(i) Let initially the edge of the cube be l.
Initial surface area = 6l2
If each edge of the cube is doubled, then it becomes \(2l.\)
New surface area =\( 6(2l)^ 2 \)
\(= 24 l\)2
= 4 \(\times\) 6l2
Clearly, the surface area will be increased by 4 times.
(ii) Initial volume of the cube = l3
When each edge of the cube is doubled, it becomes 2l.
New volume \(= (2l)^3 = 8l^ 3 = 8 \times l^ 3 \)
Clearly, the volume of the cube will be increased by 8 times.
Let the side length of the cube be a.
Volume of the cube (V) is: \(V = a^3\)
Surface area of the cube (S) is: \(S = 6a^2\)
Since the side length is doubled, the new side length is 2a.
(i) New surface area:
\(\text{New surface area} = 6 \times (2a)^2 = 6 \times 4a^2 = 24a^2 = 4S\)
So, the surface area will increase by 4 times.
(ii) New volume of the cube:
\(\text{New volume} = (2a)^3 = 8a^3 = 8V\)
So, the volume will increase by 8 times.