Physics in Depth

Image source : Physics Forums In \(_{Z}X^A\), Z is the number of protons, A is the atomic mass number (i.e., sum of number of protons and number of neutrons) and (A-Z) gives the number of neutrons. Calculate the binding energy per nucleon of 26 Fe 56 . Atomic mass of 26 Fe 56 is 55.9349 u and that of 1 H 1 is 1.00783 u. Mass of 0 n 1 = 1.00867 u and 1 u = 931 MeV. Solution: Given, mass of 26 Fe 56 , M = 55.9349 u mass of 1 H 1 i.e., mass of proton, m p = 1.00783 u mass of 0 n 1 , m n = 1.00867 u 1 u = 931 MeV binding energy per nucleon, \(\Delta E_{ben}\) = ? The relation to find the binding energy (when the mass is in atomic mass unit (u)) is, \[\Delta E_{be}=\Delta m \times 931\] Let's find \(\Delta m\) which is the mass defect, \[\begin{align*} \Delta m &=\text{theoretical mass of Fe nucleus} - \text{observed mass of Fe nucleus}\\ &=[Zm_p +(A-Z) m_n] - M\\ &=[26 \times 1.00783

Nikola Tesla ( Image source : Wikipedia ) A circuit consists of a capacitor of 2μF and a resistor of 1000 Ω. An alternating emf of 12 V and frequency 50 Hz is applied. Find the voltage across the capacitor and the phase angle between the applied emf and the current. Solution: Given, capacitance, C = 2μF = 2×10 -6 F resistnce , R = 1000 Ω alternating emf, V = 12 V frequency, f = 50 Hz voltage across capacitor, V C = ? phase angle between the applied emf and the current, φ = ? Now, the current in this circuit is given by the expression \[I=\frac{V}{\sqrt{R^2+ X_C^2}}\] Here, \[\begin{align*} X_C&=\frac{1}{\omega C}\\ &=\frac{1}{2 \pi f C}\\ &=\frac{1}{2 \times \pi \times 50 \times 2 \times 10^{-6}}\\ &=1591.55 \hspace{0.1cm} \Omega\\ \end{align*}\] Then, \[\begin{align*} I&=\frac{V}{\sqrt{R^2 + X_C^2}}\\ &=\frac{12}{\sqrt{(1000)^2 + (1591.55)^2}}\\ &=0.0064

Warning! The questions below are frequenty asked questions in NEB examination. Please follow the syllabus thoroughly. Every chapter and topic of your textbook is important. This questions are only for practice. Mechanics Define moment of inertia. Derive an expression for the moment of inertia of thin uniform rod about an axis through its center and perpendicular to its length. Derive a relation between the torque and angular acceleration in case of a rigid body. State the conservation of angular momentum. Derive the relation between torque and angular momentum. Write the physical significance of moment of inertia. Show that K.E. = \(\frac{1}{2} I \omega^2\) where the symbols have their usual meanings. Define simple harmonic motion. Show that the bob of a simple pendulum may move with simple harmonic motion and find its time period. Find an expression for the energy of particle in SH

The magnetic flux passing perpendicular to the plane of coil is given by φ = 4t 2 +5t+2 where φ is in weber and t is in second. Calculate the magnitude of instantaneous emf induced in the coil when t = 2 sec. Solution: Given, magnetic flux, φ = 4t 2 +5t+2 time, t = 2 sec induced emf, ε = ? From Faraday's law, \[\begin{align*} \epsilon& = -\frac{d\phi}{dt}\\ &=-\frac{d(4t^2+5t+2)}{dt}\\ &=-(8t+5)\\ \end{align*}\] At t = 2 sec, the induced emf is , \[\begin{align*} \epsilon&=-(8 \times 2 + 5)\\ \therefore \epsilon& = -21\hspace{0.1cm}\text{V}\\ \end{align*}\] Thus, the magnitude of induced emf is 21 V. A straight conductor of length 25 cm is moving perpendicular to its length with a uniform speed of 10 m/s making an angle of 45 0 with a uniform magnetic field of 10 T. Calcuate the emf induced across its length. Solution: Given, length, l = 25 cm = 0.25 m sp

Two galvanometers, which are otherwise identical, are fitted with different coils. One has a coil of 50 turns and resistance 10 Ω while the other has 500 turns and a resistance of 600 Ω. What is the ratio of the deflection when each is connected in turns to a cell of emf 25 V and internal resistance 50 Ω? Solution: For galvanometer coil 1, number of turns, N 1 = 50 resistance, R 1 = 10 Ω For galvanometer coil 2, number of turns, N 2 = 50 resistance, R 2 = 10 Ω emf of a cell, E = 25 V internal resistance of a cell, r = 50 Ω Let \(\alpha_1\) and \(\alpha_2\) be the deflection of the galvanometer coil 1 and 2 respectively. \(\frac{\alpha_1}{\alpha_2}\) = ? We know, the restoring torque on a coil is equal to the deflecting torque. \[\begin{align*} \tau &= BINA\\ k \alpha & = BINA\\ \end{align*}\] where k is the restoring torque per unit twist of the suspension wire. So, \[I_1 =

Photons Quantum nature of radiation Light has wave character as well as a particle character. Depending on the situation, one of the character is dominant. When light is passed through a double slit, it shows interference which can only be understood by wave theory of light. When light of sufficiently low wavelength falls on a metal surface, electrons are ejected. This phenomena is called Photoelectric effect and can be understood only in terms of the particle nature of light. According to Planck's quantum theory of radiation, energy from the radiating body is emitted in separate packets of energy, and each packet is called quantum of energy. Each quanta carries a definite amount of energy called Photons . The energy carries by each photon is given by, \[E=hf \hspace{0.1cm} \text{ ... (i)}\] where, f = frequency of radiation, h = Planck's constant (6.625×10 -34 Js) Equation (i) can also be expressed in terms of wavelength as, \[