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 Correct answer is (b) volume

Water has more density than the cooking oil.

Correct answer is (b) 1000

Correct answer is (d) area

A physical balance is more sensitive than a beam balance True.

Verifying Euler’s formula

F + V – E 

= 12 + 6 – 16 

= 18 – 16 

= 2

Yes, the polyhedron can have F = 12, E = 16 and V = 6

(a) 2 + 5 = 6 + 1 = 3 + 4 = 7

∴ It can be a die.

(b) It is a cuboid

Euler’s formula is given by F + V – E

(i) F = 4 ; V = 4; E = 6

F + V – E = 4 + 4 – 6 = 8 – 6

F + V – E = 2

∴ Euler’s formula is satisfied.

(ii) F = 10; V = 6; E = 12

F + V – E = 10 + 6 – 12

= 16 – 12 = 4 ≠ 2

∴ Euler’s formula is not satisfied.

(iii) F = 12 ; V = 20 ; E = 30

F + V – E = 12 + 20 – 30

= 32 – 30 = 2

∴ Euler’s formula is satisfied.

(iv) F = 20 ; V = 13 ; E = 30

F + V – E = 20 + 13 – 30

= 33 – 30 = 3 ≠ 2

∴ Euler’s formula is not satisfied.

(v) F = 32 ; V = 60 ; E = 90

F + V – E = 32 + 60 – 90

= 92 – 90 = 2

∴ Euler’s formula is satisfied.

Length of the arc = θ°/360° × 2πr units

Given central angle 0 = 60°

Radius of the sector r = 42 cm

l = 60°/360° × 2 × 227 × 42 = 44 cm

∴ Length of the arc = 44 cm

By Euler’s formula F + V -  E = 2 fora polyhedron.

Here F = 12, V = 17, E = 22 

F + V – E = 12 + 17 – 22 

= 29 – 22 

= 7 ≠ 2 

∴ The polyhedron cannot have 12 faces 22 edges and 17 vertices.

Area of the card = Area of the rectangle + area of 2 semicircles

Length of the rectangle l = 30 cm

Breadth b = 21 cm

Radius of the semicircle = 21/27

∴ Area of the card

= (l × b) + (2 × 1/2 πr²) sq. units

= 30 × 21 + 22/7 × 21/2 × 21/2 cm² 

= 630 + 346.5

= 976.5 cm² (approximately)

∴ Area of the Invitation card = 976.5 cm²

(i) π 

(ii) chord 

(iii) diameter 

(iv) 12 cm 

(v) an arc

Number of equal sectors ‘n’ = 8

Radius of the sector ‘r’ = 56 cm

Area of the each sector = 1/n πr² sq. units

= 1/8 × 22/7 × 56 × 56 cm² = 1232 cm²

Area of each sector = 1232 cm² (approximately)

Area of the figure = Area of 2 parallelograms with base 8 cm and height 3 cm

= 2 × (bh) sq. units

= 2 × 8 × 3 cm² = 48 cm²

∴ Area of the given figure = 48 cm²

(i) Area = Area of 6 squares of side 4 cm

= 6 × a² sq. units

= 6 × 4 × 4 cm²

= 96 cm²

(ii) Area = Area of 2 rectangles of

l = 10, b = 6 + Area of 2 rectangles of l = 6, b = 4 +

Area of 2 rectangles of l = 10,b = 4

= (10 × 6) + (6 × 4) + (10 × 4) cm²

= 60 + 24 + 40 cm²

= 124 cm²

Area of the combined shape = Area of the rectangle + Area of 2 triangles

Length of the rectangle l = 10 cm

Breadth b = 8 cm

Base of the triangle base = 8 cm

Height h = 6 cm

∴ Area of the shape = (l × b) + (2 × 1/2 × base × h) cm²

= (10 × 8) + (8 × 6) cm² 

= 80 + 48 cm² 

= 128 cm²

Area of the given shape = 128 cm²

Area of the glass painting = Area of the square + Area of the triangle

Side of the square a = 30 cm

Base of the triangle b = 30 cm

Height of the triangle h = 8 cm

∴ Area of the painting = (a)² + (12 bh) sq. units

= (30 × 30) + (12 × 30 × 8) cm² 

= 900 + 120 cm²

= 1020 cm²

Area of the glass painting = 1020 cm²

(i) length, breadth and height

(ii) vertex

(iii) six

(iv) circle

(v) cube

Area of the doormat = Area of 2 trapezium

Height of the trapezium h = 702 cm;

a = 90 cm; b = 70 cm

∴ Area of the trapezium

= 12h (a + b) sq. units

Area of the door mat

= 2 × 12 × 35 (90 + 70) cm²

= 35 × 160 cm² 

= 5600 cm²

∴ Area of the door mat = 5600 cm²

Area = Area of a rectangle + Area of a triangle + Area of a trapezium

For rectangle length l = 120 – 20 – 20 cm = 80 cm

Breadth b = 30 cm

For the triangle base = 30 cm

Height = 20 cm

For the trapezium height h = 20 cm

Parallel sided a = 50 cm

b = 30 cm

∴ Area of the figure = (l × b) + (12 × base × height) + 12 × h × (a + b) sq. units

= (80 × 30) + ( 12 × 30 × 20) + 12 × 20 × (50 + 30) cm²

= 2400 + 300 + 800 cm² 

= 3500 cm²

Area of the figure = 3500 cm²

(i) Cube – dice, building blocks, jewel box.

(ii) Cuboid – books, bricks, containers.

(iii) Cylinder – candles, electric tube, water pipe.

(iv) Cone – Funnel, cap, ice cream cone

(v) Sphere – ball, beads, lemon.

For a regular polygon all sides and all the angles must be equal. But in a rhombus all the
sides are equal. But all the angles are not equal

∴ It is not a regular polygon.

Area = 6 × Area of a square of side 6 cm

= 6 × (6 × 6) cm²

= 216 cm²

(ii) Area = Area of 2 rectangles of side (8 × 6) cm² + Area of 2 rectangles of side (8 × 4) cm² + Area of 2 rectangles of side (6 × 4) cm²

= (8 × 6) + (8 × 4) + (6 × 4)cm²

= 48 + 32 + 24 cm²

= 104 cm²

(i) Area of the sector

A = lr/2 sq. units

l = 48 m

r = 10 m

= (48 × 10)/2 m²

= 24 × 10 m²

= 240 m²

Area of the sector = 240 m²

(ii) length of the arc l = 12.5 cm

Radius r = 6 cm

Area of the sector

A = lr/2 sq. units

= (12.5×6)/2

= 12.5 × 3 cm² 

= 37.5 cm²

Area of the sector = 37.5 cm²

(iii) length of the arc l = 50 cm

Radius r = 13.5 cm

Area of the sector

A = lr/2 sq. units

= (50×13.5)/2

= 25 × 13.5 cm²

= 337.5 cm²

Area of the sector = 337.5 cm²

Area of the house = Area of a square of side 6 cm + Area of a rectangle with l = 8cm, h = 6 cm + Area of a ∆ with b = 6 cm and h = 4 cm + Area of a parallelogram with b = 8 cm, h = 4 cm

= (side × side) + (l × b) + (12 × b × h) + 6h cm²

= (6 × 6) + (8 × 6) + (12 × 6 × 4) + (8 × 4) cm²

= 36 + 48 + 12+ 32 cm²

Required Area = 128 cm²

No, the equal parts are not sectors. Because a sector is a plane surface that is enclosed between two radii and the circular arc of the circle. Here the boundaries are not radii.

Radius of the sector r = 4.2 cm

Area of the sector = 9.24 cm²

lr/2 = 9.24

(l×4.2)/2 = 9.24

l × 2.1 = 9.24

l = 9.24/2.1

l = 4.4 cm

Perimeter of the sector = 1 + 2r units 

= 4.4 + 2(4.2) cm

= 4.4 + 8.4 cm 

= 12. 8 cm

Perimeter of the sector = 12.8 cm

Length of the arc of the sector l = 44 cm

Perimeter of the sector P = 64 cm

l + 2r = 64 cm

44 + 2 r = 64 .

2 r = 64 – 44

2 r = 20

r = 20/2 = 10 cm²

Area of the sector = lr/2 sq. units 

= (44×10)/2 cm² 

= 22 × 10 cm² 

= 220 cm²

Area of the sector = 220 cm²

Given Radius of the sector = 10 cm

Perimeter of the sector P = 45 cm

l + 2r = 45

l + 2(10) = 45

l + 20 = 45

l = 45 – 20

l = 25 cm

Length of the arc l = 25 cm

Data: m = 500 g, c = 0.1 cal/(g.°C), T’= 100 °C, m₁ = 100 g, c₁ = 1 cal/(g.°C), T₁ = 20°C, m₂ = 100 g, c₂ = 0.1 cal/(g.°C), T₂ = 20 °C, T= ?

Heat lost by the sphere = heat gained by the water and the calorimeter.

∴ mc (T’ – T) = m₁ c₁ (T – T₁ ) + m₂ c₂ (T – T₂ )

∴ 500 g × o.l cal/(g.°C) × (100 °C – T)

= 100 g × 1 cal/(g.°C) × (T – 20 °C) + 100 g × 0.1 cal/(g.°C) × (T – 20 °C)

∴ 50 (100 °C – T) = 100 × (T – 20 °C) + 10 × (T – 20 °C)

∴ 50 (100 °C – T) = 110 × (T – 20 °C) 

∴ 500 °C – 5T = 11T – 220 °C 

∴ 16T = 720 °C

∴ T = \(\cfrac{720^\circ C}{16}\) = 45 °C

Maximum temperature of the mixture = 45 °C.

Correct answer is (c) FPS

Correct answer is (c) FPS

Because mercury is harmful for us, it has been replaced by alcohol in a thermometer. 

[Notes: (i) Mercury thermometers are still widely used in laboratories in schools and colleges. (ii) A thermometer is a device to measure temperature. A thermometer containing mercury in its bulb is called a mercury thermometer. There are other types of thermometer such as a thermocouple thermometer, a platinum resistance thermometer, a thermistor thermometer, etc.]

1. True

2. False - Temperature is a measure of average kinetic energy of the particle in a system.

3. False - In thermometers, boiling point of water is taken as the upper Fixed point.

4. False - One coulomb of charge flowing per second is called 'ampere'.

5. True.

6.True.

7. True

8. False - Atomic clocks are used in GPS Devices.

9. False - Candela is used to express luminous intensity.

10. True

Lumen is the SI unit of luminous flux.

The measure of the power of the emitted light, by a light source in a particular direction, per unit solid angle is called as luminous intensity.

Correct answer is (c) visible light

Accuracy in Measurements :

1. Measurement is the base of all experiments in science and technology. The value of every measurement contains some uncertainty. These uncertainties are called as 'Errors'.

2. The difference between the real value and the observed value is called an error.

• Accuracy : Accuracy is the closeness of a measured value to the actual value or true value.
• Precision : Precision is the closeness of two or more measurements to each other.

• No. It is not possible of 100°C fever. The normal temperature of human body is between 98.4°F and 98.6°F. 
• So, he should say that, he was affected by a fever of 100°F and it is not 100°C.

(i) A thermometer is used to measure temperature.

(ii) The apparatus used to measure heat is called a calorimeter.

(iii) Temperature is the measure of the average kinetic energy of the atoms in a substance.

(iv) The heat contained in a substance is the measure of the total kinetic energy of the atoms in the substance.

(i) False. (In general, objects expand on heating. There are some exceptions to this, you will learn about them in Standard X.)

(ii) False. (Atoms of a solid are bound to each other due to the forces acting between them.)

(iii) False. (The average kinetic energy of atoms in a hot object is more than the average kinetic energy of atoms in a cold object.)

Length of the field l = 76 m

Breadth of the field b = 60 m

Area of the field A = l x b sq. units 

= 76 x 60 m²

Area of the field A = 4560 m²

Length of the rope = 35 m

Radius of the land that the cow can graze = 35 m

Area of the land tha the cow can graze = circle of radius 35 m 

= πr² sq.units

π x 35 x 35 m² = \(\frac{22}{7}\) x 35 x 35 m²

= 3850 m²

Area of the land the cow cannot graze = Area of the field – Area

that the cow can graze

= (4560 – 3860) m² 

= 710 m²

Area of the land that the cow cannot graze = 710 m²

Let the length of the rectangular field = ‘L’ m

Breadth of the rectangular field = ‘B’ m

Area of the rectangular field = (L x B) m²

Also given length = 3 x Breadth

L = 3B

Width of the path (W) = 5m

Length of the inner rectangle = L – 2W 

= l – 2(5)

= 3B – 10m

Breadth of the inner rectangle = B – 2W

= B – 2(5)

= B – 10 m

Area of the inner rectangle = (3B – 10) (B – 10)

= 3B² – 10B – 30B + 100

Area of the path = Area of outer rectangle – Area of inner rectangle

= (L x B) – (3B² – 10B – 30B + 100)

3B x B – (3B² – 40B + 100)

= 3B² – 3B² + 40B – 100

Area of the path = 40B – 100

Given area of the path = 500 m²

40B – 100 = 50

40B = 500 + 100 = 600

B = \(\frac{600}{40}\)

B = 15m

Length of the field = 45 m; 

Breadth of the field = 15 m

Given outer radius R = 16 cm

Area of the circular pathway = πR² = πr²

Area of the circular pathway = 352 sq. cm

πR² – πr² = 352 cm²

π(R² – r²) = 352

16² – r² = \(\frac{352\times7}{22}\)

16² – r² = 16 x 7

16² – r² = 112

16² – 112 = r²

r² = (256 – 112) cm²

r² = 144 cm²

r = 12 cm

Inner radius r = 12 cm

Radius of the circular hall R = 120 cm 

Radius of the circular carpet r = 106 cm

Area of the hall uncovered = Area of the hall – Area of the carpet

= π(R² – r²) cm² 

= \(\frac{22}{7}\) x (120² – 106²) cm²

= \(\frac{22}{7}\) x (120 + 106) x (120 – 106) cm²

= \(\frac{22}{7}\) x 226 x 14 cm² 

= 9,944 cm² 

Area of the hall uncovered = 9, 944 cm²

If the bull is tethered by a rope then the area it can graze is a circular area of radius 

= length of the rope 

Area of the circle = 2464 m² 

πr² = 2464 m² 

\(\frac{22}{7}\) x r² = 2464 

r² = 2464 x \(\frac{7}{22}\)

= 122 x 7 

= 16 x 7 x 7 

r² = 42 x 72 

r² = 4 x 7 

= 28 m 

Length of the rope r = 28 m

Area of the outer rectangle = 48 sq.cm 

Area of the inner rectangle = 15 sq.cm 

Area of the rectangular pathway = Area of the outer rectangle – Area of the inner rectangle

= (48 – 15) cm² 

= 33 cm²

Area of the outer circle = 1386 m²

πR² = 1386m²

Area of the inner circle = 616 m²

πr² = 616m²

Area of the path = Area of outer circle – Area of the inner circle

1386 m² – 616 m²

Area of the path = 770m²

Also πR² = 1386

R² = \(\frac{1386\times7}{22}\)

R² = 63 x 7

R² = 9 x 7 x 7

R² = 3² x 7²

R = 3 x 7

Outer Radius R = 21 m

Again πr² = 616

\(\frac{22}{7}\) x r² = 616

r² = 28 x 7 

r² = 4 x 7 x 7 

r² = 2² x 7² 

r = 2 x 7 

Inner radius r = 14m 

Width of the path = Outer radius – Inner radius 

= 21 – 14 

Width of the path = 7m

Area of the rectangular garden L x B = 11 m x 8 m 

= 88 m²

Length of the inner rectangle L = L – 2 

W = 11 – 2(2) 

= 11 – 4 

= 7 m 

Breadth of the inner rectangle b = B – 2W 

= 8 – 2(2) 

= 8 – 4 

= 4 m 

Area of the inner rectangle = l x b sq. units 

= 7 x 4 m² 

= 28 m² 

Area of the path = Area of the outer rectangular garden – Area of the inner rectangle 

= 88 m² – 28 m² 

= 60 m² 

Area of the path = 60 m²