Question

Upon heating length of side of cube changes by 2% volume of cube is

Answer 1

Let the side of a cube is x.

Given that length of side of cube changes by 2%.

= > x + 2% of x

= > x + 2x/100

= > x + x/50

= > 51x/50.

We know that volume of cube = x^3.

= > (51x/50)^3

= > 132651x^3/125000

Therefore, the increase in volume:

= > (132651x^3/125000) - x^3

= > (132651x^3 - 125000x^3)/125000

= > 7651x^3/125000.

Hope this helps!

Answer 2

in case of heating length of side of cube chagnes by 2 % .
Let length of cube is L
so, change in length, ∆L = 2% of L
e.g., ∆L = 0.02L

we know,
Volume of cube = (side length)³
V = L³
now, differentiate V with respect to L
dV/dL = 3L²
dV = 3L²dL
dividing V from both sides,
dV/V = 3L²dL/V
dV/V = 3L²dL/L³ [ as you know, V = L]
dV/V = 3dL/L
so, we gain the formula for finding change in volume if change in side length of cube is given. e.g.,$\frac{\Delta{V}}{V}=3 \times\frac{\Delta{L}}{L}$
now, ∆V/V = 3 × 0.02L/L = 0.06
for finding % change , multiply 100 with (∆V/V)
so, % change in volume of cube = 100 ×(∆V/V)
= 100 × 0.06 = 6 %


[note :- you should know that of % change is less then 10% then, you should use error concept like I did above ]