Question

At sea level the atmospheric pressure is 1.04×10^5 Pa. Assuming=10m/s^2 and density of air to be uniform equal to 1.3kg/m^3. Find the height of atmosphere
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Answer 1

Explanation:

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Answer 2

$\huge\bold{\mathbb{QUESTION}}$

At sea level, the atmospheric pressure is $(1.04\times 10^5)\: Pa.$ Assuming $g = 10 m s^{-2}$ and density of air to be uniform and equal to $1.3 kg m^{-3},$ find the height of the atmosphere.

$\huge\bold{\mathbb{GIVEN}}$

Atmospheric pressure at sea level is $(1.04\times 10^5)\: Pa.$

It is assumed that $g = 10\: m/s^{2}.$

The density of air is uniform and equal to $1.3\: kg/m^{3}.$

$\huge\bold{\mathbb{TO\:FIND}}$

The height of the atmosphere.

$\huge\bold{\mathbb{CONCEPT}}$

$P=h \rho g$

$\implies h={\dfrac{P}{ \rho g}}$

Where,

$P$ is the atmospheric pressure.

$h$ is the height.

$\rho$ is the density of air.

$g$ is the gravitational acceleration.

Here,

$P=(1.04\times 10^5)\: Pa$

$g = 10\: m/s^{2}$

$\rho = 1.3\: kg/m^{3}$

$\huge\bold{\mathbb{SOLUTION}}$

Let the height be $h\:m.$

$\therefore h={\dfrac{P}{ \rho g}}$

$\implies h={\dfrac{(1.04\times 10^5)}{(10\times1.3)}}$

$\implies h={\dfrac{(1.04\times 100000)}{(10\times1.3)}}$

$\implies h={\dfrac{104000}{13}}$

$\implies h=\cancel{\dfrac{104000}{13}}$

$\implies {\boxed{\mathfrak{\pink{h=8000}}}}$

So, $h=8000.$

Height $= h\:m=8000\:m=8\:km$

$\huge\bold{\mathbb{THEREFORE}}$

The height of the atmosphere is $8\:km.$