Three solid spheres of radius 3 cm, 4 cm and 5 cm are melted and recasted into a solid sphere. What will be the percentage decrease in the surface area?

Let radius of new sphere = $$R$$ cm

Thus, volume of new sphere = Sum of volumes of all the three spheres

=> $$\frac{4}{3}\pi (R)^3=\frac{4}{3}\pi (r_1)^3+\frac{4}{3}\pi (r_2)^3+\frac{4}{3}\pi (r_3)^3$$

=> $$(R)^3=(3)^3+(4)^3+(5)^3$$

=> $$(R)^3=27+64+125=216$$

=> $$R=\sqrt[3]{216}=6$$ cm

Total surface area of new sphere = $$4\pi R^2$$

= $$4\pi (6)^2=144\pi$$ $$cm^2$$

Total surface area of the three spheres = $$4\pi(r_1)^2+4\pi(r_2)^2+4\pi(r_3)^2$$

= $$4\pi (9+16+25)=200\pi$$ $$cm^2$$

$$\therefore$$ Percentage decrease in surface area = $$\frac{200\pi-144\pi}{200\pi}\times100$$

= $$\frac{56}{2}=28\%$$

=> Ans - (D)