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Chapter 12: Surface Areas and Volumes

Overview

Practice Assertion and Reason questions on Class 10 Surface Areas and Volumes. Master relationships between radius, slant height, and volume when combining solids like cones, cylinders, and hemispheres for CBSE Board Section A.

Assertion-Reason Worksheet

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Directions:

In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:

  • (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  • (B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
  • (C) Assertion (A) is true but Reason (R) is false.
  • (D) Assertion (A) is false but Reason (R) is true.
  1. Question 1:
    Assertion (A): Two cubes each of volume 64 cm$^3$ are joined end to end. The surface area of the resulting cuboid is 160 cm$^2$.
    Reason (R): If $l, b, h$ are length, breadth and height of a cuboid, then Total Surface Area $= 2(lb+bh+hl)$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Volume $a^3 = 64 \Rightarrow a = 4$ cm.
    Step 2: For cuboid, $l = 4+4=8$, $b=4$, $h=4$.
    Step 3: TSA $= 2(32+16+32) = 2(80) = 160$ cm$^2$. A is true.
    Step 4: R is the correct formula used.
  2. Question 2:
    Assertion (A): The volume of the largest sphere that can be carved out of a cube of side $a$ is $\frac{1}{6}\pi a^3$.
    Reason (R): The diameter of the largest sphere carved out of a cube is equal to the side of the cube.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Diameter $d = a \Rightarrow$ Radius $r = a/2$.
    Step 2: Volume $= \frac{4}{3}\pi (a/2)^3 = \frac{4}{3}\pi \frac{a^3}{8} = \frac{1}{6}\pi a^3$. A is true.
    Step 3: R explains the relationship between sphere and cube.
  3. Question 3:
    Assertion (A): The volume of a cube is 64 cm$^3$ if its side is 4 cm.
    Reason (R): The total surface area of a cube of side $a$ is $6a^2$.
    (A)
    (B)
    (C)
    (D)
    Solution: (B)
    Step 1: Volume $= 4^3 = 64$. A is true.
    Step 2: R is the correct formula for TSA, but it does not explain the volume calculation in A.
  4. Question 4:
    Assertion (A): The total surface area of a top (lattu) is the sum of the curved surface area of the hemisphere and the curved surface area of the cone.
    Reason (R): The top is obtained by joining the plane surfaces of the hemisphere and cone together.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Since the plane surfaces are joined, they are internal. Only curved surfaces are exposed. A is true.
    Step 2: R correctly explains the construction of the solid.
  5. Question 5:
    Assertion (A): The number of solid spheres of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm is 5.
    Reason (R): Volume of material remains conserved during melting and recasting.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Vol(Cylinder) $= \pi (2)^2 (45) = 180\pi$.
    Step 2: Vol(Sphere) $= \frac{4}{3}\pi (3)^3 = 36\pi$.
    Step 3: Number $= 180\pi / 36\pi = 5$. A is true.
    Step 4: R is the principle used.
  6. Question 6:
    Assertion (A): If a cylinder and a cone have the same base radius and same height, then the ratio of their volumes is 3:1.
    Reason (R): Volume of cylinder is $\pi r^2 h$ and volume of cone is $\frac{1}{3}\pi r^2 h$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Ratio $= \pi r^2 h : \frac{1}{3}\pi r^2 h = 1 : 1/3 = 3:1$. A is true.
    Step 2: R gives the correct formulas used for comparison.
  7. Question 7:
    Assertion (A): If the height of a cone is 24 cm and diameter of base is 14 cm, then the slant height is 15 cm.
    Reason (R): Slant height $l = \sqrt{h^2 + r^2}$.
    (A)
    (B)
    (C)
    (D)
    Solution: (D)
    Step 1: Radius $r = 7$ cm. Height $h = 24$ cm.
    Step 2: $l = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25$ cm.
    Step 3: Assertion says 15 cm, which is false. R is true.
  8. Question 8:
    Assertion (A): The volume of a hemisphere of radius $r$ is $\frac{2}{3}\pi r^3$.
    Reason (R): The total surface area of a hemisphere is $2\pi r^2$.
    (A)
    (B)
    (C)
    (D)
    Solution: (C)
    Step 1: A is the correct formula for volume. A is true.
    Step 2: TSA of hemisphere is $3\pi r^2$ (curved + base). R is false.
  9. Question 9:
    Assertion (A): A solid metallic sphere of radius 8 cm is melted and recast into spherical balls each of radius 2 cm. The number of such balls is 64.
    Reason (R): Number of balls $= \frac{\text{Volume of big sphere}}{\text{Volume of small sphere}}$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: Ratio of volumes $= (\frac{R}{r})^3 = (\frac{8}{2})^3 = 4^3 = 64$. A is true.
    Step 2: R is the correct method.
  10. Question 10:
    Assertion (A): The volume of a cylinder is 3080 cm$^3$ and base radius is 14 cm. The height is 5 cm.
    Reason (R): Volume of cylinder $= \pi r^2 h$.
    (A)
    (B)
    (C)
    (D)
    Solution: (A)
    Step 1: $3080 = \frac{22}{7} \times 14 \times 14 \times h \Rightarrow 3080 = 616 h$.
    Step 2: $h = 3080/616 = 5$ cm. A is true.
    Step 3: R is the formula used.
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