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QuestionPhysicsClass 11

A vertical cylinder of height 100cm contians air at a constant temperatur. The top is closed by a frictionless light postpn. The atmospheric pressure pver the piston.

Given, $${P}_{{1}}={A}\to{m}{o}{s}{p}{h}{e}{r}{i}{c}{p}{r}{e}{s}{s}{u}{r}{e}$$
$$={75}\times\rho{g},$$
$${V}_{{1}}={100}\times{A}$$
$${\left({P}_{{2}}\right)}={A}{t}{m}{o}{s}{p}{h}{e}{r}{i}{c}{p}{r}{e}{s}{s}{u}{r}{e}$$
+ Mercury pressure
$$={75}\rho{g}+{h}\rho{g}$$
$${\left({\quad\text{if}\quad}{h}={h}{e}{i}{g}{h}{t}{o}{f}{m}{e}{r}{c}{u}{r}{y}\right)}$$
$${V}_{{2}}={\left({100}-{h}\right)}$$
$${\left({\left({P}_{{1}}\right)}{\left({V}_{{1}}\right)}\right)}={\left({\left({P}_{{2}}\right)}{\left({V}_{{2}}\right)}\right)}$$
$$\Rightarrow{75}\rho{g{{\left({100}{A}\right)}}}$$
$$={\left({\left({75}+{h}\right)}\rho{g{{\left({100}-{h}\right)}}}{A}\right)}$$
$$\Rightarrow{\left({75}\times{100}\right)}={\left({\left({75}+{h}\right)}{\left({100}-{h}\right)}\right)}$$
$$\Rightarrow{7500}={7500}-{75}{h}+{100}{h}-{\left({h}^{{2}}\right)}$$
$$\Rightarrow{\left({h}^{{2}}\right)}-{100}{h}+{75}{h}={0}$$
$${h}^{{2}}-{25}{h}={0}$$
$$\Rightarrow{\left({h}^{{2}}\right)}={25}{h}$$
$${h}={25}{c}{m}$$
Height of Mercury that can be poured
$$={25}{c}{m}$$ .          