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Question: A TV tower has a height \[150m\]. What is the total population covered by the TV tower, if the popul...

A TV tower has a height 150m150m. What is the total population covered by the TV tower, if the population density around the TV tower is 103km210^{3} km^{-2}? Radius of the earth 6.4×106m6.4 \times 10^{6} m

& A.60.288lakhs \\\ & B.40.192lakhs \\\ & C.100lakhs \\\ & D.20.228lakhs \\\ \end{aligned}$$
Explanation

Solution

A transmission tower is a tall structure, which provides over-head power. The current is transmitted as alternating currents. To find the population covered, we need to calculate the distance up to which transmission could be viewed, given by d=2hRd=\sqrt{2hR}, and The total area over which the transmission is viewed which is given by,A=πd2A=\pi d^{2}. Then the population covered =population  density×A=population\; density \times A

Formula used:
d=2hR,A=πd2d=\sqrt{2hR},A=\pi d^{2}

Complete step by step answer:
A transmission tower is a tall structure, which provides over-head power. The current is transmitted as alternating currents. These are used to carry high voltage from the power grids to the electrical station, where these are further divided into households. The various types of tower designs are based on the voltage transmitted, they are: one circuit, two circuits, and four circuits. The various materials used are: wood, tubular steel, lattices, and concrete. Since they carry high voltages, we must be careful around the transmission tower.
We know that the distance up to which transmission could be viewed is d=2hRd=\sqrt{2hR}, where hh is the height of the tower and RR is the radius of the earth, which is a constant.
Given that, h=150m,R=6.4×106mh=150m, R=6.4\times 10^{6}m and average population density =103km2=103m2=10^{3}km^{-2}=10^{-3}m^{-2}
Substituting, we get, d=2×150×6.4×106=192×106d=\sqrt{2\times 150 \times 6.4\times 10^{6}}=\sqrt{192\times 10^{6}}
The total area over which the transmission is viewed is given by, A=πd2=3.14×192×106=6028×106A=\pi d^{2}=3.14\times 192\times 10^{6}=6028 \times 10^{6}
Then the population covered =population  density×A=103×6028×106=60.28lakhs=population\;density \times A= 10^{-3} \times 6028 \times 10^{6}= 60.28 lakhs
Hence, the correct answer is option A.

Note:
This may seem like a tricky problem, but if the formulas are known, then it is an easy one. It is suggested to convert all the values to meters first and then proceed with the calculation, to avoid confusion of power. Be careful with calculations and the powers. Also don’t waste time calculating the square root for dd, as in AA we use d2d^{2}.