logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Ad
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
0 votes

Figure shows the arrangement , gas is thermally insulated . Having pressure $\;p_{g}\;$ ( greater than atmospheric pressure $\;p_{a}\;$ ) . Spring of force constant k is initially unstretched . Position of mass m & area S is friction less and in equilibrium position , it rises up a distance $\;x_{0}\;$ , then -

$(a)\;final\;pressure=p_{a}+\large\frac{kx_{0}}{S}+\large\frac{mg}{S}\qquad(b)\;work\;done\;by\;gas=\large\frac{kx_{0}^2}{2}\qquad(c)\;final\;pressure=p_{a}+\large\frac{kx_{0}}{S}\qquad(d)\;work\;done\;by\;gas=\large\frac{kx_{0}^2}{2}+mgx_{0}$

Can you answer this question?
 
 

1 Answer

0 votes
Answer : $\;final\;pressure=p_{a}+\large\frac{kx_{0}}{S}+\large\frac{mg}{S}$
Explanation :
FBD of pistern in equation m position
$pS=p_{a} S + mg +kx_{0}$
$p=p_{a}+\large\frac{mg}{S}+\large\frac{kx_{0}}{S}$
answered Mar 15, 2014 by yamini.v
edited Mar 15, 2014 by yamini.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...