Four identical cylindrical chalk-sticks, each of radius r = 0.5 cm and length l = 10 cm, are bound tightly together using a duct tape as shown in the following figure. The width of the duct tape is equal to the length of the chalk-stick. The area (in cm^2) of the duct tape required to wrap the bundle of chalk-sticks once is:
A. 2(4 + pi) B. 20(4 + pi) C. 10(8 + pi) D. 10(4 + pi)
GATE 2024 · General Aptitude · Geometry · medium
Answer: Area of duct tape = 10(4 + pi) cm^2.
Identify the cross-section path of the tape: In the 2x2 arrangement each pair of adjacent cylinders shares a common external tangent of length equal to the centre-to-centre spacing minus nothing (they touch), so the straight section between each pair of adjacent cylinder centres is 2r = 1 cm. At each of the four outer corners the tape follows a quarter-circle of radius r = 0.5 cm.
Compute total straight length and arc length: L_straight = 4 * 2*(0.5) = 4 cm. L_arc = 2*pi*(0.5) = pi cm. Total perimeter = 4 + pi cm.
Multiply perimeter by tape width to get area: The tape wraps once around the bundle; its area = length of tape * width of tape = (4+pi) * 10 = 10(4+pi) cm^2.