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If the diameter of cross-section of a wire is decreased by 20%. How much percent should...

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nasirjam | Student, Grade 9 | Honors

Posted December 29, 2012 at 5:38 PM via web

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If the diameter of cross-section of a wire is decreased by 20%. How much percent should the length be increased,so that the volume remains the same ?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted December 29, 2012 at 6:01 PM (Answer #1)

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Since the shape of wire is cylindrical, you may evaluate its volume using the following formula, such that:

`V = pi*d^2/4*l`

d represents the diameter of cross section of wire

l represents the length of wire

Since the new diameter of cross section of wire is `d_1 = d - ` `20/100*d` , you need to substitute in equation of volume, such that:

`V = pi*(d_1^2)/4*l_1 => pi*d^2/4*l = pi*(d_1^2)/4*l_1`

Reducing duplicate factors yields:

`pi*d^2/4*l = pi*d^2(1 - 20/100)^2/4*l_1`

Reducing duplicate factors yields:

`l = (1 - 20/100)^2*l_1 => l_1 = l/(1 - 20/100)^2`

`l_1 = l*(100/80)^2 => l_1 = l*(5/4)^2 => l_1 = 1.56*l =>l_1 = 156/100*l = 156%*l`

Hence, evaluating the number of percents the original length of wire needs to be increased, under the given conditions, yields `l_1 = 156%*l.`

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vaaruni | High School Teacher | Salutatorian

Posted December 30, 2012 at 11:56 AM (Answer #2)

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Let the original diameter be = d1 and original length be = l1 The original radious = d1/2 initial volume V1 = (pie)*(d1/2)^2*l1= (pie)*d1^2/4*l1 The new diameter d2 (after decreasing by 20% of original)= 0.8d1 [ d2=d1-(20/100)*d1= d1-0.2d1= 0.8d1] The new radious = (0.8d1)/2 = 0.4d1 Let the new length be = l2 The new volume V2= (pie)*(0.4d1)^2*l2 = (pie)*0.16(d1^2)*l2 Since the volume remained same, therefore V1 = V2 (pie)*d1^2/4*l1 = (pie)*0.16(d1^2)*l2 => l1/4 = 0.16 l2 [ pie and d1^2 get cancelled ] => l2 = l1/(4*0.16) = l1/(0.64)= l1*(100/64)= (25/16)l1 => l2 = (25/16)l1 increase in length = l2-l1 = (25/16)l1 - l1 = (9/16)l1 percentge increase in length = ((9/16)l1/l1)*100 = (9/16)*100 = 225/4 = 56.25% Hence, percentage increase in length of the wire = 56.25% <- Answer

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