1. 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?
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By: anil on 05 May 2019 01.48 am
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 = $$4pi R^2$$ = $$4pi (6)^2=144pi$$ $$cm^2$$ Total surface area of the three spheres = $$4pi(r_1)^2+4pi(r_2)^2+4pi(r_3)^2$$ = $$4pi (9+16+25)=200pi$$ $$cm^2$$ $$ herefore$$ Percentage decrease in surface area = $$frac{200pi-144pi}{200pi} imes100$$ = $$frac{56}{2}=28\%$$ => Ans - (D)
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