Class 10 Surface Areas and Volumes Mock Test (CBSE)

Topics in Surface Areas and Volumes

• Surface area of a combination of solids
• Volume of a combination of solids

Surface Areas and Volumes practice questions with answers

1. Q1Multiple choice·1 mark
   A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1\text{ cm}$ and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$.

   1. (a)$\frac{2}{3}\pi\text{ cm}^3$
   2. (b)$\pi\text{ cm}^3$
   3. (c)$\frac{4}{3}\pi\text{ cm}^3$
   4. (d)$2\pi\text{ cm}^3$
   Show answer and step marks

   1. Correct option: (b) $\pi\text{ cm}^3$. Volume = $\frac{1}{3}\pi(1)^2(1) + \frac{2}{3}\pi(1)^3 = \pi\text{ cm}^3$+1

   Answer: (b)
2. Q2Assertion and reason·1 mark
   Assertion (A): The volume of a solid formed by joining two identical solid hemispheres of radius $r$ base-to-base is $\frac{4}{3}\pi r^3$. Reason (R): Joining two hemispheres of radius $r$ base-to-base forms a solid sphere of radius $r$, whose volume is $\frac{4}{3}\pi r^3$.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true but Reason (R) is false.
   4. (d)Assertion (A) is false but Reason (R) is true.
   Show answer and step marks

   1. Correct option: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). Both Assertion and Reason are true, and the Reason correctly explains why the volume is $\frac{4}{3}\pi r^3$.+1

   Answer: (a)
3. Q3Multiple choice·1 mark
   A solid is in the shape of a cone surmounted on a hemisphere. The radius of the hemisphere is $3\text{ cm}$ and the total height of the solid is $9\text{ cm}$. Find the volume of the solid.

   1. (a)$36\pi\text{ cm}^3$
   2. (b)$54\pi\text{ cm}^3$
   3. (c)$27\pi\text{ cm}^3$
   4. (d)$45\pi\text{ cm}^3$
   Show answer and step marks

   1. Correct option: (a) $36\pi\text{ cm}^3$. Height of cone $h = 9 - 3 = 6\text{ cm}$. Total volume = $\frac{2}{3}\pi(3)^3 + \frac{1}{3}\pi(3)^2(6) = 18\pi + 18\pi = 36\pi\text{ cm}^3$.+1

   Answer: (a)
4. Q4Multiple choice·1 mark
   A solid toy is in the form of a hemisphere surmounted by a right circular cone. If the radius of the hemisphere is $7\text{ cm}$ and the total height of the toy is $31\text{ cm}$, then the total surface area of the toy is (use $\pi = \frac{22}{7}$):

   1. (a)$858\text{ cm}^2$
   2. (b)$770\text{ cm}^2$
   3. (c)$924\text{ cm}^2$
   4. (d)$1078\text{ cm}^2$
   Show answer and step marks

   1. Correct option: (a) $858\text{ cm}^2$. Height of the cone $h = 31 - 7 = 24\text{ cm}$. Slant height $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = 25\text{ cm}$. Total surface area = $\pi r l + 2\pi r^2 = \frac{22}{7} \times 7 \times 25 + 2 \times \frac{22}{7} \times 7^2 = 550 + 308 = 858\text{ cm}^2$.+1

   Answer: (a)
5. Q5Assertion and reason·1 mark
   Assertion (A): When a solid cylinder is melted and recast into a cone, the volume of the cone is equal to the volume of the cylinder. Reason (R): The total surface area of a solid remains unchanged when it is recast into a different shape.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true but Reason (R) is false.
   4. (d)Assertion (A) is false but Reason (R) is true.
   Show answer and step marks

   1. Correct option: (c) Assertion (A) is true but Reason (R) is false. Assertion is true due to conservation of volume during recasting. Reason is false because total surface area generally changes when a solid is recast into a different shape.+1

   Answer: (c)
6. Q6Long answer·5 marks
   A solid metallic toy is in the form of a right circular cylinder with a hemispherical base attached at the lower end and a conical top attached at the upper end. The radius of the common base is $7\text{ cm}$ and the height of the cylindrical part is $12\text{ cm}$. If the total height of the toy is $43\text{ cm}$, find: (i) the slant height of the conical part, (ii) the total surface area of the toy (use $\pi = \frac{22}{7}$), (iii) the total volume of the material used in the toy.
   Show answer and step marks

   1. Height of the conical part = Total height - Height of cylinder - Radius of hemisphere $= 43 - 12 - 7 = 24\text{ cm}$.+1
   2. Slant height of the cone $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\text{ cm}$.+1
   3. Total surface area = CSA of cylinder + CSA of hemisphere + CSA of cone $= 2\pi rh + 2\pi r^2 + \pi rl = 2 \times \frac{22}{7} \times 7 \times 12 + 2 \times \frac{22}{7} \times 7^2 + \frac{22}{7} \times 7 \times 25 = 528 + 308 + 550 = 1386\text{ cm}^2$.+1.5
   4. Total volume = Volume of cylinder + Volume of hemisphere + Volume of cone $= \pi r^2 h + \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h_{cone} = \frac{22}{7} \times 7^2 \times 12 + \frac{2}{3} \times \frac{22}{7} \times 7^3 + \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24 = 1848 + 718.67 + 1232 = 3798.67\text{ cm}^3$.+1.5

   Answer: (i) Slant height $= 25\text{ cm}$; (ii) total surface area $= 1386\text{ cm}^2$; (iii) volume $= 3798.67\text{ cm}^3$ (approx.)

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