NCERT Solutions Maths Class 10 Exercise 13.3

NCERT Solutions Maths Class 10 Exercise 13.3

 

Maths Class 10 Exercise 13.3

 

1. 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: Radius of the sphere (r) = 4.2 cm

Volume of the sphere = 4/3 × πr3 

                                = 4/3 × π(4.2)3 cm3

Radius of the cylinder (R) = 6 cm

Let the height of the cylinder be H cm.

Then, the volume of the cylinder = πR2H  

                                                    = π(6)2H cm3

According to the question,

Volume of the sphere = Volume of the cylinder

4/3 × π(4.2)3 = π(6)2H

H = 4(4.2)3/3(6)2 

H = 2.74 cm

 

2. 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: Let the radii of the smaller spheres be r1, r2 and r3 and let the radius of the resulting sphere be r cm.

According to the question,

Volume of the resulting sphere = Volume of the three smaller spheres

4/3 × πr3 = 4/3 × πr1+ 4/3 × πr23 + 4/3 × πr33

4/3 × πr3 = 4/3 × π(6)3 + 4/3 × π(8)3 + 4/3 × π(10)3

r3 = (6)3 + (8)3 + (10)3

r3 = 216 + 512 + 1000

r3 = 1728 

r3 = (12)3

r = 12 cm

Hence, the radius of the resulting sphere is 12 cm.

 

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

 

Solution: Diameter of the well = 7 m

Radius of the well (r) = 7/2 m

Depth of the well (h) = 20 m

Length of the platform (l) = 22 m and breadth of the platform (b) = 14 m

Let the height of the platform be H m.

According to the question,

Volume of the earth dug from well = Volume of the earth spread out on the platform

πr2h = l × b × H

22/7 × 7/2 × 7/2 × 20 = 22 × 14 × H 

H = 2.5 m

 

4. 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: Diameter of the well = 3 m

Radius of the well (r) = 3/2 m

Depth of the earth dug from the well (h) = 14 m

Width of the embankment = 4 m

Radius of the well with embankment (r1) = 3/2 + 4 = 11/2 m

Let the height of the embankment be h1 m.

According to the question,

Volume of the embankment = Volume of the earth dug from the well

π[(r1)2 – (r)2]h1 = πr2h

[(11/2)2 – (3/2)2]h1 = (3/2)2 × 14

[121/4 – 9/4]h1 = 9/4 × 14

112/4 × h1 = 9/4 × 14

h1 = 1.125 m

Hence, the height of the embankment is 1.125 m.

 

5. 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 the top. Find the number of such cones which can be filled with ice cream.

 

Solution: Diameter of the right circular cylinder = 12 cm

Radius of the right circular cylinder (r) = 12/2 = 6 cm and its height (h) = 15 cm

Diameter of the cone = 6 cm

Radius of the cone (r1) = 6/2 = 3 cm and height (h1) = 12 cm

Radius of the hemispherical shape = 3 cm

Volume of ice cream in a cone = 1/3 × πr12h1 + 2/3 × πr13

Let the number of cones that can be filled with ice cream be n.

According to question,

Volume of n cones = Volume of the right circular cylinder

n × [1/3 × πr12h1 + 2/3 × πr13] = πr2h

n × π[1/3 × (3)2 × 12 + 2/3 × (3)3] = π × (6)2 × 15

n × [4 × (3)2 + 2 × (3)2] = (6)2 × 15

n × 54 = 36 × 15

n = 540/54

n = 10

Hence, 10 cones can be filled with ice cream.

 

6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

 

Solution: Diameter of the silver coin = 1.75 cm

Radius of the silver coin (r) = 1.75/2 = 7/8 cm and its thickness (h) = 2 mm = 1/5 cm

Dimensions of the cuboid = 5.5 cm × 10 cm × 3.5 cm

Let the number of coins to be melted is n.

According to question,

Volume of n silver coins = Volume of the cuboid

n × πr2h = 5.5 × 10 × 3.5

n × π(7/8)2 × 1/5 = 5.5 × 10 × 3.5

n × 22/7 × 49/64 × 1/5 = 5.5 × 10 × 3.5

n = 400

Hence, 400 silver coins must be melted.

 

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

 

Solution: Radius of the base of the cylindrical bucket (r) = 18 cm and its height (h) = 32 cm.

Volume of the cylindrical bucket = πr2h 

= π(18)2 × 32  

= 10368π cm3

Height of the conical heap (h1) = 24 cm

Let the radius of the conical heap be r1 cm.

Volume of the conical heap = 1/3 × πr12h1  

= 1/3 × π × r12 × 24  

= 8πr12 cm3

According to question, Volume of bucket = Volume of conical heap

10368π = 8πr12

r12 = 10368/8 

r12 = 1296

r1 = 36 cm

Now, slant height of the heap (l) = 

= 

= 

=   

=

 

8. 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: Width of the canal = 6 m

Depth of the canal = 1.5 m

Speed of the flow of water = 10 km/h

= 10 × 1000 m/h

= 10000 m/h

= 10000/60 m/min

= 500/3 m/min

Distance of water flown in 30 minutes = 500/3 × 30 m

Volume of water that flows in 30 minutes = 6 × 1.5 × 5000 = 45000 m3

Volume of water that flows in 30 minutes = Area it will irrigate × height of water

45000 = A × 8/100 

A = 4500000/8

A = 562500 m3

A = 562500/10000 hectares

A = 56.25 hectares

Hence, it will irrigate 562500 m3 or 56.25 hectares.

 

9. 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: Diameter of the cylindrical tank = 10 m

Radius of the cylindrical tank (r) = 10/2 = 5 m and depth of the cylindrical tank (h) = 2 m

Volume of the cylindrical tank = πr2h 

= π(5)2 ×

= 50π m3                                                             

Rate of flow of water (h1) = 3 km/h

= 3000 m/h

= 3000/60 m/min

= 50 m/min

Internal diameter of the pipe = 20 cm

Radius of the pipe (r1) = 10 cm = 0.1 m

Volume of water that flows in the pipe per minute = πr12h1 

= π(0.1)2 × 50  

π/2 m3

Required time = 50π ÷ π/2 

= 100 minutes

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