Question

If the diameter of the wire is decreased by 5%how much %will be the length increased so that the volume remain same

Answer 1

Let the Initial Volume of the wire be V and Final volume be V'
Also, Let the Initial radius be r and Final radius be r
         Let the initial length be L and Final length be L'
         Let the initial diameter be d and Final diameter be d'

Now according to the Question,
                 Diameter is increased be 5%
So,  
                      d' = d + 5% of d
                          = d + 0.05 d
                     d'  = 1.05 d
   also we know, radius,
                                 r = d / 2
 and 
                                 r' = d' / 2

putting the above value of d',

                                r' = 1.05 d / 2 = 0.525 d
 
we know Volume of  cylinder can be calculated as.
Initially, 
                             V = pi * r^2 * l
Thus                     V = pi * 0.5d^2 l

Finally,
                             V' = pi * r'^2 * l'
                                 = pi * 0.525d^2 * l'
now we are given that the volumes are same
Thus, 

                              V = V'
Putting the values,
                     
                   pi * 0.5 d^2 * l = pi * 0.525 d^2 *l
                         0.25 d^ * l = 0.275 d^2 * l'
                           0.25 l = 0.275 l'
                                   l' = 0.90 l

Thus l' = 9 % of l