The document RD Sharma Solutions -Ex-19.2, (Part -2), Surface Area And Volume Of Right Circular Cylinder, Class 9 Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**Q12. The curved surface area of a cylinder is 1320cm ^{2} and its base had diameter 21 cm. Find the height and volume of the cylinder.**

**Solution:**

Let, r be the radius of the cylinder

h be the height of the cylinder

â‡’ 2r = 21cm

â‡’ r = 21/2

= 10.5cm

Given, Curved surface area(CSA) = 1320cm^{2}

â‡’ 2Î rh = 1320

â‡’

â‡’ h = 1320/66

â‡’ h = 20 cm

Volume of cylinder = Ï€r^{2}*h

= 22 * 1.5 * 10.5 * 20

= 6930cm^{2}

**Q13. The ratio between the radius of the base and the height of a cylinder is 2:3.Find the total surface area of the cylinder, if its volume is 1617cm ^{2}.**

**Solution:**

Let, r be the radius of the cylinder

h be the height of the cylinder

Volume of cylinder = Ï€r^{2}*h

h = 10.5 cm

from, eq 1

= 7 cm

Total surface area of cylinder = 2Ï€r(h+r)

**Q14. A rectangular sheet of paper, 44 cm*20 cm, is rolled along its length of form cylinder. Find the volume of the cylinder so formed.**

**Solution:**

Given, the dimensions of the sheet are 44cm*20cm

Here, length = 44 cm

Height = 20 cm

2Ï€r = 44

Volume of cylinder = r^{2}*h

= 154*20 = 3080cm^{3}

**Q15. The curved surface area of cylindrical pillar is 264m ^{2} and its volume is 924m^{3}. Find the diameter and the height of the pillar.**

**Solution:**

Let, r be the radius of the cylindrical pillar

h be the height of the cylindrical pillar

CSA = 264m^{2}

2Ï€rh = 264m^{2} â€”â€”â€“ 1

â‡’ Volume of the cylinder = 924m^{2}

Î *r^{2}*h = 924

Î rh(r) = 924

Substitute Ï€rh in eq 1

substitute r value in eq 1

so, the diameter = 2r = 2(7) = 14 m and height = 6 m

**Q16. Two circular cylinders of equal volumes have their heights in the ratio 1:2. Find the ratio of two radii.**

**Solution:**

Let, r_{1},r_{2} be the radii of the cylinder

h_{1},h_{2} be the height of the cylinder

v_{1},v_{2} be the volume of the cylinder

â‡’

â‡’

â‡’

â‡’

Hence, the ratio of the radii are

**Q17. The height of a right circular cylinder is 10.5 m. Three times the sum of the areas of its two circular faces is twice the area of the curved surface. Find the volume of the cylinder.**

**Solution:**

Let, r be the radius of the right circular cylinder

h be the height of the right circular cylinder

h = 10.5 cm

â‡’ 3(2Ï€r^{2}) = 2(2Ï€rh)

â‡’ 3r = 2h

â‡’

â‡’

â‡’ r = 7cm

Volume of the cylinder = r^{2}*h

= 154*10.5

= 1617cm^{3}

**Q18. How many cubic meters of earth must be dug out to sink a well 21m deep and 6 m diameter? Find the cost of plastering the inner surface as well at Rs.9.50 perm ^{2}.**

**Solution:**

Let, r be the radius

h be the height

here, h = 21m

2r = 6

â‡’ r = 6/2

= 3 m

Volume of the cylinder = r^{2}*h

= 66*9

= 594cm3

Cost of plastering = 9.5 per m^{3}

Cost of plastering inner surface = Rs.(594*9.50) = Rs. 5643

**Q19. The trunk of a tree is cylindrical and its circumference is 176 cm. If the length of the tree is 3 m. Find the volume of the timber that can be obtained from the trunk.**

**Solution:**

We know that, circumference = 2Ï€r

â‡’176 = 2Ï€r

â‡’ r = 28 cm

Here, height(h) = 3m = 300cm

Volume of timber = r^{2}*h

= 44*8400 = 739200cm3 (or) 0.7392m^{3}

**Q20. A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment.**

**Solution:**

Let, r be the radius of well

h be the height of well

here, h = 8m

2r = 14

â‡’ r = 14/2

= 7m

Volume of well = r^{2}*h

= 22*56

= 1232m^{3}

Let, r_{e} be the radius of embankment

h_{e }be the height of embankment

Volume of well = Volume of embankment

**Q21. The difference between inside and outside surfaces of a cylindrical tube is 14 cm long is 88 sq.cm. If the volume of the tube is 176 cubic cm, Find the inner and outer radii of the tube.**

**Solution:**

Let, R be the outer radius

R be the inner radius

Here, h = 14cm

2Ï€Rh â€“ 2Ï€rh = 88

â‡’ 2Ï€h(R â€“ r) = 88

â‡’ (R â€“ r) = 1cm â€”â€”â€” 1

Volume of tube = Ï€*R^{2}*h â€“ Ï€*r^{2}*h

176 = Ï€h(R^{2} â€“ r^{2})

â‡’ (R^{2} â€“ r^{2}) = 4

â‡’ (R + r)(R â€“ r) = 4

Here, (R â€“ r) = 1

â‡’ (R + r)(1) = 4

â‡’ (R + r) = 4 cm

â‡’ R = 4 â€“ r â€”â€”â€”â€” 2

Here , R â€“ r = 1

â‡’ R = 1 + r

Substitute R value in eq 2

â‡’ 1 + r = 4 â€“ r

â‡’ 2r = 3

â‡’ r = 32

= 1.5 cm

Substitute â€˜râ€™ value in eq 1

â‡’ R â€“ 1.5 = 1

â‡’ R = 1 + 1.5

â‡’ R = 2.5 cm

Hence, the value of inner radii is 1.5 cm and radius of outer radii is 2.5 cm

**Q.22. Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 meters per second into a cylindrical tank. The water is collected in a cylindrical vessel radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes?**

**Solution:**

Given data is as follows :

Internal diameter of the pipe = 2 cm

Water flow rate through the pipe = 6 m/sec

Radius of the tank = 60 cm

Time = 30 minutes

The volume of water that flows for 1 sec through the pipe at the rate of 6 m/sec is nothing but the volume of the cylinder with n = 6

Also, given is the diameter which is 2 cm. Therefore ,

R = 1 cm

Since the speed with which water flows through the pipe is in meters/second, let us convert the radius of the pipe from centimeters to meters . Therefore ,

Volume of water collected in the tank after 30 minutes = Volume of water that flows through the pipe for 30 minutes

Now , we have to find the volume of water that flows for 30 minutes .

Since , speed of water is in metres/second , let us convert 30 minutes into seconds . It will be 30Ã—60

Volume of water that flows for 30 minutes

Now , considering the tank , we have been given the radius of tank in centimeters . Let us first convert it into metres . Let radius of tank be â€˜Râ€™ .

R = 60 cm

Volume of water collected in the tank after 30 minutes = Volume of water that flows through the pipe for 30 minutes

h = 3 m

Therefore , the height of the tank is 3 metres

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