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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
Options
10 cm
\[10\sqrt{2}\] cm
\[10/\sqrt{2}\] cm
none of these
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Solution
\[\text{ Given }: \]
\[\text{ l = 10 cm }\]
\[\alpha = {45}^0 \]
\[\text{ Rise in water level after the tube is tilted = h}\]
\[ \Rightarrow \text{ l = h} \cos {45}^0 \]
\[ \Rightarrow h = \frac{l}{\cos {45}^0} = \frac{10}{\left( \frac{1}{\sqrt{2}} \right)} = 10\sqrt{2}\text{ cm }\]
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