When we press the bulb of a dropper with its nozzle kept in water, air in the dropper is seen to escape in the form of bubbles. Once we release the pressure on the bulb, water gets filled in the dropper. The rise of water in the dropper is due to
- pressure of water.
- gravity of the earth.
- shape of rubber bulb.
- atmospheric pressure.
The Correct Option is D
Solution and Explanation
The rise of water in the dropper when the bulb is released is due to atmospheric pressure.
When you press the bulb, you increase the pressure inside the dropper. As a result, the air inside the dropper is expelled, creating a low-pressure zone. When you release the bulb, the atmospheric pressure pushes the water into the dropper to equalize the pressure, causing the water to rise in the dropper. This phenomenon is based on the principle of atmospheric pressure.
Therefore, the correct option is (D): atmospheric pressure.
Top Questions on Pressure
- Pressure inside a soap bubble is greater than the pressure outside by an amount;(given : R = Radius of bubble, S = Surface tension of bubble)
- Mercury is filled in a tube of radius \(2 \, \text{cm}\) up to a height of \(30 \, \text{cm}\). The force exerted by mercury on the bottom of the tube is ___N.
(Given, atmospheric pressure = \(10^5 \, \text{N/m}^2\), density of mercury = \(1.36 \times 10^4 \, \text{kg/m}^3\), \(g = 10 \, \text{m/s}^2\), \(\pi = \frac{22}{7}\)) Two vessels A and B are of the same size and are at the same temperature. A contains 1 g of hydrogen and B contains 1 g of oxygen. \(P_A\) and \(P_B\) are the pressures of the gases in A and B respectively, then \(\frac{P_A}{P_B}\) is:
- \(T-V\) graph is given for two different pressures \(P_1\) and \(P_2\), Then :
- A spherical soap bubble inside an air chamber at pressure \(𝑃_0 = 10^5 \)Pa has a certain radius so that the excess pressure inside the bubble is Δ𝑃 = 144 Pa. Now, the chamber pressure is reduced to \(8𝑃_0/27\) so that the bubble radius and its excess pressure change. In this process, all the temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure Δ𝑃 in both the cases to be much smaller than the chamber pressure. The new excess pressure Δ𝑃 in Pa is
Questions Asked in CBSE Class VIII exam
(a) Find out the position of the image of an object situated at A in the plane mirror.
(b) Can Paheli at B see this image?
(c) Can Boojho at C see this image?
(d) When Paheli moves from B to C, where does the image of A move?- CBSE Class VIII
- Multiple Images
- A 5 m 60 cm high vertical pole casts a shadow 3 m 20 cm long. Find at the same time -
(i) the length of the shadow cast by another pole 10 m 50 cm high
(ii) the height of a pole which casts a shadow 5 m long.- CBSE Class VIII
- Direct Proportion
- Identify the part which vibrates to produce sound in the following instruments.
- Dholak
- Sitar
- Flute
- CBSE Class VIII
- Sound is Produced by a Vibrating Body
- The process that you saw in Activity 11.7 is used for purification of copper. A thin plate of pure copper and a thick rod of impure copper are used as electrodes. Copper from impure rod is sought to be transfered to the thin copper plate. Which electrode should be attached to the positive terminal of the battery and why?
- CBSE Class VIII
- Electroplating
- How many people and animals died in Yala National Park?
- CBSE Class VIII
- The Tsunami