What happens to the
Question: What happens to the inductive reactance and the current in a purely inductive circuit if the frequency is halved?Both, inductive reactance and current will be halved.Inductive reactance will be halved and current will be doubled.Inductive reactance will be doubled and current will be halved.Both, inducting reactance and current will be doubled.Correct Option: , 2 Solution: (2) $X_{L}=\omega L$ $\mathrm{i}=\frac{\mathrm{v}_{0}}{\omega \mathrm{L}}$...
Read More →Your friend is having eye sight problem.
Question: Your friend is having eye sight problem. She is not able lo see clearly a distant uniform window mesh and it appears to her as nonuniform and distorted. The doctor diagnosed the problem as :AstigmatismMyopia with AstigmatismPresbyopia with AstigmatismMyopia and hypermetropiaCorrect Option: , 2 Solution: If distant objects are blurry then problem is Myopia. If objects are distorted then problem is Astigmatism...
Read More →Two identical photocathodes
Question: Two identical photocathodes receive the light of frequencies $f_{1}$ and $f_{2}$ respectively. If the velocities of the photo-electrons coming out are $v_{1}$ and $v_{2}$ respectively, then$\mathrm{v}_{1}^{2}-\mathrm{v}_{2}^{2}=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left[\mathrm{f}_{1}-\mathrm{f}_{2}\right]$$\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left[\mathrm{f}_{1}+\mathrm{f}_{2}\right]$$v_{1}+v_{2}=\left[\frac{2 h}{m}\left(f_{1}+f_{2}\right)\right]^{\frac{...
Read More →The fractional change in the magnetic field intensity at a distance
Question: The fractional change in the magnetic field intensity at a distance ' $r$ ' from centre on the axis of current carrying coil of radius ' $a$ ' to the magnetic field intensity at the centre of the same coil is: (Take $ra)$ $\frac{3}{2} \frac{\mathrm{a}^{2}}{\mathrm{r}^{2}}$$\frac{2}{3} \frac{a^{2}}{r^{2}}$$\frac{2}{3} \frac{r^{2}}{a^{2}}$$\frac{3}{2} \frac{r^{2}}{a^{2}}$Correct Option: , 4 Solution: $\mathrm{B}_{\text {axis }}=\frac{\mu_{0} 1 R^{2}}{2\left(R^{2}+x^{2}\right)^{3 / 2}}$ $...
Read More →A conducting bar of length L is free to slide on two parallel conducting rails as shown in the figure
Question: A conducting bar of length L is free to slide on two parallel conducting rails as shown in the figure Two resistors $R_{1}$ and $R_{2}$ are connected across the ends of the rails. There is a uniform magnetic field $\vec{B}$ pointing into the page. An external agent pulls the bar to the left at a constant speed $v$. The correct statement about the directions of induced currents $I_{1}$ and $I_{2}$ flowing through $R_{1}$ and $\mathrm{R}_{2}$ respectively is :Both $\mathrm{I}_{1}$ and $\...
Read More →The atomic hydrogen
Question: The atomic hydrogen emits a line spectrum consisting of various series.Which series of hydrogen atomic spectra is lying in the visible region ?Brackett seriesPaschen seriesLyman seriesBalmer seriesCorrect Option: , 4 Solution: Conceptual...
Read More →An AC source rated 220V , 50Hz is connected to a resistor.
Question: An AC source rated $220 \mathrm{~V}, 50 \mathrm{~Hz}$ is connected to a resistor. The time taken by the current to change from its maximum to the rms value is :$2.5 \mathrm{~ms}$$25 \mathrm{~ms}$$2.5 \mathrm{~s}$$0.25 \mathrm{~ms}$Correct Option: 1 Solution: $\mathrm{i}=\mathrm{i}_{0} \cos (\omega \mathrm{t})$ $\mathrm{i}=\mathrm{i}_{0}$ at $\mathrm{8}=0$ $\mathrm{i}=\frac{\mathrm{i}_{0}}{\sqrt{2}}$ at $\omega \mathrm{t}=\frac{\pi}{4}$ $\mathrm{t}=\frac{\pi}{4 \omega}=\frac{\pi}{4(2 \p...
Read More →Two identical blocks
Question: Two identical blocks $A$ and $B$ each of mass $m$ resting on the smooth horizontal floor are connected by a light spring of natural length $L$ and spring constant $\mathrm{K}$. A third block $\mathrm{C}$ of mass $\mathrm{m}$ moving with a speed $v$ along the line joining A and B collides with A.The maximum compression in the spring is $v \sqrt{\frac{M}{2 K}}$$\sqrt{\frac{m V}{2 K}}$$\sqrt{\frac{\mathrm{mv}}{\mathrm{K}}}$$\sqrt{\frac{m}{2 K}}$Correct Option: 1 Solution: (1) $\mathrm{C}$...
Read More →Match List-I with List-II
Question: Match List-I with List-II Choose the most appropriate answer from the options given below :(a)-(i),(b)(iii),(c)-(iv),(d)-(ii)(a)(ii),(b)(iv),(c)(iii),(d)(i)(a)-(ii),(b)-(iii),(c)-(iv),(d)-(i)(a)-(ii),(b)-(iii),(c)-(i),(d)-(iv)Correct Option: , 4 Solution: (d) $\tan \phi=\frac{\mathrm{V}_{\mathrm{L}}-\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{R}}}=\frac{\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}$...
Read More →In a scries LCR resonance circuit,
Question: In a scries LCR resonance circuit, if we change the resistance only, from a lower to higher value :The bandwidth of resonance circuit will increase.The resonance frequency will increase.The quality factor will increase.The quality factor and the resonance frequency will remain constant.Correct Option: 1 Solution: Bandwidth $=\mathrm{R} / \mathrm{L}$ Bandwidth $\propto \mathrm{R}$ So bandwidth will increase...
Read More →A block of 200 g mass moves with a uniform speed in a horizontal circular groove,
Question: A block of 200 g mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius 20 cm. If the block takes 40 s to complete one round, the normal force by the side walls of the groove is :0.0314 N$9.859 \times 10^{-2} \mathrm{~N}$$6.28 \times 10^{-3} \mathrm{~N}$$9.859 \times 10^{-4} \mathrm{~N}$Correct Option: , 4 Solution: $\mathrm{N}=m \omega^{2} \mathrm{R}$ $\mathrm{N}=\mathrm{m}\left[\frac{4 \pi^{2}}{\mathrm{~T}^{2}}\right] \mathrm{R}$ .....(1) ...
Read More →In the experiment of Ohm's law,
Question: In the experiment of Ohm's law, a potential difference of $5.0 \mathrm{~V}$ is applied across the end of a conductor of length $10.0 \mathrm{~cm}$ and diameter of $5.00 \mathrm{~mm}$. The measured current in the conductor is $2.00$ A. The maximum permissible percentage error in the resistivity of the conductor is :-$3.9$$8.4$$7.5$$3.0$Correct Option: 1 Solution: $\mathrm{R}=\frac{\rho \ell}{\mathrm{A}}=\frac{\mathrm{V}}{\mathrm{I}}$ $\rho=\frac{\mathrm{AV}}{\mathrm{I} \ell}=\frac{\pi \...
Read More →Two particles A and B
Question: Two particles A and B of equal masses are suspended from two massless springs of spring constants $K_{1}$ and $K_{2}$ respectively.If the maximum velocities during oscillations are equal, the ratio of the amplitude of $A$ and $B$ is$\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}$$\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}$$\sqrt{\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}}$$\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}$Correct Option: , 4 Solution: (4) $\mathrm{A}_{1} \omega_{1}=\mathrm{A}_{2} \omega_{...
Read More →The four arms of a Wheatstone
Question: The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of $15 \Omega$ resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of $10 \mathrm{~V}$ is maintained across AC. $2.44 \mu \mathrm{A}$$2.44 \mathrm{~mA}$$4.87 \mathrm{~mA}$$4.87 \mu \mathrm{A}$Correct Option: , 3 Solution: $\frac{x-10}{100}+\frac{x-y}{15}+\frac{x-0}{10}=0$ $53 x-20 y=30$..(1) $\frac{y-10}{60}+\frac{y-x}{15}+\frac{y-0}{5}=...
Read More →The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation :
Question: The stopping potential in the context of photoelectric effect depends on the following property of incident electromagnetic radiation :PhaseIntensityAmplitudeFrequencyCorrect Option: , 4 Solution: Stopping potential changes linearly with frequency of incident radiation....
Read More →The position, velocity and acceleration of a particle
Question: The position, velocity and acceleration of a particle moving with a constant acceleration can be represented by :Correct Option: , 2 Solution: Option (2) represent correct graph for particle moving with constant acceleration, as for constant acceleration velocity time graph is straight line with positive slope and $x$-t graph should be an opening upward parabola....
Read More →A particle is travelling 4 times as fast as an electron.
Question: A particle is travelling 4 times as fast as an electron. Assuming the ratio of de-Broglie wavelength of a particle to that of electron is $2: 1$, the mass of the particle is :-$\frac{1}{16}$ times the mass of $\mathrm{e}^{-}$8 times the mass of $e^{-}$16 times the mass of $\mathrm{e}^{-}$$\frac{1}{8}$ times the mass of $\mathrm{e}^{-}$Correct Option: , 4 Solution: $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\frac{\lambda_{p}}{\lambda_{e}}=\frac{p_{c}}{p_{p}}=\frac{m_{e} v_{e}}{m_{p} v_{p}...
Read More →An RC circuit as shown in the figure is driven by a AC source generating a square wave.
Question: An RC circuit as shown in the figure is driven by a AC source generating a square wave. The output wave pattern monitored by CRO would look close to : Correct Option: , 3 Solution: For $t_{1}-t_{2}$ Charging graph $\mathrm{t}_{2}-\mathrm{t}_{3}$ Discharging graph...
Read More →A hairpin like shape
Question: A hairpin like shape as shown in figure is made by bending a long current carrying wire. What is the magnitude of a magnetic field at point $P$ which lies on the centre of the semicircle ? $\frac{\mu_{0} \mathrm{I}}{4 \pi \mathrm{r}}(2-\pi)$$\frac{\mu_{0} \mathbf{I}}{4 \pi \mathrm{r}}(2+\pi)$$\frac{\mu_{0} \mathbf{I}}{2 \pi \mathrm{r}}(2+\pi)$$\frac{\mu_{0} \mathbf{I}}{2 \pi r}(2-\pi)$Correct Option: , 2 Solution: (2) $\mathrm{B}=2 \times \mathrm{B}_{\text {st.wire }}+\mathrm{B}_{\text...
Read More →Two cells of emf
Question: Two cells of emf $2 \mathrm{E}$ and E with internal resistance $r_{1}$ and $r_{2}$ respectively are connected in series to an external resistor $R$ (see figure). The value of $R$, at which the potential difference across the terminals of the first cell becomes zero is $r_{1}+r_{2}$$\frac{\mathrm{r}_{1}}{2}-\mathrm{r}_{2}$$\frac{r_{1}}{2}+r_{2}$$\mathrm{r}_{1}-\mathrm{r}_{2}$Correct Option: , 2 Solution: $i=\frac{3 E}{R+r_{1}+r_{2}}$ $\mathrm{TPD}=2 \mathrm{E}-\mathrm{ir}_{1}=0$ $2 \mat...
Read More →For an electromagnetic wave travelling in free space,
Question: For an electromagnetic wave travelling in free space, the relation between average energy densities due to electric (Ue ) and magnetic (Um) fields is :$\mathrm{U}_{\mathrm{e}}=\mathrm{U}_{\mathrm{m}}$$U_{e}U_{m}$$\mathrm{U}_{\mathrm{e}}\mathrm{U}_{\mathrm{m}}$$\mathrm{U}_{\mathrm{e}} \neq \mathrm{U}_{\mathrm{m}}$Correct Option: 1 Solution: In EMW, Average energy density due to electric $\left(U_{\rho}\right)$ and magnetic $\left(U_{m}\right)$ fields is same....
Read More →A carrier signal
Question: A carrier signal $\mathrm{C}(\mathrm{t})=25 \sin \left(2.512 \times 10^{10} \mathrm{t}\right)$ is amplitude modulated by a message signal $\mathrm{m}(\mathrm{t})=5 \sin \left(1.57 \times 10^{8} \mathrm{t}\right)$ and transmitted through an antenna. What will be the bandwidth of the modulated signal?$8 \mathrm{GHz}$$2.01 \mathrm{GHz}$$1987.5 \mathrm{MHz}$$50 \mathrm{MHz}$Correct Option: , 4 Solution: (4) Band width $=2 \mathrm{f}_{\mathrm{m}}$ $\omega_{\mathrm{m}}=1.57 \times 10^{8}=2 \...
Read More →Solve this following
Question: The zener diode has a $\mathrm{V}_{\mathrm{z}}=30 \mathrm{~V}$. The current passing through the diode for the following circuit is ......... mA. Solution: $i=\frac{60}{4000} A$ $\mathrm{i}_{1}=\frac{30}{5000} \mathrm{~A}$ $i-i_{1}=\frac{60}{4000}-\frac{30}{5000}=\frac{9}{1000} \mathrm{~A}$ current from zener diode $\mathrm{i}_{\mathrm{z}}=\mathrm{i}-\mathrm{i}_{1}=9 \mathrm{~mA}$...
Read More →A 25 m long antenna is mounted on an antenna tower. The height of the antenna tower is 75 m.
Question: A 25 m long antenna is mounted on an antenna tower. The height of the antenna tower is 75 m. The wavelength (in meter) of the signal transmitted by this antenna would be :300400200100Correct Option: 4, Solution: Length of Antena $=25 \mathrm{~m}=\frac{\lambda}{4}$ $\Rightarrow \lambda=100 \mathrm{~m}$...
Read More →In Young's double slit arrangement,
Question: In Young's double slit arrangement, slits are separated by a gap of $0.5 \mathrm{~mm}$, and the screen is placed at a distance of $0.5 \mathrm{~m}$ from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of 5890A is :-$1178 \times 10^{-9} \mathrm{~m}$$1178 \times 10^{-6} \mathrm{~m}$$1178 \times 10^{-12} \mathrm{~m}$$5890 \times 10^{-7} \mathrm{~m}$Correct Option: 2, Solution: $\beta=\frac{\lambda \mathrm{D}}{...
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