Question

Water rises to a height of 2 cm in a capillary tube held vertically. When the tube is tilted ${60^0}$ from the vertical, the length of the water column in the tube will be
$
  (a){\text{ 1cm}} \\
  (b){\text{ 2cm}} \\
  (c){\text{ 3cm}} \\
  (d){\text{ 4cm}} \\
 $

Answer 1

Hint: In this question let the length of the water column in the titled tube by 60 degree from vertical be H cm. Use the concept that if we rotate the component of the titled tube in vertical direction the component of the titled tube length up to which water is filled is equal to the original height of the water in a capillary tube when kept in a vertical position. This will help approaching the problem.

Complete step-by-step solution -

Given data:
Water rises in a capillary tube when held vertically by 2 cm. as shown in the figure 1.
Let x = 2cm.
Now we have to find out the length of the water column in the tube when the tube is tilted 60 degrees from the vertical as shown in figure 2.
Let the length of the water column in the titled tube by 60 degrees from vertical be H cm.
Now if we rotate the component of the titled tube in vertical direction the component of the titled tube length up to which water is filled is equal to the original height of the water in a capillary tube when kept in a vertical position.
Therefore,
Component of H in vertical direction = height of the water rises in a capillary tube when kept in vertical position.
Now the component of H in the vertical direction is the product of the titled tube length up to which water is filled to the cosine of the vertical angle i.e. the angle by which the tube is titled.
Therefore, component of H in the vertical direction = $H\cos {60^o}$
$ \Rightarrow H\cos {60^o} = 2$
Now as we know that $\cos {60^o} = \dfrac{1}{2}$ so use this property in above equation we have,
$ \Rightarrow H \times \dfrac{1}{2} = 2$
Now simplify this we have,
$ \Rightarrow H = 2 \times 2 = 4$ cm.
So the length of the column in the tube when titled by 60 degrees from the vertical is 4cm.
So this is the required answer.
Hence option (D) is the correct answer.

Note: In this question we have simply taken the vertical height of the titled tube that will be the component of H in vertical direction see figure 2 that is $H\cos {60^o}$ equal to the water rise level in case of figure 1 that is 2cm because the original tube is the one that is being altered to attain this new configuration thus the vertical height of water must be same in both of them as the water is not dripping outside thus the total amount of water remains same in both the configurations.