Physics - Fluid Question with Solution | TestHub

Heights if only single material tubes are used of sufficient length, For table$T_1\Rightarrow h_1=\frac{2T\mathrm{c}\mathrm{o}\mathrm{s}\theta }{prg}=\frac{2\times 0.075\times \mathrm{c}\mathrm{o}\mathrm{s}0Kd}{1000\times 2\times {10}^{-4}\times 10}=7.5\mathrm{c}\mathrm{m}$For table$T_2\Rightarrow h_2=\frac{2T\mathrm{c}\mathrm{o}\mathrm{s}{\theta }^{\prime }}{prg}=\frac{2\times 0.075\times \mathrm{c}\mathrm{o}\mathrm{s}60°}{1000\times 2\times {10}^{-4}\times 10}=3.75\mathrm{c}\mathrm{m}$Weight of water in meniscus will be different in two cases because angle of contact is different$\Rightarrow \left(\mathrm{C}\right)$is correct. Liquid will rise only upto height of$5\mathrm{ }\mathrm{c}\mathrm{m}$and meniscus will adjust by changing is radius of curvature. If the liquid goes up in tube$2$then it will not be able to support the weight of the liquid$\Rightarrow \left(\mathrm{D}\right)$is incorrect. For option$\left(\mathrm{B}\right)$Refer diagram below -As$K_4$maximum height upto which liquid may rise in capillary is$h_1=7.5\mathrm{c}\mathrm{m}$And length of tube available is$8\mathrm{ }\mathrm{c}\mathrm{m}$$\Rightarrow$Water will rise upto$7.5\mathrm{ }\mathrm{c}\mathrm{m}$$\Rightarrow \left(\mathrm{B}\right)$is correct. For option$\left(\mathrm{A}\right)$, Refer figure below -Here maximum height upto which water can rise in case-$2$,$h_2=3.75\mathrm{c}\mathrm{m}$. Also the length of tube available is$5\mathrm{ }\mathrm{c}\mathrm{m}$.$\Rightarrow$Water will rise upto$3.75\mathrm{c}\mathrm{m}$$\Rightarrow \left(\mathrm{A}\right)$is correct.