Admin Office of CHENNAI campus and DELHI Head Office.

CBSE Previous Year Solved  Papers  Class 12 Physics Delhi 2016

Time allowed : 3 hours                                                                                           Maximum Marks: 70

General Instructions:

1. 1.  All questions are compulsory. There are 26
      questions in all.
   2.  This question paper has five sections : Section A, Section B, Section C, Section D and Section E.
   3. Section A contains five questions of one mark each, Section B contains five questions of two marks each, Section C contains twelve questions of three marks each, Section D contains one value based question of four marks and Section E contains three questions of five marks each.
   4.  There is no overall choice. However, an internal choice has been provided in one question of two marks, one question of three marks and all the three questions of five marks weight age. You have to attempt only one of the choices in such questions.
   5. You may use the following values of physical constants wherever necessary:

SET I

SECTION-A

Question.1. A point charge +Q is placed at point O as show in the figure. Is the potential difference V_A — V_B positive, negative or zero ?

Question.2. How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased ?
Answer : According to Gauss’s law

Flux depends only on the charge enclosed.
Hence, the electric flux remains constant.

Motion in Straight Line Class 11 NCERT Solutions are very helpful in scoring more marks in class 11 exams.

 

Question.3. Write the underlying principle of a moving coil galvanometer.
Answer : When a current carrying coil is placed in magnetic field then it experiences a torque.

Question.4. Why are microwaves considered suitable for radar systems ,used in aircraft navigation ?
Answer : Microwaves of frequency 1 GHz to 300 GHz bounces from even the smallest aircraft so that they are suitable to avoid getting bombed. Microwaves can penetrate through clouds also.

Question.5. Define ‘quality factor ‘ of resonance in series LCR circuit. What is its SI unit ?
Answer : The Q factor of series resonance circuit is defined as the ratio of the voltage developed across the inductor or capacitor at resonance to the impressed voltage, which is the voltage across R.

It is dimensionless hence, it has no units.

SECTION -B

Question.6. Explain the terms (i) Attenuation and (ii) Demodulation used in Communication System.
Answer: (i) Attenuation: The loss of strength of a signal while . propagating through a medium is known as attenuation.
(ii) Demodulation : The process of retrieval of information from the carrier wave at the receiver is termed demodulation. This is the reverse process of modulation.

Question.7.

Question.8. A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into two fragments each of A = 120 with BE/A = 8.5 MeV Calculated the released energy.

Answer : Gain in binding energy for nucleon is about 0.9 MeV.

Question.9. Two cell of emfs 1.5 V and 2.0 V having internal resistances 0.2 Ω and 0.3 Ω respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.

Question.10. State Brewster’s law.
The value of Brewster angle for a transparent medium is different for light of different colour. Give reason.
Answer : Brewster’s law : The law states that the tangent of the polarising angle of incidence of a transparent medium is equal to its refractive index. The light incident at this angle. When reflects back is perfectly polarised.
M = tan i_p
The refractive index of a material depends on the colour or wavelength of light. As the polarising angle depends on refractive index (M = tan ip), so it also depends on wavelength of light.

SECTION – C

Question.11. A charge is distributed uniformly over a ring of radius ‘a’. Obtain an expression for the electric intensity E at a point on the axis of the ring. Hence show that for points at large distances from the ring, it behaves like a point charge.
Answer : Suppose we have a ring of radius a that carries a uniformly distributed positive charge q.

(i) The axial components dE cos θ and
(ii) The perpendicular component dE sin θ.
Since the perpendicular component of any two diametrically opposite elements are equal and opposite, they cancel out in pairs. Only the axial components will add up to produce the resultant field.


This expression is similar to electric field due to point charge.

Question.12. Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only using Einstein’s equation.
Answer: (i) Existence of threshold frequency: According to wave theory, there should not exist any threshold frequency but Einstein’s theory explains the existence of Threshold frequency.
(ii) Dependence of kinetic energy on frequency of incident light: According to wave theory, the maximum kinetic energy of emitted electrons should depend on intensity of incident light and not on frequency whereas Einstein’s equation explains that it depends on frequency and not on intensity.
(iii) Instantaneous emission of electrons : According to wave theory there should be time lag between emission of electrons and incident of light whereas Einstein’s equation explains why there is no time lag between incident of light and emission of electrons.

Question.13. (a) Write the expression for the magnetic force acting on a charged particle moving with velocity v in the presence of magnetic field B.
(b) A neutron, an electron and an alpha particle moving with equal velocities, enter a uniform magnetic field going into the plane of the paper as shown. Trace their paths in the field and justify your

Question.14. (i) Define mutual inductance.
(ii) A pair of adjacent coils has a mutual inductance of 1.3 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil ?
Answer : (i) Mutual induction is the phenomenon of production of induced emf in one coil due to change of current in the neighbouring coil. The coil in which the current changes is called primary coil and the coil in which emf is induced is called the secondary coil.

Question.15. Two parallel plate capacitors X and Y have the same area of plates and same separation between them. X has air between the plates while Y contains a dielectric medium of ε_r= 4.

(i) Calculate capacitance of each capacitor if equivalent capacitance of the combination is 4 µF.
(ii) Calculate the potential difference between the plates of X and Y.
(iii) Estimate the ratio of electrostatic energy stored in X and Y

Question.16. Two long straight parallel conductors carry steady current I₁ and I₂ separated by a distance d. If the currents are flowing in the same direction, show how the magnetic field set up in one produces an attractive force on the other. Obtain the expression for this force. Hence define one ampere.

Question.17. How are e.m. waves produced by oscillating charges ?
Draw a sketch of linearly polarized e.m. waves propagating in the Z-direction. Indicate the directions of the oscillating electric and magnetic fields.

Answer: These waves are constituted by varying or oscillating electric and magnetic fields. The electric and magnetic fields are perpendicular to each other and are also perpendicular to the direction of propagation of the wave E is the envelope of electric intensity vector and B is the envelope of magnetic intensity vector.

OR
Correction in Amperes Circuital law (Modified Ampere’s law) : Maxwell removed the problem of current continuity and inconsistency observed in Ampere’s Circuital law by introducing the concept of displacement current, Displacement current arises due to change in electric flux

Conduction current is because of How of charges but displacement current is not because of How of charges but because of change in electric flux.

Question.18. (a) Explain any two factors which justify the need of modulating a low frequency signal.
(b) Write two advantages of frequency modulation over amplitude modulation.

Question.19. (i) Write the functions of three segments of a transistor.
(ii) Draw the circuit diagram for studying the input and output characteristics of n-p-n transistor in common emitter configuration. Using the circuit, explain how input, output characteristics are obtained.
Answer : (i) Three segments of transistor are :
(i) Emitter (ii) Base
(iii) Collector
Emitter : It is of moderate size and heavily doped, it supplies a large number of majority carriers which flow through the transistor.
Base : It is very thin and lightly doped and it separates emitter and collector region of transistor and controls the flow of charge carriers.
Collector : This segment is moderately doped and larger in size as compared to emitter. It collects a major portion of majority carriers supplied by the emitter.

Question.20. (a) Calculate the distance of an object of height h from a concave mirror of radius of curvature 20 cm, so as to obtain a real image of magnification 2. Find the location of image also,
(b) Using mirror formula, explain why does a convex mirror always produce a virtual image.

Question.21. (i) State Bohr’s quantization condition for defining stationary orbits. How does de-Broglie hypothesis explain the stationary orbits ?
(ii) Find the relation between the three wavelengths λ₁, λ₂ and λ₃ from the energy level diagram shown below.

Question.22. Draw a schematic ray diagram of reflecting telescope showing how rays coming from a distant object are received at the eyepiece. Write its two important advantages over a refracting telescope.
Answer : Reflecting Telescope : The reflecting telescope makes us of a concave mirror as objective. The rays of light coming from distant object are incident on the objective (parabolic reflective). After reflection the rays of light meet at a point where another convex mirror is placed. This mirror focusses light inside the telescope tube. The final image is seen through the eye piece. The images produced by the reflecting telescope is very bright and its resolving power is high.

Advantages:
(i) The resolving power (the ability to observe two object distinctly) is high, due to the large diameter of the objective.
(ii) There is no chromatic aberration as the objective is a mirror.

SECTION – D

Question.23. Meeta’s father was driving her to the school. At the traffic signal she noticed that each traffic light was made of many tiny lights instead of a single bulb. When Meeta asked this question to her father, he explained the reason for this.
Answer the following questions based on above information:
(i) What were the values displayed by Meeta and her father ?
(ii) What answer did Meeta’s father give ?
(iii) What are the tiny lights in traffic signals called and how do these operate ?
Answer : (i) Awareness for energy conservation, power saving and knowledge about traffic rules.
(ii) Meeta’s father said that these are LED light which consume less power and high reliability.
(iii) The tiny lights in traffic signals are Light Emitting Diode. These are operated by connecting the P-N junction diode in forward biased condition.

SECTION – E

Question.24. (i) An a. c. source of voltage V = V₀ sin tor is connected to a series combination of L, C and R. Use the phasor diagram to obtain expressions for impedance of the circuit and phase angle between voltage and current. Find the condition when current will be in phase with the voltage. What is the circuit in this condition called ? (ii) In a series LR circuit X_L = R and power factor of the circuit is P₁. When capacitor with capacitance C such that X_L= X_C is put in series, the power factor becomes P₂. Calculate P₁/P₂.
OR
(i) Write the function of a transformer. State is principle of working with the help of a diagram. Mention various energy losses in this device.
(ii) The primary coil of an ideal step up transformer has 100
turns and transformation ratio is also 100. The input voltage and power are respectively 220V and 1100 W Calculate:
(a) number of turns in secondary.
(b) current in primary.
(c) voltage across secondary.
(d) current in secondary.
(e) power in secondary.
Answer : (i) Let a, series LCR circuit is connected to an ac source V (Fig). We take the voltage of the source to be V = V₀ sin ωt.



OR
(i) A transformer is an electrical device for converting an alternating current at low voltage into that at high voltage or vice-versa.

• If it increases the input ac voltage, it is called step up transformer.
•  If it decreases the input ac voltage, it is called step down transformer.

Principle : It works on the principle of mutual induction i.e., When a changing current is passed through one of the tow inductively coupled coils, an induced emf is set up in the other coil.


Working Theory : As the AC flows through the primary, it generate an alternating Magnetic flux in the core which passes through the secondary coil.
Let N1 = No. of turns in primary coils
N2 = No. of turns in secondary coils
This changing flux set up an induced emf in the secondary,/ also a self induced emf in the primary.
If there is no leakage of magnetic flux, then flux linked with each turn of the primary will be equal to that linked with each of the secondary. According to Faraday’s law of induction.

• Copper loss : Some energy is lost due to the heating of copper wires used in the primary and secondary winding’s. This power less (P = 12R) can be minimised by using thick copper wires of low resistance.
• Eddy current loss : The alternating magnetic flux induces eddy current in the iron core which leads to some energy loss in the form of heat. This loss can be reduced by using laminated iron core.
• Hysteresis loss : The alternating current carries the iron core through cycles of magnetisation and demagnetisation. Work done in each of these cycles and is lost as heat. This is called hysteresis loss and can be magnetised by using c.ore material having narrow hysteresis loop.
• Flux leakage: The magnetic flux produced by the primary may not fully pass through the secondary. Some of the flux may leak into air. This loss can be minimised by winding the primary and secondary coils over one another.

Question.25. (i)In Young’s double slit experiment, deduce the condition for (a) constructive, and (b) destructive interference at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position V on the screen.
(ii) Compare the interference pattern observed in Young’s double slit experiment with single slit diffraction pattern, pointing out three distinguishing features.
OR
(i) Plot a graph to show variation of the angle of deviation as a function of angle of incidence for light passing through a prism. Derive ah expression for refractive index of the prism in terms of angle of minimum deviation and angle of prism.
(ii) What is dispersion of light ? What is its cause ?
(iii) A ray of light incident normally on one face of a right isosceles prism is totally reflected as shown in fig. What must be the minimum value of refractive index of glass? Give relevant calculations.




(ii) Compose of interference pattern observed in Young’s double slits and the single slits diffraction :

OR
(i) It is a transparent medium having two planes and non-parallel refracting surface inclined to each other and three surfaces are not participating in refraction.




(ii) Dispersion of light : These colours are often observed as light passes Through a triangular prism upon passing through the prism, the white light is separated into its component colours : red, orange, yellow, green, blue, and violet. The separation of visible light into its different colours is known as dispersion. Dispersion occurs because for different colour of light a transparent medium will have different refractive indices (µ).

Question.26. (i) Define the term drift velocity.
(ii) On the basis of electron drift, derive an expression for resistivity of a conductor in terms of number density of free electrons and relaxation time. On what factors does resistivity of a conductor depend ?
(iii) Why alloys like constantan and manganin are used for making standard resistors ?
OR
(i) State the principle of working of a potentiometer.
(ii) In the following potentiometer circuit AB is a uniform wire of length 1 m and resistance 10 Ω. Calculate the potential gradient along the wire and balance length AO (=l).

Answer : (i) Drift velocity is defined as the average velocity which the electrons are drifted towards the positive terminal under the effect of applied electric field. Thermal velocities are randomly distributed and average thermal velocity is zero.



Where P is the specific resistance or resistivity of the material of the wire. It depends on number of free electron per unit volume and temperature.
(iii) They are used to make standard resistors because :
(a) They have high value of resistivity
(b) Temperature coefficient of resistance is less.
(c) They are least affected by temperature.
OR
(i) Principle of potentiometer : The basic principle of potentio meter is that when a constant current flows through a wire of uniform cross-section area and the composition of the potential drop across any length of the wire is directly proportional to that length.
A potentiometer is a device used to measure an unknown emf or potential difference and internal resistance of a cell accurately.
Construciton :

1. A potentiometer consists of a long uniform cross-section of wire generally made of manganin or constantan.
2. Usually, 1 m long separate pieces of wire are fixed on a wooden board parallel to each other.
3. The wire are joined in series by thick copper strips.
4. The ends A and B are connected to a battery (called driving all), a plug key and rheostat.
5. A jockey J is provided with the help of which contact can be made at any point on the wire.
6. This circuit sends a constant current I through the wire AB.

Principle : When a constant current flows through a wire of uniform cross sectional area and composition the potential drop across any length of the wire is directly proportional to that length.
Let V be the potential difference across the portion of the wire of length l whose resistance is R
By Ohm’s law,

Looking for an easy way to Learning of new elementary english grammar and composition for class 7 answers, Solutions. You have to learn basic English Grammar topics like Tenses Verbs, Nouns, etc… In this article, we will review the best English Grammer Topics and compare them against each other.

Noun Phrase And Noun Clause Exercises For Class 7 CBSE Pdf

“Main (or independent) clause is a clause that expresses a complete thought and , can stand as a sentence. But subordinate (or dependent) clauses cannot stand by themselves and are further divided into three types,

1. Noun Phrase
2. Adjective Phrase
3. Adverb Phrase

Noun Phrase:
A phrase that acts as a noun in a sentence is called a noun phrase. It consists of a noun and other words that give information about the noun. It answers the question what or who. The noun phrase can be replaced by a pronoun.
Examples:

• The coffee – table books looked beautiful. (noun: the coffee – table books)
• I like going for a swim. (noun: going for a swim)
• Mona met the newly appointed chief. (noun: the newly appointed chief)

Remember, a noun phrase must have a noun.

Noun Clause:
The clause that tells you what someone says or does. It can also be the object of a verb.
A noun clause is a dependent clause that acts as a noun. To identify the noun clause in a sentence, we can ask the question ‘What’ after the verb. Noun clauses begin with words such as how, that, what, whatever, when, where, whether, which, whichever, who, whoever, whom, whomever, and why.
Noun clauses can act as subjects, direct objects, indirect objects.

Noun clauses as direct objects:
Examples:

• They know what my name is.
• I noticed where they were going.

Noun clauses as a subject:
Examples

• What she is doing is not known.
• What you said is not clear.
• Where she lives is not considered.

Noun clauses as an indirect object:
Examples:

• I will do whatever it takes to score good marks.
• Nancy will whoever finds the purse for her.

Noun Phrase Vs Noun Clause
Examples

• No one knows when he will come. (Noun-Clause)
• No one knows the time of his coming. (Noun-Phrase)
• She heard that he had succeeded. (Noun-Clause)
• She heard of his success. (Noun-Phrase)
• They will never know why he succeeded. (Noun-Clause)
• They will never know the reason for his success. (Noun-Phrase)
• The law will punish whosoever is guilty. (Noun-Clause)
• The law will punish the guilty. (Noun)
• Her friends want to know where she is living. (Noun-Clause)
• Her friends want to know her residence. (Noun)

Focus:
A noun clause can be a complement after a ‘be’ verb.
E.g.

• Why she was working so late is something we need to find out

A noun clause may sometimes be placed first in a sentence for emphasis.
E.g.

• How we’re going to deal with the issue is the question.

Presentation:
Read the picture story.

Noun Phrase And Noun Clause Worksheet for Class 7 CBSE

A. Underline the noun phrases in the following sentences.
1. He wished to talk to his manager.
2. The wicked man loves getting poor people into trouble,
3. The poor man wanted to pay back every penny he owed.
4. He hates to punish his children.
5. Horses prefer living in dark stables.
6. I will never do such a thing.
7. Have you ever tried to climb a tree?
8. He refused to answer my question.
9. He promised to get me something.
10. Why do you want to meet him?

B. In each of the following sentences replace the words in italics by suitable noun clauses.

1. No one can guess the time of his arrival.
___________________________________________
2. I predict a change in the weather
___________________________________________
3. I heard of his success.
___________________________________________
4. Nobody knows the reason of his failure.
___________________________________________
5. The jury believed the man to be guilty.
___________________________________________
6. The doctors expect an improvement in his health.
___________________________________________
7. I know him to be trustworthy.
___________________________________________
8. He confessed his guilt.
___________________________________________
9. I hope to be there in time.
___________________________________________
10. He is confident of winning the match.
___________________________________________
11. He seems to be a spy.
___________________________________________
12. His arrival was totally unexpected.
___________________________________________

C. Combine the following sentences using noun clauses.

1. I don’t know. What kind of a battery is this?
___________________________________________
2. My brother doesn’t know. How tall is he?
___________________________________________
3. I am not sure. Which chapter are we on?
___________________________________________
4. Scientists claim. They will cure cancer.
___________________________________________
5. Who is she? No one knows.
___________________________________________
6. Could you tell me? Where is the bookstore?
___________________________________________
7. Do you remember? How long did we stay in Qatar?
___________________________________________
8. Please tell us… What did you hear?
___________________________________________

D. Complete the following sentences by adding suitable noun clauses. The first one has been done for you.

1. I cannot understand why she is angry with me.
2. They said _______________________.
3. I think _______________________.
4. He told me _______________________.
5. hurts me. _______________________.
6. You can eat _______________________.
7. Can you guess _______________________.
8. I feel certain _______________________.
9. I do not know _______________________.
10. No one knows _______________________.