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Indian Navy SSR Surface Areas and Volumes — practice questions

Explore 8 free Surface Areas and Volumes questions in Indian Navy SSR 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 1–8 of 8.

The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in three-dimensional space is:Read the answer and worked solution →Angle between line and plane: if direction of line is L and normal to plane is N, sin(angle line-plane):Read the answer and worked solution →The cosine of the angle between two lines with direction ratios $(1,2,2)$ and $(2,2,1)$ is:Read the answer and worked solution →A cone of height 12 cm and radius 5 cm has volumeRead the answer and worked solution →A frustum of a cone has radii 4 cm and 8 cm and height 3 cm. Its slant height 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 →Match each solid to its shapeRead the answer and worked solution →True or false: solidsRead the answer and worked solution →

Worked practice questions

The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in three-dimensional space is:

  1. $\sqrt{(x_2 + x_1)^2 + (y_2 + y_1)^2 + (z_2 + z_1)^2}$
  2. $(x_2 - x_1)(y_2 - y_1)(z_2 - z_1)$
  3. $|x_2 - x_1| + |y_2 - y_1| + |z_2 - z_1|$
  4. $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$
Show answer and solution

Correct answer: D. $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

3D distance formula extends Pythagoras into three dimensions: the distance is the square root of the sum of squared coordinate differences. Derivation: project the segment $PQ$ onto each axis to get the leg lengths $|x_2 - x_1|$, etc. Apply Pythagoras twice. Option B is the **Manhattan** (taxicab) distance, useful in some lattice problems but not the Euclidean answer.

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Angle between line and plane: if direction of line is L and normal to plane is N, sin(angle line-plane):

  1. |L · N| / (|L| |N|) — since 90° - angle(line, normal)
  2. 0
  3. tan
  4. cos
Show answer and solution

Correct answer: A. |L · N| / (|L| |N|) — since 90° - angle(line, normal)

Angle between line and plane (measured to surface, not normal) = complement of angle between line and normal. **So sin(line-plane) = cos(line-normal) = |L · N|/(|L||N|).**

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The cosine of the angle between two lines with direction ratios $(1,2,2)$ and $(2,2,1)$ is:

  1. $\tfrac{4}{9}$
  2. $\tfrac{2}{3}$
  3. $\tfrac{1}{9}$
  4. $\tfrac{8}{9}$
Show answer and solution

Correct answer: D. $\tfrac{8}{9}$

We have cosθ = (1·2+2·2+2·1)/(√9·√9). So cosθ = 8/9.

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More Mathematics topics

SequencesBinomial TheoremSequences and SeriesProbability & StatisticsGeometryPermutations and CombinationsLinear ProgrammingVectors & 3D Geometry