# A cylindrical can whose base is horizontal and of radius 3.5cm contains sufficient water so that if a sphere is placed in the can, the water just covers the sphere. given that the can just fits into the can, calculate:i.) the surface area of can in contact with water when the sphere is in it,ii.) the depth of water in the can before the sphere was put into the can.give your answers as proper fractions and take =22/7

$\\ \text{(1). we can see the figure height of cylinder is equal to diameter of sphere} (h= 2 r) \\ When sphere is in the can then total sur face area of the can =\pi r^{2}+2 \pi r h \Rightarrow \frac{22}{7} \times 3.5 \times(3.5+14)=192.5 \mathrm{~cm}^{2} \\ (2). Let the height of water be fore putting sphere in it be x volume of can when sphere not in can = volume of sphere+volume of water \pi r^{2} h=\pi r^{2} x+\frac{4}{3} \pi r^{3} \\ \Rightarrow h=x+\frac{4}{3} r \\ x=\frac{4}{3} \times 3.5-7=\frac{7}{3}=2 \frac{1}{3} \mathrm{~cm} \\ Hence, the sur face area of can in contact with water when the sphere is in it is 192.5 \mathrm{~cm}^{2} and height upto which can was filled before we put sphere in it is 2 \frac{1}{3} \mathrm{~cm}$

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