Electricity

Q.No.1. What does an electric circuit mean ?

Answer

An electric circuit means a continuous and closed path of an electric current.

Q.No. 2. Define the unit of current.

Answer

The electric current is expressed by a unit called ampere (A), named after the French scientist, Andre-Marie Ampere . One ampere is constituted by the flow of one coulomb of charge per second, that is, 1 A = 1 C/1 s.

Q.No. 3. Calculate the number of electrons constituting one coulomb of charge.

Answer

We know that

1.6×10-¹⁹ coulomb charge = 1 Electron.

or, 1 coulomb charge = 1÷(1.6×10-¹⁹) Electrons.

or, 1 coulomb charge = 6×10¹⁸ Electrons.

Q.No.1. Name a device that helps to maintain a potential difference across a conductor.

Answer 

Battery or a combination of cells helps to maintain a potential difference across a conductor. 

2. What is meant by saying that the potential difference between two points is 1 V?

Answer 

One volt is the potential difference between two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.

Therefore,

1 volt = 1 joule/1 coulomb

1 volt = 1joule per coulomb. 

1 V = 1 J/C.

3. How much energy is given to each coulomb of charge passing through a 6 V battery?

Answer 

Work done = Potential difference×Charge 

or, Work done = 6×1

or, Work done = 6 joule.

Continue .....

प्रकाश – परावर्तन तथा अपवर्तन

प्रश्न

1. अवतल दर्पण के मुख्य फोकस की परिभाषा लिखिए।

उत्तर

दर्पण पर मुख्य अक्ष के समांतर कुछ किरणें आपतित हो रही हैं। सभी दर्पण की मुख्य अक्ष के एक बिंदु पर मिल रही/प्रतिच्छेदी हैं। यह बिंदु अवतल दर्पण का मुख्य फोकस कहलाता है।

2. एक गोलीय दर्पण की वक्रता त्रिज्या 20  cm है। इसकी फोकस दूरी क्या होगी?

उत्तर

f = R/2

or, f = ±20/2

or, f = ±10 cm.

3. उस दर्पण का नाम बताइए जो बिंब का सीधा तथा आवर्धित प्रतिबिंब बना सके।

उत्तर

अवतल दर्पण

4. हम वाहनों में उत्तल दर्पण को पश्च-दृश्य दर्पण के रूप में वरीयता क्यों देते हैं?

उत्तर

उत्तल दर्पणों को इसलिए भी प्राथमिकता देते हैं, क्योंकि ये सदैव सीधा प्रतिबिंब बनाते हैं यद्यपि वह छोटा होता है। इनका दृष्टि-क्षेत्र भी बहुत अधिक है क्योंकि ये बाहर की ओर वक्रित होते हैं।

1. उस उत्तल दर्पण की फोकस दूरी ज्ञात कीजिए जिसकी वक्रता-त्रिज्या  32  cm है।

उत्तर

f = R÷2

or, f = 32÷2

or, f = 16 cm.

2. कोई अवतल दर्पण आपने सामने  10  cm दूरी पर रखे किसी बिंब का तीन गुणा आवर्धित (बड़ा) वास्तविक प्रतिबिंब बनाता है। प्रतिबिंब दर्पण से कितनी दूरी पर है।

उत्तर

दर्पण के अवर्धन सूत्र से

प्रतिबिंब की ऊंचाई/बिंब की ऊंचाई = प्रतिबिंब दूरी/बिंब दूरी

or, प्रतिबिंब दूरी = -3×10

or, प्रतिबिंब दूरी = -30 cm

1. वायु में गमन करती प्रकाश की एक किरण  जल  में तिरछी प्रवेश करती है। क्या प्रकाश किरण अभिलंब की ओर झुकेगी अथवा अभिलंब से दूर हटेगी? बताइए क्यों?

उत्तर

प्रकाश की किरण अभिलंब की ओर झुकेगी क्योंकि प्रकाश की किरण विरल से सघन माध्यम में प्रवेश कर रही है।

2. प्रकाश वायु से  1.50 अपवर्तनांक की काँच की प्लेट में प्रवेश करता है। काँच में प्रकाश की चाल कितनी है? निर्वात में प्रकाश की चाल  3  ×  108  m/s है।

उत्तर

कांच का अपवर्तनंक = हवा में प्रकाश की चाल/कांच में प्रकाश की चाल

or, कांच में प्रकाश की चाल = 3×10⁸/1.5

or, प्रकाश में कांच की चाल = 2×10⁸ m/s

3. सारणी 10.3 से  अधिकतम प्रकाशिक घनत्व के माध्यम को ज्ञात कीजिए। न्यूनतम प्रकाशिक  घनत्व के माध्यम को भी ज्ञात कीजिए।

4. आपको किरोसिन,  तारपीन का तेल तथा जल दिए गए हैं। इनमें से  किसमें प्रकाश सबसे अधिक तीव्र गति से चलता है? सारणी  10.3 में दिए गए आँकड़ों का उपयोग कीजिए।

उत्तर

जल का अपवर्तनांक सबसे कम है अतः जल में प्रकाश सबसे तीव्र गति में चलता है।

5. हीरे  का अपवर्तनांक  2.42 है। इस कथन का क्या अभिप्राय है?

इसका अर्थ है कि वायु में प्रकाश का वेग तथा हीरे में प्रकाश के वेग का अनुपात 2.42 है।

10. Light Reflection and Refraction

QUESTIONS

1. Define the principal focus of a concave mirror.

Answer:

 A number of rays parallel to the principal axis are falling on a concave mirror. They are all meeting/intersecting at a point on the principal axis of the mirror. This point is called the principal focus of the concave mirror.

2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Answer:

Focal length = Radius of curvature ÷ 2

or, Focal length = ±20÷2

or, Focal length = ±10 cm.

3. Name a mirror that can give an erect and enlarged image of an object.

Answer:

Concave Mirror.

4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?

Answer:

Convex mirrors are preferred because they always give an erect, though diminished, image. Also, they have a wider field of view as they are curved outwards. Thus, convex mirrors enable the driver to view much larger area than would be possible with a plane mirror.

1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.

Answer:

Focal length = Radius of curvature = Radius of curvature÷2

or, Focal length = 32÷2

or, Focal length = 16 cm.

2. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?

Answer:

Height of image÷Height of object = -(Image distance÷Object distance)

or, 3÷1 = -(Image distance÷10)

or, Image distance = -(3x10)

or, Image distance = 30 cm.

1. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?

Answer:

The ray of light bends towards the normal.

Water is denser with respect to air.

2. Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 108 m s–1.

Answer:

Refractive index of glass with respect to air or vacuum = Speed of light in vacuum ÷ speed of light in glass.

or, 1.5 = 3×10⁸÷speed of light in glass.

or, Speed of light in glass = 3×10⁸÷1.5.

or, Speed of light in glass = 2×10⁸ m/s Ans.

3. Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.

Answer:

Diamond having highest optical density whereas air with lowest optical density.

4. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 10.3.

Answer:

Water having lowest refractive index out of these three. So, the light travels fastest in the water.

5. The refractive index of diamond is 2.42. What is the meaning of this statement?

Answer:

This means that the ratio of the speed of light in air and the speed of light in diamond is equal to 2.42.

1. Define 1 dioptre of power of a lens.

Answer:

The power of a lens is the reciprocal of its focal length. The unit of power of lens is dioptre when focal length is in metre. 1 dioptre means the focal length is 1 metre.

2. A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.

Answer:

From question it is clear that

v = 50 cm.

Height of the image = Height of the object

So, v÷u = 1

or, u = -50 cm.

As per image formation property

Convex lens forms real and inverted image of eual size at 2f.

So, 2f = 50 cm.

or, f = 50÷2

or, f = 25 cm.

or, f = 25÷100 m.

or, Power of lens = 1÷f

or, Power of lens = 100÷25

or, Power of lens = 4 dioptre.

3. Find the power of a concave lens of focal length 2 m.

Answer:

From question it is clear that

f = -2 m.

Power of lens = 1÷f

Power of lens = 1÷(-2)

or, Power of lens = -0.5 dioptre.

EXERCISES

1. Which one of the following materials cannot be used to make a lens?

(a) Water (b) Glass (c) Plastic (d) Clay

Answer

Clay cannot be used to make a lens.

2. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?

(a) Between the principal focus and the centre of curvature

(b) At the centre of curvature

(c) Beyond the centre of curvature

(d) Between the pole of the mirror and its principal focus.

Answer

The correct option will be (d)

3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?

(a) At the principal focus of the lens

(b) At twice the focal length

(c) At infinity

(d) Between the optical centre of the lens and its principal focus.

Answer

The correct option will be (b)

4. A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be

(a) both concave.

(b) both convex.

(c) the mirror is concave and the lens is convex.

(d) the mirror is convex, but the lens is concave.

Answer

The correct option will be (a)

5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

(a) only plane.

(b) only concave.

(c) only convex.

(d) either plane or convex.

Answer

The correct option will be (d)

6. Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

(a) A convex lens of focal length 50 cm.

(b) A concave lens of focal length 50 cm.

(c) A convex lens of focal length 5 cm.

(d) A concave lens of focal length 5 cm.

Answer

The correct option will be (b)

7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

Answer

The range of distance should be less than 15 cm.

The nature of image will be virtual.

The image is larger than object.

The ray diagram is given below

8. Name the type of mirror used in the following situations.

(a) Headlights of a car.

(b) Side/rear-view mirror of a vehicle.

(c) Solar furnace.

Support your answer with reason.

Answer

(a)

Concave mirror is used for headlights of car because it converges the ray of light in its focal length.

(b)

Convex mirror is used for side/rear-view mirror of a vehicle because it has large view range and erract and virtual image is formed by it.

9. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Answer

This lens will produce a complete image when object is placed between optical centre and main focus of it.

10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Answer

From question it is clear that

u = -25 cm.

Length of object = 5 cm

f = 10 cm.

From lens formula we know that

1/v = 1/f + 1/u

1/v = 1/10 + 1/-25

or, 1/v = (5 - 2)÷50

or, 1/v = 3/50

or, v = 50/3 cm.

Length of image = Length of object×v÷u

or, Length of image = 5×50÷(25×3)

or, Length of image = 10/3 cm.

11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Answer

From question it is clear that

f = -15 cm

v = -10 cm

u = ?

1/u = 1/v - 1/f

or, 1/u = -1/10 + 1/15

or, 1/u = (-3+ 2)/30

or, 1/u = -1/30

u = -30 cm.

12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Answer

From question it is clear that

u = -10 cm

f = 15 cm

v = ?

From mirror formula

1/v = 1/f - 1/u

or, 1/v = 1/10 + 1/15

or, 1/v = (3 + 2)/30

or, 1/v = 5/30

or, 1/v = 1/6

v = 6 cm.

Nature of image will be virtual.

13. The magnification produced by a plane mirror is +1. What does this mean?

Answer

Plane mirror makes virtual image of equal height.

14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Answer

From question it is clear that

u = -20 cm

f = 30/2 = 15 cm

v = ?

From mirror formula

1/v = 1/f - 1/u

or, 1/v = 1/15 + 1/20

or, 1/v = 7/60

or, v = 60/7 cm.

Image will be virtual.

Size of image = Size of object × v/u

or, Size of image = 5×60÷20×7

or, Size of image = 15/7 cm.

15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.

Answer

Size of object = 7 cm

From mirror formula

1/v = 1/f - 1/u

or, 1/v = -1/18 + 1/27

or, 1/v = (-3+2)/54

or, 1/v = -1/54

or, v = -54 cm

Nature of image will be real.

Size of image = Size of object×image distance÷object distance

or, Size of image = 7×54÷27

or, Size of image = 14 cm.

16. Find the focal length of a lens of power – 2.0 D. What type of lens is this?

Answer

Focal length = 1/Power of lens

or, f = 1/-2

or, f = -0.5 m

This is type of a concave lens.

17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Answer

f = 1/P

or, f = 1/1.5

or, 0.66 m

The prescribed lens is converging.

10. Circles

Learning Points

1.Tangent

2. Point of contact

3. Theorem 10.1

4. Theorem 10.2

Tangent

A line which passes through a certain point of a circle is said to be tangent whereas the point is said to be point of contact.

Theorem 10.1

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given

There is a circle of centre O. A tangent XY passes through the point P. OP is the radius through the point of contact P.

Prove that

XY perpendicular on OP.

Construction

Take a point Q on XY.

Proof

We know that the shortest distance between a point and line is perpendicular to the line.

For all points on XY, OP is the shortest distance between O and XY.

So, XY is perpendicular on the radius OP. 

Proved.

Theorem 10.2

The lengths of tangents drawn from an external point to a circle are equal.

Solution:

Given:

PQ and PR are the tangents from the external point P.

To Prove that:

PQ = PR

Construction:

Join O to P, Q and R.

Proof:

Consider in ∆OPQ and ∆OPR.

OP = OP (Common)

OQ = OR (Radius of same circle)

<OQP = <ORP (Each right angle)

∆OPQ ≈ ∆OPR (RHS)

Therefore,

PQ = PR Proved.

12. Areas Related to the Circle Class 10

Exercise 12.1

Unless stated otherwise, use π = 22/7

1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Solution:

Circumference of new circle = 2π(AC + AB)

= 2π(19+9)

= 2π×28

Radius of new circle = Circumference of new circle ÷2π

Radius of new circle = 2π×28÷2π

Radius of circle = 28 cm. Ans.

2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Solution:

Area of new circle = π(AB² + AC²)

Now putting the value of AB and AC

Area of new circle = π(6² + 8²)

Area of new circle = π(36+64)

Area of new circle = π×100

Radius of new circle = √Area of new circle ÷π

Radius of new circle = √2π×100÷π

Radius of new circle = √100

Radius of new circle = 10 cm Ans.

3. Fig. 12.3 depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Solution:

Radius of Gold scoing= 21/2 cm. = 10.5 cm

Area of Gold scoring = π(10.5)2

or, Area of Gold scoring = 22×10.5×10.5÷7

or, Area of Gold scoring = 346.5 cm2 Ans.

Radius of Red scoring = 10.5+10.5 = 21 cm.

Area of Red scoring = π(21)2

or, Area of Red scoring = 22×441÷7

or, Area of Red scoring = 1386

Radius of Blue scoring = 21 + 10.5 = 31.5 cm.

Area of Blue scoring = π(31.5)2

Area of Blue scoring = 22×31.5×31.5÷7

or, Area of Blue scoring = 3118.5 cm2 Ans.

Radius of Black scoring = 31.5 + 10.5 = 42 cm.

Area of Black scoring = π(42)2

Area of Black scoring = 22×42×42÷7

= 22×42×6 = 132×42 = 5544 cm2

Radius of white scoring = 42 + 10.5 = 52.5 cm.

Area of White scoring = π(52.5)2

or, Area of White scoring = 22×52.5×52.5÷7 = 22×52.5×7.5 = 8662.5 cm2

4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Solution:

Distance covered by car in 10 minutes = speed×time

or, Distance = 60×10/60.   (1 minute = 1/60 hour)

Distance = 11 km.

or, Distance = 11×1000×100 cm.

Circumference of the wheel = π×Diameter

or, Circumference = 80×22/7

No. of revolutions done by wheel = Distance/Circumference

or, No. of revolution = 11×1000×10×7÷80×22

No. of revolution = 4375 Ans.

5. Tick the correct answer in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

(A) 2 units  (B) π units (C) 4 units (D) 7 units

Solution:

From question it is clear that

Circumference of the circle = Area of the circle

2πr = πr2

2 = r

r = 2 unit.

Hence correct answer or option = (A)

Exercise 12.2

Unless stated otherwise, use π = 22/7

1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

From question it is clear that

Radius of the circle = r = 6 cm.

Central Angle = θ = 60°

We know that

Area of sector of a circle = πr²θ÷360°

Now putting the value of r, θ and π

Area of sector = 22×6×6×60°÷360°×7

= 22×6÷7

= 132÷7

= 18.88 cm² Ans.

2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Solution:

From question it is clear that

2πr = 22 cm

r = 22÷2π

r = 22×7÷22×2

= 3.5 cm.

θ = 90°

Area of the quadrant = πr²θ÷360°

Now putting the value of π, r and θ

Area of the quadrant = 22×3.5×3.5×90°÷360°

= 11.5×3.5×3.5

= 140.875 cm² Ans.

3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Solution:

From question it is clear that

Angle subtended by minute hand in 5 minutes = 360°×5÷60°

= 30°

Length of minute hand = r = 14 cm.

Area swept by minute hand in 5 minutes = πr²θ÷360

Now putting the value of r, π and θ

Area swept by minute hand in 5 minutes = 22×14×14×30°÷360°

= 359.33 cm² Ans.

4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14)

From question it is clear that

AB = AC = r = 10 cm.

θ = 90°

Area of sector = πr²θ÷360°

Now putting the value of r, π and θ.

Area of sector = 3.14×10×10×90°÷360°

= 314÷4

= 78.57 cm²

Area of right triangle ABC = AB×AC÷2

= 10×10÷2

= 50 cm²

Area of minor segment = Area of corresponding sector - Area of corresponding triangle = 78.57 - 50 = 28.57 cm² Ans.

Area of Circle = πr²

or, Area of Circle = 3.14×10×10

= 314 cm²

Area of major sector = Area of Circle - Area of minor sector

or, Area of major sector = 314 - 78.57 = 235.43 cm² Ans.

5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

(i) the length of the arc (ii) area of the sector formed by the arc

(iii) area of the segment formed by the corresponding chord 

Solution:

From question it is clear that

r = 21 cm.

θ = 60°

Length of arc = Circumference of the circle×θ÷360°

= 2πr×60°÷360° = 2×22×21×60°÷360°×7 = 22 cm.

Area of the sector = πr²θ÷360° = 22×21×21×60°÷360°×7 = 231 cm².

Area of segment = Area of the sector-Area of corresponding triangle = 231 -√3×21×21÷4 = 231-190.953 = 40.047 cm².

6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.

(Use π = 3.14 and √3= 1.73)

Solution:

From question it is clear that

AB = AC = r = 15 cm.

θ = 60°

Triangle inscribed with centre will be equilateral triangle.

 Area of sector = πr²θ÷360°

= 3.14×15×15×60°÷360°

= 1.57×15×5

= 117.75 cm²

Area of corresponding triangle = √3r²÷4

= √3×15×15÷4

= 1.73×225÷4

= 97.3125

Area of minor segment = 117.75 - 97.3125 = 20.4375 cm²

Area of the Circle = πr²

3.14×15×15 = 706.5 cm²

Area of major segment = Area of the Circle - Area of minor segment = 706.5 - 20.4375 = 686.0625 cm².

7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.

(Use π = 3.14 and √3= 1.73)

Solution:

From question it is clear that

r = AB = AC = 12 cm.

θ = 120°

Consider in right triangle ADB.

AD = AB×cos60°

AD = 12×1÷2

or, AD = 6.

Similarly,

BD = AB×sin60°

or, BD = 12×√3÷2

or, BD = 6×√3

BC = 2×BD = 12√3.

Area of sector ABC = πr²θ÷360°

Now putting the value

or, Area of sector ABC = 3.14×12²×120°÷360°

or, Area of sector ABC = 150.72 cm².

Area of triangle ABC = AD×BC÷2

or, Area of triangle ABC = 6×12×√3÷2

or, Area of triangle ABC = 62.28 cm².

Area of the corresponding segment of the Circle = Area of sector ABC - Area of triangle ABC

or, Area of the corresponding segment of the Circle = 150.72 - 62.28 = 88.44 cm² Ans.

8. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig. 12.11). Find

(i) the area of that part of the field in which the horse can graze.

Fig. 12.11

(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)

Solution:

From question it is clear that

(i) 

r = 5 m.

θ = 90°

Grazing area of the field = πr²θ÷360°

or, Grazing area of the field = 3.14×5×5×90°÷360°

or, Grazing area of the field = 19.625 m².

(ii)

When

r = 10 m.

Grazing area of the field = πr²θ÷360°

or, Grazing area of the field = 3.14×10×10×90°÷360°

or, Grazing area of the field = 78.5 cm²

The increase in the grazing area = 78.5 - 19.625 = 58.875 m².

9. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 12.12. Find :

(i) the total length of the silver wire required.

(ii) the area of each sector of the brooch.

Fig. 12.12

Solution:

(i)

Total length of the silver wire = π×diameter + 5×diameter

or, total length of the wire = (22×35÷7) + 5×35

or, total length of the wire = 110 + 175 

or, total length of the wire = 285 mm.

(ii)

The area of each sector of the brooch = π(d/2)²(360°/10)÷360°

or, The area of the sector of the brooch = 22×35×35×360°÷7×2×2×10×360°

or, The area of the sector of the brooch = 96.25 mm².

10. An umbrella has 8 ribs which are equally spaced (see Fig. 12.13). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Fig. 12.13

Solution:

From question it is clear that

r = 45 cm.

θ = 360°÷8 = 45°

Area between two consecutive ribs = πr²θ÷360°

or, Area between two consecutive ribs = 3.14×45²×45°÷360°

or, Area between two consecutive ribs = 3.14×45×45÷8

or, Area between two consecutive ribs = 794.8125 cm² Ans.

11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Solution:

From question it is clear that

r = 25 cm.

θ = 115°

Total area cleaned at each sweep of the blades = 2×πr²θ÷360°

or, Total area cleaned at each sweep of the blades = 2×3.14×25×25×115°÷360°

or, Total area cleaned at each sweep of the blades = 1,253.82 cm² Ans.

12. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)

Solution:

From question it is clear that

r = 16.5 km.

θ = 80°

The area of the sea over which the ships are warned = πr²θ÷360°

or, The area of the sea over which the ships are warned = 3.14×16.5²×80°÷360°

or, The area of the sea over which the ships are warned = 3.14×16.5×16.5×2÷9

or, The area of the sea over which the ships are warned = 189.97 km² Ans.

13. A round table cover has six equal designs as shown in Fig. 12.14. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of

 0.35 per cm². (Use √3 = 1.7)

Fig. 12.14

From question it is clear that

r = 28 cm.

θ = 360°÷6 = 60°

Area of the sector = πr²θ÷360°

or, Area of the sector = 22×28²×60°÷360°×7

or, Area of the sector = 22×4×28÷6

or, Area of the sector = 410.67 cm²

Area of the corresponding equilateral triangle = √3×28²÷4 (side = r = 28 cm)

or, Area of the corresponding equilateral triangle = √3×7×28

or, Area of the corresponding equilateral triangle = 1.7×7×28

or, Area of the corresponding equilateral triangle = 332.2 cm²

Area of design = Area of the sector - Area of the corresponding equilateral triangle

or, Area of design = Area of the sector - Area of the corresponding equilateral triangle

or, Area of design = 410.67 - 332.2

or, Area of design = 78.47 cm².

Total area of design = 6×78.47

Cost of design = Rs. 0.35×6×78.47

or, Cost of design = Rs. 164.79 Ans.

14. Tick the correct answer in the following :

Area of a sector of angle p (in degrees) of a circle with radius R is

Solution:

The correct answer is (D)

Which is just equal to πr²θ÷360°