Surface Areas and Volumes CBQs — CBSE Class 10 (2027) with step-wise solutions

Sample Surface Areas and Volumes CBQs

1. Question 1MCQ1 mark
   The amount of mixture required to make one biscuit is 18 cm³. After cooking, the biscuit becomes a cylinder of radius 3 cm and height 0.7 cm and has some air trapped inside it. Find the volume of air trapped in the biscuit.
   • A. 0.7 cm³
   • B. 1.5 cm³
   • C. 1.8 cm³
   • D. 3.2 cm³
   Show step-wise solution

   1. [1 mark] Volume of air trapped = Volume of cooked biscuit − Volume of mixture = 19.8 − 18 = 1.8 cm³. Answer: C.

   Answer: C. 1.8 cm³

2. Question 2Assertion-Reason1 mark
   A water tank is a cylinder of radius 2 m and height 3 m standing on a hemispherical bottom of radius 2 m. Leave answers in terms of π. Assertion (A): The capacity of the tank is 52π/3 m³. Reason (R): The volume of a solid made by joining two solids is the sum of the volumes of the two solids.
   • (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
   • (B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.
   • (C) Assertion is true but Reason is false.
   • (D) Assertion is false but Reason is true.
   Show step-wise solution

   1. [1 mark] (A): Capacity = π(2)²(3) + (2/3)π(2)³ = 12π + 16π/3 = 52π/3 m³, found by adding the two volumes, so A and R are true and R explains A.

   Answer: (A)

3. Question 3Short answer3 marks
   Ramesh has recently built his house and installed a cylindrical water tank of radius 50 cm and height 175 cm. If water is filled in the tank at the rate of 11 litres per minute, how long will it take for the tank to be completely filled?
   Show step-wise solution

   1. [1 mark] Volume of tank = πr²h = (22/7) × 50 × 50 × 175 = 1,375,000 cm³ = 1375 litres.
   2. [1 mark] Identify rate of filling = 11 litres per minute.
   3. [1 mark] Time = volume / rate = 1375 / 11 = 125 minutes.

   Answer: 125 minutes

4. Question 4Case study4 marks
   Tennis balls are sold in a cylindrical tin that holds exactly three balls placed one on top of another. Each ball is a sphere of diameter 7 cm; each ball just touches the curved wall of the tin, and the stack of three balls just touches the lid and the bottom. (Take π = 22/7.) Based on the above information, answer the following questions: (i) Write the radius and the height of the tin. [1 mark] (ii) Find the volume of one tennis ball. [1 mark] (iii) Find the volume of the empty space inside the tin, and the fraction of the tin's volume that is empty. [2 marks]
   Show step-wise solution

   1. [1 mark] (i) Radius of tin = radius of a ball = 3.5 cm; height of tin = 3 × 7 = 21 cm.
   2. [1 mark] (ii) Volume of one ball = (4/3)πr³ = (4/3) × (22/7) × 3.5 × 3.5 × 3.5 = 539/3 cm³ ≈ 179.67 cm³.
   3. [1 mark] (iii) Volume of tin = πr²h = (22/7) × 3.5 × 3.5 × 21 = 808.5 cm³; volume of three balls = 3 × 539/3 = 539 cm³.
   4. [1 mark] (iii) Empty space = 808.5 − 539 = 269.5 cm³, which is 269.5/808.5 = 1/3 of the tin.

   Answer: (i) r = 3.5 cm, h = 21 cm (ii) 539/3 cm³ ≈ 179.67 cm³ (iii) 269.5 cm³; 1/3 of the tin

5. Question 5Long answer5 marks
   How many cubic centimetres of iron is required to construct an open box whose external dimensions are 36 cm, 25 cm and 16.5 cm provided the thickness of the iron is 1.5 cm. If one cubic cm of iron weighs 7.5 g, find the weight of the box.
   Show step-wise solution

   1. [1 mark] External volume = 36 × 25 × 16.5 = 14850 cm³.
   2. [1 mark] Internal length = 36 − 2(1.5) = 33, width = 25 − 2(1.5) = 22.
   3. [1 mark] Internal height = 16.5 − 1.5 = 15 (open box, only the base removes thickness once), so internal volume = 33 × 22 × 15 = 10890 cm³.
   4. [1 mark] Volume of iron = 14850 − 10890 = 3960 cm³.
   5. [1 mark] Weight = 3960 × 7.5 g = 29700 g = 29.7 kg.

   Answer: Iron required = 3960 cm³; weight = 29.7 kg.

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