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Question: A balloon whose total weight uninflected is \(150kg\) is moored to the ground and then filled with h...

A balloon whose total weight uninflected is 150kg150kg is moored to the ground and then filled with hydrogen of density 0.090gm/litre.0.090gm/litre. If the capacity of the envelope is 512kilolitres512kilolitres and the density of the air is 1.293gm/litre1.293gm/litre, find the tension in the mooring rope.

Explanation

Solution

In order to solve the given question, first of all we need to find the total mass of the balloon when it is filled with hydrogen. After that we need to find the mass of the envelope where air is inside it. Then we can calculate the total tension in the mooring rope by putting the equation of tension. After that we can finally conclude with the correct solution.

Complete step by step solution:
The mass of the balloon is given as, m=150kgm = 150kg
And the density of hydrogen is given as, ρh=0.090{\rho _h} = 0.090gm/litre
Also the volume of the envelope is given as, V=512×103V = 512 \times {10^3}litres
We know that density of a substance is mass per unit volume of the substance.
Therefore, ρh=mV{\rho _h} = \dfrac{m}{V}
m=ρhV\Rightarrow m = {\rho _h}V
Now let us calculate the total mass of the balloon including the mass of the hydrogen.
mb=150+0.090×512×103103\Rightarrow {m_b} = 150 + \dfrac{{0.090 \times 512 \times {{10}^3}}}{{{{10}^3}}}
mb=150+46.08=196.08kg\Rightarrow {m_b} = 150 + 46.08 = 196.08kg
Now, let us find the mass of the envelope with air filled inside it.
me=ρaV\Rightarrow {m_e} = {\rho _a}V
me=1.293×512103=0.66kg\Rightarrow {m_e} = \dfrac{{1.293 \times 512}}{{{{10}^3}}} = 0.66kg
Now, we need to calculate the total tension in the mooring rope.
We know the formula of tension is written as, T=mg±maT = mg \pm ma
Here, in this case the acceleration will be only due to gravity. So, the above equation can be written as,
T=mbgmeg\Rightarrow T = {m_b}g - {m_e}g
T=196.08×100.66×10\Rightarrow T = 196.08 \times 10 - 0.66 \times 10
T=1960.86.6=1954.2N\therefore T = 1960.8 - 6.6 = 1954.2N

Hence, the required value of tension in the mooring rope is 1954.2N1954.2N.

Note: We define tension as the pulling force which is transmitted axially by a flexible medium. This flexible medium can be a rope, string cable etc. It is also defined as the action reaction pair of forces that acts at the each end of the flexible medium.