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Similarity between both the figures is that both have the same heights.

The difference between the two figures is that one is a cylinder and the other is a cube.

Lateral surface area of the cube = 4l2 = 4 (7 cm)² = 196 cm²

Lateral surface area of the cylinder = 2πrh = 2 x (22/7) x (7/2) x 7 cm² = 154 cm²

Hence, the cube has larger lateral surface area.

Given,

Ratio of dimensions of cuboid = 5:3:1

Total surface area of cuboid = 414 m²

Let dimensions are = 5x x 3x x x

So,

= 2(lb x bh x hl) = 414

= 2(15x² + 3x² + 5x²) = 414

= 2 x 23x² = 414

= x² = \(\frac{414}{46}\) = 9

= x = \(\sqrt9\) = 3

So,

Dimensions are = 5x = 5 x 3 = 15 m

= 3x = 3 x 3 = 9 m

= x = 3 m

Correct option is: D) 9

Given that radius of base of the cone is r = 7 cm.

Let h be the height of the cone.

Then volume of the cone = \(\frac 13 \pi r^2h\)

\(\therefore\) \(\frac 13 \pi r^2h\) = 462

\(\Rightarrow\) h = \(\frac {462 \times 3 \times 7}{22 \times 7 \times 7}\) (\(\because\) r = 7 cm)

= \(\frac {21\times 3}{7} = 3\times 3 = 9 \, cm\)

Hence, height of the cone is 9 cm

Correct option is: D) 9

Correct option is: (D) 792

We have 

r = 6 cm & h = 7 cm 

\(\therefore\) Volume of cylinder = \(\pi r^2h\) 

= \(\frac {22}7\) \(\times\) 6 \(\times\) 6 \(\times\) 7

= 22 \(\times\) 36

= 792 \(cm^3\)

Correct option is: (D) 792

Correct option is: (B) 9

We have l = length of the cuboid = 8,

b = breadth of the cuboid = 4 and 

h = height of the cuboid = 1

Maximum length of the stick that can be placed in a cuboid  = length of diagonal of cuboid

= \(\sqrt {l^2+b^2+h^2} = \sqrt {8^2+4^2+1^2} \)

\(= \sqrt {64 + 16 +1} = \sqrt {81} = 9\)

Correct option is: (B) 9

Correct option is: (A) 300 cm³

Given that base area of the prism = 30 \(cm^2\) and height of the prism = 10 cm

\(\therefore\) Volume of the prism = Base area of the prism \(\times\) Height of the prism

= 30 \(\times\) 10 = 300 \(cm^3\)

Correct option is: (A) 300 cm³

Correct option is: B) 4 : 4 : √5

Given that 

Radius of cylinder = Radius of cone = Radius of sphere = r

\(\because\) Height of the sphere = Diameter of the sphere = 2 r

\(\therefore\) Height of cylinder = Height of cone  = 2r

Now, curved surface area of sphere \(4 \pi r^2\)

curved surface are of cylinder = \(2 \pi r h\)

= 2\(\pi r (2 r) \) (\(\because\) h = 2r)

= \(4 \pi r^2\) 

curved surface area of cone = \(\pi rl\)

= \(\pi r\sqrt{r^2+h^2}\)

= \(\pi r\sqrt {r^2 + 4r^2}\) (\(\because\) h = 2r)

= \(\sqrt5 \pi r^2\)

Now, the ratio of their curved surface areas

= \((C.S.A)_S : (C.S.A)_{cylinder} : (C.S.A)_{cone}\)

= \(4 \pi r^2\) : \(4 \pi r^2\) : \(\sqrt5 \pi r^2\) 

= 4 : 4 : \(\sqrt5 \) (On dividing by \(\pi r^2\))

Hence, the ratio for their curved surface areas = 4 : 4 : \(\sqrt5 \).

Correct option is: B) 4 : 4 : √5

Correct option is: (B) 400π

Given that radius of the sphere is r = 10 cm

\(\therefore\) Curved surface area of the sphere = 4\(\pi r^2\)

= 4\(\pi\) \(\times\) \(10^2\) = 400 \(\pi\)\(cm^2\) 

Correct option is: (B) 400π

Area of rhombus = \(\frac{1}{2}\)x d₁ x d₂

Area of rhombus = \(\frac{1}{2}\)x 7.5 x 12

Area of rhombus = 6 x 7.5 = 45.0

Hence, area of rhombus is 45 cm²

Area of rhombus = \(\frac{1}{2}\)x d₁ x d₂

240 = \(\frac{1}{2}\)x 16 x d₂

d₂ = \(\frac{240}{8}\) = 30

Hence, other diameter is 30 cm

Given,

Depth of tube well = 280 m

Diameter of tube well = 3 m

Radius of well = \(\frac{3}{2}\) = 1.5 m

so,

Volume = πr²h = \(\frac{22}{7}\) x 1.5 x 1.5 x 280 = 1980 m³

Cost of sinking tube well at rate Rs.3.60/m³ = 1980×3.60 = Rs.7128

Curved surface area = 2πrh = 2 x \(\frac{22}{7}\) x 1.5 x 280 = 2640 m²

Cost of cementing its inner curved surface at rate Rs.2.50/m² = 2.50×2640 =Rs.6600

Given,

Total surface area of cylinder = 231 cm²

Curved surface area = \(\frac{2}{3}\) total surface area = \(\frac{2}{3}\) x 231 = 154 cm²

so,

= 2πrh = \(\frac{2}{3}\) 2πr(r + h)

= 3 h = 2h + 2r

= h = 2r.............(i)

And,

= 2πr(r + h) = 231

= 2 x \(\frac{22}{7}\) x r x 3r = 231(putting h = 2r)

= r² = \(\frac{231\times 3\times 7}{44}\) = \(\frac{49}{4}\)

= r = \({\sqrt\frac{49}{4}}\) = \(\frac{7}{2}\) = 3.5 cm

= h = 2r = 2×3.5 = 7 cm

Volume of cylinder = πr²h = \(\frac{22}{7}\) x 3.5 x 3.5 x 7 = 269.5 cm³

Correct option is: B) cube

If in a cuboid l = b = h then it becomes cube.

Correct option is: B) cube

Given,

Diameter of metallic cylinder = 1 cm

Radius of cylinder = \(\frac{1}{2}\) cm

Length of cylinder = 5 cm

Diameter of wire drawn from it = 1 mm = 0.1 cm

Let length of wire = h cm

so,

Length of wire drawn from metal = \(\frac{volume\,of\,metal}{volume\,of\,wire}\) = \(\frac{π\times \frac{1}{4}\times 5}{π\times 0.1^2}\) = 500 cm = 5 m

Volume of the cylinder is

V =  π r² h = π 5/4 cm³

The wire is drawn into cylinder of radius 0.05cm and length L  (1mm = 0.1cm = 2*radius)

Therefore

L π  (0.05cm)² = π 5/4 cm³ ⇒ L = 500cm

Given,

Dimensions of rectangular sheet = 30 cm× 18 cm

Case (i)

when paper is rolled along its length:

= 2πr = 30

= r = \(\frac{30}{2π}\) cm

= height = 18 cm

Volume of cylinder thus formed = πr²h = π x (\(\frac{30}{2π}\))² x 18 cm³

Case (ii)

When paper is rolled along its breadth:

= 2πr = 18

=  r = \(\frac{18}{2π}\) cm

= Height = 30 cm

Volume of cylinder thus formed = πr²h = π(\(\frac{18}{2π}\))² x 30 

Hence,

= \(\frac{volume\,of\,cylinder\,1}{volume\,of\,cylinder\,2}\) = \(\cfrac{[π(\frac{30}{2π})^2\times 18]}{[π(\frac{18}{2π})^2\times 30]}\) = \(\frac{30}{18}\) = \(\frac{5}{3}\)

As we roll the sheet, one of the dimensions is the height of the cylinder, the other is circumference of the base circle.

Therefore:

C₁ = 2 π r₁ = 18cm ⇒ r₁ = 9cm/π

h₁ = 30cm

V₁ = π (r₁)² h₁ = 81cm*30cm/π 

C₂ = 2 π r₂ = 30cm ⇒ r = 15cm/π

h₂ = 18cm

V₂ = π (r₂)² h₂ = 225cm*18cm/π 

V₁/V₂ = (81cm*30cm)/(225cm*18cm) = 0.6

Given,

Length of field = 150 m

Width of field = 100 m

Area of field = 150 m × 100 m = 15000 m²

Length of plot = 50 m

Breadth = 30 m

Depth upto which it dug = 8 m

So volume of earth taken out from it = 50 × 30 × 8 = 12000 m³

let raise in earth level of field on which it spread = h metre

so,

= 15000 x h = 12000

= h = \(\frac{12000}{15000}\) = 0.8 m

The level of field raised by 0.8 metre

Volume of cuboid = l × b × h

= 20 × 20 × 30

= 12000 cm³

\(=\frac{12000}{1000}\) litres

= 12 litres

∴ The oil can will contain 12 litres of oil.

The correct answer is option (c) 96 cm²

Explanation:

Let “a” be the side of the cube

Given that, the volume of cube is 64 cm³

It means that a³ = 64 cm³

Hence, a = 4 cm

Therefore, the surface area of a cube = 6 x 4² = 6 x 16 = 96

The correct answer is option (c) 4000

Explanation:

Given that, the dimensions of the godown are 40 m, 25 m and 10 m

Volume = 40 m × 25 m × 10 m = 10000 m³

Given that, volume of each cuboidal box is 2 m × 1.25 m × 1 m = 2.5 m³

Hence, the total number of boxes to be filled in the godown is

= 10000/2.5 = 4000

The correct answer is option (b) 1/2 of the original square

Explanation:

Let “a” be the side of the square sheet

Thus, the area of the bigger square sheet = a² …. (1)

Now, the circle of maximum possible size from it is given as:

The radius of the circle = a/2 … (2)

Then the diameter = a

We know that, any square in the circle of maximum size should have the length of the diagonal which is equal to the diameter of the circle.

It means that, the diagonal of a square formed inside a circle is “a”

Hence, the square side = a / √2

Thus, the area of square = a² / 2

By equating the equations (1) and (2), we will get:

Area of the resultant square is ½ of the original square.

The correct answer is option (c) 16 cm

Explanation:

Given that, the area of a quadrilateral = 20 cm²

We know that, the area of a quadrilateral = (1/2) (diagonal) (sum of the altitudes)

20 = (1/2) (1+1.5) BD

20 = (1/2) (2.5) BD

20×2 = 2.5 BD

40 = 2.5 BD

BD = 16 cm

Area of rhombus = 1/2 (Product of its diagonals)

Therefore, area of the given rhombus

= 1/2 x 7.5cm x 12cm

= 45 cm²

Correct option is: (B) vertical height

In right circular cone, we have 

r = Radius of base of the cone, 

h = Vertical height of the cone

l = slant height of the cone

Now, \(\sqrt{(l + r)(l - r)}\) = \(\sqrt {l^2-r^2}\) (\(\because\) (a+b) (a-b) = \(a^2-b^2\))

= \(\sqrt {h^2}\) (\(\because\) \(l^2 = r^2 + h^2 = l^2 -r^2 = h^2\))

= h which is vertical height of the right circular cone.

Correct option is: (B) vertical height

Perimeter of new figure is-

(D) 50 cm.

We know that,

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

Total surface area of the cube = 6 (l)²

Total surface area of cuboid (a) = [2{(60) (40) + (40) (50) + (50) (60)}] cm²

= [2(2400 + 2000 + 3000)] cm²

= (2 × 7400) cm²

= 14800 cm²

Total surface area of cube (b) = 6 (50 cm)² = 15000 cm²

Thus, the cuboidal box (a) will require lesser amount of material.

½ (h₁ +h₂) d

Explanation:

Assume that ABCD be a quadrilateral, h₁ and h₂ are the heights on the diagonal BD = d, then, the area of a quadrilateral be

= (1/2)(h₁ +h₂) BD

Since the diagonal is doubled and the heights are halved, we will get

= (1/2) [ (h₁/2) +(h₂/2) ] 2d

= ½ (h₁ +h₂) d

volume is h³ and surface area is 6h²

Explanation:

Each side of a cube = h

Thus, volume of cube = h³

Surface area of a cube = 6 (h³)

10a²

Explanation:

Let “a” be the side of two cubes.

When the two cubes are joined face to face, the figure obtained should be a cuboid having the same breadth and height. As the combined cube has a length twice of the length of a cube.

It means that l = 2a, b = a and h = a

Hence, the total surface area of cuboid = 2(lb + bh + hl)

= 2(2a × a + a × a + a × 2a)

Simplify the above expression, we get

= 2[2a² + a² + 2a²]

= 10a²

The correct answer is option (d) None of the above.

Explanation:

We know that, the total surface area of a cylinder is 2π r(h + r), when the radius is “r” and height is “h”.

If the radius is 2r and the height is (1/4)h, then the total surface area becomes,

= 2π (2r) ((1/4)h + 2r)

= 4 πr [(h+8r)/4]

= πr (h+8r)

Twice

Explanation:

Rectangular sheet dimension is 20 cm × 10 cm

When a cylinder is folded along its length, which is 20 cm, then the resultant cylinder is with height 10 cm.

Again, if a cylinder is folded along its breadth, which is 10 cm, then the resultant cylinder is with height 20 cm

When the above conditions are applied in the volume of cylinder formula,

Then we get v = 2V

Volume of cuboidal reservoir = 108 m³ = (108 × 1000) L = 108000 L It is given that water is being poured at the rate of 60 L per minute. That is, (60 × 60) L = 3600 L per hour

Required number of hours = 108000/3600

= 30 hours

Thus, it will take 30 hours to fill the reservoir.

Rate of flow of water = 60 litres/minute

Volume of the tank = 108 m³

1 m³ = 1000 litres

108 m³= 108000 litres

Time taken to fill the complete tank = 108000/60 = 1800 minutes = 1800/60 hours = 30 hours

The tank will be filled completely in 30 hours.