Question - 12 : - Three cubes of a metal whose edges are in the ratio 3: 4: 5 are melted and converted into a single cube whose diagonal is 12√3 cm. Find the edges of the three cubes.

Answer - 12 : -

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

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

= 9x³ +64x³ + 125x³ = 216x³

Now, let a be the edgeof 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 thethree cubes are 6 cm, 8 cm and 10 cm respectively.

Question - 13 : - A solid metallic sphere of radius 10.5 cm is melted and recast into a number of smaller cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed.

Answer - 13 : -

Given,

Radius of metallicsphere = R = 10.5 cm

So, its volume = 4/3πR³ = 4/3 π(10.5)³

We also have,

Radius of each cone =r = 3.5 cm

Height of each cone =h = 3 cm

And, its volume = 1/3πr²h = 1/3 π(3.5)²(3)

Thus,

The number of cones =Volume of metallic sphere/ Volume of each cone

= 4/3 π(10.5)³ /1/3 π(3.5)²(3)

= 126

Question - 14 : - The diameter of a metallic sphere is equal to 9 cm. It is melted and drawn into a long wire of diameter 2 mm having uniform cross-section. Find the length of the wire.

Answer - 14 : -

Given,

Radius of the sphere =9/2 cm

So, its volume = 4/3πr³ = 4/3 π(9/2)³

And, the radius of thewire = 2 mm = 0.2 cm

Let the length of thewire = h cm

So, the volume of wire= πr²h = π(0.2)²h

Then, according to thequestion we have

Volume of wire =Volume of sphere

π(0.2)²h =4/3 π(9/2)³

h = 4 x 729/ (3 x 8 x0.01) = 12150 cm

Therefore, the lengthof the wire = 12150 cm

Question - 15 : - An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is 1/4 of the radius of the original ball, how many such balls are made? Compare the surface area, of all the smaller balls combined together with that of the original ball.

Answer - 15 : -

Let the radius of thebig ball be x cm

The, the radius of thesmall ball = x/4 cm

And, let the number ofballs = n

Then according to thequestion, we have

Volume of n smallballs = Volume of the big ball

n x 4/3 π(x/4)³ =4/3 πx³

n x (x³/64) = x³

n = 64

Therefore, the numberof small balls = 64

Next,

Surface area of allsmall balls/ surface area of big ball = 64 x 4π(x/4)²/ 4π(x)²

= 64/16 = 4/1

Thus, the ratio of thesurface area of the small balls to that of the original ball is 4:1

Question - 16 : - A copper sphere of radius 3 cm is melted and recast into a right circular cone of height 3 cm. Find the radius of the base of the cone?

Answer - 16 : -

Given,

Radius of the coppersphere = 3 cm

We know that,

Volume of the sphere =4/3 π r³

= 4/3 π × 3³ …..(i)

Also, given that thecopper sphere is melted and recasted into a right circular cone

Height of the cone = 3cm

We know that,

Volume of the rightcircular cone = 1/3 π r²h

= 1/3 π × r²×3 ….. (ii)

On comparing equation(i) and (ii) we have,

4/3 π × 3³ =1/3 π × r²× 3

r² =36

r = 6 cm

Therefore, the radiusof the base of the cone is 6 cm.

Question - 17 : - . A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire?

Answer - 17 : -

Given,

Diameter of the copperwire = 1 cm

So, radius of thecopper wire = 1/2 cm = 0.5 cm

Length of the copperrod = 8 cm

We know that,

Volume of the cylinder= π r²h

= π × 0.5²×8 ……. (i)

Length of the wire =18 m = 1800 cm

Volume of the wire = πr²h

= π r² ×1800 ….. (ii)

On equating both theequations, we have

π × 0.5²×8 = π r² × 1800

r² = 2/1800 = 1/900

r = 1/30 cm

Therefore, thediameter of the wire is 1/15 cm i.e. 0.67 mm which is the thickness of thewire.

Question - 18 : - The diameters of internal and external surfaces of a hollow spherical shell are 10cm and 6 cm respectively. If it is melted and recast into a solid cylinder of length of 8/3, find the diameter of the cylinder?

Answer - 18 : -

Given,

Internal diameter ofthe hollow sphere = 6 cm

So, the internalradius of the hollow sphere = 6/2 cm = 3 cm = r

External diameter ofthe hollow sphere = 10 cm

So, the externalradius of the hollow sphere = 10/2 cm = 5 cm = R

We know that,

Volume of the hollowspherical shell = 4/3 π × (R³ – r³)

= 4/3 π × (5³ –3³) ….. (i)

And given, the lengthof the solid cylinder = 8/3 cm

Let the radius of thesolid cylinder be r cm

We know that,

Volume of the cylinder= π × r² × h

= π × r² ×8/3 ….. (ii)

Now equating both (i)and (ii), we have

4/3 π × 5³ –3³ = π × r² × 8/3

4/3 x (125 – 27) = r² ×8/3

98/2 = r²

r² =49

r = 7

So, d = 7 x 2 = 14 cm

Therefore, thediameter of the cylinder is 14 cm

Question - 19 : -
How many coins 1.75 cm in diameter and 2 mm thick must be melted to forma cuboid 11 cm x 10 cm x 7 cm?

Answer - 19 : -

Given,

Diameter of the coin =1.75 cm

So, its radius =1.74/2 = 0.875 cm

Thickness or theheight = 2 mm = 0.2 cm

We know that,

Volume of the cylinder(V₁) = πr²h

= π 0.875² ×0.2

And, the volume of thecuboid (V₂) = 11 × 10 × 7 cm³

Let the number ofcoins needed to be melted be n.

So, we have

V₂ = V₁ ×n

11 × 10 × 7 = π 0.875² ×0.2 x n

11 × 10 × 7 = 22/7 x0.875² × 0.2 x n

On solving we get, n =1600

Therefore, the numberof coins required are 1600

Question - 20 : -
The surface area of a solid metallic sphere is 616 cm². It ismelted and recast into a cone of height 28 cm. Find the diameter of the base ofthe cone so formed.

Answer - 20 : -

Given,

The height of the cone= 28 cm

Surface area of thesolid metallic sphere = 616 cm³

We know that,

Surface area of thesphere = 4πr²

So, 4πr² =616

r² =49

r = 7

Radius of the solidmetallic sphere = 7 cm

Let’s assume r to bethe radius of the cone

We know that,

Volume of the cone =1/3 πr²h

= 1/3 πr²(28)….. (i)

Volume of the sphere =4/3 πr³

= 4/3 π7³ ………. (ii)

On equating equations(i) and (ii), we have

1/3 πr²(28)= 4/3 π7³

Eliminating the commonterms, we get

r²(28)= 4 x 7³

r²=49

r =7

So, diameter of thecone = 7 x 2 = 14 cm

Therefore, thediameter of the base of the cone is 14 cm

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