# RD Sharma Solutions Class 8 Mathematics Solutions for Surface Area and Volume of Right Circular Cylinder Exercise 22.1 in Chapter 22 - Surface Area and Volume of Right Circular Cylinder

Question 12 Surface Area and Volume of Right Circular Cylinder Exercise 22.1

A cylindrical vessel open at the top has diameter 20 cm and height 14 cm. Find the cost of tin-plating it on the inside at the rate of 50 paise per hundred square centimetre.

We have,

Diameter of base of cylinder = 20 cm

Radius of cylinder = 20/2 = 10cm Height of cylinder = 14 cm

Total surface area of cylinder = 2πrh + πr2

= (2 × 22/7 × 10 × 14) + (22/7 × 102)

= 880 + 2200/7

= (6160+2200)/7

= 8360/7 cm2

We know that cost per 100cm2 = 50paise So, cost per 1cm2 = Rs 0.005

∴ Cost of tin painting the area inside the vessel = 8360/7 × 0.005 = Rs 5.97

Video transcript
[Music] hello dear student i am sunita anaya from lido learning i am here to help you with this problem which goes like this a cylindrical vessel open at the top has a diameter 20 centimeters and height 14 centimeters find the cost of tin plating it on the inside at the rate of 50 paisa per 100 square centimeters so what is this tin plating it is a process of applying a thin coat of tin to any vessel to prevent it from rusting right so here we have a cylindrical vessel open at the top so the surface area in question is let me draw it and show you so supposing this is my vessel [Music] a cylindrical vessel which is open at the top right so this vessel would have the curved surface area on the inside as well as the base right so the area to be tin plated is the curved surface area and the base of the vessel so we shall write it here area to be plated is equal to the curved surface area csa plus area of base now the formula for curved surface area is pi d that is a circumference into the height of the vessel and the area of the base is given by pi r squared okay now let's substitute the values pi d h is 22 over 7 into d is 20 and h is 14 plus pi r squared is 22 over 7 into now what is our r if diameter is 20 the radius is half of 20 which is 10 [Music] all right so that is pi r squared let's try and reduce this so here we have [Music] 22 into 40 which is 880 right [Music] 22 into 40 is 880 plus two thousand two hundred or twenty two hundred divided by seven [Music] all right now if i take the lcm of these two terms it will it will become 7 into 880 plus 2200 the whole divided by seven which works out to six one six zero seven into eight hundred and eighty is six one six zero plus two thousand two hundred the whole over seven now six one six zero plus two thousand two hundred is eight three six zero upon 7 all right so this is the area to be plated now let's look at the rate of plating i will work on the rate over here [Music] rate of plating is given as 50 pesa per 100 square centimeters right now if it is 50 pi sub 100 square centimeters the rate for one square centimeter would be rate per square centimeter would be equal to 50 divided by 100 [Music] which is equal to 0.5 right 0.5 pisa now if i want to convert this into rupees i will have to divide it by 100 so 0.5 pisa is equal to 0.5 divided by 100 right so we shift the decimal place to 2 places to the left so that becomes 0.005 all right so this is 0.005 rupee per centimeter squared this is our rate of tinning it tin plating it so let's multiply this area which we have found a 3 6 0 by 7 into the rate right so into 0.005 will be equal to now let's multiply 5 by 8360 so that is zero five sixes are thirty zero carry three five threes are fifteen and three eighteen five eighths are forty and one forty 41 all right and three decimal places correct so we have three decimal places so that's 41.8 upon seven now this works out to be seven fives are 35 a remainder of six seven nines are 63 a remainder of five seven sevens are 49 yes that would that should that should be enough so it will cost us rupees 5.97 so this is our answer this is the cost of tin plating the cylindrical vessel i hope you understood the solution please drop a comment in the comment section if you have any doubt visit our channel regularly for more homework solutions and subscribe to our channel for updates as well thank you
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