Question

If the radius of wire is increased by one third of its original the volume remains same then the height is changed to

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

The height becomes 9/16th of its original value.

Step-by-step explanation:

CASE 1 : Let the height & radius of the wire be h & r respectively initially.

⇒ Volume of wire ( V₁ ) = πr²h

CASE 2 : Let the changed height & radius be  H & R respectively .

R = r + r/3 = 4r / 3

⇒ Volume of wire ( V₂ ) = πR²H

⇒ πR²H = π$(\frac{4r}{3} )^2$H ------> (1)

From the question we know that in CASE 1 & CASE 2 volumes of the wire remain same.

⇒π$(\frac{4r}{3} )^2$H = πr²h

⇒16πr²H/9 = πr²h

⇒ 16H / 9 = h

⇒ H = $\frac{9}{16}$(h)

∴ The changed height becomes 9/16th of the original value.