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√2 cm
(c) 10/√2 cm
(d) none of these

Answer 1

Answer:

Length of water risen in the tube = (b), (c) 10/√2

Answer 2

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 10√2 cm.

Explanation:

Step 1:

When the tube is bent the vertical rise remains the same.

Given data,  

I $=10 \mathrm{cm}$

l is length  

$\alpha=45^{\circ}$

Step 2:

Rise in level of water after the conduit has been named = h

Thus the water in the inclined tube goes up to

$\mathrm{I}=\mathrm{h} \cos \alpha$

$10 \mathrm{cm}=\mathrm{h} \cos 45^{\circ}$

$\mathrm{h}=\frac{10}{\cos 45^{\circ}}\left(\mathrm{or} 10 \mathrm{sec} 45^{\circ}=10 \sqrt{2} \mathrm{cm}\right.$

$h=\frac{10}{\frac{1}{\sqrt{2}}}$

$h=10 \sqrt{2} \mathrm{cm}$