Answer:
We know that the rate of decrease of density \[\rho \] of
air with height y is directly proportional to density \[\rho \] i.e.
\[\text{p=10 atm}=10\times 1\cdot 013\times
{{10}^{5}}\]
where
K is a constant of proportionality. Here ve sign shows that \[\rho \]
decreases as y increases.
\[pa;B=37\times
{{10}^{\text{9}}}\text{N }{{\text{m}}^{\text{-2}}}\] \[=\frac{\vartriangle
V}{V}=\frac{P}{B}=\frac{10\times 1\cdot 013\times {{10}^{5}}}{37\times
{{10}^{9}}}\]
Integrating
it within the conditions, as y changes from 0 to y, density changes from \[=2\cdot
74\times {{10}^{-5}}\],we have
\[\therefore
\]
\[\frac{\vartriangle
V}{V}\]
or
\[=2\cdot 74\times {{10}^{-5}}\]
Here
K is a constant. Suppose \[7\times {{10}^{6}}Pa.\]is a constant such that \[\text{L=10
cm}=0\cdot \text{10 m; p}=7\times {{10}^{6}}\text{Pa;}\] then \[\text{ B=140
GPa}=140\times {{10}^{9}}Pa\]
(b)
The balloon will rise to a height, where its density becomes equal to the air
at that height.
Density
of balloon,
\[B=\frac{pV}{\vartriangle
V}=\frac{p{{L}^{3}}}{\vartriangle V}\text{ or }\vartriangle
V=\frac{p{{L}^{3}}}{B}\]
\[=\frac{\left(
7\times {{10}^{6}} \right)\times {{\left( 0\cdot 10 \right)}^{3}}}{140\times
{{10}^{9}}}\]\[=5\times {{10}^{-8}}{{m}^{3}}\]
As,
\[=5\times {{10}^{-2}}m{{m}^{3}}\] \[0\cdot 10%\] \[=2\cdot 2\times {{10}^{9}}N{{m}^{-2}}.\]
or
\[V=1\] or \[{{10}^{-3}}{{m}^{3}};\]
Taking
log on both the sides \[\vartriangle V/V=0\cdot 10/100={{10}^{-3}}\]
\[B=\frac{pV}{\vartriangle
V}\]\[p=B\frac{\vartriangle V}{V}=\left( 2\cdot 2\times {{10}^{9}} \right)\]
\[\times
{{10}^{-3}}=2\cdot 2\times {{10}^{6}}Pa\]s
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