CBSE Class 10 Mathematics Surface Areas and Volumes Assignment Set 20

Exercise-1 Subjective Questions: Surface Area and Volume of Solid Figures

Question A-1. Find the volume of a cube whose surface area is 150 m².
Answer:
6a² = 150 ⇒ a² = 25 ⇒ a = 5 m.
Volume = a³ = 125 m³.

Teacher's Note:
1. Find the edge from the surface area first.

Question A-2. Three cubes each of side 5 cm are joined end to end. Find the surface area of the resulting cuboid.
Answer:
The cuboid is 15 cm × 5 cm × 5 cm.
T.S.A. = 2 (15 × 5 + 5 × 5 + 5 × 15) = 2 (75 + 25 + 75) = 350 cm².

Teacher's Note:
1. Add the three edges that are joined into one length.

Question A-3. The difference between the outer and inner curved surface areas of a hollow right circular cylinder, 14 cm long, is 88 cm². If the volume of metal used in making the cylinder is 176 cm³, find the outer and inner diameters of the cylinder. [Use π = 22/7]
Answer:
Let R and r be the outer and inner radii. Difference of curved areas: 2π(R − r)(14) = 88 ⇒ R − r = 1.
Volume of metal: π (R² − r²)(14) = 176 ⇒ R² − r² = 4 ⇒ R + r = 4.
So R = 2.5 cm and r = 1.5 cm. Outer diameter = 5 cm, inner diameter = 3 cm.

Teacher's Note:
1. Use R² − r² = (R − r)(R + r) to solve the two equations.

Question A-4. Two cylinders of same volume have their heights in the ratio 1 : 3. Find the ratio of their radii.
Answer:
\( \pi r_1^2 h_1 = \pi r_2^2 h_2 \Rightarrow \dfrac{r_1^2}{r_2^2} = \dfrac{h_2}{h_1} = \dfrac{3}{1} \)
\( r_1 : r_2 = \) √3 : 1.

Teacher's Note:
1. For equal volumes, the radius squared varies inversely with the height.

Question A-5. If the diameter of a cylindrical jar is increased by 25% without altering the volume, then by what percent must its height decrease?
Answer:
New radius = 1.25 r. Since volume is unchanged: r²h = (1.25r)² h′ ⇒ h′ = \( \dfrac{h}{1.5625} \) = 0.64h.
Decrease = 100 − 64 = 36%.

Teacher's Note:
1. Radius changes by a factor, volume depends on its square.

Question A-6. A right-angled triangle whose sides are 15 cm and 20 cm is made to revolve about its hypotenuse. Find the volume and the surface area of the double cone so formed. [Take π ≈ 3.14]
Answer:
Hypotenuse = 25 cm. The radius of the double cone = altitude on the hypotenuse = \( \dfrac{15 \times 20}{25} \) = 12 cm.
Volume = \( \dfrac{1}{3}\pi (12)^2 (25) \) = 1200π = 3768 cm³.
Surface area = π(12)(15) + π(12)(20) = 420π = 1318.8 cm². (With π = 22/7 these are 3771.43 cm³ and 1320 cm².)

Teacher's Note:
1. Two cones share the same base circle, whose radius is the altitude to the hypotenuse.

Conversion of Solids

Question B-1. Three cubes with sides in the ratio 3 : 4 : 5 are melted to form a single cube whose diagonal is \( 12\sqrt{3} \) cm. Find the sides of the cubes.
Answer:
Diagonal of cube = a√3 = 12√3 ⇒ a = 12 cm, so a³ = 1728.
Let the sides be 3k, 4k, 5k: 27k³ + 64k³ + 125k³ = 216k³ = 1728 ⇒ k³ = 8 ⇒ k = 2.
Sides of the cubes: 6 cm, 8 cm, 10 cm.

Teacher's Note:
1. A cube of edge a has diagonal a√3.

Question B-2. A wall 10 m long, 1 m broad and 4 m high is to be constructed with bricks 20 cm long, 10 cm broad and 10 cm high. How many bricks are required to build the wall?
Answer:
Volume of wall = 1000 × 100 × 400 = 4 × 10⁷ cm³. Volume of a brick = 20 × 10 × 10 = 2000 cm³.
Number of bricks = \( \dfrac{4 \times 10^7}{2000} \) = 20000.

Teacher's Note:
1. Use the same unit (cm) for the wall and the brick.

Question B-4. Water flows out through a circular pipe whose internal radius is 1 cm, at the rate of 80 cm/second into an empty cylindrical tank, the radius of whose base is 40 cm. By how much will the level of water rise in the tank in half an hour?
Answer:
Volume flowing in 1800 s = π × 1² × (80 × 1800) = 144000π cm³.
π × 40² × h = 144000π ⇒ h = 90 cm.

Teacher's Note:
1. Length of water column = speed × time.

Question B-5. A solid sphere of radius 3 cm is melted and then cast into small spherical balls each of diameter 0.6 cm. Find the number of balls thus obtained.
Answer:
Number = \( \dfrac{\frac{4}{3}\pi (3)^3}{\frac{4}{3}\pi (0.3)^3} = \left( \dfrac{3}{0.3} \right)^3 = 10^3 \) = 1000.

Teacher's Note:
1. The radius of each ball is 0.3 cm.

Question B-6. The diameters of external and internal surfaces of a hollow spherical shell are 10 cm and 6 cm respectively. If it is melted and recast into a solid cylinder of length \( 2\dfrac{2}{3} \) cm, find the diameter of the cylinder.
Answer:
Volume of shell = \( \dfrac{4}{3}\pi (5^3 - 3^3) = \dfrac{392\pi}{3} \) cm³.
π R² × \( \dfrac{8}{3} \) = \( \dfrac{392\pi}{3} \) ⇒ R² = 49 ⇒ R = 7 cm.
Diameter of the cylinder = 14 cm.

Teacher's Note:
1. Subtract the inner sphere from the outer sphere to get the volume of the shell.

Question B-7. A hemispherical bowl of internal radius 36 cm is full of liquid. The liquid is to be filled into cylindrical shaped small bottles each of diameter 3 cm and height 6 cm. How many bottles are needed to empty the bowl?
Answer:
Volume of bowl = \( \dfrac{2}{3}\pi (36)^3 \) = 31104π cm³. Volume of a bottle = π (1.5)² (6) = 13.5π cm³.
Number of bottles = \( \dfrac{31104}{13.5} \) = 2304.

Teacher's Note:
1. Divide the volume of the bowl by the volume of one bottle.

Question B-9. The surface area of a solid metallic sphere is 616 cm². It is melted and recast into a cone of height 28 cm. Find the diameter of the base of the cone so formed. [Use π = 22/7]
Answer:
4πr² = 616 ⇒ r² = 49 ⇒ r = 7 cm. Volume of sphere = \( \dfrac{4}{3}\pi (7)^3 \).
\( \dfrac{1}{3}\pi R^2 (28) = \dfrac{4}{3}\pi (343) \Rightarrow R^2 = 49 \Rightarrow R = 7 \)
Diameter of the base = 14 cm.

Teacher's Note:
1. Find the radius of the sphere from its surface area first.

Question B-10. A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of water displaced out of the cylindrical vessel.
Answer:
Volume of water displaced = volume of the cone = \( \dfrac{1}{3} \times \dfrac{22}{7} \times 3.5^2 \times 6 \)
= 77 cm³.

Teacher's Note:
1. A fully immersed solid displaces water equal to its own volume.

Combination of Solids

Question C-2. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter ℓ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Answer:
Radius of hemisphere = \( \dfrac{\ell}{2} \).
Surface area = 6ℓ² − π \( \dfrac{\ell^2}{4} \) + 2π \( \dfrac{\ell^2}{4} \) = 6ℓ² + \( \dfrac{\pi \ell^2}{4} \)
= \( \dfrac{\ell^2}{4} \) (24 + π) sq. units.

Teacher's Note:
1. Remove the circle from the face and add the curved surface of the hemisphere.

Question C-3. A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and slant height of the conical portion is 53 m, find the area of the canvas needed to make the tent.
Answer:
Canvas = C.S.A. of cylinder + C.S.A. of cone = 2π(52.5)(3) + π(52.5)(53)
= \( \dfrac{22}{7} \times 52.5 \times (6 + 53) \) = 165 × 59 = 9735 m².

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

Teacher's Note:
1. The hemisphere hides a circle of the top face but adds its curved surface.

Teacher's Note:
1. Actual capacity = apparent capacity − volume of the raised hemisphere.