Allied & ICAR Exam Bank • ALLIED_2436 MCQ • 2 Marks

Section 8: General Aptitude

Analytical & Spatial Aptitude

Subtopic: Spatial Reasoning & 3D Representations

Question Statement:

An ant crawls on the outer surface of a cylindrical grain bin of radius R = 1 m and height H = 4 m. Starting from the bottom at theta = 0, it travels to a point directly opposite at the top (theta = pi, z = H). What is the minimum crawling distance along the surface?

Options:

  • (A) sqrt(pi^2 * R^2 + H^2) = sqrt(pi^2 + 16) approx 5.09 m
  • (B) pi * R + H = 3.14 + 4 = 7.14 m
  • (C) H = 4 m
  • (D) 2*pi*R + H = 10.28 m

Answer Key:

A

Step-by-Step Solution & Explanation

On the unrolled development, the shortest path is the straight hypotenuse between (0, 0) and (pi*R, H): d = sqrt((pi*R)^2 + H^2) = sqrt(pi^2 + 16) approx 5.09 m.
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