Physics in Depth

Worked Problems Class 12 Physics · NEB 1. The graph of a wave motion is given below. Observe the graph and answer the following questions. (a) What properties of the medium are relevant for the propagation of this type of wave? The given wave is a transverse mechanical wave . For its propagation, the medium should have: Elasticity Inertia In addition, the frictional/resistive force of medium should be minimum to minimize energy loss. (b) What do PQ and PR indicate with respect to the wave? PQ represents the amplitude (i.e. maximum displacement of a particle from mean position) and PR represents the wavelength (i.e. distance between two consecutive troughs/crests). (c) What do you mean by phase in a wave? Why are particles at 'S' and 'R' in anti-phase? Phase in a wave refers to the state of motion of a vibrating particle at a given instant, which defines both the position and direction of motion. Particles S and R are in anti-phase ...

Model Question Solution — Physics DM State the principle of homogeneity. Solution The principle of homogeneity states that, 'the dimensions of all term in an equation must be identical.i.e. dimension in L.H.S. of an equation is equal to the dimension in R.H.S. of an equation.' DM A student writes the formula for speed of longitudinal waves as \( v=\sqrt{\frac{\gamma P}{D}} \) where P is the pressure, D is density and \(\gamma\) is the ratio of specific heats which is dimensionless. Check the correctness of the formula. Solution The given equation is \[ v=\sqrt{\frac{\gamma P}{D}} \]. The dimensional formula of each of the quantity here are: Dimensional formula of velocity i.e., Dimension of L.H.S. = \([LT^{-1}]\) Dimensional formula of pressure = \(\frac{[MLT^{-2}]}{[L^2]}\) = \([ML^{-1}T^{-2}]\)...

Differential Expansion Differential Expansion Differential expansion is the difference in the amounts by which two objects (rods or layers) of different materials expand when subjected to the same temperature difference $\Delta \theta$. Derivation Consider two parallel metallic rods rigidly connected to a common base at one end. Let their initial lengths be $l_1$ and $l_1'$ with corresponding coefficients of linear expansion $\alpha$ and $\alpha'$. Let the initial temperature of the setup be $\theta_1^\circ\text{C}$, where the initial difference between their free ends is $d_1 = l_1 - l_1'$. When the temperature increases uniformly to $\theta_2^\circ\text{C}$, the rods expand to new lengths $l_2$ and $l_2'$. The new separation distance between their free ends becomes $d_2 = l_2 - l_2'$. The individual final lengths are giv...

Simple harmonic motion — circle and sine curve Circle and simple harmonic motion projection A particle travels around a circle at constant angular speed. A dashed line drops from the particle to the vertical diameter, and that foot point's height is plotted over time on the right as a sine curve. time particle on circle height of foot over time (SHM) Pause Period 3.0 s For a particle moving in a uniform circular motion, its foot of perpendicular in either X axis or Y axis is at simple harmonic motion (SHM).

Kirchhoff's circuit laws — worked problem Student observation in a potentiometer A student sets up a circuit as shown in the figure given below to measure the emf of a test cell. (i) Explain why he is unable to find a balance point and state the change he must make in order to achieve the balance. (ii) State how he would recognize the balance point. (iii) He obtained the balance point for distance 37.5 cm using standard cell of emf 1.50 V. And for the test cell, the balance distance AB was 25.0 cm. Calculate the emf of the test cell. (iv) He could have used an ordinary voltmeter to measure the emf of the test cell directly. The student, however, argues that the above instrument is more precise than an ordinary voltmeter. Justify his logic. (i)Problem in circuit The positive terminals of both the driving cell and the test cell must be connected to the same terminal A. If their polarities oppose each other at the joint (A...

Simple Pendulum Simulation Explore how length & temperature affect the period of oscillation Parameters Pendulum Length 1.00 m Temperature 25 °C Initial Angle 15° Live Measurements Period T 2.00 s Frequency 0.50 Hz Eff. Length 1.000 m Δ Length +0.000 m Physics T = 2π √(L / g) L eff = L₀ (1 + α·ΔT) α (steel) = 12×10⁻⁶ /°C ·   g = 9.81 m/s² Controls ⏸ Pause ...