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A container made of a metal sheet open at the top is of the form of a frustum of cone, whose height is 16 cm and the radii of its lower and upper circular edges are 8 cm and 20 cm respectively. Find(i) The cost of metal sheet used to make the container if it costs Rs. 10 per 100 cm2.(ii) The cost of milk at the rate of Rs. 35 per liter which can fill it completely. |
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Answer» Radius of lower end = r = 8 cm Radius of upper end = R = 20 cm Height of container frustum = h = 16 cm Cost of 100 cm2 metal sheet = Rs 10 So, Cost of 1 cm2 metal sheet = 10/100 = Rs 0.1 Let l be the slant height. l2 = (R-r)2 + h2 = (20-8)2 + 162 = 256 + 144 = 400 or l = 20 cm Surface area of frustum of the cone = πr2 + π(R + r)l cm2 = π[ (20 + 8 )20 + 82] = π[560 + 64] = 624 X ( 22/7 ) = 1961.14 cm2 (i) Find cost of metal sheet used to make the container if it costs Rs. 10 per 100 cm2 Cost of metal sheet per 100 cm2 = Rs. 10 Cost of metal for Rs. 1961.14 cm2 = (1961.14 x10)/100 = Rs. 196.114 (ii) Find the volume of frustum: Volume of frustum = 1/3 πh(r2 + R2 + rR) = 1/3 x 22/7 x 16(82 + 202 + 8×20) = 1/3 x 22/7 x 16(64 + 400 + 160) = 10345.4208 cm3 = 10.345 liters Using 1000 cm3 = 1 litre 1 cm3 = 1/1000 litre Cost of 1 liter milk = Rs. 35 Cost of 10.345 liter milk = Rs. 35 x 10.345 = Rs. 363 (approx) |
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