Question

The figure shows a capillary tube of radius r dipped into water. The atmospheric pressure is${{P}_{o}}$ and the capillary rise of water is h. S is the surface tension for water-glass. Initially, $h=10\,cm$. If the capillary tube is now inclined at $45{}^\circ$, the length of water rising in the tube will be

& A\,.\,\,10\,cm \\\ & B.\,10\sqrt{2}\,cm \\\ & C\,.\,\dfrac{10}{\sqrt{2}}\,cm \\\ & D.\,\,\text{None of}\,\text{these} \\\ \end{aligned}$$

Answer 1

We will make use of the height formula, that is, the capillary rise of liquid to compute the value of the rise in the liquid. The rise of the liquid formula is the initial height of the liquid present in the cube by the cosine of the angle by which the tube is tilted.
Formulae used:

& h=\dfrac{2S\cos \theta }{\rho gr} \\\ & l=\dfrac{h}{\sin \theta } \\\ \end{aligned}$$ **Complete step-by-step solution:** The diagram representing the before and after positioning of the capillary tube.  The formula for computing the capillary rise of liquid is given as follows. $$h=\dfrac{2S\cos \theta }{\rho gr}$$ Where $$h$$is the height of the liquid, $$S$$is the surface tension, $$\theta $$is the angle of inclination, $$\rho $$is the density, $$g$$is the acceleration due to gravity and $$r$$is the radius of curvature. Firstly, we will compute the value of the constants when the tube was not inclined. Consider the formula $$h=\dfrac{2S\cos \theta }{\rho gr}$$ Substitute the values in the above formula $$\begin{aligned} & 10=\dfrac{2S\cos 0{}^\circ }{\rho gr} \\\ & 10=\dfrac{2S}{\rho gr} \\\ \end{aligned}$$ Consider the above equation for further computation. Now, we will compute the value of the length of water rising in the tube when the tube is inclined by an angle of $$45{}^\circ $$ . $$h=\dfrac{2S\cos 45{}^\circ }{\rho gr}$$ Continue further computation. $$\begin{aligned} & h=\dfrac{2S\times {}^{1}/{}_{\sqrt{2}}}{\rho gr} \\\ & \Rightarrow h=\dfrac{1}{\sqrt{2}}\left( \dfrac{2S}{\rho gr} \right) \\\ \end{aligned}$$ Now substitute the value of the constant in the above equation. $$h=10\sqrt{2}$$ **$$\therefore $$The value of the length of water rising in the tube will be, $$10\sqrt{2}\,cm$$, thus, the option (B) is correct.** **Note:** Sometimes the other parameters, such as, the density of the liquid will be given to create confusion. But, while computing the rise in liquid for the tilted tube, there will be a need of only the initial height of the liquid and the angle by which the tube is tilted.