GATE ECE 2024 | GA | Question: 7

Question

​​​​​Four identical cylindrical chalk-sticks, each of radius $r=0.5 \mathrm{~cm}$ and length $l=10 \mathrm{~cm}$, 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 $\mathrm{cm}^{2}$ ) of the duct tape required to wrap the bundle of chalk-sticks once, is

• A. $20(4+\pi)$
• B. $20(8+\pi)$
• C. $10(8+\pi)$
• D. $10(4+\pi)$

Comments

The Correct Answer is Option D) 

Solution

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Answer 1

Area of the duct tape required to wrap the bundle of chalk-sticks once (A) = length of duct tape required (L) $\times$ width of duct required (W)

$A = L \times W$

$A = L \times 10$  (width of duct tape = length of chalk-stick = 10cm.)

$A = 10L$                  .............................(1)

Now, $L = $  perimeter of outer black square - total lenght of perimeter of black region + total length of circular region of duct tape that is toucing each circle

$L = (4 \times 4r) - (4 \times 2r) + \Big( 4 \times \dfrac{2 \pi r}{4} \Big)$

$L = 16r - 8r + 2 \pi r$

$L = 8r + 2 \pi r$

$L = 8 \times 0.5 + 2 \pi \times 0.5$

$L = 4 + \pi$

Now, putting value of $L$ in eqn. (1),

$\boxed{A = 10 (4 + \pi)}$

PS: don't do this mistake that you first unroll the given diagram and then calculate required duct tape because as you can see in the unrolled diagram the surface of duct tape increased due to increase in surface open to duct tape.

Comments

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399

 

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Answer 2

H ( L x B + 2π r )

10(2 x 2  + 2 x π x 0.5)

10(4+π )