The density of the atmosphere at sea level is 1.3 kg/m3. Assume it does not change with altitude and g = 10 ms–2, how high would the atmosphere extend ?
A tube 1 cm2 in cross-section is attached to the top of a vessel 1 cm high and of cross-section 100 cm2. Water is poured into the system filling it to a depth of 100 cm above the bottom of the vessel as shown in figure. Take g = 10 ms–2. Now,
A body with a volume V neither sinks nor floats in a liquid. If the vessel containing the liquid falls with an acceleration g/3 , then the volume of the solid inside the liquid in the falling condition is:
A sample of metal weights 210 gram in air, 180 gram in water and 120 gram in an unknown liquid. Then:
The reading of spring balance when a block is suspended from it in air, is 60 N. This reading is changed to 40 N when the block is immersed in water. The specific gravity of the block is :
Solution:
Loss of weight in water = $$\text{Weight in air} - \text{Weight in water} = 60\text{ N} - 40\text{ N} = 20\text{ N}$$
Specific gravity = $$\frac{\text{Weight in air}}{\text{Loss of weight in water}} = \frac{60}{20} = 3$$
Therefore, the specific gravity of the block is 3 (Option 1).
An open U-tube contains mercury. When 11.2 cm of water is poured into one of the arms of the tube, how high does the mercury rise in the other arm from its initial level ?
When 11.2 cm of water is poured into one arm, it balances a mercury column of height 2x, where x is the height the mercury rises in the other arm.
Using the pressure balance equation (density of water * height of water = density of mercury * 2x), we get 1 * 11.2 = 13.6 * 2x. Solving for x yields 0.41 cm, making Option 3 the correct answer.
A simple pendulum oscillating in air has a period of \(\sqrt{3}\text{ s}\). If it is completely immersed in non-viscous liquid, having density \((\frac{1}{4})^{\text{th}}\) of the material of the bob, the new period will be
The effective acceleration due to gravity in the liquid is \(g' = g\left(1 - \frac{\rho_L}{\rho_B}\right) = g\left(1 - \frac{1}{4}\right) = \frac{3}{4}g\). Since \(T \propto \frac{1}{\sqrt{g}}\), the new period is \(T' = T\sqrt{\frac{g}{g'}} = \sqrt{3}\sqrt{\frac{4}{3}} = 2\text{ s}\).
A wooden cube is floating in water with some part inside water. When a stone of mass \(4.5\text{ kg}\) is placed on cube then it further sinks by \(5\text{ cm}\). Then side of cube is:
The additional weight of the stone is balanced by the extra buoyant force: \(mg = a^2 \Delta x \rho_w g\). Substituting the values: \(4.5 = a^2 (0.05)(1000)\) gives \(a^2 = 0.09\text{ m}^2\), which yields a side length of \(a = 30\text{ cm}\).
Assertion (A): Weight of an empty balloon measured in air is \(W_1\). If air at atmospheric pressure is filled inside balloon and again weight of the balloon is measured. Weight of balloon in second case is equal to \(W_1\).
Reason (R): Upthrust is equal to weight of the fluid displaced by the body.
Concept: Archimedes' Principle and apparent weight. When air at atmospheric pressure is filled into a balloon, the weight of the air inside is equal to the upthrust exerted by the surrounding air on the volume displaced by this internal air. Thus, the net change in apparent weight due to the air inside is zero. Both Assertion and Reason are true, and Reason explains Assertion by defining upthrust as per Archimedes' principle.