1. A gold ingot in the shape of a cylinder is melted and the resulting molten metal molded into a few identical conical ingots. If the height of each cone is half the height of the original cylinder and the area of the circular base of each cone is one fifth that of the circular base of the cylinder, then how many conical ingots can be made?
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By: anil on 05 May 2019 02.34 am
Let the radius of the cylinder be $$r$$ and height be $$h$$ (as shown). Let the dimensions of the cone be $$r_c$$ and $$h_c$$ (as shown). It is given that $$h_c = frac{h}{2}$$ Area of base of cone $$A_{c}=pi imes r_{c}^{2}$$ Area of base of cylinder $$A=pi imes r^{2}$$ Given, $$A_{c}=frac{1}{5} imes A $$ $$ Rightarrow pi imes r_{c}^{2} =frac{1}{5} imes pi imes r^{2} $$ $$ Rightarrow r_{c}^{2} = frac{1}{5} imes r^{2} $$ $$ Rightarrow frac{r^{2}}{r_{c}^{2}} = 5 $$
Now we know that the volume of cylinder = total volume of cones
Let the number of cones be n. So, volume of cylinder = n x volume of each cone $$ Rightarrow V = n imes V_{c} $$
$$ Rightarrow pi imes r^{2} imes h = n imes frac{1}{3} imes pi imes r_{c}^{2} imes h_c $$
$$ Rightarrow r^{2} imes h = n imes frac{1}{3} imes r_{c}^{2} imes h_c $$
$$ Rightarrow frac{r^{2}}{r_{c}^{2}} imes frac{h}{h_c} imes 3 = n $$
$$ Rightarrow 5 imes 2 imes 3 = n $$
$$ Rightarrow 30 = n $$
Thus, 30 conical ingots can be made.
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Now we know that the volume of cylinder = total volume of cones
Let the number of cones be n. So, volume of cylinder = n x volume of each cone $$ Rightarrow V = n imes V_{c} $$
$$ Rightarrow pi imes r^{2} imes h = n imes frac{1}{3} imes pi imes r_{c}^{2} imes h_c $$
$$ Rightarrow r^{2} imes h = n imes frac{1}{3} imes r_{c}^{2} imes h_c $$
$$ Rightarrow frac{r^{2}}{r_{c}^{2}} imes frac{h}{h_c} imes 3 = n $$
$$ Rightarrow 5 imes 2 imes 3 = n $$
$$ Rightarrow 30 = n $$
Thus, 30 conical ingots can be made.