Question

A metal sheet $27$ cm long, $8$ cm broad and $1$ cm thick is melted into a cube so that volume is constant. The difference between surface areas of two solids is

• A. $284c{m}^2$
• B. $286c{m}^2$
• C. $296c{m}^2$
• D. $300c{m}^2$

Answer 1

Answer: (B)

$286c{m}^2$

We know that,
Total surface area of cuboid = 2 (lb + bh + hl)
SurfWe know that,

Total surface area of cuboid = 2 (lb + bh + hl)

Surface area of given cuboid

= 2 (27 $\times$ 8 + 8 $\times$ 1 + 27 $\times$ 1) = 502 $c{m}^2$

Volume of cube = 27 $\times$ 8 $\times$ 1 = 216 $c{m}^3$

$\therefore$ Edge of the cube a = $\sqrt[3]{3216}$ = 6 cm

We know that,

Total surface area of cube = $6a^2$

Surface area of this cube = 6 $\times (6{)}^2$ = 216 $c{m}^2$

Hence, the difference of surface areas = 502 - 216 = 286 $c{m}^2$ace area of given cuboid
= 2 (27 $\times$ 8 + 8 $\times$ 1 + 27 $\times$ 1) = 502 $c{m}^2$
Volume of cube = 27 $\times$ 8 $\times$ 1 = 216 $c{m}^3$
$\therefore$ Edge of the cube a = $\sqrt[3]{3216}$ = 6 cm
We know that,
Total surface area of cube = $6a^2$
Surface area of this cube = 6 $\times (6{)}^2$ = 216 $c{m}^2$
Hence, the difference of surface areas = 502 - 216 = 286 $c{m}^2$