The document RD Sharma Solutions -Ex-19.2, (Part -1), 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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**Q1. A soft drink is available in two packs- (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm, Which container has greater capacity and by how much?**

**Solution:**

Given,

The tin can will be cubical in shape.

Length (L) of tin can = 5 cm

Breadth (B) of tin can = 4 cm

Height (H) of tin can = 15 cm

Capacity of the tin can = l x b x h = (5 x 4 x 15) cm^{3}

Radius (R) of the circular end of the plastic cylinder = 7/2 cm = 3.5 cm

Height (H) of plastic cylinder = 10 cm

Capacity of plastic cylinder

Therefore, the plastic cylinder has greater capacity.

Difference in capacity = (385 – 300) cm^{3 }= 85 cm^{3}

**Q2. The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 20 cm and height 10 m, how much concrete mixture would be required to build 14 such pillars?**

**Solution:**

Given,

The concrete mixture is used to build up the pillars is required for the entire space of the pillar i.e, we need to find the volume of the cylinders.

Radius of the base of a cylinder = 20 cm

Volume of the cylindrical pillar = πR^{2}H

Therefore, volume of 14 pillars

**Q3. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 gm**.

**Solution:**

Given,

Inner radius (r_{1}) of a cylindrical pipe = 24/2 = 12 cm

Outer radius (r_{2}) of a cylindrical pipe = 24/2 = 14 cm

Height of pipe (h) = length of pipe = 35 cm

Mass of pipe = volume x density =

Mass of 1 cm^{3} wood = 0.6 gm

Therefore, mass of 5720 cm^{3 }wood = 5720 x 0.6 = 3432 gm = 3.432 kg

**Q4. If the lateral surface of a cylinder is 94.2 cm ^{2} and its height is 5 cm, find:**

**(i) radius of its base (ii) volume of the cylinder [Use pi = 3.141]**

**Solution:**

(i) Given,

Height of the cylinder = 5 cm

Let radius of cylinder be ‘r’

Curved surface of the cylinder = 94.2 cm^{2}

2 πrh = 94.2 cm^{2}

r = 3 cm [π = 31.4, h = 5 cm]

(ii) Volume of the cylinder = πr2h = (3.14 x 3^{2} x 5) cm^{3} = 141.3 cm^{3}

**Q5. The capacity of a closed cylindrical vessel of height 1 m is 15.4 liters. How many square meters of the metal sheet would be needed to make it?**

**Solution:**

Given,

Height of the cylindrical vessel = 15.4 litres = 0.0154 m^{3} [1m^{3} = 1000 litres]

Let the radius of the circular ends of the cylinders be ‘r’

πr2h = 0.0154 m^{3}

r = 0.07 m [π = 31.4, h =1m]

Total surface area of a vessel = 2πr(r+h)

= 2(3.14(0.07)(0.07+1)) m^{2 }= 0.4703 m^{2}

**Q6. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?**

**Solution:**

Given,

Radius (R) of cylindrical bowl = 7/2 cm = 3.5 cm [Diameter = 7 cm]

Height up to which the bowl is filled with soup = 4 cm

Volume of soup in 1 bowl = πr^{2}h

Volume of soup n 250 bowls = (250 x 154) cm^{3} = 38500 cm^{3} = 38.5 liters.

The hospital has to prepare 38.5 liters of soup daily in order to serve 250 patients.

**Q7. A hollow garden roller, 63 cm wide with a girth of 440 cm, is made of 4 cm thick iron. Find the volume of the iron.**

**Solution:**

Given, h = 63cm

The outer circumference of the roller = 440cm

Thickness of the roller = 4cm

Let, R be the external radius

We know that,

2πR = 440

R = 70

The thickness is given as 4cm, so the inner radius ‘r’ is given as

⇒ r = R – 4

⇒ = 70 – 4

⇒ = 66cm

Here, we know the inner and outer radii

So, the volume is given as

⇒ V = π(R^{2}−r^{2})h

**Q8. A solid cylinder has a total surface area of 231cm ^{2}. Its curved surface area is 3 of the total surface area. Find the volume of the cylinder.**

**Solution:**

Given,

Total surface area = 231cm^{2}

Curved surface area

= 154

We know that,

2πrh + 2πr^{2} = 231 —– 1

Here 2πrh is the curved surface area, so substitute tha value of CSA in eq 1

⇒ 154 + 2πr^{2} = 231

⇒ 2πr^{2} = 231- 154

⇒ 2πr^{2} = 77

We need to find the value of h

CSA = 154cm^{2}

⇒ 2πrh = 154

So the volume of the cylinder is,

Volume = πr^{2}h

The volume of the cylinder is 269.5cm^{3}

**Q9. The cost of painting the total outside surface of a closed cylindrical oil tank at 50 paise per square decimetre is Rs 198. The height of the tank is 6 times the radius of the base of the tank. Find the volume corrected to 2 decimal places.**

**Solution:**

Let the radius of the tank be r dm

Then, height = 6r dm

Cost of painting for 50 paisa per dm^{2} = Rs 198

Therefore, h =(6 x 3) dm = 18 dm

Volume of the tank

**Q10. The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. Calculate the ratio of their volumes and the ratio of their curved surfaces.**

**Solution:**

Let the radius of the cylinders be 2x and 3x and the height of the cylinders be 5y and 3y.

**Q11.The ratio between the curved surface area and the total surface area of a right circular cylinder is 1:2. Find the volume of the cylinder, if its total surface area is 616cm ^{2}.**

**Solution:**

Let, r be the radius of cylinder

h be the radius of cylinder

Total surface area (T.S.A) = 616cm^{2}

Now,

TSA = 2πrh + 2πr^{2}

⇒ 616 = CSA + 2πr^{2}

⇒ 616 = 308 + 2πr^{2}

⇒ 2πr^{2} = 616 – 308

⇒ 2πr^{2} = 308

⇒ r = 7 cm

Since,CSA = 308cm^{2}

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