JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Semiconductors

WhatsApp Email

The required height of a TV tower which can cover the population of 6.03 lakh is h . If the average population density is 100 per square km and the radius of earth is 6400 km , then the value of

h

will be

____

Answer: 150

Solution

The distance to the horizon from a tower of height

h

is:

d = 2 R h

where

R = 6400

km is Earth's radius.

The coverage area is:

A = π d^{2} = π \times 2 R h = 2 π R h

The total population covered is:

Population = Density \times Area = 100 \times 2 π R h

Given population =

6.03 \times 1 0^{5}

6.03 \times 1 0^{5} = 100 \times 2 π \times 6400 \times 1 0^{- 3} \times h

(converting

h

to meters and

R

to km)

6.03 \times 1 0^{5} = 100 \times π \times 12.8 \times h

h = \frac{6.03 \times 1 0 ^{5}}{100 \times π \times 12.8} = \frac{6.03 \times 1 0 ^{5}}{4021.2} \approx 150 m

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Integer answer

Practise more: all JEE Main PYQs on Semiconductors →

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Semiconductors

Solution

More JEE Main questions on Semiconductors