A graph is shown between the maximum kinetic energy (\( E_k \)) of emitted photoelectrons and frequency (\( \nu \)) of the incident light in an experiment of the photoelectric effect. Find:
(i) Threshold frequency
(ii) Work function (in eV)
(iii) Planck's constant
From the graph: The threshold frequency (\( \nu_0 \)) is the x-intercept of the graph: \[ \nu_0 = 2.5 \times 10^{14} \, \mathrm{Hz}. \] The work function (\( \phi \)) is given by: \[ \phi = h \nu_0 = (6.6 \times 10^{-34}) (2.5 \times 10^{14}) = 1.65 \times 10^{-19} \, \mathrm{J}. \] Converting to eV: \[ \phi = \frac{1.65 \times 10^{-19}}{1.6 \times 10^{-19}} = 1 \, \mathrm{eV}. \] \item From the slope of the graph, Planck's constant \( h \) is: \[ h = \frac{\Delta E_k}{\Delta \nu} = \frac{5 \times 10^{-19}}{7 \times 10^{14} - 2.5 \times 10^{14}} = 6.6 \times 10^{-34} \, \mathrm{Js}. \]
The switch (S) closes at \( t = 0 \) sec. The time, in sec, the capacitor takes to charge to 50 V is _________ (round off to one decimal place).
The op-amps in the following circuit are ideal. The voltage gain of the circuit is __________(round off to the nearest integer).
In the system shown below, the generator was initially supplying power to the grid. A temporary LLLG bolted fault occurs at \( F \) very close to circuit breaker 1. The circuit breakers open to isolate the line. The fault self-clears. The circuit breakers reclose and restore the line. Which one of the following diagrams best indicates the rotor accelerating and decelerating areas?
The transformer connection given in the figure is part of a balanced 3-phase circuit where the phase sequence is “abc”. The primary to secondary turns ratio is 2:1. If \( I_a + I_b + I_c = 0 \), then the relationship between \( I_A \) and \( I_{ad} \) will be:
In the circuit shown below, if the values of \( R \) and \( C \) are very large, the form of the output voltage for a very high frequency square wave input is best represented by:
(b) Order of the differential equation: $ 5x^3 \frac{d^3y}{dx^3} - 3\left(\frac{dy}{dx}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^4 + y = 0 $