Today i am solving the important question of ncert of class 10 chapter surface areas and volumes of exercise 13.4 question number 4 . Please help me to solve the tricky question of this question . Give me the best solution of this question. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs. 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm2.
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A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs. 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm2.Q.4
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Given,
r1 = 20 cm,
r2 = 8 cm and
h = 16 cm
∴ Volume of the frustum = (⅓)×π×h(r12+r22+r1r2)
It is given that the rate of milk = Rs. 20/litre
So, Cost of milk = 20×volume of the frustum
= Rs. 209
Now, slant height will be
So, CSA of the container = π(r1+r2)×l
= 1758.4 cm2
Hence, the total metal that would be required to make container will be = 1758.4 + (Area of bottom circle)
= 1758.4+201 = 1959.4 cm2
∴ Total cost of metal = Rs. (8/100) × 1959.4 = Rs. 157