Question

A solid sphere is cut into identical pieces by three mutually perpendicular planes passing through its centre. Increase in total surface area of all the pieces with respect to the total surface area of the original sphere is

• A. $250\%$
• B. $175\%$
• C. $150\%$
• D. $125\%$

Answer 1

The correct option is A $250\%$

Every cut igin a sphere adds two identical surfaces or two disks to the original and surface area of each disk $=\pi \times r^2$. Hence, each cut adds $2\times \pi \times r^2$ to the surface area

3 cuts, therefore add $6\times \pi \times r^2$ to the original surface area, make the total surface area after the cut to be $10\times \pi \times r^2$.

Hence, increase in surface area w.r.t to original surface area $=10\times \pi \times \frac{r^2}{4}\times \pi \times r^2=2.5=250\%$.