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Q.

The cylindrical vessel shown in the figure has two pistons in it. The piston on the left touches a spring attached to the wall of the vessel. The wall has a hole in it. The volume of the air between the pistons is  2000cm3 and its pressure is initially equal to the external atmospheric pressure of  105N/m2. The piston on the right is slowly pressed inwards, maintaining constant temperature, until its inner surface is at the position where the inner surface of the piston on the left was initially. What will be the final volume of the air between the pistons? 

The cross–sectional area of the cylinder is  100  cm2, and a force of 10 N compresses the spring by 1 cm. If  V=x×104m3  find x6  (nearest integer)

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Detailed Solution

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p2=p1+DxA=105Pa+1000N/m102m2.x,

Where D is the spring constant. Then the volume of the enclosed gas is    V2=A.x=102m2.x.
Since temperature is constant, Boyle’s law can be applied:  p1V1=p2V2=(p1+DxA)Ax.

Rearranged by the powers of x :   Dx2+p1Axp1V1=0
and the solution of the equation is  x=p1A±p12A2+4Dp1V12D
Numerically, the equation (1) is  103Nm.x2+103N.x2.10h2Nm=0
Which simplifies to  5x2+5x1=0, and the solution is  x=5±25+2010=0.1708m0.171m

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