problems on surface areas and volumes ! plz help . provide wid full sol '?

Question

if the diameter of cross section of a wire is decreased by 5 % , how much percent will the length be increased so dat volume remains same ??????????????????????????
for ur reference
vol of cylinder = Pi*r^2*h

Answer 1

Easy once you figure out how to set up the equation. To make things simpler, let's assume the radius = 1. Since

V = pi(r^2)h'

Then V = pi(1^2)h = pi(h)

Now we're going to reduce the radius by 5%, the new volume will be

pi(0.95^2)h =

pi(.9025)h

Now we need the multiplier x (for the change in height) such that:

x(pi(0.9025)h) = V = pi(h)

Solving for x gives

x = pi(h)/[pi(0.9025)h

x = 1/(0.9025)

This is about 1.108 and is the multiplier for the height to restore the volume, right? So you would subtract 1 (the original radius) from that, then multiple by 100 to get the percent change in h needed which yields about 10.80%. The exact answer would be

100((1/0.9025)-1), again approximately equal to 10.80

Let's test and make sure. If the original wire is 100mm long with a radius of 1mm, its volume is:

pi(1^2)(100) = 100pi.

Reducing radius 5% and increasing the height by the proposed exact amount gives a new volume of:

pi(.95^2)(100 * (1/(0.9025)) =

100pi, exactly restoring the volume.

Gary H

Answer 2

In terms of diameter, Vol = (1/4)*Pi*d^2*h
Here, height h corresponds to the length of the wire.

Let d and h, as above, be the dimensions of the wire in question before its diameter is reduced.

We'll write D = (something) * d to denote the diameter after the wire is shrunk; decreasing it by 5% means it is now 95% of what it used to be, so this means the new diameter is

D = (.95)*d

We're going to increase the length, too, so let H be the new length. We want to find the relative sizes of h and H, that's the goal we've been given. So if we write

H = A*h,

Then we're just looking for A. Whatever A turns out to be, we subtract 1 from it to get the fractional increase, then convert that to a percentage to get the percent increase.

New vol = (1/4)*Pi*D^2*H

Substitute for D and H, to get the new volume in terms of the old cylinder's dimensions:

New vol = (1/4)*Pi*[(.95)*d]^2*A*h = (1/4)*Pi*(.95)^2*d^2*A*h

Now we set New vol = Old vol to find A. To take advantage of cancelling, we rewrite this as

(New vol)/(Old vol) = 1

When we substitute and cancel we end up with simply

(.95)^2*A = 1

Solve for A, then do with A as I described above. Should come out to something a little over 10%.

Answer 3

if the diameter is decreased 5%, it is now .95 of the original. The area is proportional to d^2 so it is now .9025 of the original. to keep the same Volume, the length must increase to
1/.9025=1.108 or be 10.8% longer

Answer 4

Let us assume that length increases by x%
Initial volume
=Pi*d^2*h/4....1
Volume with changes
=Pi*[.95d]^2*[1+x/100]*h......2
We have to equate 1 and 2
Pi*d^2*h/4=Pi*[.95d]^2[1+x/100]*h/4
1=[.95]^2*[1+x/100]
1=.9025+.9025*x/100
.9025x/100=.0975
x
=100*.0975/.9025
=10.803%
We have to increase the length
by 10.803% to keep same volume.

Answer 5

v = 2(pi)rl where l=length
so
let x be the percentage increase of length, then:
2(pi)(0.95r)(1+(x/100))l = 2(pi)rl
so (0.95r)(1+(x/100)) = r
so (1/0.95)-1 = x/100
so x = (100/0.95)-100 = 5.26% (2 decimal places)

Answer 6

=Pi*[.95d]^2*[1+x/100]*h
Pi*d^2*h/4=Pi*[.95d]^2[1+x/100
1=[.95]^2*[1+x/100]
1=.9025+.9025*x/100
.9025x/100=.0975
=100*.0975/.9025
=10.803%