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Chapter 12: Surface Areas and Volumes

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This page provides comprehensive Chapter 12: Surface Areas and Volumes - Standard Worksheet - SJMaths. Standard level practice worksheet for Class 10 Surface Areas and Volumes. Practice for CBSE Board Exams.

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  1. Question 1: Two cubes each of volume 64 cm$^3$ are joined end to end. Find the surface area of the resulting cuboid.
    Solution:
    Step 1: Volume of cube $= a^3 = 64 \Rightarrow a = 4$ cm.
    Step 2: When joined end to end, length $L = 4+4=8$ cm, breadth $B=4$ cm, height $H=4$ cm.
    Step 3: Surface Area $= 2(LB + BH + HL) = 2(32 + 16 + 32) = 2(80) = 160$ cm$^2$.
    Answer: 160 cm$^2$.
  2. Question 2: A decorative block is made of two solids - a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Take $\pi = 22/7$)
    Solution:
    Step 1: TSA of block = TSA of Cube + CSA of Hemisphere - Area of Base of Hemisphere.
    Step 2: TSA $= 6a^2 + 2\pi r^2 - \pi r^2 = 6a^2 + \pi r^2$.
    Step 3: $a=5, r=2.1$. TSA $= 6(25) + \frac{22}{7}(2.1)(2.1) = 150 + 22(0.3)(2.1) = 150 + 13.86 = 163.86$ cm$^2$.
    Answer: 163.86 cm$^2$.
  3. Question 3: A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
    Solution:
    Step 1: Radius $r = 3.5$ cm. Height of cone $h = 15.5 - 3.5 = 12$ cm.
    Step 2: Slant height $l = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5$ cm.
    Step 3: TSA $= \text{CSA(Cone)} + \text{CSA(Hemisphere)} = \pi rl + 2\pi r^2 = \pi r(l + 2r)$.
    Step 4: TSA $= \frac{22}{7} \times 3.5 (12.5 + 7) = 11 \times 19.5 = 214.5$ cm$^2$.
    Answer: 214.5 cm$^2$.
  4. Question 4: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.
    Solution:
    Step 1: $r = 1$ cm, $h = 1$ cm.
    Step 2: Volume $= \text{Vol(Cone)} + \text{Vol(Hemisphere)} = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$.
    Step 3: Volume $= \frac{1}{3}\pi (1)^2 (1) + \frac{2}{3}\pi (1)^3 = \frac{1}{3}\pi + \frac{2}{3}\pi = \pi$ cm$^3$.
    Answer: $\pi$ cm$^3$.
  5. Question 5: 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.
    Solution:
    Step 1: Volume of Sphere = Volume of Cylinder. $\frac{4}{3}\pi R^3 = \pi r^2 h$.
    Step 2: $\frac{4}{3}(4.2)^3 = (6)^2 h \Rightarrow h = \frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 36}$.
    Step 3: $h = \frac{296.352}{108} = 2.744$ cm.
    Answer: 2.744 cm.
  6. Question 6: From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm$^2$.
    Solution:
    Step 1: $h=2.4, r=0.7$. Slant height $l = \sqrt{2.4^2 + 0.7^2} = \sqrt{5.76 + 0.49} = \sqrt{6.25} = 2.5$ cm.
    Step 2: TSA $= \text{CSA(Cylinder)} + \text{CSA(Cone)} + \text{Area(Base)}$.
    Step 3: TSA $= 2\pi rh + \pi rl + \pi r^2 = \pi r(2h + l + r)$.
    Step 4: TSA $= \frac{22}{7} \times 0.7 (4.8 + 2.5 + 0.7) = 2.2 \times 8 = 17.6$ cm$^2$.
    Answer: 18 cm$^2$.
  7. Question 7: 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.
    Solution:
    Step 1: Volume of earth dug $= \pi r^2 h = \frac{22}{7} \times (3.5)^2 \times 20 = 22 \times 0.5 \times 3.5 \times 20 = 770$ m$^3$.
    Step 2: Volume of platform $= L \times B \times H = 22 \times 14 \times H$.
    Step 3: $22 \times 14 \times H = 770 \Rightarrow H = \frac{770}{308} = 2.5$ m.
    Answer: 2.5 m.
  8. Question 8: A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm$^3$ of iron has approximately 8g mass. (Use $\pi = 3.14$)
    Solution:
    Step 1: Vol(Big Cylinder) $= \pi (12)^2 (220) = 3.14 \times 144 \times 220 = 99475.2$ cm$^3$.
    Step 2: Vol(Small Cylinder) $= \pi (8)^2 (60) = 3.14 \times 64 \times 60 = 12057.6$ cm$^3$.
    Step 3: Total Volume $= 111532.8$ cm$^3$.
    Step 4: Mass $= 111532.8 \times 8 = 892262.4$ g $= 892.26$ kg.
    Answer: 892.26 kg.
  9. Question 9: A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm$^3$. Check whether she is correct, taking the above as the inside measurements, and $\pi = 3.14$.
    Solution:
    Step 1: Vol(Sphere) $= \frac{4}{3} \pi (4.25)^3 = \frac{4}{3} \times 3.14 \times 76.7656 = 321.39$ cm$^3$.
    Step 2: Vol(Neck) $= \pi (1)^2 (8) = 3.14 \times 8 = 25.12$ cm$^3$.
    Step 3: Total Volume $= 321.39 + 25.12 = 346.51$ cm$^3$.
    Answer: She is incorrect. Correct volume is 346.51 cm$^3$.
  10. Question 10: 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.
    Solution:
    Step 1: Vol(Container) $= \pi (6)^2 (15) = 540\pi$.
    Step 2: Vol(One Cone + Hemisphere) $= \frac{1}{3}\pi (3)^2 (12) + \frac{2}{3}\pi (3)^3 = 36\pi + 18\pi = 54\pi$.
    Step 3: Number of cones $= \frac{540\pi}{54\pi} = 10$.
    Answer: 10 cones.
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