Question

If the radius of a wire is decreased to one fourth of its original and it's volume remain the same then how many time will the new length become it's original length

Answer 1

Answer:

Initial radius is R

length is L

volume of wire = π R²L

now new radius is R - R/4 = 3R/4

new length be L'

new volume = initial volume

⇒π (3R/4)²× L' = π R²L

⇒ (9/16 ) L' = L

⇒L' = 16/9 L

new length is 16/9 times the original length

Answer 2

Question:

If the radius of a wire is decreased to one fourth of its original and it's volume remain the same then how many time will the new length become it's original length

Answer:

16 times

Step-by-step explanation:

Let R be the original radiusg of the wire and r be the new one.

Let L be the original length of the wire and l be the new one.

Let V be the original volume of the wire and v be the new one.

Now, R = r/4 (Given)

Therefore, v = πr²l

We know, V = πR²L

But, V = v

So, πR²L = πr²l

→ r²L/16 = r²l

→ L = 16l

Hence, if the radius of a wire is decreased to one fourth of its original and it's volume remain the same then 16 times will the new length become it's original length.

HOPE IT WAS HELPFUL.