What is the simplest way for solving the question of class 9^{th} of exercise 13.4 of math of question no 5 give me the best and simple way for solving this question in simple way of Surface areas and Volumes chapter Give me the best and simple way for solving the question of class 9^{th} of Surface areas and Volumes chapter of math of question no.1 how i solve this question in easy and simple way because it is very important type of question

AnilSinghBoraGuru

# A hemispherical bowl made of brass has inner diameter 10.5cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2. (Assume π = 22/7) Q.5

Share

Inner radius of hemispherical bowl, say r = diameter/2 = (10.5)/2 cm = 5.25 cm

Formula for Surface area of hemispherical bowl = 2πr

^{2}= 2×(22/7)×(5.25)

^{2}= 173.25Surface area of hemispherical bowl is 173.25 cm

^{2}Cost of tin-plating 100 cm

^{2}area = Rs 16Cost of tin-plating 1 cm

^{2}area = Rs 16 /100Cost of tin-plating 173.25 cm

^{2 }area = Rs. (16×173.25)/100 = Rs 27.72Therefore, the cost of tin-plating the inner side of the hemispherical bowl at the rate of Rs 16 per 100 cm

^{2}is Rs27.72.