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पहिये का व्यास = 60 सेमी

पहिये की त्रिज्या r = 60/2 = 30 सेमी

साइकिल द्वारा 1 चक्कर में चली दूरी = पहिये की परिधि

= 2πr = 2 x 22/7 x 30 = 1320/7 सेमी

∵ लड़के द्वारा 1 मिनट में लगाये चक्करों की संख्या = 140

तब, 140 चक्करों में चली दूरी = 1320/7 x 140 = 26400 सेमी

= 26400/1000 x 100 किमी = 0.264 किमी

तथा समय = 1 घण्टा = 60 मिनट

∵ 1 मिनट में चली गई दूरी = 0.264 किमी

∴ 60 मिनट में चली गई दूरी = 0.264 x 60 = 15.84 किमी

अतः साइकिल की चाल = 15.84 किमी/घण्टा

We know that volume of cuboid = length × breadth × height

Volume of the cistern = 11.2 × 6 × 5.8

= 389.76 m³

= 389.76 × 1000

= 389760 liters.

Area of the sheet that required to make the cistern = total surface area of the cistern we know that total surface area of cuboid= 2(l b + b h + h l)

= 2 (11.22 × 6 + 6 × 5.8 + 5.8 × 11.2)

= 2 (67.2 + 64.96 + 34.8)

= 333.92 cm²

We know that water in a canal forms a cuboid

The dimensions are

Breadth = 30dm = 3m

Height = 12dm = 1.2m

We know that

Length = distance covered by water in 3 minutes = velocity of water in m/hr × time in hours

By substituting the values

Length = 20000 × (30/60)

So we get

Length = 10000m

We know that

Volume of water flown in 30 minutes = l × b × h

By substituting the values

Volume of water flown in 30 minutes = 10000 × 3 × 1.2 = 36000 m³

Consider A m² as the area irrigated

So we get

A × (9/100) = 36000

On further calculation

A = 400000 m²

Therefore, the area to be irrigated is 400000 m².

It is given that

Diameter of a cylinder = 28cm

We know that radius = diameter/2 = 28/2 = 14cm

Height of a cylinder = 40cm

We know that

Curved surface area = 2 πrh

By substituting the values

Curved surface area = 2 × (22/7) × 14 × 40

So we get

Curved surface area = 3520 cm²

We know that

Total surface area = 2 πrh + 2 πr²

By substituting the values

Total surface area = (2 × (22/7) × 14 × 40) + (2 × (22/7) × 14²)

On further calculation

Total surface area = 3520 + 1232 = 4752 cm²

We know that

Volume of cylinder = πr²h

By substituting the values

Volume of cylinder = (22/7) × 14² × 40

So we get

Volume of cylinder = 24640 cm³

Therefore, the curved surface area, total surface area and the volume of cylinder are 3520 cm², 4752 cm² and 24640cm³.

The dimensions of the room is

Length = 9m

Breadth = 8m

Height = 6.5m

We know that

Area of four walls of the room = 2 (l + b) × h

By substituting the values

Area of the four walls of the room = 2 (9 + 8) × 6.5

On further calculation

Area of the four walls of the room = 34 × 6.5

So we get

Area of the four walls of the room = 221 m²

The dimensions of the door are

Length = 2m

Breadth = 1.5m

We know that

Area of one door = l × b

By substituting the values

Area of one door = 2 × 1.5

So we get

Area of one door = 3m²

The dimensions of the window are

Length = 1.5m

Breadth = 1m

We know that

Area of two windows = 2 (l × b)

By substituting the values

Area of two windows = 2 (1.5 × 1)

On further calculation

Area of two windows = 2 × 1.5 = 3m²

So the area to be whitewashed = Area of four walls of the room – Area of one door – Area of two windows

By substituting the values

Area to be whitewashed = (221 – 3 – 3)

So we get

Area to be whitewashed = 215m²

It is given that the cost of whitewashing = ₹ 25 per square metre

So the cost of whitewashing 215m² = ₹ (25 × 215)

Cost of whitewashing 215m² = ₹ 5375

Therefore, the cost of whitewashing 215m² is ₹ 5375.

It is given that

Radius of the sphere = 10.5cm

Radius of smaller cone = 3.5cm

Height = 3cm

We know that

Number of cones = Volume of the sphere/ Volume of one small cone

So we get

Number of cones = (4/3 × (22/7) × 10.5³)/ (1/3 × (22/7) × 3.5² × 3)

On further calculation

Number of cones = 4851/ 38.5 = 126

Therefore, 126 cones are obtained from the metallic sphere.

The dimensions of cuboid are

Length = 9m

Breadth = 8m

Height = 2m

We know that

Volume of cuboid = l × b × h

By substituting the values

Volume of cuboid = 9 × 8 × 2

So we get

Volume of cuboid = 144 m³

We know that

Volume of each cube of edge 2m = a³

So we get

Volume of each cube of edge 2m = 2³ = 8 m³

So the number of cubes formed = volume of cuboid / volume of each cube

By substituting the values

Number of cubes formed = 144/8 = 18

Therefore, the number of cubes formed is 18.

It is given that

Length of the cuboidal water tank = 6m

Breadth of the cuboidal water tank = 5m

Height of the cuboidal water tank = 4.5m

We know that

Volume of a cuboidal water tank = l × b × h

By substituting the values

Volume of a cuboidal water tank = 6 × 5 × 4.5

By multiplication

Volume of a cuboidal water tank = 135 m³

We know that 1m³ = 1000 litres

So we get

Volume of a cuboidal water tank = 135 × 1000 = 135000 litres

Therefore, the cuboidal water tank can hold 135000 litres of water.

Given details are,

Dimensions of metal block = 2.25m × 1.5m × 27cm = 2.25m × 1.5m × 0.27m

Side of each cube formed = 45cm = 0.45 m

We know that,

Number of cubes can formed = volume of metal block / volume of one cube

= (2.25×1.5×0.27) /(0.45×0.45×0.45)

= 0.91125 / 0.091125

= 10 cubes

∴ 10 cubes are formed.

Given:-  

r₁ = 16 cm 

r₂ = 10 cm 

h = 800 cm 

Let l be the slant height of the friction clutch, then 

l = √(h² + {r₁ - r₂}²) 

l =  √(640000 + 36) 

l = √640036 = 2√160009 cm  

L.S.A of friction clutch  

= π(r₁ + r₂)l 

= 52π√160009 cm  

Volume of friction clutch  

= πh/3 (r₁² + r₂² + r₁r₂) 

= 800π/3 (256 + 100 + 160) 

= 800π × 516/3 

= 800π × 172 

= 137600π cm³

We know that, 

1 m = 100 cm 

Also, 

Volume of a cuboid = Length × Breadth × Height 

Therefore, 

Volume of the original block = 225 × 150 × 27 

= 911250 cm³ 

Given that, 

Length of the edge of the cube = 45 cm 

Therefore, 

Volume of one cube = a³ = (45)³ 

= 91125 cm³ 

Hence, 

Total number of blocks that can be cast = \(\frac{Volume\,of\,the\,block}{Volume\,of\,the\,cube}\)

= \(\frac{911250}{91125}\)

= 10

Solution:

False

Since, The volume of the frustum of a cone is 1/3 πh[r₁² + r₂² + r₁r₂] where h is vertical height of the frustum and r₁, r₂ are the radii of the ends.

Let the length of the pipe be h cm.

Then, Volume of cuboid = (53 x 40 x 15) cm³

Internal radius of the pipe = 7/2 cm = r

External radius of the pipe = 8/2 = 4 cm = R

So, the volume of iron in the pipe = (External Volume) – (Internal Volume)

= πR²h – πr²h

= πh(R²– r²)

= πh(R – r) (R + r)

= π(4 – 7/2) (4 + 7/2) x h

= π(1/2) (15/2) x h

Then from the question it’s understood that,

The volume of iron in the pipe = volume of iron in cuboid

π(1/2) (15/2) x h = 53 x 40 x 15

h = (53 x 40 x 15 x 7/22 x 2/15 x 2) cm

h = 2698 cm

Therefore, the length of the pipe is 2698 cm.

Let the edges of three cubes (in cm) be 3x, 4x and 5x respectively.

So, the volume of the cube after melting will be = (3x)³ + (4x)³ + (5x)³

= 9x³ + 64x³ + 125x³ = 216x³

Now, let a be the edge of the new cube so formed after melting

Then we have,

a³ = 216x³

a = 6x

We know that,

Diagonal of the cube = √(a² + a² + a²) = a√3

So, 12√3 = a√3

a = 12 cm

x = 12/6 = 2

Thus, the edges of the three cubes are 6 cm, 8 cm and 10 cm respectively.

Given,

Radius of each spherical lead shot = r = 4.2/ 2 = 2.1 cm

The dimensions of the rectangular lead piece = 66 cm x 42 cm x 21 cm

So, the volume of a spherical lead shot = 4/3 πr³

= 4/3 x 22/7 x 2.1³

And, the volume of the rectangular lead piece = 66 x 42 x 21

Thus,

The number of spherical lead shots = Volume of rectangular lead piece/ Volume of a spherical lead shot

= 66 x 42 x 21/ (4/3 x 22/7 x 2.1³)

= 1500

Given,

The radius of each spherical lead shot = r = 4/2 = 2 cm

Volume of each spherical lead shot = 4/3 πr³ = 4/3 π 2³ cm³

Edge of the cube = 44 cm

Volume of the cube = 44³ cm³

Thus,

Number of spherical lead shots = Volume of cube/ Volume of each spherical lead shot

= 44 x 44 x 44/ (4/3 π 2³)

= 2541

It is given that

Diameter of the bucket = 28cm

Radius = 28/2 = 14cm

Height of the bucket = 72cm

Length of the tank = 66cm

Breadth of tank = 28 cm

We know that

Volume of tank = volume of cylindrical bucket

l × b × h = πr²h

By substituting the values

66 × 28 × h = (22/7) × (14)² × 72

On further calculation

h = (22 × 2 × 14 × 72)/ (66 × 28)

So we get

h = 24 cm

Therefore, the height of the water level in the tank is 24 cm.

Total surface area of solid hemisphere = 3πr²

⇒ 3πr² = 462 

⇒ 3 × (\(\frac{22}7\)) × r² = 462 

⇒ r² = \(\frac{{462}\times{7}}{{22}\times{3}}\)

⇒ r² =49 

⇒ r = 7 cm 

Volume of solid hemisphere = \(\frac{2}3πr^3\)

= \(\frac{2}3π7^3\) cm³ 

= \(\frac{2}{3}\times\frac{22}7\times7\times7\times7\)

= 718.67 cm³

Diameter of the cylindrical base = 3 m 

∴ Radius of cylindrical tank \(\frac{3}2\) m 

= 1.5 m 

Height of the tank = 3.5 m 

Volume of the tank = πr²h

= π x 1.5² x 3.5 = \((\frac{22}{7})\) x 1.5 x 1.5 x 3.5

= 24.75 m³ 

Now, 

1 m³ = 1000 liters 

24.75 m³ = 1000 × 24.75 liters 

= 24750 liters 

Full quantity of the water when it is full = 24750 m³ 

Quantity of water when it is half filled = \(\frac{24750}2\) liters 

= 12375 liters 

Time taken by it to empty 225 liters 0f water = 1 minute 

∴ Time taken by it empty 12375 litersof water = \(\frac{12375}{225}\) minutes 

= 55 minutes

Diameter of spherical marbles = \(\frac{1.4}2\)cm 

= 0.7 cm 

Diameter of cylinder vessel \(\frac{7}2\) cm

= 3.5 cm 

Volume of 150 spherical balls =150 × \(\frac{4}3\)πR³

= \(\frac{{150}\times{4}\times{22}\times{0.7}\times{0.7}\times{0.7}}{{3}\times{7}}\) cm³

= 215.6 cm³ 

Let the rise in level of water be h cm 

Volume of rise in level of water (Volume of cylinder) = πr²h

= \(\frac{22}7\times3.5\times3.5\times{h}\)

Volume of Rise in level of water in the vessel = volume of 150 spherical balls 

⇒ (\(\frac{22}7\)) × 3.5 × 3.5 × h = 215.6 

⇒ h = \(\frac{{215.6}\times{7}}{{22}\times{3.5}\times{3.5}}\) cm

⇒ h = 5.6 cm

(A) 142296

According to the question,

Volume of cube =22³=10648cm³

Volume of cube that remains unfilled =1/8×10648=1331cm³

volume occupied by spherical marbles =10648−1331=9317cm³

Radius of the spherical marble = 0.5/2=0.25cm=1/4cm

Volume of 1 spherical marble = 4/3×22/7 × (1/4)³=11/168cm³

Numbers of spherical marbles, n = 9317 × (11/168) =142296

Radius of a hemisphere = Radius of a solid hemisphere = 10 cm

Surface area of the hemisphere = 2πr² 

= 2 x 3.14 x (10)² cm² 

= 628 cm² 

And, surface area of solid hemisphere = 3πr² 

= 3 x 3.14 x (10)² cm² 

= 942 cm²

Volume of a sphere = \(\frac{4}{3}\)πr³ Cubic Units 

Where, r = radius of a sphere 

(i) diameter =14 cm 

So, radius = \(\frac{diameter}{2}\) 

= \(\frac{14}{2}\) 

= 7 cm 

Volume = \(\frac{4}{3}\) x \(\frac{22}{7}\) x (7)³ 

= 1437.33 

Volume = 1437.33 cm³ 

(ii) diameter = 3.5 dm 

So, radius = \(\frac{diameter}{2}\) 

= \(\frac{3.5}{2}\) 

= 1.75 dm 

Volume = \(\frac{4}{3}\) x \(\frac{22}{7}\) x (1.75)³ 

= 22.46 

Volume = 22.46 dm³ 

(iii) diameter = 2.1 m 

So, radius = \(\frac{diameter}{2}\) 

= \(\frac{2.1}{2}\) 

= 1.05 m 

Volume = \(\frac{4}{3}\) x \(\frac{22}{7}\) x (1.05)³ 

= 4.851 

Volume = 4.851 m³

Surface area of a sphere = 4πr² 

Where, r = radius of a sphere 

(i) Diameter = 14 cm 

So, Radius = \(\frac{Diameter}{2}\) 

= \(\frac{14}{2}\) cm 

= 7 cm 

Surface area = 4 x \(\frac{22}{7}\) x (7)² 

= 616 

Surface area is 616 cm² 

(ii) Diameter = 21 cm 

So, Radius =  \(\frac{Diameter}{2}\)  

= \(\frac{21}{2}\) cm 

= 10.5 cm 

Surface area = 4 x \(\frac{22}{7}\) x (10.5)² 

= 1386 

Surface area is 1386 cm² 

(iii) Diameter= 3.5 cm 

So, Radius =  \(\frac{Diameter}{2}\) 

= \(\frac{3.5}{2}\) cm 

= 1.75 cm 

Surface area = 4 x \(\frac{22}{7}\) x (1.75)² 

= 38.5 

Surface area is 38.5 cm²

Surface area of a sphere is 5544 cm² 

Surface area of a sphere = 4πr² 

So, 4πr² = 5544

4 x \(\frac{22}{7}\) x (r)² = 5544 

r² = \(\frac{(5544 \times 7)}{88}\) 

r² = 441 

or r = 21 cm

Now, Diameter=2(radius) = 2(21) = 42 cm

Data: E is any point on Median AD of an ∆ABC. 

To Prove: ar.(∆ABE) = ar. (∆ACE) 

Proof: In ∆ABC, AD is the median. 

∴ ∆ABD = ∆ACD ……….. (i) 

In ∆EBC, DE is the median. 

∴ ∆EBC = ∆ECD …………. (ii)

Subtracting (ii) from (i), 

∆ABD – ∆EBD = ∆ACD – ∆ECD 

∴ ar. (∆ABE) = ar. (∆ACE).

Volume of a cube = side³

Volume of a sphere = \(\frac{4}{3}\)πr³ 

Given, 

spherical bullets are to be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter. 

Let the number of bullets be ‘a’. 

⇒ 44³ = a × \(\frac{4}{3}\)× \(\frac{22}{7}\) × 2³ 

⇒ a = 2541

Surface area of a sphere = 154 cm² 

We know, Surface area of a sphere = 4πr² 

So, 4πr² = 154 

4 x \(\frac{22}{7}\) x r² = 154 

or r² = \(\frac{49}{4}\) 

or r = \(\frac{7}{2}\) 

= 3.5 

Radius (r) = 3.5 cm 

Now, Volume of sphere = \(\frac{4}{3}\) π r³ 

= (\(\frac{4}{3}\)) π × 3.5³ 

= 179.66 

Therefore, Volume of sphere is 179.66 cm³.

Diameter of hollow sphere = 7 m 

So, radius of hollow sphere = \(\frac{7}{2}\) m = 3.5 cm 

Now, 

Area available to the motorcyclist for riding = Surface area of a sphere = 4πr² 

= 4 × (\(\frac{22}{7}\)) × 3.5² m² 

= 154 m²

Surface area of a sphere = 154 cm² 

We know, Surface area of a sphere = 4πr² 

So, 4πr² = 154 

4 x \(\frac{22}{7}\) x r² = 154 

r² = \(\frac{49}{4}\) 

or r = \(\frac{7}{2}\) 

= 3.5 

Radius of a sphere is 3.5 cm.

Radius of a sphere (R)= 4 cm

Height of the cylinder = \(\frac{2}{3}\) diameter

We know, 

Diameter = 2(Radius) 

Let h be the height and r be the base radius of a cylinder, then 

h = \(\frac{2}{3}\) x (2r) 

= \(\frac{4r}{3}\)

Volume of the cylinder = Volume of the sphere 

πr²h = \(\frac{4}{3}\)πR³ 

π × r² × (\(\frac{4r}{3}\)) = \(\frac{4}{3}\)π(4)³ 

(r)³ = (4)³ 

or r = 4 

Therefore, radius of the base of the cylinder is 4 cm.

Option : (A)

Volume of sphere = \(\frac{4}{3}\)πr³ 

Given, 

solid sphere of radius 10 cm is moulded into 8 spherical solid balls of equal radius 

⇒ \(\frac{4}{3}\)π × 10³ = 8 × \(\frac{4}{3}\)π × r³ 

⇒ r= \(\frac{10}{2}\)= 5cm 

Surface area of a sphere = 4πr² 

Thus, 

the surface area of each sphere = 4 × π × 5² 

= 100π

Volume of a sphere = \(\frac{4}{3}\)πr³
Volume of a cylinder = πr²h

Given, 

Diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm.

Long wire can be assumed to be a cylinder.

⇒ \(\frac{4}{3}\)π x 3³ x h = π x 0.1² x l

⇒ l = 3600 cm = 36 m

Option : (C)

Volume of a sphere = \(\frac{4}{3}\)πr³ 

Volume of a cylinder = πr²h 

Given, 

cylindrical rod whose height is 8 times of its radius is melted and recast into spherical balls of same radius. Let the number of such balls be ‘a’. 

⇒ π × r² × 8r = a × \(\frac{4}{3}\)π × r³ 

⇒ a = 6

Given that, 

radius of cylindrical jar, r = 6 cm 

Depth/height of the jar, h = 2 cm 

Volume of the jar, V’ = πr² h 

⇒ V’ = π (6)² (2) cm³ 

Radius of the sphere = 1.5 cm 

So, 

volume of the sphere = \(\frac{4}{3}πr^3\)

= \(\frac{4}{3}π(1.5)^3\)

Volume of oil in cylindrical jar = volume of n spheres needed to raise the level by 2 cm 

∴ π (6)² (2) = \(n\times\frac{4}{3}π(1.5)^3\)

⇒ n = 16 

∴ number of iron sphere needed = 16

Volume of a sphere = \(\frac{4}{3}\)πr³
Volume of a cylinder = πr²h

Given,

Diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. The length of the wire is 108 m.

Long wire can be assumed to be a cylinder.

⇒ \(\frac{4}{3}\)π x 9³ = π x r² x 10800

⇒ r = 0.6 cm

Volume of a sphere = \(\frac{4}{3}\)πr³
Volume of a cylinder = πr²h

Given, 

Measuring jar of internal diameter 10 cm is partially filled with water. Four equal spherical balls of diameter 2 cm each are dropped in it and they sink down in water completely.

Let the rise in level of water be ‘h’ cm.

⇒ π x 5² x h = 4 x \(\frac{4}{3}\)π x 1³

⇒ h = \(\frac{16}{75}\) cm

Volume of a hemisphere = (\(\frac{2}{3}\))πr³ 

Volume of a cone = (\(\frac{1}{3}\))πr²h 

Given, 

Cone and a hemisphere have equal bases which implies they have the same radius. 

Height of the hemisphere is its radius. 

Let the base radius be ‘r’ and the height of cone be ‘h’. 

Given, 

Cone and hemisphere have equal volume. 

(\(\frac{2}{3}\))πr³ = (\(\frac{1}{3}\))πr²h 

⇒ h : r = 2 : 1

Radius of sphere = 5 cm

Let ‘r’ be the radius of cylinder. 

We know, Volume of sphere = \(\frac{4}{3}\)πr³  

= \(\frac{4}{3}\) x π x (5)³ 

Height (h) of water rises is \(\frac{5}{3}\) cm

Volume of water rises in cylinder = πr²h 

Therefore, Volume of water rises in cylinder = Volume of sphere 

So, πr²h = \(\frac{4}{3}\)πr³ 

πr² x \(\frac{5}{3}\) = \(\frac{4}{3}\) x π x (5)³ 

or r² = 100 

or r = 10 

Therefore, radius of the cylinder is 10 cm.

Volume of a sphere = \(\frac{4}{3}\)πr³
Volume of a cylinder = πr²h

Given, 

Cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil.

Level of the oil has to rise by 2 cm. 

Let the number of spheres required be ‘n’.

⇒ n x \(\frac{4}{3}\) x π x 1.5³ = π x 6² x 2

⇒ n = 16

Volume of a sphere = \(\frac{4}{3}\)πr³ 

Volume of a cone = \(\frac{1}{3}\)πr²h

Given,

metallic sphere of radius 10.5 cm is melted and thus recast into small cones, each of radius 3.5 cm and height 3 cm.

Let the number of cones be ‘n’.

⇒ n × (1/3)π × 3.5² × 3 = (4/3) × π × 10.5³

⇒ n = 126

Volume of a hemisphere = \(\frac{2}{3}\)πr³ 

Volume of a right circular cone = \(\frac{1}{3}\)πr²h 

Given,

a cone, a hemisphere and a cylinder stand on equal bases and have the same height.

Height of a hemisphere is the radius and equal bases implies equal base radius.

Thus, 

height of cone = height of cylinder = base radius = r

Ratio of volumes = \(\frac{1}{3}\)πr²h :  \(\frac{2}{3}\)πr³ : πr²h

⇒ Ratio of volumes = r³ : 2r³ : 3r³ = 1 : 2 : 3

We have,

Total surface area of cylinder = 231 cm²

Curved surface area = 2/3 total surface area = 2/3 × 231 = 154 cm²

2πrh = 2/3 2πr(h + r)

3h = 2(h+r)

3h = 2h + 2r

h = 2r ……… (i)

And,

2πr(h + r) = 231

2 × 22/7 × r × (2r+r) =231

2 × 22/7 × r × 3r = 231

3r² = 231×7 / 2×22

= 1617 / 44

= 36.75

r² = 36.75 / 3

= 12.25

r = √12.25

= 3.5 cm

Since, h = 2r = 2×3.5 = 7cm

∴ Volume of cylinder = πr²h

= 22/7 × 3.5 × 3.5 × 7

= 269.5 cm³

Volume of a sphere = (\(\frac{4}{3}\))πr³ 

Volume of a cylinder = πr²h

Let the radius of the cylinder be r cm. 

Given, 

Sphere of radius 5 cm is immersed in water filled in a cylinder, 

the level of water rises \(\frac{5}{3}\)cm 

Volume of the sphere = Volume of the water in the cylinder.

⇒ \(\frac{4}{3}\)π x 5³ = π x r² x \(\frac{5}{3}\)

⇒ r² = 4 × 5²

⇒ r = 10 cm

Volume of a sphere = \(\frac{4}{3}\)πr³ 

Volume of a right circular cone = \(\frac{1}{3}\)πr²h 

Given, 

diameter of the internal and external surfaces of a hollow spherical shell are 4 cm and 8 cm respectively. It is melted into a cone of base diameter 8 cm.

⇒ Volume of material in sphere = Volume of the cone

⇒ \(\frac{4}{3}\) x π x (4³ - 2³) =\(\frac{1}{3}\)π x 4² x h 

⇒ h = 14 cm

Volume of a sphere = \(\frac{4}{3}\)πr³ 

Volume of a cylinder = \(\frac{1}{3}\)πr²h

Given,

radius of the internal and external surfaces of a hollow spherical shell are 2 cm and 4 cm respectively. It is melted into a cone of base radius 4 cm.

⇒ Volume of material in sphere = Volume of the cone

⇒ \(\frac{4}{3}\)x π x (4³ - 2³) = \(\frac{1}{3}\)π x 4² x h

⇒ h = 14 cm

L² = h² + r²

⇒ l = \(\sqrt{14^2+4^2}\)

⇒ l = \(\sqrt{212}\) = 14.56 cm

Volume of a sphere = \(\frac{2}{3}\)πr³ 

Volume of a cylinder = \(\frac{1}{3}\)πr²h

Given,

hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm

⇒ \(\frac{2}{3}\)π x 7³ = \(\frac{1}{3}\) x π x 49 x r²

⇒ r² = 14 

⇒ r = 3.74 cm

Volume of a cube = side³ 

Volume of a sphere = \(\frac{4}{3}\)πr³ 

Given, 

cube of side 4 cm contains a sphere touching its side 

Radius of the sphere = \(\frac{4}{2}\) = 2 cm 

Volume of the gap in between = 4³ - \(\frac{4}{3}\)π × 2³ 

⇒ Volume of the gap in between = 30.48 cm³

Largest sphere that can be carved out of a cube will have its diameter as the side of the cube. 

Radius of the largest sphere carved out of a cube of side 10.5 cm = \(\frac{10.5}{2}\) = 5.25 cm

Volume of a cube = side³

Volume of a sphere =  \(\frac{4}{3}\)πr³ 

Ratio of their volumes = \(\frac{\frac{4}{3}\pi\times5.25^3}{(10.5)^3}\)

= \(\frac{4}{3}\) x \(\frac{22}{7}\) x \(\frac{1}{8}\)

= 11 : 21

We have,

Sum of base radius and height of cylinder = 37m

(r + h) = 37m

Total surface area = 1628 m²

By using the formula,

Total surface area = 2πr(h+r)

So,

2πr(h+r) = 1628

2 × 22/7 × r (37) = 1628

r = 1628×7 / 2× 22×37

= 11396 / 1628

= 7m

∴ Circumference of its base = 2πr

= 2 × 22/7 × 7

= 44 m