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Correct answer is

(b) condensation

Correct answer is

(b) Glyptal

(a) As the poet is with her mother in a speeding car, the tree on the roadside seems to be sprinting in the opposite direction.

(b) After the security check, the poet looked at her mother again as she was going to part from her and wanted to reassure herself before her departure.

(c) She observed her mother’s pale and lifeless face which seemed dull and colourless like the late winter’s moon.

(d) Personification; Trees sprinting.

Trees of Eugenia, Michelia, Rodenadrasa are found in the high altitude of mountainous regions.

The Autumn Forests are found in the regions where the annual average rainfall is 100 – 200 cm.

Correct answer is (a) 1, -6

f(x) = ax³ + bx² + 11x – 6 satisfies the conditions of Rolle’s theorem in [1, 3]

∴ f(1) = f(3)

a(1)³ + b(1)² + 11(1) – 6 = a(3)³ + b(3)³ + 11(3) – 6 

a + b + 11 = 27a + 9b + 33 

26a + 8b = -22 

13a + 4b = -11 

Only a = 1, b = -6 satisfy this equation.

Government Policy is to make 33 per cent of Indian land forested.

(c) 740, 261 sq. km

(c) 80 per cent

(c) 22.02 per cent

The given pair of equations is

\(\frac{x}4+\frac{y}3= -\frac{15}{12}\)........(i) 

\(\frac{x}2+y = 1\)......(ii)

 Multiplying (i) by 12 and (ii) by 4, we have 

3x + 4y = 5 ……….(iii) 

2x + 4y = 4 ………(iv) 

Now, subtracting (iv) from (iii), we get 

x = 1 

Putting x = 1 in (iv), we have 

2 + 4y =4 

⇒ 4y = 2 

⇒ y = 1/2 

∴ x + y = 1 + 1/2 = 3/2 

Hence, the value of x + y is 3/2 .

(a) 1 cm

Let the internal radius of the tube be r cm. 

External radius = \(\frac82\) cm = 4 cm, 

Height = 20 cm 

∴ Volume of iron used in the tube = π(4² - r²) x 20 cm

Given, π x (16 - r²) x 20 = 440

⇒ \(\frac{22}{7}\times20\times(16-r^2) = 440\)

⇒ 16 – r² = 7 ⇒ r² = 16 – 7 = 9 ⇒ r = 3

∴ Thickness of tube = 4 cm – 3 cm = 1 cm.

(c) 44.0625 cm

Let h be the length (or height) of the solid cylinder 2 cm in diameter.

∴ Volume of solid cylinder = π (1)² h = πh

External radius of hollow cylinder = 6 cm Internal radius of hollow cylinder = 6 cm – 0.25 cm = 5.75 cm 

∴ Volume of the hollow cylinder = π(6² - 5.75²) x 15 

Given,π(6² - 5.75²) x 15 = πh

⇒ (6 + 5.75) (6 – 5.75) × 15 = h 

⇒ h = 11.75 × 0.25 × 15 = 44.0625 cm.

Given: Curved surface area = 660 sq.cm, and height = 21 cm 

To find: area of base and radius of a cylinder 

i. Curved surface area of cylinder = 2πrh 

∴ 660 = 2 x (22/7) x r x 21 

∴ 660 = 2 x 22 x r x 3

∴ 660/2 x 22 x 3 = r

660/2 x 66 = r

∴ 5 = r 

i.e., r = 5 cm

ii. Area of a base of the cylinder = πr² 

= (22/7) x 5 x 5 

= 550/7

= 78.57 sq.cm 

∴The radius of the cylinder is 5 cm and the area of its base is 78.57 sq.cm.

Given

Length of cuboid (l) = 12 cm

Breadth (b) =9 cm

and Height (h) = 5 cm

Total surface area of cuboid = 2(lb + bh + hl)

= 2[(12 × 9) + (9 × 5) +(5 × 12)]

= 2[108 + 45 + 60]

2 × 213 = 426 cm²

Volume = l × b × h

= 12 × 9 × 5 = 540 cm³

Hence, total surface area of cuboid 426 cm²

And volume of the cuboid 540 cm³

Given:

Length of water tank (l) = 48 m

Breath (b) = 36 m

Height (h) = 28 m

∴ Volume of the water tank = l × b × h

= 48 × 36 × 28

= 48384 m³.

Hence, volume of water tank = 48384 m³

Given,

Radius of hemisphere (r) = 4 cm

Curved surface area = 2πr²

= 2 × π × 4²

= 32π cm²

Volume of the cube with core of 8 cm = (core)³

= 8³ = 512 cm³

Volume of the cube with core 6 cm = (core)³

= (6)³ = 216 cm³.

Volume of the cube with core 1 cm = (core)³

= (1)³ = 1 cm³.

The total volume of three cubes 512 + 216 + 1 = 729 cm³.

Having melted these cubes, a new cube is recasted

∴ The volume of the cube recasted = 729 cm³

⇒ (core)³ = 729

⇒ core = \(\sqrt [ 3 ]{ 729 } \)

= (9³)^1/3 = 9 cm

Total surface area of cuboid recasted =6 (core)²
6 × 9 × 9 = 486 cm².

Hence the surface area of the new cube = 486 cm².

Given,

Diameter of cylindrical pipe = 7 cm

∴ Radius (r) = \(\frac { 7 }{ 2 }\) = 3.5 cm = 0.035 m and the volume of flowing water in one minute = 192.5 liter

∴ The volume of flowing water in one hour 

= 192.5 × 60 liter

= \(\frac { 192.5\times 60 }{ 1000 } \) m³

= 11.55 m³ …….(i)

Let the speed of the water in pipe be x km./h

∴ Length of flowing water in one hour (h) = x m

∴ Volume of water flowing in one hour = πr²h 

= \(\frac { 22 }{ 7 }\) × (0.035)² × x

= 0.00385 × x m³ …….(ii)

From equation (i) and (ii) we get

0.00385 × x = 11.55

⇒ x = \(\frac { 11.55 }{ 0.00385 }\)

⇒ x = \(\frac { 1155000 }{ 385 }\)

⇒ x = 3000 m

= \(\frac { 3000 }{ 1000 }\) km = 3 km

Hence the speed of water in the pipe = 3 km/h

Given,

Diameter of well = 7 m

∴ Radius of well (r) = \(\frac { 7 }{ 2 }\) m

Depth of well (h) = 20m

The volume of the soil taken out from well = πr²h

= \(\frac { 22 }{ 7 }\) × \({ \left( \frac { 7 }{ 2 } \right) }^{ 2 }\) × 20

= 770 cm³.

The length of plate form (L) = 22 m

The breadth of plate form (B) = 14 m

Let the height of plate form be H.

Volume of plate form = L x B x H m³.

= 22 × 14 × H m³

According to question

Volume of plate form = volume of earth digut 22 × 14 × H = 770

H = \(\frac { 770 }{ 22\times 14 }\)

= 2.5 m.

Hence the height of plate form is 2.5 m.

Given,

Volume of cylinder =30π cm³

and base area = 6π cm²

:. Volume of cylinder = base area × height

⇒ 30π = 6π × height

⇒ height = \(\frac { 30\pi }{ 6\pi } \)

= 5 cm.

Hence, the height of the cylinder 5 cm.

Given: Height (h) = 12 cm, length (l) = 13 cm 

To find: Radius of the base of the cone (r) 

Solution:

l² = r² + h² 

∴ 13² = r² + 12² 

∴ 169 = r² + 144 

∴169 – 144 = r² 

∴ r² = 25 

∴ r = √25 … [Taking square root on both sides] 

= 5 cm 

∴ The radius of base of the cone is 5 cm.

Given,

Length of the tank (l) = 4 m

Breadth (b) = 3 m

Depth of water into tank (h) = 2 m

Volume of the water poured into tank = l × b × h

= 4 × 3 × 2

= 24 m³

Hence, 24 m³ water should be poured.

Given: Curved surface area of cylinder = 1980 sq.cm., 

radius (r) = 15 cm 

To find: Height of the cylinder (h)

Solution:

Curved surface area of cylinder = 2πrh 

∴ 1980 = 2 x \(\frac{22}7\) x 15 x h

∴ h = \(\frac{1980\, \times\,7}{2\,\times\,22\,\times\,5}\)

∴ h = 21 cm

∴ The height of the cylinder is 21 cm.

Given: Radius (r) = 20 cm, height (h) = 13 cm 

To find: Curved surface area and the total surface area of the cylinder

 Solution:

i. Curved surface area of cylinder = 2πrh

= 2 x 3.14 x 20 x 13 

= 1632.8 sq.cm 

ii. Total surface area of cylinder = 2πr(r + h) 

= 2 x 3.14 x 20(20 + 13) 

= 2 x 3.14 x 20 x 33 = 4144.8 sq.cm 

∴ The curved surface area and the total surface area of the cylinder are 1632.8 sq.cm and 4144.8 sq.cm respectively.

Given

Length of box (l) = 50 cm

Breadth of box (b) = 36 cm

Height of box (h) = 25 cm

The necessary cloth to make its cover = surface area of the box

= 2(lb + bh + hl)

= 2[(50 × 36) + (36 + 25) + (25 + 50)]

= 2[1800 + 900 + 1250]

= 2 × 3950 = 7900 cm².

Given,

Base radius (r) = 3 cm

Height (h) = 8 cm

Volume of right cylindrical cup = πr²h

= π × (3)² × 8

= 72 π cm³

Volume of water in the cup = \(\frac { 72\pi }{ 2 } \)

= 36 π cm³

i. Given: r = 7 cm and h = 10 cm 

To find: Curved surface area of cylinder and total surface area 

Curved surface area of the cylinder = 2πrh

= 2 x (22/7) x 7 x 10 

= 2 x 22 x 10

= 440 sq.cm 

Total surface area of the cylinder: = 2πr(h + r) 

= 2 x (22/7) x 7(10 + 7) 

= 2 x (22/7) x 7 x 17 

= 2 x 22 x 17 = 748 sq.cm 

The curved surface area of the cylinder is 440 sq.cm and its total surface area is 748 sq.cm

ii. Given: r = 1.4 cm and h = 2.1 cm 

To find: Curved surface area of cylinder and total surface area 

Curved surface area of the cylinder = 2πrh 

= 2 x (22/7) x 1.4 x 2.1 

= 2 x 22 x 0.2 x 2.1 

= 18.48 sq.cm 

Total surface area of the cylinder = 2πr (h + r)

= 2 x (22/7) x 1.4 (2.1 + 1.4) 

= 2 x (22/7) x 1.4 x 3.5 

= 2 x 22 x 0.2 x 3.5 

= 30.80 sq.cm 

∴ The curved surface area of the cylinder is 18.48 sq.cm and its total surface area is 30.80 sq.cm.

iii. Given: r = 2.5 cm and h = 7 cm 

To find: Curved surface area of cylinder and total surface area 

Curved surface area of the cylinder = 2πrh 

= 2 x (22/7) x 2.5 x 7 

= 2 x 22 x 2.5 

= 110 sq.cm 

Total surface area of the cylinder = 2πr(h + r) 

= 2 x (22/7) x 2.5 (7+ 2.5)

= 2 x (22/7) x 2.5 x 9.5 

= 1045/7

= 149.29 sq.cm 

∴ The curved surface area of the cylinder is 110 sq.cm and its total surface area is 149.29 sq.cm.

iv. Given: r = 70 cm and h = 1.4 cm 

To find: Curved surface area of cylinder and total surface area 

Curved surface area of the cylinder = 2πrh 

= 2 x (22/7) x 70 x 1.4 = 2 x 22 x 10 x 1.4 

= 616 sq.cm 

Total surface area of the cylinder = 2πr(h + r) 

= 2 x (22/7) x 70(1.4 + 70) 

= 2 x (22/7) x 70 x 71.4

= 2 x 22 x 10 x 71.4

= 2 x 22 x 714 

= 31416 sq.cm 

∴ The curved surface area of the cylinder is 616 sq.cm and its total surface area is 31416 sq.cm.

v. Given: r = 4.2 cm and h = 14 cm 

To find: Curved surface area of cylinder and total surface area 

Curved surface area of the cylinder = 2πrh 

= 2 x (22/7) x 4.2 x 14 

= 2 x 22 x 4.2 x 2 

= 369.60 sq.cm 

Total surface area of the cylinder = 2πr (h + r)

= 2 x (22/7) x 4.2 (14+ 4.2)

= 2 x (22/7) x 4.2 x 18.2 

= 2 x 22 x 0.6 x 18.2 

= 480.48 sq.cm 

∴ The curved surface area of the cylinder is 369.60 sq.cm and its total surface area is 480.48 sq.cm.

i. Given: r = 10.5 cm and h = 8 cm 

To find: Volume of the cylinder 

Volume of the cylinder = πr²h

= (22/7) x 10.5 x 10.5 x 8 

= 22 x 1.5 x 10.5 x 8 

= 2772 cc 

∴ The volume of the cylinder is 2772 cc.

ii. Given: r = 2.5 m and h = 7 m 

To find: Volume of the cylinder 

Volume of the cylinder = πr²h 

= (22/7) x 2.5 x 2.5 x 7 

= 22 x 2.5 x 2.5 

= 137.5 cu.m

∴ The volume of the cylinder is 137.5 cu.m.

iii. Given: r = 4.2 cm and h = 5 cm 

To find: Volume of the cylinder 

Volume of the cylinder = πr²h 

= (22/7) x 4.2 x 4.2 x 5 = 22 x 0.6 x 4.2 x 5 

= 277.2 cc 

∴ The volume of the cylinder is 277.2 cc.

iv. Given: r = 5.6 cm and h = 5 cm 

To find: Volume of the cylinder 

Volume of the cylinder = πr²h

= (22/7) x 5.6 x 5.6 X 5 7 

= 22 x 0.8 x 5.6 x 5 

= 492.8 cc 

∴ The volume of the cylinder is 492.8 cc.

Given: For cylindrical drum: 

Diameter (d) = 50 cm and height (h) = 45 cm

To find: Total surface area of the cylindrical drum

Diameter (d) = 50 cm 

∴ radius (r) 

= d/2 = 50/2

= 25 cm 

Total surface area of the cylindrical drum = 2πr (h + r) 

= 2 x 3.14 x 25 (45 + 25) 

= 2 x 3.14 x 25 x 70 

= 10,990 sq.cm

∴ The total surface area of the cylindrical drum is 10,990 sq.cm.

Given: interior diameter of the tank (d) = 1.6 m and depth (h) = 0.7 m 

To find: Capacity of the tank interior diameter of the tank (d) = 1.6 m 

∴ Interior radius (r) = d/2 = 16/2 = 0.8 m 

= 0.8 x 100 …[∵ 1m = 100cm] 

= 80cm 

h = 0.7 m = 0.7 x 100 = 70 cm 

Capacity of the tank = Volume of the tank = πr²h

= (22/7) x 80 x 80 x 70 = 22 x 80 x 80 x 10 

= 1408000 cc 

= 1408000/1000

…[∵1 litre = 1000 cc] = 1408 litre 

∴ The tank can hold 1408 litre of water.

Given: For cylindrical rod: length of rod (h) = 90 cm, and diameter (d) = 1.4 cm 

To find: Iron required to make a rod diameter (d) = 1.4 cm 

∴ radius (r) = d/2 = 1.4/2 = 0.7 cm 

Volume of rod = πr²h 

= (22/7) x 0.7 x 0.7 x 90 = 22 x 0.1 x 0.7 x 90 

= 138.60 cc

∴ 138.60 cc of iron is required to make the rod

Given: Circumference of the base of cylinder = 132 cm and height (h) = 25 cm 

To find: Volume of the cylinder 

i. Circumference of base of cylinder = 2πr

∴ 132 = 2 x (22/7) x r

∴ (132 x 7/2 x 22) = r

(6 x 7)/2 = r

∴ 3 x 7 = r 

∴ r = 21 cm 

ii. Volume of the cylinder = πr²h 

= (22/7) x 21 x 21 x 25 = 22 x 3 x 21 x 25 

= 34650 cc 

∴ The volume of the cylinder is 34650 cc.

Given: For a cuboidal tank, 

Length (l) = 10 m, breadth (b) = 6 m, depth (h) = 3 m 

To find: Capacity of the tank and litre of water tank can hold.

i. l = 10m = 10 x 100 …[∵ 1m = 100cm] = 1000 cm, 

b = 6 m = 6 x 100 = 600 cm, 

h = 3 m = 3 x 100 = 300 cm 

Volume of the tank = l x b x h 

= 1000 x 600 x 300 

= 18,00,00,000 cc

ii. Capacity of the tank = Volume of the tank 

= 18,00,00,000 cc = 18,00,00,000/1000 …[∵ 1 litre =1000 cc] = 1,80,000 litre

∴ The capacity of the tank is 18,00,00,000 cc and it can hold 1,80,000 litre of water.

Given: For cylindrical container: diameter (d) = 28 cm, height (h₁) = 20 cm 

For cylindrical lid: height (h₂) = 2 cm 

diameter (d) = 28 cm 

∴ radius (r) = d/2 = 28/2 = 14 cm 

i. Surface area of the cylinder with one side open = Curved surface area + Area of a base 

= 2πrh₁ + πr² 

= πr (2h₁ + r) = (22/7) x 14 x (2 x 20 + 14) 

= 22 x 2 x (40 + 14) 

= 22 x 2 x 54 

= 2376 sq.cm

ii. Area of sheet required to made a lid = Curved surface area of lid + Area of upper surface 

= 2πrh + πr² = πr (2h + r) 

= (22/7) x 14 x (2 x 2 + 14)

= 22 x 2 x (4 + 14) 

= 22 x 2 x 18 

= 792 sq cm 

∴ The area of the sheet required to make the cylindrical container is 2376 sq. cm and the approximate area of a sheet required to make the lid is 792 sq. cm.

Given: For the cuboidal shape brick: 

length (l ) = 25 cm, 

breadth (b ) = 15 cm,

height (h ) = 10 cm 

For the cuboidal shape wall: 

length (l ) = 6 m, 

height (h ) = 2.5 m, breadth (b ) = 0.5 m 

To find: Number of bricks required 

When all the bricks are arranged to build a wall, the volume of all the bricks is equal to volume of wall.

∴ Number of bricks

= volume of the wall/volume of a brick

i. Volume of a brick = l₁ x b₁ x h₁ 

= 25 x 15 x 10 cc 

ii. l₂ = 6m = 6 x 100 …[∵ 1m = 100cm] = 600 cm 

h₂ = 2.5 m = 2.5 x 100 = 250 cm 

b₂ = 0.5 m = 0.5 x 100 = 50 cm

Volume of the wall = l₂ x b₂ x h₂ 

= 600 x 50 x 250 cc

iii. Number of bricks = volume of the wall/volume of a brick

= 600 × 50 × 250/25 × 15 × 10

= 40 x 2 x 25

= 2000 bricks 

∴ 2000 bricks are required to build the wall.

Given: For cuboid shaped box, 

length (l) = 20 cm, breadth (b) = 10.5 cm and height (h) = 8cm 

To find: Volume of a box

Volume of a box = l x b x h

= 20 x 10.5 x 8 

= 1680 cc 

∴ The volume of the box is 1680 cc.

Given: Length of edge of cube (l) = 7.5 cm 

To find: Volume of a cube 

Volume of a cube = l³

= (7.5)³ 

= 421.875 ≈ 421.88 cubic cm 

∴The volume of the cube is 421.88 cubic cm.

Given: Total surface area of cube = 5400 sq.cm. 

To find: Surface area of all vertical faces of the cube 

i. Total surface area of cube = 6l²

∴ 5400 = 6l²

∴ 5400/6 = l² 

∴ l² = 900 

ii. Area of vertical faces of cube = 4l² 

= 4 x 900 = 3600 sq.cm. 

∴ The surface area of all vertical faces of the cube is 3600 sq.cm.

Given: For cuboid shaped soap bar,

length (l) = 10 cm, breadth (b) = 5 cm and volume = 150 cc

To find: Thickness of the soap bar (h)

Volume of soap bar = l x b x h 

∴ 150 = 10 x 5 x h 

∴ 150 = 50 h

∴ \(\cfrac{150}{50}\) = h

∴ 3 = h 

i.e., h = 3 cm 

∴ The thickness of the soap bar is 3 cm.

Given: For cuboid shape box, 

breadth (b) = 6 unit, height (h) = 5 unit Total surface area = 500 sq. unit.

To find: Length of the box (l) 

Solution:

Total surface area of the box = 2 (lb + bh + lh) 

∴ 500 = 2 (6l + 6 x 5 + 5l)

∴ \(\frac{500}{2}\) = (11l + 30) 

∴ 250 = 11l + 30 

∴ 250 – 30= 11l 

∴ 220 = 11l 

∴ 220 = l 

∴\(\frac{220}{11}\) = l 

∴ l = 20 units 

∴ The length of the box is 20 units.

Let ABCD is a rectangle 

AC and BD are the diagonals of rectangle 

In ΔABC and ΔBCD, we have 

AB = CD 

(Opposite sides of rectangle are equal) 

∠ABC = ∠BCD 

(Each equal to 90°) 

BC = BC 

(Common) 

Therefore, 

ΔABC ≅ ΔBCD 

(By SAS congruence criterion) 

AC = BD (c.p.c.t) 

Hence, 

The diagonals of a rectangle are equal.

Correct option is B) square

Correct option is A) rectangle

Correct option is C) Cone

Correct option is C) net

Correct option is A) Parallelogram

Correct option is D) 8

Correct option is C) 8

A ball -Circle 

A cylindrical pipe – Rectangle 

A book – Rectangle