Chapter 12: Surface Areas and Volumes

For hemisphere,

Radius, r = 8 cm

We know that,

Volume of hemisphere = 2/3 πr³, where, r = radius of hemisphere

So, we get,

Volume of given hemisphere = 2/3 × π × 8³ = (1024/3) π cm³

Now,

For the cone that is recast from a hemisphere,

Base radius, r = 6 cm

We also know that,

Volume of cone = 1/3 πr²h, where, r is base radius and h is the height of the cone.

So, we get,

Volume of cone = 1/3 π(6)²h = 12πh

According to the question, we know that,

The volume remains same, when a body is reformed to another body

Volume of cylinder = Volume of cone

12πh = 1024π /3

h = 28.44 cm

Volume of water in tank = volume of cuboidal tank up to a height of 5 m

According to the question,

For cuboidal tank

Length, l = 11 m

Breadth, b = 6 m

Height, h = 5m

We know that the equation to find the volume of the tank,

Volume of tank = lbh, where, l, b and h are the length, breadth and height of tank respectively

Volume of water = 11(6)(5) = 330 m³

We also know that,

Base radius of cylindrical tank, r = 3.5 m

Let the height till which the cylindrical tank is filled = h m

Hence, using the formula,

Volume of a cylinder = πr²h, where r is base radius and h is the height of cylinder

Volume of water in cylindrical tank = π(3.5)²h

330 m ³ = 22/7 × 3.5 ×3.5 × h

330 m ³ = h × 38.5

h = 8.57 m

Hence, the height till which the cylindrical tank is filled = 8.57 m

Let the length (l), breath (b), and height (h) be the external dimension of an open box and thickness be x.

The volume of metal used in box = Volume of external box – Volume of internal box

Consider external box,

Length, l = 36 cm

Breadth, b = 25 cm

Height, h = 16.5 cm

We know that the equation of the volume of cuboid is given by,

Volume of cuboid = lbh, where, l, b and h are the length, breadth and height of tank respectively

Volume of external box = 36(25)(16.5) = 14850 cm³

Since the box is open from top,

Consider internal box,

The thickness of two sides is reduced as follows,

Length, l’ = Length of external box – 2(thickness of box) = 36 – 2(1.5) = 33 cm

Breadth, b’ = Breadth of external box – 2(thickness of box) = 25 – 2(1.5) = 22 cm

Height, h’ = Height of external box – thickness of box = 16.5 – 1.5 = 15 cm

Volume of internal box = 33(22)(15) = 10890

And,

Volume of metal in box = 14850 – 10890 = 3960 cm³

We know that,

1 cm³ weighs 7.5 g

So, 3960 cm³ weighs 3960(7.5) = 29,700 g

Therefore, the weight of box is 29,700 g i.e. 29.7 kg

Let us first calculate the volume of barrel of pen that is of cylindrical shape

Consider barrel,

Since 1cm = 10 mm

Base diameter = 5 mm = 0.5 cm

Base radius, r = 0.25 cm

Height, h = 7 cm

We know that,

Volume of a cylinder = πr²h

Volume of barrel = π (0.25)²7

Volume of barrel = 22/7 × 0.25 × 0.25 × 7 = 1.375 cm³

Hence, according to the question,

1.375 cm³ of ink can write 3300 words

No of words that can be written by 1 cm³ of ink = 3300/1.375 = 2400 words

1/5^th of a liter = 0.2 L

We know that,

1 L = 1000 cm³

0.2 L = 200 cm³

So, no of words that can be written by 200 cm³ = 2400(200) = 480000 words

Therefore, 1/5^th of a liter ink can write 480000 words.

Let the time taken by pipe to fill vessel = t minutes

Since water flows 10 m in 1 minute, it will flow 10t meters in t minutes.

According to the question,

Volume of conical vessel = Volume of water that passes through pipe in t minutes

Consider conical pope

Base Diameter = 40 cm

Base radius, r = 20 cm

Height, h = 24 cm

We know that the volume of cone = 1/3πr²h

Volume of conical vessel = 1/3π(20)²(24) = 3200 π cm³

Consider cylindrical pipe

Base diameter = 5 mm = 0.5 cm

Base radius, r = 0.25 cm

Water covers 10t m distance in pipe,

Hence, we get,

Height, h = 10t m = 1000t cm

We also know that,

Volume of a cylinder = πr²h

Volume of water passed in pipe = π(0.25)²(1000t) = 62.5tπ cm³

So, we have

62.5tπ = 3200

62.5t = 3200

t = 51.2 minutes

We know that,

0.2 minutes = 0.2(60) seconds = 12 seconds

Therefore, t = 51 minutes 12 seconds

According to the question,

Consider conical heap,

Base Diameter = 9 cm

So, base radius, r = 4.5 cm

Height, h = 3.5 cm

We know that,

Slant height,

The equation of volume of cone = 1/3πr²h

We know that,

Volume of rice = Volume of conical heap

Volume of rice = 1/3π(4.5)²(3.5)

We also know that,

Canvas requires to just cover heap = Curved surface area of conical heap

And curved surface area of a cone = πrl

Therefore, the canvas required = π(4.5)(5.7) = 80.61 cm² [appx]

The shape of pencil = cylinder.

Let the radius of base = r cm

Circumference of base = 1.5 cm

Circumference of circle is 2πr = 1.5 cm

r = 1.5/2π cm

According to the question,

Height, h = 25 cm

We know that,

Curved surface area of cylinder = 2πrh

Curved surface area of pencil = 2π (1.5/2π) 25 = 37.5 cm²

1 cm = 0.1 dm

1 cm² = 0.01 dm²

37.5 cm² = 0.375 dm²

Cost for coloring 1 dm² = Rs. 0.05

Cost for coloring 0.375 dm² (i.e. 1 pencil) = Rs. 0.01875

Cost for coloring 120000 pencils = 120000 ×0.01875 = Rs. 2250

Let the time taken by pipe to fill pond = t hours

Water flows 15 km in 1 hour, so, it will flow 15t meters in t hours.

We know that,

Volume of cuboidal pond up to height 21 cm = Volume of water that passes through pipe in “t” hours

Considering cuboidal pond,

Length, l = 50 m

Breadth, b = 44 m

Height, h = 21 cm = 0.21 m

We know that,

Volume of tank = lbh

Volume of water = 50(44)(0.21) = 462 m³

Considering cylindrical pipe

Base diameter = 14 cm

Base radius, r = 7 cm = 0.07 m

Height, h = 15t km = 15000t m

We also know that,

Volume of a cylinder = πr²h

Volume of water passed in pipe = π(0.07)²(15000t)

=22/7 × 0.07 × 0.07 × 15000t

= 231t cm³

So, we have

231t = 462

t = 2 hours

Time required to fill tank up to a height of 25 cm is 2 hours.

Considering cuboidal block

Length, l = 4 m

Breadth, b = 2.6 m

Height, h = 1 m

We know that,

Volume of tank = lbh

Volume of cuboid = 4.4(2.6)(1) = 11.44 m³

We know that,

The volume remains same when a body is recast to another body.

According to the question,

Volume of cylindrical pipe = 11.44 m³

Considering pipe or the hollow cylinder

Internal radius, r₂ = 30 cm = 0.3 m

Thickness = 5 cm

External radius, r₁ = Internal radius + thickness = 30 + 5 = 35 cm = 0.35 m

Let the length of pipe = h

We know that,

Volume of a hollow cylinder = πh(r₁² – r₂²)

Hence,

Volume of pipe = πh((0.35)² – (0.3)²)

So, the length of pipe is 112 m.

According to the question,

Average displacement by a person = 0.04 m³

Average displacement by 500 persons = 500 × 0.04 = 20 m³

Hence, the volume of water raised in pond = 20 m³

It is also given that,

Length of pond, l = 80 m

Breadth of pond, b = 50 m

Height = h

Volume of water raised in pond = 80(50)(h)

20 m³ = 4000h

h = 0.005 m = 0.5 cm

Therefore, raise in the height of water = 0.5 cm.