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Chapter 12: Surface Areas and Volumes

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This page provides comprehensive Chapter 12: Surface Areas and Volumes - Assertion Reason Worksheet - SJMaths. Assertion and Reason type questions for Class 10 Surface Areas and Volumes. Practice for CBSE Board Exams.

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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