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Indian Navy MR Surface Areas and Volumes — practice questions

Explore 6 free Surface Areas and Volumes questions in Indian Navy MR Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Surface Areas and Volumes questions and exercises with answers

Showing questions 1–6 of 6.

A line passes through the points $(1, 2, 3)$ and $(4, 6, 9)$. The direction ratios of the line are:Read the answer and worked solution →The volume of a frustum of a cone with radii 6 cm and 3 cm and height 4 cm isRead the answer and worked solution →A cylindrical glass tube of radius 1 cm is filled with water to depth 14 cm. A sphere of radius 1 cm is dropped in. The new water level rises byRead the answer and worked solution →The volume of a frustum of a cone with radii 7 cm and 14 cm and height 12 cm isRead the answer and worked solution →Match each solid to its shapeRead the answer and worked solution →True or false: solidsRead the answer and worked solution →

Worked practice questions

A line passes through the points $(1, 2, 3)$ and $(4, 6, 9)$. The direction ratios of the line are:

  1. $(4, 6, 9)$
  2. $(3, 4, 6)$
  3. $(5, 8, 12)$
  4. $(1, 2, 3)$
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Correct answer: B. $(3, 4, 6)$

Direction ratios of the line joining two points are the differences in coordinates: $(x_2 - x_1, y_2 - y_1, z_2 - z_1)$. Here: $(4 - 1, 6 - 2, 9 - 3) = (3, 4, 6)$. Direction ratios are not unique. $(6, 8, 12)$ or $(1.5, 2, 3)$ would also be valid since multiplying by any non-zero scalar preserves the direction. The standard answer just uses the raw differences. Direction cosines (the normalised version) are $(3, 4, 6) / \sqrt{9 + 16 + 36} = (3, 4, 6) / \sqrt{61}$.

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The volume of a frustum of a cone with radii 6 cm and 3 cm and height 4 cm is

  1. 36 pi cu cm
  2. 63 pi cu cm
  3. 84 pi cu cm
  4. 12 pi cu cm
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Correct answer: C. 84 pi cu cm

1. V = (1/3) × pi × h × (R squared + r squared + R r). 2. = (1/3) × pi × 4 × (36 + 9 + 18) = (1/3) × pi × 4 × 63. 3. = 84 pi cu cm. _Source: syllabus Mathematics, Ch 12 "Surface Areas and Volumes", §12.4_

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A cylindrical glass tube of radius 1 cm is filled with water to depth 14 cm. A sphere of radius 1 cm is dropped in. The new water level rises by

  1. 2 cm
  2. 2/3 cm
  3. 1 cm
  4. 4/3 cm
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Correct answer: D. 4/3 cm

1. Sphere volume = (4/3) × pi × 1 cubed = (4/3) pi cu cm. 2. Cross-section = pi × 1 squared = pi. 3. Rise = (4/3) pi / pi = 4/3 cm. _Source: syllabus Mathematics, Ch 12 "Surface Areas and Volumes", §12.3_

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