Q 25. An open container made up of metal sheet is in the form of a frustum of a cone of height 8 cm with radii of its lower and upper ends as 4 cm and 10 cm respectively. Find the cost of oil which can completely fill the container at the rate of Rs. 50 per litre. Also, find the cost of metal used, if it costs Rs. 5 per 100 sq. cm

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Solution

Dear student,

$G i v e n : h e i g h t o f t h e b u c k e t (h) = 8 c m r a d i i o f l o w e r (r_{2}) a n d u p p e r (r_{1}) e n d s = 4 c m a n d 10 c m \Rightarrow S l a n t h e i g h t o f b u c k e t (l) = \sqrt{h^{2} + {(r_{1} - r_{2})}^{2}} = \sqrt{8^{2} + {(10 - 4)}^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10 c m A r e a o f u p p e r b a s e A_{1} = {πr}_{1}^{2} = \frac{22}{7} \times 10 \times 10 = \frac{2200}{7} = 314.28 {cm}^{2} Area of lower base A_{2} = {πr}_{2}^{2} = \frac{22}{7} \times 4 \times 4 = \frac{352}{7} = 50.28 {cm}^{2} Volume of frustum = \frac{1}{3} h (A_{1} + A_{2} + \sqrt{A_{1} A_{2}}) = \frac{1}{3} \times 8 \times (314.28 + 50.28 + \sqrt{314.28 \times 50.28}) = \frac{1}{3} \times 8 \times (314.28 + 50.28 + \sqrt{15801.99}) = \frac{1}{3} \times 8 \times (364.56 + 125.70) = \frac{1}{3} \times 8 \times 490.26 = 1307.36 {cm}^{3} Thus the capacity of frustum is 1.3 L - - (as 1000 {cm}^{3} = 1 L) Cost of oil for 1 L = Rs 50 Cost of oil for 1.3 L = Rs 50 \times 1.3 = Rs 65 Area of metal sheet used = Area of lower base + C . S . A of frustum = {πr}_{2}^{2} + π (r_{1} + r_{2}) l = π [{r_{2}}^{2} + (r_{1} + r_{2}) l] = \frac{22}{7} [4^{2} + (4 + 10) \times 10] = \frac{22}{7} [16 + 14 \times 10] = \frac{22}{7} [16 + 140] = \frac{22}{7} \times 156 = \frac{3432}{7} = 490.28 {cm}^{2} Cost of metal sheet for 100 {cm}^{2} = Rs 5 Cost of metal sheet for 490.28 {cm}^{2} = Rs \frac{5}{100} \times 490.28 = Rs 24.514$
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Q. A container, open from the top, made up of a metal sheet is in the form of a frustum of a cone of height

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with radii of its lower and upper ends as

4 c m

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per litre and the cost of metal sheet used, if the costs Rs.

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