Answer:
If T is the temperature of body and Ty is the temperature
of surroundings, then according to Newton's law of cooling
\[{{10}^{-3}}\] or \[=6\cdot 9\times {{10}^{7}}Pa;\]
If the temperature of the body falls from \[=6\cdot
9\text{ }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{7}}}\text{
}\!\!\times\!\!\text{ }\left( \text{22/7} \right)\text{ }\!\!\times\!\!\text{
}{{\left( \text{3 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-3}}}
\right)}^{\text{2}}}\]and \[\therefore \]in time t, then integrating the above
relation, we have
\[=4\left[ 6\cdot 9\text{
}\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{7}}}\text{ }\!\!\times\!\!\text{
}\frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ 9 }\!\!\times\!\!\text{
1}{{\text{0}}^{\text{-6}}} \right]=7\cdot 8\times
\text{1}{{\text{0}}^{\text{3}}}\text{N}\] or \[1\cdot 1\times {{10}^{8}}Pa.\]
or
\[0\cdot 32{{m}^{3}}\]or \[=11\cdot 6\times {{10}^{11}}N{{m}^{-2}}\]or\[p=1\cdot
1\times {{10}^{8}}pa;\]
or
\[V=0\cdot 32{{m}^{3}};\]
As per question condition (i),
\[B=1\cdot 6\times
{{10}^{11}}Pa\]\[\vartriangle V=\frac{\text{pV}}{\text{B}}=\frac{\left( 1\cdot
1\times {{10}^{8}} \right)\times 0\cdot 32}{1\cdot 6\times {{10}^{11}}}\]\[=2\cdot
2\times \text{1}{{\text{0}}^{\text{-4}}}{{\text{m}}^{\text{3}}}\]\[\because
\]\[{{\text{S}}_{\text{LA}}}\text{,}{{\text{S}}_{\text{SA}}}\text{ and
}{{\text{S}}_{\text{SL}}}\]
\[\theta \] ..(2)
As per question, condition (ii),
\[{{\text{S}}_{\text{SL}}}\text{+}{{\text{S}}_{\text{LA}}}\text{cos
}\!\!\theta\!\!\text{ =}{{\text{S}}_{\text{SA}}}\]
\[\text{cos }\!\!\theta\!\!\text{
=}\frac{{{\text{S}}_{\text{SA}}}\text{-}{{\text{S}}_{\text{SL}}}}{{{\text{S}}_{\text{LA}}}}\].(3)
Dividing (3) by (2), we get
\[{{\text{S}}_{\text{SL}}}\text{}{{\text{S}}_{\text{SL}}}\text{,}\]
\[\theta \]
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