NCERT Solutions For Class 10 Math Chapter – 13 Exercise – 13.3

NCERT Solutions For Class 10 Math Chapter – 13 Exercise – 13.3

Q1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Solution:-

Volume of sphere = Volume of cylinder

4/3∏r³ = ∏r²h

4/3r³ = r²h

H = 4r³/3r₂²

= 4 x 4.2 x 4.2 x 4.2/3 x 6 x 6

= 2.74 cm

The height of the cylinder so formed will be 2.74 cm.

Q2. Metallic spheres of radii 6 cm , 8 cm and 10 cm , respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Solution:-

Volume of the resulting sphere = Sum of the volumes of three spheres

4/3 ∏r³ = 4/3∏r₁³ + 4/3∏r₁³ + 4/3∏r₁³

4/3∏r³ = 4/3∏(r₁³ + r₂³ + r₃³]

R³ = [r₁³ + r₂³ + r₃³]

R³ = [ (6 cm)³ + (8 cm)³ + (10 cm)³]

R³ = 1728 cm³

R = 12 cm

The radius of the sphere so formed will be 12 cm.

Q3. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to a form of platform 22 m by 14 m. Find the height of the platform.

Solution:-

Volume of the cylinder well = Volume of the cuboidal platform

∏r²h = = lbh

H = ∏r²h/lb

= 22/7 x 7/2 x 7/2 x 20 /22 x 14

= 5/2 m

= 2.5 m

The height of such platform will be 2.5 cm

Q4. A well of diameter 3 m is dug 14 m deep The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Solution:-

Volume of the cylindrical well = Volume of the hollow cylindrical embankment

∏r²h₁ = ∏r²h – ∏R²h

∏r²h₁ = ∏h(R² – r²)

R²h₁ = h(r – R)(R+r)

H = r²h₁/(R-r)(R+r)

= (1.5m)² x 14 m / (5.5m – 1.5m)(5.5m + 1.5)

= 1.125 m

The height of the embankment will be 1.125 m.

Q5. A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm , having a hemispherical shape on top Find the number of such cones which can be filled with ice – cream.

Solution:-

Volume of ice cream in container = n x Volume of the ice cream in each one

∏R²H = n x {1/3∏r²h + 2/3∏r³}

R²H = 1/3nr² ( h + 2r)

N = 3R²H/r²(h + 2r)

= 3 x (6 cm)² x 15 cm/(3 cm)² x (12 cm + 2 x 3 cm)

= 10 cm

Therefore 10 ice – cream cones can be filled with ice – cream in the container.

Q6. How many silver coin, 1.75 cm in diameter and of thickness 2 mm , must be melted to form a cuboid of dimensions 5.5 cm x 10 cm c 3.5 cm ?

Solution:-

Volume of n coins = Volume of cuboids

n x ∏r²h = l x b x h

n = l x b x h / ∏r²h₁

= 5.5 cm x 10 cm x 3.5 cm/22/7 x (0.875 cm) x 0.2 cm

= 400

Therefore The number of coins melted to form such a cuboid is 400.

Q7. A cylindrical bucket , 32 cm high and with radius of base 18 cm , is filled with sand. This bucket is empited on the ground and a conical heap sand is formed. If the height of the conical heap is 24 cm. Find the radius and slant height of the heap.

Solution:-

The diagram will be as-

Given,

Height (h₁) of cylindrical part of the bucket = 32 cm

Radius (r₁) of circular end of the bucket = 18 cm

Height of the conical heap ((h₂) = 24 cm

Now, let “r₂” be the radius of the circular end of the conical heap.

We know that volume of the sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.

∴ Volume of sand in the cylindrical bucket = Volume of sand in conical heap

π×r₁²×h₁ = (⅓)×π×r₂²×h₂

π×18²×32 = (⅓)×π ×r₂²×24

Or, r₂= 36 cm

And,

Slant height (l) = √(36²+24²) = 12√13 cm.

Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Solution:

It is given that the canal is the shape of a cuboid with dimensions as:

Breadth (b) = 6 m and Height (h) = 1.5 m

It is also given that

The speed of canal = 10 km/hr

Length of canal covered in 1 hour = 10 km

Length of canal covered in 60 minutes = 10 km

Length of canal covered in 1 min = (1/60)x10 km

Length of canal covered in 30 min (l) = (30/60)x10 = 5km = 5000 m

We know that the canal is cuboidal in shape. So,

Volume of canal = lxbxh

= 5000x6x1.5 m³

= 45000 m³

Now,

Volume of water in canal = Volume of area irrigated

= Area irrigated x Height

So, Area irrigated = 56.25 hectares

∴ Volume of canal = lxbxh

45000 = Area irrigatedx8 cm

45000 = Area irrigated x (8/100)m

Or, Area irrigated = 562500 m²= 56.25 hectares.

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Solution:

Ncert solutions class 10 chapter 13-21