An open container made up of a 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 cm2.

The open container in the shape of frustum so , so we need its volume and curbed surface area .is open so need bottom area also. VOLUME of Frustum =1/3 πh ( R2 +Rr +r2 ) . CURBED surface area =π (R+r)L . bottom area = π r2 . L2 = h2 +(R - r)2 . L2 = 82 +(10 – 4 )2 =64 +62 =64+36 =100=102. L =10 . Given , R =10 cm, r =4 cm, h =8 cm, L =slant hight= L =10 cm.. to fill oil is Rs.50 /litre , and metal sheet is Rs. 5 /100 cm2. Now Volume1/3 π h(R2 +Rr +r2) = 1/3 π 8 (102 +10 × 4 + 42 ) cm3 = 1/3 ×22/7× 8(100 +40 +16) cm3 =( 1/3×22/7× 8×156) cm3 now 1000 cm3 = 1litre . oil rate is Rs 50/litre . so, oil cost = 1/3×22/7×8×156×50/1000 RS. = 65.38 RS. Answer . for metal Area = curved surface area + bottom area of frustum = π(R+ r)L + π r2=π (10+4) 10 + π 42 cm2. =π (14× 10+16)cm2 = 22/7 × 156 cm2 . so cost of metal =22/7 ×156 ×5/100 Rs .= 24 .5 Rs The open container in the shape of frustum so , so we need its volume and curbed surface area .is open so need bottom area also. VOLUME of Frustum = πh ( R2 +Rr +r2 ) . CURBED surface area =π (R+r)L . bottom area = π r2 . L2 = h2 +(R - r)2 . L2 = 82 +(10 – 4 )2 =64 +62 =64+36 =100=102. L =10 . Given , R =10 cm, r =4 cm, h =8 cm, L =slant hight= L =10 cm.. to fill oil is Rs.50 /litre , and metal sheet is Rs. 5 /100 cm2. Now Volume π h(R2 +Rr +r2) = π 8 (102 +10 × 4 + 42 ) cm3 = ×× 8(100 +40 +16) cm3 =( 156) cm3 now 1000 cm3 = 1litre . oil rate is Rs 50/litre . so, oil cost = RS.= 65.38 RS. Answer . for metal Area = curved surface area + bottom area of frustum = π(R+ r)L + π r2=π (10+4) 10 + π 42 cm2. =π (14× 10+16)cm2 = × 156 cm2 . so cost of metal = ×156 ×Rs .=24 .5 Rs read less

Volume of the frustrum = Volume of the big cone - volume of the small cone = 1/3 x 22/7 x 10^2 x 13.3 - 1/3 x 22/7 x 4^2 x 5.3 = 1304.5 cmcube Cost of oil = 1304.5 x 50/1000 = Rs 65.2 Ans We can calculate the slant heights as 16.62 for the big cone and 6.64 for the small cone So cost of plate required = 5/100 x 22/7 x [10 x 16.62 -4 x 6.64 + 4^2] = Rs 24.43 read less