91Ó°ÊÓ

The potential energy is defined as the work done in lifting a mass against gravity.

Mathematically,

\({\rm{PE}} = mgh\)

Here, m is the mass, g is the acceleration due to gravity, and h is the height.

(a)

The center of mass of the pyramid is,

\({h_{cm}} = \frac{h}{4}\)

Here, h is the height of pyramid \(\left( {h = 146{\rm{ m}}} \right)\).

Putting all known values,

\(\begin{aligned} {h_{cm}} = \frac{{146{\rm{ m}}}}{4}\\ = 36.5{\rm{ m}}\end{aligned}\)

The gravitational potential energy of the pyramid is,

\({\rm{PE}} = mg{h_{cm}}\)

Here, m is the mass of the pyramid \(\left( {m = 7 \times {{10}^9}{\rm{ kg}}} \right)\), g is the acceleration due to gravity \(\left( {g = 9.8{\rm{ m}}/{{\rm{s}}^2}} \right)\), and \({h_{cm}}\) is the height of the center of mass of pyramid \(\left( {{h_{cm}} = 36.5{\rm{ m}}} \right)\).

Putting all known values,

\(\begin{aligned} {\rm{PE}} &= \left( {7 \times {{10}^9}{\rm{ kg}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {36.5{\rm{ m}}} \right)\\ &= 2.50 \times {10^{12}}{\rm{ J}}\end{aligned}\)

Therefore, the required gravitational potential energy stored in a pyramid is \(2.50 \times {10^{12}}{\rm{ J}}\).

(b)

Since, it takes 20 years working \(330{\rm{ days}}/{\rm{year}}\) and \(12{\rm{ hr}}/{\rm{day}}\) to build the pyramid. The time taken to build the pyramid is,

\(\begin{aligned} t &= \left( {20{\rm{ years}}} \right) \times \left( {\frac{{330{\rm{ days}}}}{{1{\rm{ year}}}}} \right) \times \left( {\frac{{12{\rm{ hr}}}}{{1{\rm{ day}}}}} \right)\\ &= 79200{\rm{ hr}}\end{aligned}\)

The work input of a worker is,

\({W_{in}} = Rt\)

Here, R is the rate at which the energy is consumed \(\left( {R = 300{\rm{ kcal}}/{\rm{hr}}} \right)\), and t is the time taken to build the pyramid.

Putting all known values,

\(\begin{aligned} {W_{in}} &= \left( {300{\rm{ kcal}}/{\rm{hr}}} \right) \times \left( {79200{\rm{ hr}}} \right)\\ &= 2.376 \times {10^7}{\rm{ kcal}} \times \left( {\frac{{4184{\rm{ J}}}}{{1{\rm{ kcal}}}}} \right)\\ &= 9.94 \times {10^{10}}{\rm{ J}}\end{aligned}\)

Since 1000 workers were engaged, the total work input is,

\(W = 1000{W_{in}}\)

Putting all known values,

\(\begin{aligned} W &= 1000 \times \left( {9.94 \times {{10}^{10}}{\rm{ J}}} \right)\\ &= 9.94 \times {10^{13}}{\rm{ J}}\end{aligned}\)

The efficiency of the workers is,

\(\begin{aligned} \eta &= \frac{{{\rm{Useful work done}}}}{{{\rm{total work input}}}} \times 100\% \\ &= \frac{{{\rm{PE}}}}{{{W_{in}}}} \times 100\% \end{aligned}\)

Putting all known values,

\(\begin{aligned} \eta &= \frac{{2.50 \times {{10}^{12}}{\rm{J}}}}{{9.94 \times {{10}^{13}}{\rm{ J}}}} \times 100\% \\ &= 2.5\% \end{aligned}\)

Therefore, the required efficiency of the workers is \(2.5\% \).

(c)

The energy content of 1 g of the diet is,

\(E = {E_p} \times p + {E_c} \times c + {E_f} \times f\)

Here, \({E_p}\) is the energy content of the protein \(\left( {{E_p} = 4.0{\rm{ kcal}}/{\rm{g}}} \right)\), p is amount of protein in diet \(\left( {p = 5\% } \right)\), \({E_c}\) is the energy content of the carbohydrate \(\left( {{E_c} = 4{\rm{ kcal}}/{\rm{g}}} \right)\), c is the amount of carbohydrates in diet \(\left( {c = 60\% } \right)\), \({E_f}\) is the energy content of fat \(\left( {{E_f} = 9{\rm{ kcal}}/{\rm{g}}} \right)\), and f is the amount of fat in diet \(\left( {f = 9{\rm{ kcal}}/{\rm{g}}} \right)\).

Putting all known values,

\(\begin{aligned} E &= \left( {4{\rm{ kcal}}/{\rm{g}}} \right) \times 5\% + \left( {4{\rm{ kcal}}/{\rm{g}}} \right) \times 60\% + \left( {9{\rm{ kcal}}/{\rm{g}}} \right) \times 35\% \\ &= \left( {4{\rm{ kcal}}/{\rm{g}}} \right) \times \frac{5}{{100}} + \left( {4{\rm{ kcal}}/{\rm{g}}} \right) \times \frac{{60}}{{100}} + \left( {9{\rm{ kcal}}/{\rm{g}}} \right) \times \frac{{35}}{{100}}\\ &= 5.75{\rm{ kcal}}/{\rm{g}}\end{aligned}\)

The mass of the food required to meet demand of \(3600{\rm{ kcal}}\) for one worker is,

\(\begin{aligned} m &= \frac{{3600{\rm{ kcal}}}}{{5.75{\rm{ kcal}}/{\rm{g}}}}\\ &= 626{\rm{ g}}\end{aligned}\)

As a result, the total mass of the food required to fulfil the energy demand of 20000 workers is,

\(\begin{aligned} M &= 20000 \times \left( {626{\rm{ g}}} \right)\\ &= \left( {1.252 \times {{10}^7}{\rm{ g}}} \right) \times \left( {\frac{{1{\rm{ kg}}}}{{{{10}^3}{\rm{ g}}}}} \right)\\ &= 12520{\rm{ kg}}\end{aligned}\)

Therefore, the required mass of food to be supplied each day is \(12520{\rm{ kg}}\).