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 ?
1. 0.82 cm
2. 1.35 cm
3. 0.41 cm
4. 2.32 cm
View Answer
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
1. 2 s
2. \(\frac{\sqrt{3}}{2}\text{ s}\)
3. \(2\sqrt{3}\text{ s}\)
4. \(\frac{2}{\sqrt{3}}\text{ s}\)
View Answer
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 \(T' = T\sqrt{\frac{g}{g'}} = \sqrt{3}\sqrt{\frac{4}{3}} = 2\text{ s}\).
Given below are two statements:
Assertion (A): A hydrogen-filled balloon stops rising after it has attained a certain height in the sky.
Reason (R): The atmospheric pressure decreases with height and becomes zero when maximum height is attained by balloon.
1. Both (A) and (R) are true and (R) is the correct explanation of (A).
2. Both (A) and (R) are true but (R) is not the correct explanation of (A).
3. (A) is true but (R) is false.
4. Both (A) and (R) are false.
View Answer
As the balloon rises, the density of air decreases, leading to a decrease in buoyant force until it equals the weight of the balloon, so it stops rising. Thus, Assertion is true. However, atmospheric pressure does not become zero at this height, making Reason false.
The atmospheric pressure at a place is \(10^5\text{ Pa}\). If liquid of specific gravity equal to 2, be employed as the barometric liquid, the barometric height will be (\(g = 10\text{ m/s}^2\))
1. 5 m
2. 3.2 m
3. 7 m
4. 4.5 m
View Answer
Using the relation \(P = \rho g h\), where density \(\rho = 2 \times 10^3\text{ kg/m}^3\) (specific gravity is 2). Substituting the values: \(10^5 = 2 \times 10^3 \times 10 \times h\). Solving for \(h\) gives \(h = 5\text{ m}\).
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.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
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.