 Hello and welcome to the session. Let us discuss the following question. It says a hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin plating it on the inside at the rate of rupees 16 per 100 cm square. We have to find the cost of tin plating it on the inside at the rate of rupees 16 per 100 cm square. That means we have to find the cost of tin plating this area that is the curved surface area. So we need to find the curved surface area of the hemisphere and the curved surface area of hemisphere equal to 2 pi r square where r is the radius value of pi is 22 upon 7. So this knowledge will work as k idea. Let us now move on to the solution. We are given that the diameter which is in this shape of hemisphere is equal to 10.5 cm and we know that the radius is half of the diameter. So this implies the radius of bowl is equal to 10.5 upon 2 cm that is 5.25 cm. Now we find the curved surface area of bowl which is in the shape of hemisphere. So it is 2 pi r square cm square since the unit of radius is cm. So the unit of surface area will be cm square. Now substitute the values of pi and r in this. So it is 2 into 22 upon 7 into 5.25 cm square which is equal to 2 into 22 upon 7 into 5.25 into 5.25 cm square and on simplifying we get the area as 173.25 cm square. Now we are given the cost of tin plating at the rate of rupees 16 per 100 cm square. Now cost of tin plating at 100 cm square is rupees 16. So the cost of tin plating the area of 1 cm square is rupees 16 upon 100. Therefore cost of tin plating the area of 173.25 cm square is rupees 16 upon 100 into 173.25 and it is equal to rupees 27.72. So the curved surface area of the hemispherical bowl is 173.25 cm square. Therefore the cost of tin plating the area of 173.25 cm square is rupees 27.72. The answer is rupees 27.72. So this completes the question. Bye for now. Take care. Have a good day.