1,055 practice questions for CBSE Class 12 Physics, organised by NCERT chapter (14 chapters). Every question includes a written explanation. Pick a chapter below, or try the samples first.
Semiconductor Electronics: Materials, Devices and Simple Circuits
75 questions in timed practice
Sample questions
Q1
easy
Among radio waves, infrared radiation, ultraviolet radiation and gamma rays, which has the highest frequency?
AUltraviolet radiation
BInfrared radiation
CRadio waves
DGamma rays
Show answer and explanation
Correct answer: D - Gamma rays
Gamma rays sit at the short-wavelength end of the spectrum, above about $10^{18}\ \text{Hz}$, so their frequency is the highest of the four. Ultraviolet comes next, at roughly $10^{15}$ to $10^{17}\ \text{Hz}$, then infrared near $10^{13}\ \text{Hz}$. Radio waves have the lowest frequency of the four, from a few hundred kilohertz upwards.
Q2
medium
A circuit draws an rms current of $10\ \text{A}$ from an AC source, with the current lagging the voltage by $60^\circ$. What is the wattless component of this current?
A$5\ \text{A}$
B$7.07\ \text{A}$
C$8.66\ \text{A}$
D$10\ \text{A}$
Show answer and explanation
Correct answer: C - $8.66\ \text{A}$
The wattless component is the part at right angles to the voltage, $I\sin\phi = 10\times\sin 60^\circ = 8.66\ \text{A}$. The value $5\ \text{A}$ is $I\cos\phi$, which is the power-carrying component instead; $7.07\ \text{A}$ comes from dividing by $\sqrt{2}$; and $10\ \text{A}$ would mean the whole current is wattless, which needs a phase angle of $90^\circ$.
Q3
medium
A coil of $100$ turns and area $0.020\ \text{m}^2$ rotates at $50$ revolutions per second about an axis at right angles to a uniform magnetic field of $0.10\ \text{T}$. What is the peak emf generated?
A$0.63\ \text{V}$
B$10\ \text{V}$
C$31\ \text{V}$
D$63\ \text{V}$
Show answer and explanation
Correct answer: D - $63\ \text{V}$
The angular speed is $\omega = 2\pi f = 2\pi(50) = 314\ \text{rad s}^{-1}$, so $\varepsilon_0 = NBA\omega = 100\times0.10\times0.020\times314 = 63\ \text{V}$. Slipping in an extra factor of $\tfrac12$ gives $31\ \text{V}$, putting the $50$ revolutions per second straight in as $\omega$ gives $10\ \text{V}$, and leaving out the $100$ turns gives $0.63\ \text{V}$.
Q4
easy
At a point on the screen in an interference experiment, two waves of equal amplitude arrive with a phase difference of $\pi\ \text{rad}$. What is seen at that point?
ABrightness that changes rapidly with time
BExactly half the maximum brightness
CDarkness, because the waves cancel each other
DMaximum brightness, because the waves reinforce
Show answer and explanation
Correct answer: C - Darkness, because the waves cancel each other
A phase difference of $\pi$ means one wave has a crest where the other has a trough, so equal amplitudes cancel completely and the point is dark. Reinforcement needs a phase difference of zero or a whole multiple of $2\pi$. Half the maximum brightness occurs at a phase difference of $\frac{\pi}{2}$, not $\pi$. Coherent sources give a steady pattern, so the brightness at a point does not flicker.
Q5
hard
An ideal transformer converts $2200\ \text{V}$ to $220\ \text{V}$. Its secondary carries a current of $40\ \text{A}$ and has $50$ turns. What is the primary current, and how many turns does the primary have?
A$400\ \text{A}$ and $500$ turns
B$4\ \text{A}$ and $5$ turns
C$40\ \text{A}$ and $5$ turns
D$4\ \text{A}$ and $500$ turns
Show answer and explanation
Correct answer: D - $4\ \text{A}$ and $500$ turns
The voltage falls by a factor of $10$, so the current rises by the same factor on the low-voltage side: $I_p = 40/10 = 4\ \text{A}$. The turns follow the voltages, so the high-voltage primary needs ten times as many turns as the secondary, $N_p = 50\times 10 = 500$. Giving the primary $5$ turns or a current of $400\ \text{A}$ applies the ratio the wrong way round.
Q6
hard
An AC generator has a coil of area $5.0\times10^{-2}\ \text{m}^2$ rotating at $25$ revolutions per second in a uniform magnetic field of $0.20\ \text{T}$, about an axis at right angles to the field. How many turns must the coil have if the peak emf is to be $157\ \text{V}$?
A$5$
B$50$
C$100$
D$628$
Show answer and explanation
Correct answer: C - $100$
With $\omega = 2\pi(25) = 157\ \text{rad s}^{-1}$, each turn contributes $BA\omega = 0.20\times5.0\times10^{-2}\times157 = 1.57\ \text{V}$, so $N = 157/1.57 = 100$ turns. Using the $25$ revolutions per second in place of $\omega$ gives $628$, slipping in an extra factor of $\tfrac12$ gives $50$, and leaving the area out of the calculation altogether gives $5$.
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