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 m.

Question

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 m. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

d =√2 Rh d =√2 × 6400 × h × 10-3( h in m ) Area =π d 2 =(π × 2 × 6400 × h × 10-3) km 2 6.03 × 100000=100 × π × 2 × 6400 × 10-3 h h =(6.03 × 105/10 × π × 128) h =150 m

Q. 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 ______ $m$ .

$d = 2 R h$
$d = 2 \times 6400 \times h \times 1 0^{- 3} (h$ in $m)$
Area $= π d^{2}$
$= (π \times 2 \times 6400 \times h \times 1 0^{- 3}) k m^{2}$
$6.03 \times 100000 = 100 \times π \times 2 \times 6400 \times 1 0^{- 3} h$
$h = \frac{6.03 \times 1 0 ^{5}}{10 \times π \times 128}$
$h = 150 m$

Q. 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 6400km, then the value of h will be ______m.

d=2Rh​ d=2×6400×h×10−3​(h in m) Area =πd2 =(π×2×6400×h×10−3)km2 6.03×100000=100×π×2×6400×10−3h h=10×π×1286.03×105​ h=150m

Q. 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 ______ $m$ .

$d = 2 R h$
$d = 2 \times 6400 \times h \times 1 0^{- 3} (h$ in $m)$
Area $= π d^{2}$
$= (π \times 2 \times 6400 \times h \times 1 0^{- 3}) k m^{2}$
$6.03 \times 100000 = 100 \times π \times 2 \times 6400 \times 1 0^{- 3} h$
$h = \frac{6.03 \times 1 0 ^{5}}{10 \times π \times 128}$
$h = 150 m$