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Question 5 Surface Areas and Volumes - Exercise 13.4

**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 cm ^{2}.**

Answer:

iven,

r_{1} = 20 cm,

r_{2} = 8 cm and

h = 16 cm

∴ Volume of the frustum = (⅓)×π×h(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})

=\frac{314 \times 16 \times 208}{100000} \text { litres }

It is given that the rate of milk = Rs. 20/litre

So, Cost of milk = 20×volume of the frustum

=\operatorname{Rs} .20 \times \frac{314 \times 16 \times 208}{100000}

= Rs. 209

Now, slant height will be

I=\sqrt{h^{2}+\left(r_{1}-r_{2}\right)^{2}}=\sqrt{16^{2}+(20-8)^{2}}=\sqrt{16^{2}+12^{2}}

So, CSA of the container = π(r_{1}+r_{2})×l

=\frac{314}{100}(20+8) \times 20 \mathrm{~cm}^{2}

= 1758.4 cm^{2}

Hence, the total metal that would be required to make container will be = 1758.4 + (Area of bottom circle)

= 1758.4+201 = 1959.4 cm^{2}

∴ Total cost of metal = Rs. (8/100) × 1959.4

**= Rs. 157**

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