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): 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.
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 (\(g = 10\text{ m/s}^2\))
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.
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.
The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length \(100 \text{cm}\) to stretch it by \(1 \text{mm}\) is (if Young’s modulus of the wire \(= 2.0 \times 10^{11} \text{N} \text{m}^{-2}\)
Energy density is \(u = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{1}{2} Y \left(\frac{\Delta l}{l}\right)^2 = \frac{1}{2} \times (2.0 \times 10^{11}) \times \left(\frac{10^{-3} \text{m}}{1 \text{m}}\right)^2 = 10^5 \text{J/m}^3\).
Two copper vessels A and B have the same base area but of different shapes. Vessel A takes twice the volume of water as that vessel B requires to fill upto common particular height. Then correct statement among the following is
Liquid pressure at the base depends only on height and density of the liquid, \(P = \rho g h\). Since height is the same, pressure at the base of both vessels is the same.
The Young’s modulus of wire of length L and radius r is Y. If the length and radius are reduced to \(\frac{L}{3}\) and \(\frac{r}{2}\), then its Young’s modulus will be
Young's modulus is an intrinsic property of the material of the wire and is independent of its geometrical dimensions.
Experimental observations show that for given solid material, the magnitude of strain produced is same whether the stress is tensile or compressive. The ratio of tensile stress to longitudinal strain is defined as Young’s modulus and is denoted by \(Y = \sigma/e\). The length of a metal wire is \(l_A\) when the tension in it is \(T_A\) and is \(l_B\) when tension is \(T_B\). The natural length of wire is
Let the natural length be \(L\). Using Hooke's law, \[l_A = L(1 + T_A/AY)\] and \[l_B = L(1 + T_B/AY)\]. Eliminating \(AY\) gives \[L = \frac{T_B l_A - T_A l_B}{T_B - T_A}\].