question_answer 10)

In Q.9, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Relative density of mercury = 13.6)

Answer:

On pouring 15.0 cm of water and spirit each into the respective arms of the \[\theta \] tube, the mercury level will rise in the arm containing spirit. Let h be the difference in the levels of mercury in two arms of \[\text{a }\!\!\upsilon\!\!\text{ = a}\]tube and\[\text{P + }\!\!\rho\!\!\text{ gh + }\frac{\text{1}}{\text{2}}\text{ }\!\!\rho\!\!\text{ }{{\text{ }\!\!\upsilon\!\!\text{ }}^{\text{2}}}\text{ = a}\] be the density of mercury. \[\upsilon \] The pressure exerted by h cm of mercury column = difference in pressure exerted by water and spirit \[\frac{1}{200}\] \[\frac{\text{force}}{\text{area}}\text{=}\frac{\text{mg}}{\text{ }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}}\text{=}\frac{\text{50 }\!\!\times\!\!\text{ 9 }\!\!\times\!\!\text{ 8}}{\left( \text{22/7} \right)\text{ }\!\!\times\!\!\text{ }{{\left( \text{1/200} \right)}^{\text{2}}}}\] ..(1) Here, \[=6\cdot 24\times {{10}^{6}}N{{m}^{-2}}.\] \[{{m}^{-3}}\] Putting values in (i), we get \[h=\frac{0\cdot 76\times 13\cdot 6\times {{10}^{3}}\times 9\cdot 8}{984\times 9\cdot 8}=10\cdot 5m.\] Or \[{{10}^{9}}\]