Question

If the diameter of the cross – section of a wire is decreased by 20%, how much per cent will the
length be increased so that the volume remains the same?
​

Answer 1

Answer:

Let r be the radius of cross-section of wire and h be its length.

Then volume =πr

2

h....(i)

5% of diameter of cross-seciton =

100

5

×2r=

10

r

New diameter =2r−

10

r

=

10

19r

New radius =

20

19r

Let new length be h

1

.

Volume =π(

20

19r

)

2

h

1

...(ii)

From (i) and (ii) we have

πr

2

h=(

20

19r

)

2

h

1

⇒h=

400

361

h

1

⇒h

1

=

361

400

h

Increase in length =h

1

−h=

361

400

−h=

361

39h

Percentage increase in length =

h

h

1

−h

×100=

h

39h

×100

=

361

3900

=10.8%