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The mercury manometer consists if two unequal arms of equal cross section 1 cm^(2) and lenghths 100cm and 50cm. The two open ends are sealed with air in the tube at a pressure of 80cm of mercurey. Some amount of mercury is now introduced in the manometer through the stopcock connected to it. If mercury rises in the shorter tube to a lenght 10cm in steady state, find the length of the mercury column risen in the longer tube.

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Answer Text

Solution :
Let `p_(1)` and `p_(2)` be the pressures in centimetre of mercury in the two arms ater introducing mercury in the tube. Suppose the mercury column rises in the second arm to` l_(0)cm`. <br> using `pV= constant for the shorter arm, <br> `(80cm) (50cm)= p_(1)(50cm-10cm) <br> or, p_(1)=100cm`. ...(i)` Using pV=constant fir the longer arm, <br> `(80cm) (100cm)=p_(2)(100-l_(0))cm` ....(ii) <br> From the figure, <br> `p_(1)=p_(2)+(l_(0)-10)cm`. <br> Thus by (i) <br> `100cm=p_(2)+(l_(0)-10)cm` <br> `p_(2)=100cm-l_(0)cm`. <br> Putting in (ii), <br> `(110-l_(0))(100-l_(0))=8000` <br> or,`l_(0)^(2)-210l_(0)+3000=0` <br> or,`l_(0)=15.5`. <br> The requrired length is `15.5 cm. ` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_imagesHCV_VOL2_C24_S01_008_Q01.png" width="80%">

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