Question

Suppose that 8 turns of a wire are wrapped around a pipe with a length of 60 inches and a circumference of 4​ inches, so that the wire reaches from the bottom of the pipe to the top. What is the length of the​ wire? (Hint: Picture cutting the pipe along its length and pressing it​ flat.)

Answer 1

Explanation:

The given data is as follows.

          Length of pipe = 60 inches

          Circumference of pipe = 4 inches

           No. of turns of wire = 8

As the wire reaches from bottom to the top of the pipe it means that the wire has traveled 60 inches along the length of the pipe and 32 inches around the pipe. We assume that sides of the pipe are wound around each other to form a big rectangle.

Hence, length of rectangle = 60

Breadth of rectangle = $8 \times 4$

                                   = 32

Now, let x be the diagonal of rectangle therefore, by using Pythagoras theorem length the wire is calculated as follows.

           $x^{2} = (60)^{2} + (32)^{2}$

           $x^{2}$ = 3600 +1024

               x = $\sqrt{4624}$

                  = 68

Thus, we can conclude that length of the wire is 68 inches.

Answer 2

The length of the wire is 68 inches.

The given parameters;

• number of turns of the wire, n = 8 turns

• length of the pipe, L = 60 inches

• circumference of the pipe, C = 4

When the pipe is cut, it forms a rectangle with breadth and length;

Length, L = 60

Breadth, B = C x h =  2πr x h

where;

• 2πr is the circumference of the pipe

• h is the number of turns of wire on the pipe

Breadth, B = C x h  = 4 x 8 = 32

Since the wire reaches form one bottom to the top of the wire, the length of the wire is equal to the diagonal of the rectangle.

Apply Pythagoras theorem to determine the length of the wire;

$d^2 = 60^2 + 32^2\\\\d^2 = 3600 + 1024\\\\d^2 = 4624\\\\d = \sqrt{4624} \\\\d = 68 \ inches$

Thus, the length of the wire is 68 inches.

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