Question

Water rises in a vertical capillary tube up to a length of 10 cm. If the tube is inclined at 45°, the length of water risen in the tube will be
(a) 10 cm
(b) $10\sqrt{2}$ cm
(c) $10/\sqrt{2}$ cm
(d) none of these

Answer 1

Correct option: (b)


$$\begin{gathered} \mathrm{Given}: \\ \mathrm{l}=10\mathrm{cm} \\ \mathrm{\alpha }={45}^0 \\ \mathrm{Rise}\mathrm{in}\mathrm{water}\mathrm{level}\mathrm{after}\mathrm{the}\mathrm{tube}\mathrm{is}\mathrm{tilted}=h \\ \Rightarrow \mathrm{l}=\mathit{}h\cos {45}^0 \\ \Rightarrow h=\frac{\mathrm{l}}{\cos {45}^0}=\frac{10}{\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.}\right)}=10\sqrt{2}\mathrm{cm} \end{gathered}$$