Students can Download Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 10 Maths Chapter 15 Surface Areas and Volumes Ex 15.3

Question 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.

Answer:

Given: Sphere

r_{1} = 4.2 cm.

Cylinder

r_{2} = 6 cm; h = ?

Volume of sphere = volume of cylinder

h = 2.744 cm

h = 2.74 cm

∴ height of cylinder is 2.74 cm

Question 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.

Answer:

r_{1} = 6cm, r_{2} = 8 cm and r_{3} = 10 cm.

Let the radius of resulting sphere be ‘r’

Volume of resulting sphere = sum of volumes of 3 spheres.

Question 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.

Answer:

l = 22 m and b = 14 m

Diameter of well = d= 7 m and radius = r = 7/2 m.

Depth of well = height of well = h = 20 m

Well is in the shape of cylinder

Volume of solid form well = πr^{2}h

Let height of the platform is H.

∴ Volume of soil dug from the well = Volume of soil spread out from Plat form

770 = l × b × H

770 = 22 × 14 × H

H = \(\frac{770}{22 \times 14}\) = 2.5m.

Question 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.

Answer:

Shape of well is in cylindrical diameter

= 3m and r = \(\frac{d}{2}=\frac{3}{2}\)m

Depth = h = 14 m.

Now

Outer radius of embankment

= r_{1} = \(\frac{3}{2}\) + 4 = \(\frac{11}{2}\)m.

inner radius of embankment = r_{2} = \(\frac{3}{2}\) m

Let the height of embankment = H

Volume of soil dug from well = volume of the embankment

Therefore height of embankment = 1.125 m

Question 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.

Answer:

Cylinder:

Diameter = d_{1} = 12 cm, r_{1} = \(\frac{12}{2}\) = 6cm and height = h_{1} = 15 cm.

Cone:

Diameter = d_{2} = 6 cm, r_{2} = \(\frac{6}{2}\) = 3 cm and height = h_{2} = 12 cm

Number of cones filled with ice cream

∴ Number of cones filled ice cream is 10.

Question 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?

Answer:

radius of coin = \(\frac{1.75}{2}\) = 0.875 cm and thickness

height = 2 mm = \(\frac{2}{10}\) cm = 0.2 cm

Cuboid: l = 5.5 cm, b = 10 cm and h = 3.5 cm

Number of coins

400 coins

Question 7.

A cylindrical bucket, 32 cm high and with a 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.

Answer:

Cylindrical bucket:

radius = r_{1} = 18 cm and height = h = 32 cm.

Conical heap:

radius = r_{2} = r and height = h_{2} = 24 cm.

volume of sand = volume of the cylindrical bucket

∴ volume of the conical heap will be equal to the volume of sand.

volume of cylindrical bucket = volume of the conical heap.

∴ radius = 36 cm and slant height = 12\(\sqrt{13}\) cm.

Question 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?

Answer:

Area of the cross-section of the canal

= 6 × 1.5 = 9m^{2}.

Speed of water = 10 km/hr

= \(\frac{10 \times 1000}{60}=\frac{1000}{6}\)m/sec

Now, The area of water flow from the canal in one minute

= area of cross section × speed of water

= 9 × \(\frac{1000}{6}\) = 1500 m^{2}

Hence, water flow in 30 minutes = 1500 × 30 = 45000 m^{3}

Volume of water flows in 30 minutes = volume of water irrigating the required area.

Now

∴ therefore area irrigated in 30 minutes is 562500 m^{2} = 56.25 hectares.

Question 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?

Answer:

Diameter of tank = 10 m

radius = \(\frac{d}{2}=\frac{10}{2}\) = 5m

Depth of tank = h = 2 m

Diameter of pipe = 20 cm

radius = \(\frac{20}{2}\) = 10cm = 0.1m

cross section Area of pipe = πr^{2 }= π(0.1)^{2 }= 0.01πm^{2}

Now, Speed of water 3 km/hr = \(\frac{3 \times 1000}{60}\) = 50 m/min.

Water flows in one minute from the pipe = 0.01π × 50 = \(\frac{50 \times 0.01 \times \pi}{100}\) = 0.5πm^{3}

Let the tank will be filled in t minutes water flows in ‘t’ minute from pipe = t × 0.5 π m^{3}

Now, volume of water filled in tank in ‘t’ minutes is equal to the volume of water flows from pipe in ‘f minutes.

Hence, volume of cylindrical tank = volume of water flows in ‘f minute.

= \(\frac{50}{0.5}=\frac{500}{5}\)

t = 100 minute

therefore tank will be filled in 100 minutes.