Question

S To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V , calculate this ratio for (c) a parallelepiped of dimensions × 2 a .

Answer 1

To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V ,  ratio for (c) a parallelepiped of dimensions × 2 a will be $\frac{6.29961}{c^\frac{1}{3} }$

What will be the ratio of surface area and the volume of the parallelepiped

We will proceed by stating the equations of the surface area and the volume of each shape, and then identifying each of the surface area and the volume, we will take their ratios, and as it is asking for the shape which has the minimum surface area to volume ratio, therefore at some given volume, $V=const$.

The parallelepiped whose dimension is given as

$2a,a,a$

the surface area and the volume is given respectively by:

$A=2(2a \times a)+2(a \times 2a)+2(a \times a)=10a^{2}$

$V=2a^{3}$

Let the volume be equal to the constant c, we get:

$V=c=2a^{3}\\a=(\frac{c}{2} )^\frac{1}{3}$

Hence the ratio is given by:

$\frac{A}{V} =\frac{10a^{2}}{2a^{3}}=\frac{5}{a}$

Calculating the constants, by substituting the value of $a$ in (3), we get:

$\frac{A}{V}=\frac{6.29961}{c^\frac{1}{3} }$

To learn more about parallelepiped, refer to:

brainly.com/question/27975136

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