CBSE Class 10 Mathematics Surface Areas and Volumes Assignment Set 19

Previous Year Problems

Question 2 [CBSE 2012, 4M]. Water is flowing through a cylindrical pipe of internal diameter 2 cm into a cylindrical tank of base radius 40 cm at the rate of 0.4 m/s. Determine the rise in level of water in the tank in half an hour.
Answer:
Speed = 0.4 m/s = 40 cm/s. In 30 minutes (1800 s) the length of water that flows = 40 × 1800 = 72000 cm.
Volume of water = π × 1² × 72000 cm³. Let the rise in the tank be h: π × 40² × h = π × 72000.
h = \( \dfrac{72000}{1600} \) = 45 cm.

Teacher's Note:
1. Volume through the pipe = cross-section × length of water column.

Question 3 [CBSE 2012, 3M]. A wooden toy was made by scooping out a hemisphere of the same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the volume of wood in the toy. [Use π = 22/7]
Answer:
Volume of cylinder = \( \dfrac{22}{7} \times 3.5^2 \times 10 \) = 385 cm³.
Volume of two hemispheres = \( 2 \times \dfrac{2}{3} \times \dfrac{22}{7} \times 3.5^3 \) = \( \dfrac{539}{3} \) cm³.
Volume of wood = 385 − \( \dfrac{539}{3} \) = 205.33 cm³.

Teacher's Note:
1. Scooped-out parts are subtracted from the cylinder.

Question 4 [CBSE 2013, 3M]. A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder of the same diameter. The diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the total surface area of the vessel. [Use π = 22/7]
Answer:
r = 7 cm. Height of the cylinder = 13 − 7 = 6 cm.
The vessel is open at the top, so T.S.A. = C.S.A. of hemisphere + C.S.A. of cylinder = 2πr² + 2πrh = 2 × \( \dfrac{22}{7} \) × 7 × (7 + 6)
= 572 cm².

Teacher's Note:
1. A hollow cylinder has no top face, and the hemisphere is joined to it at the bottom.

Question 5 [CBSE 2013, 3M]. A hemispherical depression is cut out from one face of a cubical block of side 7 cm, such that the diameter of the hemisphere is equal to the edge of the cube. Find the surface area of the remaining solid. [Use π = 22/7]
Answer:
r = 3.5 cm. Surface area = 6 × 7² − πr² + 2πr² = 294 + πr² = 294 + \( \dfrac{22}{7} \times 3.5^2 \)
= 294 + 38.5 = 332.5 cm².

Teacher's Note:
1. The circle removed from the top face is replaced by the curved surface of the hemisphere.

Question 7 [CBSE 2014, 1M]. A rectangular sheet of paper 40 cm × 22 cm is rolled to form a hollow cylinder of height 40 cm. Find the radius of the cylinder (in cm).
Answer:
Height is 40 cm, so the 22 cm side forms the circumference: 2πr = 22.
r = \( \dfrac{22 \times 7}{2 \times 22} \) = 3.5 cm.

Teacher's Note:
1. The side that is not the height becomes the circumference of the base.

Question 8 [CBSE 2014, 3M]. A 5 m wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used at the rate of Rs. 25 per metre. [Use π = 22/7]
Answer:
r = 7 m, ℓ = \( \sqrt{7^2 + 24^2} \) = 25 m. C.S.A. = \( \dfrac{22}{7} \times 7 \times 25 \) = 550 m².
Length of cloth = \( \dfrac{550}{5} \) = 110 m. Cost = 110 × 25 = Rs. 2750.

Teacher's Note:
1. Length of cloth = area ÷ width.

Question 10 [CBSE 2014, 4M]. Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m, and the canvas to be used costs Rs. 100 per sq. m, find the amount the associations will have to pay. What values are shown by these associations? [Use π = 22/7]
Answer:
r = 2.1 m. Slant height of cone = \( \sqrt{2.1^2 + 2.8^2} \) = 3.5 m.
Canvas for one tent = 2πrh + πrℓ = \( \dfrac{22}{7} \times 2.1 \times (8 + 3.5) \) = 75.9 m². Cost of one tent = 75.9 × 100 = Rs. 7590.
Cost of 100 tents = Rs. 759000. The associations pay 50%: Rs. 379500.
Values shown: helpfulness, concern for the flood victims and social responsibility.

Teacher's Note:
1. Find the canvas for one tent first, then scale to 100 tents and take half.

Question 11 [CBSE 2014, 4M]. A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of each bottle, if 10% liquid is wasted in this transfer.
Answer:
Volume of liquid = \( \dfrac{2}{3}\pi (18)^3 \) = 3888π cm³. Liquid used = 90% of it = 3499.2π cm³.
72 × π × 3² × h = 3499.2π ⇒ 648 h = 3499.2
h = 5.4 cm.

Teacher's Note:
1. Only 90% of the liquid reaches the bottles.

Question 12 [CBSE 2015, 3M]. A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs. 5 per 100 sq. cm. [Use π = 3.14]
Answer:
The hemisphere sits on a face of the cube, so its largest diameter is the edge of the cube: 10 cm, r = 5 cm.
T.S.A. = 6 × 10² − πr² + 2πr² = 600 + 3.14 × 25 = 678.5 cm².
Cost = 678.5 × \( \dfrac{5}{100} \) = Rs. 33.93 (approx).

Teacher's Note:
1. The hemisphere covers a circle of the top face and adds its own curved surface.

Question 13 [CBSE 2015, 3M]. 504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area. [Use π = 22/7]
Answer:
Volume of 504 cones = \( 504 \times \dfrac{1}{3}\pi (1.75)^2 (3) \) = 1543.5π cm³.
\( \dfrac{4}{3}\pi r^3 = 1543.5\pi \Rightarrow r^3 = 1157.625 \Rightarrow r = 10.5 \) cm.
Diameter = 21 cm. Surface area = 4πr² = 4 × \( \dfrac{22}{7} \) × 10.5² = 1386 cm².

Teacher's Note:
1. Add the volumes of all cones, then equate to the sphere volume.

Question 14 [CBSE 2015, 4M]. From each end of a solid metal cylinder, metal was scooped out in hemispherical form of same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire. [Use π = 22/7]
Answer:
Volume of the remaining solid = π(4.2)²(10) − 2 × \( \dfrac{2}{3}\pi (4.2)^3 \) = π (176.4 − 98.784) = 77.616π cm³.
Wire radius = 0.7 cm. π (0.7)² L = 77.616π ⇒ L = \( \dfrac{77.616}{0.49} \) = 158.4 cm.

Teacher's Note:
1. Thickness of a wire means its diameter, so the radius is half of 1.4 cm.

Teacher's Note:
1. The tent has no floor, so only curved surfaces are counted.

Question 16 [CBSE 2016, 3M]. A conical vessel, with base radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel. [Use π = 22/7]
Answer:
Volume of water = \( \dfrac{1}{3}\pi (5)^2 (24) \) = 200π cm³.
π (10)² h = 200π ⇒ h = 2 cm.

Teacher's Note:
1. Equal volumes of water in both vessels.

Question 17 [CBSE 2016, 3M]. A sphere of diameter 12 cm is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by \( 3\dfrac{5}{9} \) cm. Find the diameter of the cylindrical vessel.
Answer:
Volume of sphere = \( \dfrac{4}{3}\pi (6)^3 \) = 288π cm³. Rise = \( \dfrac{32}{9} \) cm.
π r² × \( \dfrac{32}{9} \) = 288π ⇒ r² = 81 ⇒ r = 9 cm.
Diameter of the vessel = 18 cm.

Teacher's Note:
1. The volume of water displaced is the volume of the sphere.

Question 18 [CBSE 2016, 4M]. Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively offered to the state government to provide place and the canvas for 1500 tents to be fixed by the government and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 m and height 3.5 m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs. 120 per sq. m, find the amount shared by each school to set up the tents. What value is generated by the above problem? [Use π = 22/7]
Answer:
r = 2.8 m. Slant height of cone = \( \sqrt{2.8^2 + 2.1^2} \) = 3.5 m.
Canvas for one tent = 2πrh + πrℓ = \( \dfrac{22}{7} \times 2.8 \times (7 + 3.5) \) = 92.4 m². For 1500 tents = 138600 m².
Total cost = 138600 × 120 = Rs. 16632000. Share of each of the 50 schools = Rs. 332640.
Value: concern for the homeless, helpfulness and cooperation.

Teacher's Note:
1. Divide the total cost equally among the 50 schools.

Question 20 [CBSE 2017, 3M]. The dimensions of a solid iron cuboid are 4.4 m × 2.6 m × 1.0 m. It is melted and recast into a hollow cylindrical pipe of 30 cm inner radius and thickness 5 cm. Find the length of the pipe.
Answer:
Volume of cuboid = 4.4 × 2.6 × 1.0 = 11.44 m³ = 11440000 cm³. Outer radius of pipe = 35 cm.
Cross-sectional area of the pipe wall = \( \dfrac{22}{7} (35^2 - 30^2) \) = \( \dfrac{22}{7} \times 325 \) cm².
Length = \( \dfrac{11440000 \times 7}{22 \times 325} \) = 11200 cm = 112 m.

Teacher's Note:
1. Cross-section of a hollow pipe is a ring: π(R² − r²).

Question 21 [CBSE 2017, 4M]. In a rain water harvesting system, the rain-water from a roof of 22 m × 20 m drains into a cylindrical tank having diameter of base 2 m and height 3.5 m. If the tank is full, find the rainfall in cm. Write your views on water conservation.
Answer:
Volume of tank = \( \dfrac{22}{7} \times 1^2 \times 3.5 \) = 11 m³.
Rainfall × (22 × 20) = 11 ⇒ rainfall = \( \dfrac{11}{440} \) m = 0.025 m = 2.5 cm.
Views: harvesting rain water recharges ground water and saves water for future use.

Teacher's Note:
1. Water collected on the roof equals the volume of the full tank.

Question 22 [CBSE 2017, 4M]. Water in a canal, 5.4 m wide and 1.8 m deep, is flowing with a speed of 25 km/hour. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation?
Answer:
Length of water flowing in 40 minutes = 25 × \( \dfrac{40}{60} \) km = \( \dfrac{50}{3} \) km = \( \dfrac{50000}{3} \) m.
Volume = 5.4 × 1.8 × \( \dfrac{50000}{3} \) = 162000 m³.
Area irrigated = \( \dfrac{162000}{0.1} \) = 1620000 m².

Teacher's Note:
1. Area irrigated = volume of water ÷ depth of standing water.