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1ldpm229z | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>If $$\mathrm{R}, \mathrm{X}_{\mathrm{L}}$$, and $$\mathrm{X}_{\mathrm{C}}$$ represent resistance, inductive reactance and capacitive reactance. Then which of the following is dimensionless :</p> | [{"identifier": "A", "content": "$$\\frac{R}{X_{L} X_{C}}$$"}, {"identifier": "B", "content": "$$R X_{L} X_{C}$$"}, {"identifier": "C", "content": "$$\\frac{R}{\\sqrt{X_{L} X_{C}}}$$"}, {"identifier": "D", "content": "$$R \\frac{X_{L}}{X_{C}}$$"}] | ["C"] | null | $R=$ Resistance
<br/><br/>$$
\begin{aligned}
& {\left[X_{L}\right]=[R]} \\\\
& {\left[X_{C}\right]=[R]}
\end{aligned}
$$
<br/><br/>So, $\frac{R}{\sqrt{X_{L} X_{C}}}$ is dimensionless. | mcq | jee-main-2023-online-31st-january-morning-shift | 8,912 |
1ldpng9k5 | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>An inductor of $$0.5 ~\mathrm{mH}$$, a capacitor of $$20 ~\mu \mathrm{F}$$ and resistance of $$20 ~\Omega$$ are connected in series with a $$220 \mathrm{~V}$$ ac source. If the current is in phase with the emf, the amplitude of current of the circuit is $$\sqrt{x}$$ A. The value of $$x$$ is ___________</p> | [] | null | 242 | $$
X_L=X_C
$$
<br/><br/>So, $\mathrm{Z}=\mathrm{R}=20 \Omega$
<br/><br/>$$
\begin{aligned}
& \mathrm{i}_{\mathrm{rms}}=\frac{220}{20}=11 \\\\
& \mathrm{i}_{\max }=11 \sqrt{2}=\sqrt{242}
\end{aligned}
$$ | integer | jee-main-2023-online-31st-january-morning-shift | 8,913 |
ldqv7t89 | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | In the given circuit, rms value of current $\left(I_{\mathrm{rms}}\right)$ through the resistor $R$ is:<br/><br/>
<img src="data:image/png;base64,UklGRloNAABXRUJQVlA4IE4NAACQuQCdASoAA7UBP4G61WY2LaunIXCZ0sAwCWlu+F6oMCEpNlvV6X73+I/aB9wf7zxucHjymL5nmP/709ffQk/JhUdCiOA84GRDc5OX85zyGrZeM/6lAVHQoh5oRY9wYg5clRTgZThbSjZR1hTYDI... | [{"identifier": "A", "content": "$ 2 \\sqrt{2} \\mathrm{~A}$"}, {"identifier": "B", "content": "$2 \\mathrm{~A}$"}, {"identifier": "C", "content": "$\\frac{1}{2} \\mathrm{~A}$"}, {"identifier": "D", "content": "$20 \\mathrm{~A}$"}] | ["B"] | null | <p>$${I_{rms}} = {{{V_{rms}}} \over z} = {{200\sqrt 2 } \over {\sqrt {{{100}^2} + {{(200 - 100)}^2}} }}$$</p>
<p>$$ = {{200\sqrt 2 } \over {100\sqrt 2 }}$$</p>
<p>= 2 A</p> | mcq | jee-main-2023-online-30th-january-evening-shift | 8,914 |
1ldr2rai3 | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>In a series LR circuit with $$\mathrm{X_L=R}$$, power factor P<sub>1</sub>. If a capacitor of capacitance C with $$\mathrm{X_C=X_L}$$ is added to the circuit the power factor becomes P<sub>2</sub>. The ratio of P<sub>1</sub> to P<sub>2</sub> will be :</p> | [{"identifier": "A", "content": "1 : $$\\sqrt2$$"}, {"identifier": "B", "content": "1 : 3"}, {"identifier": "C", "content": "1 : 2"}, {"identifier": "D", "content": "1 : 1"}] | ["A"] | null | <p>$${X_L} = R$$</p>
<p>$$ \Rightarrow {P_1} = {R \over {\sqrt {X_L^2 + {R^2}} }} = {1 \over {\sqrt 2 }}$$</p>
<p>Now, $${X_L} = {X_C} = R$$</p>
<p>$$ \Rightarrow {P_2} = {R \over {\sqrt {{R^2} + {{({X_L} - {X_C})}^2}} }} = 1$$</p>
<p>$$ \Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over {\sqrt 2 }}$$</p> | mcq | jee-main-2023-online-30th-january-morning-shift | 8,915 |
1ldsad6ef | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>For the given figures, choose the correct options :</p>
<p><img src="data:image/png;base64,UklGRqAMAABXRUJQVlA4IJQMAAAQoACdASoAA3gBP4G41WY2LSunIXBZosAwCWlu4WsBG/Pz80f6+09s9/7jwj/c+Mtfod4qfger5LkXyXIvkuRfJci0ijxSBoTBZ9E3YZdNGWP2owveHvU8Uo6Fvt6T6PQHgMT/ZdrCaTzSV3U1MgqRX2J3r6IpAzRVcuVyuVyuVyuVyuVyuVyViRO1EvZIxH/cazmO58... | [{"identifier": "A", "content": "The rms current in circuit (b) can be larger than that in (a)"}, {"identifier": "B", "content": "The rms current in figure (a) is always equal to that in figure (b)"}, {"identifier": "C", "content": "The rms current in circuit (b) can never be larger than that in (a)"}, {"identifier": "... | ["C"] | null | <p>For (a), $$i = {V \over R} = {{220} \over {40}} = 5.5\,A$$</p>
<p>for (b), $${X_C} = {1 \over {2\pi fC}} = {1 \over {2\pi \times 50 \times 0.5 \times {{10}^{ - 6}}}} = {{{{10}^6}} \over {50\pi }}\Omega $$</p>
<p>$${X_L} = 2\pi fL = 2\pi \times 50 \times 50 \times {10^{ - 3}} = 50\pi \,\Omega $$</p>
<p>$${X_C} > {X... | mcq | jee-main-2023-online-29th-january-evening-shift | 8,916 |
1ldsbiyqa | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>An inductor of inductance 2 $$\mathrm{\mu H}$$ is connected in series with a resistance, a variable capacitor and an AC source of frequency 7 kHz. The value of capacitance for which maximum current is drawn into the circuit $$\frac{1}{x}\mathrm{F}$$, where the value of $$x$$ is ___________.</p>
<p>(Take $$\pi=\frac{... | [] | null | 3872 | <p>Current drawn is maximum when circuit is in resonance.</p>
<p>$$\omega = {1 \over {\sqrt {LC} }}$$</p>
<p>$$2\pi (7000) = {1 \over {\sqrt {2 \times {{10}^{ - 6}}C} }}$$</p>
<p>$$ \Rightarrow C = {1 \over {3872}}F$$</p> | integer | jee-main-2023-online-29th-january-evening-shift | 8,917 |
1ldtzw3xk | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>A series LCR circuit is connected to an AC source of 220 V, 50 Hz. The circuit contains a resistance R = 80$$\Omega$$, an inductor of inductive reactance $$\mathrm{X_L=70\Omega}$$, and a capacitor of capacitive reactance $$\mathrm{X_C=130\Omega}$$. The power factor of circuit is $$\frac{x}{10}$$. The value of $$x$$ ... | [] | null | 8 | $$
\begin{aligned}
& \cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^2}} \\\\
& \cos \phi=\frac{80}{\sqrt{(80)^2+(60)^2}} \\\\
& \cos \phi=\frac{80}{100} \Rightarrow \frac{8}{10}
\end{aligned}
$$<br/><br/>
So, $x=8$ | integer | jee-main-2023-online-25th-january-evening-shift | 8,918 |
1ldui18qn | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes $$x$$ times its initial resonant frequency $$\omega_0$$. The value of $$x$$ is :</p> | [{"identifier": "A", "content": "1/4"}, {"identifier": "B", "content": "1/16"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "16"}] | ["A"] | null | The resonance frequency of LC oscillations circuit is<br/><br/>
$$
\begin{aligned}
& \omega_0=\frac{1}{\sqrt{\mathrm{LC}}} \\\\
& \mathrm{L} \rightarrow 2 \mathrm{~L} \\\\
& \mathrm{C} \rightarrow 8 \mathrm{C} \\\\
& \omega=\frac{1}{\sqrt{2 \mathrm{~L} \times 8 \mathrm{C}}}=\frac{1}{4 \sqrt{\mathrm{LC}}} \\\\
& \omega=... | mcq | jee-main-2023-online-25th-january-morning-shift | 8,919 |
1ldujqk2f | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>An LCR series circuit of capacitance 62.5 nF and resistance of 50 $$\Omega$$, is connected to an A.C. source of frequency 2.0 kHz. For maximum value of amplitude of current in circuit, the value of inductance is __________ mH.</p>
<p>(Take $$\pi^2=10$$)</p> | [] | null | 100 | $\because$ For maximum amplitude of current, circuit should be at resonance.
<br/><br/>
$$
\begin{aligned}
& \therefore X_{L}=X_{C} \\\\
& \omega L=\frac{1}{\omega C} \\\\
& L=\frac{1}{\omega^{2} C} \\\\
& =\frac{1}{\left(2 \pi \times 2 \times 10^{3}\right)^{2} \times 62.5 \times 10^{-9}} \\\\
& =100 ~\mathrm{mH}
\end... | integer | jee-main-2023-online-25th-january-morning-shift | 8,920 |
1lgp0cep6 | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>Given below are two statements:</p>
<p>Statement I : An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor.</p>
<p>Statement II : An AC circuit containing a pure capacitor or a pure inductor consumes high power due to its non-zero power factor.</p>
<p>In the light of above sta... | [{"identifier": "A", "content": "Both Statement I and Statement II are false"}, {"identifier": "B", "content": "Statement I is true but statement II is false"}, {"identifier": "C", "content": "Statement I is false but statement II is true"}, {"identifier": "D", "content": "Both Statement I and Statement II are true"}] | ["A"] | null | <p>Statement I: An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor.</p>
<p>This statement is incorrect. Electrical resonance occurs in an AC circuit when the capacitive reactance and inductive reactance are equal, causing the impedance of the circuit to be minimum. This typica... | mcq | jee-main-2023-online-13th-april-evening-shift | 8,921 |
1lgyf3l3z | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>Given below are two statements:</p>
<p>Statement I : Maximum power is dissipated in a circuit containing an inductor, a capacitor and a resistor connected in series with an AC source, when resonance occurs</p>
<p>Statement II : Maximum power is dissipated in a circuit containing pure resistor due to zero phase diffe... | [{"identifier": "A", "content": "Statement I is false but Statement II is true"}, {"identifier": "B", "content": "Both Statement I and Statement II are false"}, {"identifier": "C", "content": "Statement I is true but Statement II is false"}, {"identifier": "D", "content": "Both Statement I and Statement II are true"}] | ["D"] | null | <p>In a series LCR circuit connected to an AC source, resonance occurs at a particular frequency at which the inductive reactance is equal to the capacitive reactance, resulting in the minimum impedance of the circuit. At this frequency, the circuit draws maximum current from the source, and thus, the maximum power is ... | mcq | jee-main-2023-online-10th-april-morning-shift | 8,923 |
1lh2zyj8b | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>A capacitor of capacitance $$150.0 ~\mu \mathrm{F}$$ is connected to an alternating source of emf given by $$\mathrm{E}=36 \sin (120 \pi \mathrm{t}) \mathrm{V}$$. The maximum value of current in the circuit is approximately equal to :</p> | [{"identifier": "A", "content": "$$\\frac{1}{\\sqrt{2}} A$$"}, {"identifier": "B", "content": "$$2 \\sqrt{2} A$$"}, {"identifier": "C", "content": "$$\\sqrt{2} A$$"}, {"identifier": "D", "content": "$$2 A$$"}] | ["D"] | null | <p>For a capacitor connected to an AC source, the maximum current $$I_\text{max}$$ can be calculated using the formula:</p>
<p>$$I_\text{max} = E_\text{max} \cdot \omega C$$</p>
<p>where $$E_\text{max}$$ is the maximum voltage, $$\omega$$ is the angular frequency, and $$C$$ is the capacitance.</p>
<p>Given the emf equa... | mcq | jee-main-2023-online-6th-april-evening-shift | 8,924 |
jaoe38c1lscpsrwh | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>A series LCR circuit with $$\mathrm{L}=\frac{100}{\pi} \mathrm{mH}, \mathrm{C}=\frac{10^{-3}}{\pi} \mathrm{F}$$ and $$\mathrm{R}=10 \Omega$$, is connected across an ac source of $$220 \mathrm{~V}, 50 \mathrm{~Hz}$$ supply. The power factor of the circuit would be ________.</p> | [] | null | 1 | <p>$$\begin{aligned}
& \mathrm{X}_{\mathrm{c}}=\frac{1}{\omega \mathrm{C}}=\frac{\pi}{2 \pi \times 50 \times 10^{-3}}=10 \Omega \\
& \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=2 \pi \times 50 \times \frac{100}{\pi} \times 10^{-3} \\
& =10 \Omega \\
& \because \mathrm{X}_{\mathrm{C}}=\mathrm{X}_{\mathrm{L}}, \text { Henc... | integer | jee-main-2024-online-27th-january-evening-shift | 8,927 |
jaoe38c1lsd7vbge | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>An AC voltage $$V=20 \sin 200 \pi t$$ is applied to a series LCR circuit which drives a current $$I=10 \sin \left(200 \pi t+\frac{\pi}{3}\right)$$. The average power dissipated is:</p> | [{"identifier": "A", "content": "21.6 W"}, {"identifier": "B", "content": "200 W"}, {"identifier": "C", "content": "173.2 W"}, {"identifier": "D", "content": "50 W"}] | ["D"] | null | <p>$$\begin{aligned}
& <\mathrm{P}>=\mathrm{IV} \cos \phi \\
& =\frac{20}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \times \cos 60^{\circ} \\
& =50 \mathrm{~W}
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift | 8,928 |
jaoe38c1lsflvkm7 | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>In an a.c. circuit, voltage and current are given by:</p>
<p>$$V=100 \sin (100 t) V$$ and
$$I=100 \sin \left(100 t+\frac{\pi}{3}\right) \mathrm{mA}$$ respectively.</p>
<p>The average power dissipated in one cycle is:</p> | [{"identifier": "A", "content": "5 W"}, {"identifier": "B", "content": "25 W"}, {"identifier": "C", "content": "2.5 W"}, {"identifier": "D", "content": "10 W"}] | ["C"] | null | <p>$$\begin{aligned}
& P_{\text {avg }}=V_{\text {rms }} I_{r m s} \cos (\Delta \phi) \\
& =\frac{100}{\sqrt{2}} \times \frac{100 \times 10^{-3}}{\sqrt{2}} \times \cos \left(\frac{\pi}{3}\right) \\
& =\frac{10^4}{2} \times \frac{1}{2} \times 10^{-3} \\
& =\frac{10}{4}=2.5 \mathrm{~W}
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift | 8,929 |
1lsgd2r5a | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>A series L.R circuit connected with an ac source $$E=(25 \sin 1000 t) V$$ has a power factor of $$\frac{1}{\sqrt{2}}$$. If the source of emf is changed to $$\mathrm{E}=(20 \sin 2000 \mathrm{t}) \mathrm{V}$$, the new power factor of the circuit will be :</p> | [{"identifier": "A", "content": "$$\\frac{1}{\\sqrt{3}}$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{\\sqrt{2}}$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{\\sqrt{5}}$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{\\sqrt{7}}$$"}] | ["C"] | null | <p>$$\begin{aligned}
& E=25 \sin (1000 t) \\\\
& \cos \theta=\frac{1}{\sqrt{2}}
\end{aligned}$$</p>
<p>LR circuit</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqmaruy/fe41a022-2b85-4305-a918-b331d99425f6/41e7dea0-cddf-11ee-a0d3-7b75c4537559/file-6y3zli1lsqmaruz.png?format=png" ... | mcq | jee-main-2024-online-30th-january-morning-shift | 8,931 |
luxwco8e | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>A capacitor of reactance $$4 \sqrt{3} \Omega$$ and a resistor of resistance $$4 \Omega$$ are connected in series with an ac source of peak value $$8 \sqrt{2} \mathrm{~V}$$. The power dissipation in the circuit is __________ W.</p> | [] | null | 4 | <p>To calculate the power dissipation in the circuit, we follow a systematic approach. We're provided with the reactance of the capacitor ($$X_C = 4 \sqrt{3} \Omega$$), the resistance ($$R = 4 \Omega$$), and the peak value of the AC voltage source ($$V_{peak} = 8 \sqrt{2} V$$). The power dissipated in an AC circuit is ... | integer | jee-main-2024-online-9th-april-evening-shift | 8,932 |
luz2u3rd | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb :</p> | [{"identifier": "A", "content": "becomes zero\n"}, {"identifier": "B", "content": "remains same\n"}, {"identifier": "C", "content": "increases\n"}, {"identifier": "D", "content": "decreases"}] | ["C"] | null | <p>To understand the impact of placing a dielectric between the plates of the capacitor on the glow of the bulb, we need to consider the properties and behavior of capacitors in an AC circuit.</p>
<p>When a capacitor is connected in series with a bulb in an AC circuit, the impedance of the capacitor plays a significan... | mcq | jee-main-2024-online-9th-april-morning-shift | 8,933 |
luz2umql | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>When a coil is connected across a $$20 \mathrm{~V}$$ dc supply, it draws a current of $$5 \mathrm{~A}$$. When it is connected across $$20 \mathrm{~V}, 50 \mathrm{~Hz}$$ ac supply, it draws a current of $$4 \mathrm{~A}$$. The self inductance of the coil is __________ $$\mathrm{mH}$$. (Take $$\pi=3$$)</p> | [] | null | 10 | <p>Let's first determine the resistance of the coil when connected to a DC supply. The current drawn by the coil in a DC circuit can be used to calculate its resistance using Ohm's law:</p>
<p>$$R = \frac{V}{I}$$</p>
<p>Given the DC supply voltage $$V_{DC} = 20 \, \mathrm{V}$$ and the current $$I_{DC} = 5 \, \mathrm{... | integer | jee-main-2024-online-9th-april-morning-shift | 8,934 |
lv0vxltl | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>In an ac circuit, the instantaneous current is zero, when the instantaneous voltage is maximum. In this case, the source may be connected to :</p>
<p>A. pure inductor.</p>
<p>B. pure capacitor.</p>
<p>C. pure resistor.</p>
<p>D. combination of an inductor and capacitor.</p>
<p>Choose the correct answer from the opti... | [{"identifier": "A", "content": "A, B and C only\n"}, {"identifier": "B", "content": "A and B only\n"}, {"identifier": "C", "content": "A, B and D only\n"}, {"identifier": "D", "content": "B, C and D only"}] | ["C"] | null | <p>To understand which answer is correct, we must consider the relationship between voltage and current in various types of circuits, namely circuits with pure inductors, pure capacitors, pure resistors, and a combination of inductors and capacitors (LC circuits).</p>
<p>In a <strong>purely resistive</strong> circuit,... | mcq | jee-main-2024-online-4th-april-morning-shift | 8,935 |
lv0vy1s1 | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>A alternating current at any instant is given by $$i=[6+\sqrt{56} \sin (100 \pi t+\pi / 3)]$$ A. The $$r m s$$ value of the current is ______ A.</p> | [] | null | 8 | <p>The given alternating current (AC) can be represented as $$i=6+\sqrt{56} \sin (100 \pi t+\pi / 3)$$ A, where $$6$$ is the DC component and $$\sqrt{56} \sin (100 \pi t+\pi / 3)$$ is the AC component of the current. The RMS (Root Mean Square) value of an alternating current is a measure of the equivalent direct curren... | integer | jee-main-2024-online-4th-april-morning-shift | 8,936 |
lv2es39u | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>Match List I with List II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;bo... | [{"identifier": "A", "content": "A-IV. B-I, C-III, D-II\n"}, {"identifier": "B", "content": "A-I. B-IV, C-II, D-III\n"}, {"identifier": "C", "content": "A-IV. B-I, C-II, D-III\n"}, {"identifier": "D", "content": "A-I. B-IV, C-III, D-II"}] | ["B"] | null | <p>For pure capacitive circuit, $$I$$ lead by $$90^{\circ}$$ to $$\mathrm{V}$$</p>
<p>For pure inductive circuit, $$V$$ lead by $$90^{\circ}$$ to $$I$$</p>
<p>At series LCR resonance, $$I$$ and $$V$$ are in same phase.</p>
<p>For LCR series circuit, $$V$$ and I may suffer some phase difference.</p> | mcq | jee-main-2024-online-4th-april-evening-shift | 8,937 |
lv3ve9g6 | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>A coil of negligible resistance is connected in series with $$90 \Omega$$ resistor across $$120 \mathrm{~V}, 60 \mathrm{~Hz}$$ supply. A voltmeter reads $$36 \mathrm{~V}$$ across resistance. Inductance of the coil is :</p> | [{"identifier": "A", "content": "0.91 H"}, {"identifier": "B", "content": "0.76 H"}, {"identifier": "C", "content": "2.86 H"}, {"identifier": "D", "content": "0.286 H"}] | ["B"] | null | <p>To find the inductance of the coil, we need to analyze the given circuit and use the information provided. The circuit consists of a resistor and an inductor in series, connected to an AC supply. Here are the given values:</p>
<p>1. Resistance, $$R = 90 \Omega$$</p>
<p>2. Supply voltage, $$V_{\text{total}} = 120 \... | mcq | jee-main-2024-online-8th-april-evening-shift | 8,938 |
lv3veaqd | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>An alternating emf $$\mathrm{E}=110 \sqrt{2} \sin 100 \mathrm{t}$$ volt is applied to a capacitor of $$2 \mu \mathrm{F}$$, the rms value of current in the circuit is ________ $$\mathrm{mA}$$.</p> | [] | null | 22 | <p>To determine the RMS (Root Mean Square) value of the current in the circuit, we start by analyzing the given emf and the capacitive reactance.</p>
<p>The given alternating emf is:</p>
<p>
$$\mathrm{E} = 110 \sqrt{2} \sin 100 \mathrm{t} \, \text{volts}$$
</p>
<p>Here, the peak voltage (or maximum voltage) $$\mathr... | integer | jee-main-2024-online-8th-april-evening-shift | 8,939 |
lv5gst30 | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now:</p> | [{"identifier": "A", "content": "halved\n"}, {"identifier": "B", "content": "same\n"}, {"identifier": "C", "content": "Zero\n"}, {"identifier": "D", "content": "double"}] | ["D"] | null | <p>To solve this problem, we need to understand the relationship between the current amplitude in a series LCR circuit at resonance and the resistance $ R $. At resonance, the impedance $ Z $ of the series LCR circuit is equal to the resistance $ R $, and thus:</p>
<p>$$ Z = R $$</p>
<p>The amplitude of the current $... | mcq | jee-main-2024-online-8th-april-morning-shift | 8,940 |
lv7v3ofn | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>An alternating voltage of amplitude $$40 \mathrm{~V}$$ and frequency $$4 \mathrm{~kHz}$$ is applied directly across the capacitor of $$12 \mu \mathrm{F}$$. The maximum displacement current between the plates of the capacitor is nearly :</p> | [{"identifier": "A", "content": "10 A"}, {"identifier": "B", "content": "8 A"}, {"identifier": "C", "content": "13 A"}, {"identifier": "D", "content": "12 A"}] | ["D"] | null | <p>Let's calculate the maximum displacement current between the plates of the capacitor when an alternating voltage is applied directly across it.</p>
<p>The formula for the capacitive reactance $$X_C$$ of a capacitor is given by:</p>
<p>$X_C = \frac{1}{2\pi fC}$</p>
<p>where</p>
<ul>
<li>$$f$$ is the frequenc... | mcq | jee-main-2024-online-5th-april-morning-shift | 8,941 |
lv7v4rcl | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>An ac source is connected in given series LCR circuit. The rms potential difference across the capacitor of $$20 \mu \mathrm{F}$$ is __________ V.</p>
<p><img src="data:image/png;base64,UklGRtoRAABXRUJQVlA4IM4RAABQ5gCdASoAA60BP4G+2GO2MCymo5FqQsAwCWlu+F+7jdsLRCN+ef6a3LbJuNzcO7o/4XjboJCunmff+Yrr0T///6ufv0bM6E82rze7/eO... | [] | null | 50 | <p>$$\begin{aligned}
& \left(V_{r m s}\right)_c=\frac{V_{r m s}}{Z} X_c \\
& =\frac{50}{\sqrt{(500-100)^2+300^2}} \times 500 \\
& =50 \mathrm{~V}
\end{aligned}$$</p> | integer | jee-main-2024-online-5th-april-morning-shift | 8,942 |
lv9s20yo | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>A series LCR circuit is subjected to an ac signal of $$200 \mathrm{~V}, 50 \mathrm{~Hz}$$. If the voltage across the inductor $$(\mathrm{L}=10 \mathrm{~mH})$$ is $$31.4 \mathrm{~V}$$, then the current in this circuit is _______.</p> | [{"identifier": "A", "content": "10 A"}, {"identifier": "B", "content": "10 mA"}, {"identifier": "C", "content": "68 A"}, {"identifier": "D", "content": "63 A"}] | ["A"] | null | <p>$$\begin{aligned}
&V_L=I(\omega L)=31.4\\
&\begin{aligned}
\Rightarrow \quad I & =\frac{31.4}{2 \times 3.14 \times 50 \times 10 \times 10^{-3}} \\
& =10 \mathrm{~A}
\end{aligned}
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift | 8,943 |
lvb29eno | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is $$2.5 \mathrm{~nF}$$. If resistance of $$200 \Omega$$ and $$100 \mathrm{~mH}$$ inductor is being used in the given circuit. The frequency of ac source is _________ $$\times 10^3 \mathrm{~Hz}$$ (given $$\mat... | [] | null | 10 | <p>To solve this problem, we need to use the concept of resonance in an LCR (inductor-capacitor-resistor) circuit. At resonance, the inductive reactance and capacitive reactance cancel each other out. The condition for resonance in an LCR circuit is given by:</p>
<p>$$\omega L = \frac{1}{\omega C}$$</p>
<p>Where:
<u... | integer | jee-main-2024-online-6th-april-evening-shift | 8,944 |
lvc57cv4 | physics | alternating-current | ac-circuits-and-power-in-ac-circuits | <p>Given below are two statements :</p>
<p>Statement I : In an LCR series circuit, current is maximum at resonance.</p>
<p>Statement II : Current in a purely resistive circuit can never be less than that in a series LCR circuit when connected to same voltage source.</p>
<p>In the light of the above statements, choose t... | [{"identifier": "A", "content": "Statement I is true but Statement II is false\n"}, {"identifier": "B", "content": "Statement I is false but Statement II is true\n"}, {"identifier": "C", "content": "Both Statement I and Statement II are true\n"}, {"identifier": "D", "content": "Both Statement I and Statement II are fal... | ["C"] | null | <p>Statement I : True</p>
<p>Statement II : True</p>
<p>Current in purely resistive circuit is equal to current in LCR circuit with same resistance at resonance, otherwise more.</p> | mcq | jee-main-2024-online-6th-april-morning-shift | 8,945 |
4607qdKnqCzVxh96 | physics | alternating-current | ac-generator-and-transformer | In a transformer, number of turns in the primary coil are $$140$$ and that in the secondary coil are $$280.$$ If current in primary coil is $$4A,$$ then that in the secondary coil is | [{"identifier": "A", "content": "$$4A$$ "}, {"identifier": "B", "content": "$$2A$$ "}, {"identifier": "C", "content": "$$6A$$ "}, {"identifier": "D", "content": "$$10A$$ "}] | ["B"] | null | $${N_p} = 140,\,\,{N_s} = 280,\,\,{I_p} = 4A,\,\,{I_s} = ?$$
<br><br>For a transformer $${{{I_s}} \over {{I_p}}} = {{{N_p}} \over {{N_s}}}$$
<br><br>$$ \Rightarrow {{{I_s}} \over 4} = {{140} \over {280}} \Rightarrow {I_s} = 2A$$ | mcq | aieee-2002 | 8,947 |
6Ec6OxBh8BJj2TNa | physics | alternating-current | ac-generator-and-transformer | The core of any transformer is laminated so as to | [{"identifier": "A", "content": "reduce the energy loss due to eddy currents"}, {"identifier": "B", "content": "make it light weight"}, {"identifier": "C", "content": "make it robust and strong"}, {"identifier": "D", "content": "increase the secondary voltage "}] | ["A"] | null | Laminated core provide less area of cross-section for the current to flow. Because of this, resistance of the core increases and current decreases thereby decreasing the eddy current losses. | mcq | aieee-2003 | 8,948 |
miSvleiDv9GBrw0o | physics | alternating-current | ac-generator-and-transformer | In an $$AC$$ generator, a coil with $$N$$ turns, all of the same area $$A$$ and total resistance $$R,$$ rotates with frequency $$\omega $$ in a magnetic field $$B.$$ The maximum value of $$emf$$ generated in the coil is | [{"identifier": "A", "content": "$$N.A.B.R.$$$$\\omega $$ "}, {"identifier": "B", "content": "$$N.A.B$$"}, {"identifier": "C", "content": "$$N.A.B.R.$$ "}, {"identifier": "D", "content": "$$N.A.B.$$$$\\omega $$ "}] | ["D"] | null | $$e = - {{d\phi } \over {dt}} = - {{d\left( {N\overrightarrow B .\overrightarrow A } \right)} \over {dt}}$$
<br><br>$$ = - N{d \over {dt}}\left( {BA\,\cos \,\omega t} \right) = NBA\omega \sin \,\omega t$$
<br><br>$$ \Rightarrow {e_{\max }} = NBA\omega $$ | mcq | aieee-2006 | 8,949 |
2kCrUXgD1yzCfozAGg18hoxe66ijvzt10aj | physics | alternating-current | ac-generator-and-transformer | A transformer consisting of 300 turns in the
primary and 150 turns in the secondary gives
output power of 2.2 kW. If the current in the
secondary coil is 10A, then the input voltage
and current in the primary coil are : | [{"identifier": "A", "content": "220 V and 20 A"}, {"identifier": "B", "content": "220 V and 10A"}, {"identifier": "C", "content": "440 V and 5A"}, {"identifier": "D", "content": "440 V and 20 A"}] | ["C"] | null | Given N<sub>P</sub> = 300, N<sub>s</sub> = 150, P<sub>0</sub> = 2200W<br><br>
I<sub>s</sub> = 10 A<br>
P<sub>0</sub> = V<sub>0</sub>I<sub>0</sub> $$ \Rightarrow $$ 2200 = V<sub>0</sub> × 10 $$ \Rightarrow $$ V<sub>0</sub> = 220 V<br><br>
$$ \because $$ $${{{V_i}} \over {{V_0}}} = {{{N_P}} \over {{N_S}}} \Rightarrow {V_... | mcq | jee-main-2019-online-10th-april-morning-slot | 8,952 |
bjXPIxd9tlzVZDL10y1kmipnmkp | physics | alternating-current | ac-generator-and-transformer | For the given circuit, comment on the type of transformer used.<br/><br/><img src="data:image/png;base64,UklGRh4LAABXRUJQVlA4IBILAAAQRgCdASp9AagAPm00lkikIqIhInM7CIANiWlu4WeRG/Of8U/zD8T/Bz+y/jP2HvlX2s9W34A6oLNL9X/un8y/r3/B/Mj75fuXgn7j9QL1F/df6L/VPF37sKrXoF+xPzv+3f0T+Zf8z+s+ej+3fz39yf6N8G/Ll7gH8s/kH+H/l/7tf3X5p8ICgH/J/6J... | [{"identifier": "A", "content": "Auxiliary transformer"}, {"identifier": "B", "content": "Step down transformer"}, {"identifier": "C", "content": "Step-up transformer"}, {"identifier": "D", "content": "Auto transformer"}] | ["C"] | null | <p>Voltage across secondary coil,</p>
<p>= $${{Power\,across\,load} \over {Current\,pas\sin g\,through\,load}}$$</p>
<p>$$ \Rightarrow {V_2} = {P \over {{I_L}}} = {{60} \over {0.11}} \Rightarrow {V_2} = 545.45$$ V</p>
<p>Voltage across primary coil, V<sub>1</sub> = 220V</p>
<p>$$\Rightarrow$$ V<sub>2</sub> > V<sub>1</s... | mcq | jee-main-2021-online-16th-march-evening-shift | 8,954 |
1l5akkd0b | physics | alternating-current | ac-generator-and-transformer | <p>Match List-I with List-II.</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;b... | [{"identifier": "A", "content": "(A) - (II), (B) - (I), (C) - (IV), (D) - (III)"}, {"identifier": "B", "content": "(A) - (II), (B) - (I), (C) - (III), (D) - (IV)"}, {"identifier": "C", "content": "(A) - (III), (B) - (IV), (C) - (II), (D) - (I)"}, {"identifier": "D", "content": "(A) - (III), (B) - (I), (C) - (II), (D) -... | ["A"] | null | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial,... | mcq | jee-main-2022-online-25th-june-morning-shift | 8,955 |
1l6nsrt1d | physics | alternating-current | ac-generator-and-transformer | <p>A transformer operating at primary voltage $$8 \,\mathrm{kV}$$ and secondary voltage $$160 \mathrm{~V}$$ serves a load of $$80 \mathrm{~kW}$$. Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be</p> | [{"identifier": "A", "content": "$$800 \\,\\Omega$$ and $$1.06 \\,\\Omega$$"}, {"identifier": "B", "content": "$$10 \\,\\Omega$$ and $$500 \\,\\Omega$$"}, {"identifier": "C", "content": "$$800 \\,\\Omega$$ and $$0.32 \\,\\Omega$$"}, {"identifier": "D", "content": "$$1.06 \\,\\Omega$$ and $$500 \\,\\Omega$$"}] | ["C"] | null | <p>$${V_1}{i_1} = {V_2}{i_2} = 80$$ kW</p>
<p>$$ \Rightarrow {i_1} = 10\,A$$ and $${i_2} = {{80 \times 1000} \over {160}} = 500\,A$$</p>
<p>$$ \Rightarrow {R_1} = {{{V_1}} \over {{i_1}}} = 800\,\Omega $$ and $${R_2} = {{160} \over {500}} = 0.32\,\Omega $$</p> | mcq | jee-main-2022-online-28th-july-evening-shift | 8,956 |
1ldnybebc | physics | alternating-current | ac-generator-and-transformer | <p>A square shaped coil of area $$70 \mathrm{~cm}^{2}$$ having 600 turns rotates in a magnetic field of $$0.4 ~\mathrm{wbm}^{-2}$$, about an axis which is parallel to one of the side of the coil and perpendicular to the direction of field. If the coil completes 500 revolution in a minute, the instantaneous emf when the... | [] | null | 44 | Area $(\mathrm{A})=70 \mathrm{~cm}^2=70 \times 10^{-4} \mathrm{~m}^2$
<br/><br/>$$
\mathrm{B}=0.4 \mathrm{~T}
$$
<br/><br/>$f=\frac{500 \text { revolution }}{60 \text { minute }}=\frac{500}{60} \frac{\text { rev. }}{\mathrm{sec} .}$
<br/><br/>Induced emf in rotating coil is given by
<br/><br/>$$
\begin{aligned}
& e=N \... | integer | jee-main-2023-online-1st-february-evening-shift | 8,957 |
ldqw2z2d | physics | alternating-current | ac-generator-and-transformer | In an ac generator, a rectangular coil of 100 turns each having area $14 \times 10^{-2} \mathrm{~m}^{2}$ is rotated at $360 ~\mathrm{rev} / \mathrm{min}$ about an axis perpendicular to a uniform magnetic field of magnitude $3.0 \mathrm{~T}$. The maximum value of the emf produced will be ________ $V$.
<br/><br/>
$\left(... | [] | null | 1584 | <p>$$\phi=B.A$$</p>
<p>$$\phi=\mathrm{BNA}\cos\omega t$$</p>
<p>So, $$Emf = {{ - d\phi } \over {dt}} = NBA\omega \sin \omega t$$</p>
<p>So maximum value of emf is</p>
<p>$${E_{\max }} = NBA\omega $$</p>
<p>$$ = 100 \times 3 \times 14 \times {10^{ - 2}} \times {{360 \times 2\pi } \over {60}} = 1584$$</p> | integer | jee-main-2023-online-30th-january-evening-shift | 8,959 |
jaoe38c1lscp19dg | physics | alternating-current | ac-generator-and-transformer | <p>Primary side of a transformer is connected to $$230 \mathrm{~V}, 50 \mathrm{~Hz}$$ supply. Turns ratio of primary to secondary winding is $$10: 1$$. Load resistance connected to secondary side is $$46 \Omega$$. The power consumed in it is :</p> | [{"identifier": "A", "content": "11.5 W"}, {"identifier": "B", "content": "12.5 W"}, {"identifier": "C", "content": "10.0 W"}, {"identifier": "D", "content": "12.0 W"}] | ["A"] | null | <p>$$\begin{aligned}
& \frac{V_1}{V_2}=\frac{N_1}{N_2} \\
& \frac{230}{V_2}=\frac{10}{1} \\
& V_2=23 \mathrm{~V}
\end{aligned}$$</p>
<p>Power consumed $$=\frac{\mathrm{V}_2^2}{\mathrm{R}}$$</p>
<p>$$=\frac{23 \times 23}{46}=11.5 \mathrm{~W}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift | 8,961 |
1lsg6y28g | physics | alternating-current | ac-generator-and-transformer | <p>A power transmission line feeds input power at $$2.3 \mathrm{~kV}$$ to a step down transformer with its primary winding having 3000 turns. The output power is delivered at $$230 \mathrm{~V}$$ by the transformer. The current in the primary of the transformer is $$5 \mathrm{~A}$$ and its efficiency is $$90 \%$$. The w... | [] | null | 45 | <p>$$\begin{aligned}
& P_i=2300 \times 5 \text { watt } \\
& P_0=2300 \times 5 \times 0.9=230 \times I_2 \\
& I_2=45 A
\end{aligned}$$</p> | integer | jee-main-2024-online-30th-january-evening-shift | 8,962 |
1lsgcytzp | physics | alternating-current | ac-generator-and-transformer | <p>Primary coil of a transformer is connected to $$220 \mathrm{~V}$$ ac. Primary and secondary turns of the transforms are 100 and 10 respectively. Secondary coil of transformer is connected to two series resistances shown in figure. The output voltage $$\left(V_0\right)$$ is :</p>
<p><img src="data:image/png;base64,Uk... | [{"identifier": "A", "content": "7 V"}, {"identifier": "B", "content": "44 V"}, {"identifier": "C", "content": "22 V"}, {"identifier": "D", "content": "15 V"}] | ["A"] | null | <p>$$\begin{aligned}
& \frac{\varepsilon_1}{\varepsilon_2}=\frac{N_1}{N_2}=\frac{100}{10} \Rightarrow \varepsilon_2=22 \mathrm{~V} \\
& I=\frac{22}{22 \times 10^3}=1 \mathrm{~mA}, V_0=7 \mathrm{~V}
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift | 8,963 |
8CMiNsI8xNXYas3C | physics | alternating-current | growth-and-decay-of-current | In an oscillating $$LC$$ circuit the maximum charge on the capacitor is $$Q$$. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is | [{"identifier": "A", "content": "$${Q \\over 2}$$ "}, {"identifier": "B", "content": "$${Q \\over {\\sqrt 3 }}$$ "}, {"identifier": "C", "content": "$${Q \\over {\\sqrt 2 }}$$ "}, {"identifier": "D", "content": "$$Q$$"}] | ["C"] | null | When the capacitor is completely charged, the total energy in the $$L.C$$ circuit is with the capacitor and that energy is $$E = {1 \over 2}{{{Q^2}} \over C}$$
<br><br>When half energy is with the capacitor in the form of electric field between the plates of the capacitor we get
$${E \over 2} = {1 \over 2}{{Q{'^2}} \o... | mcq | aieee-2003 | 8,964 |
9dRlbRiHU2IJPiTo | physics | alternating-current | growth-and-decay-of-current | An inductor of inductance $$L=400$$ $$mH$$ and resistors of resistance $${R_1} = 2\Omega $$ and $${R_2} = 2\Omega $$ are connected to a battery of $$emf$$ $$12$$ $$V$$ as shown in the figure. The internal resistance of the battery is negligible. The switch $$S$$ is closed at $$t=0.$$ The potential drop across $$L$$ as ... | [{"identifier": "A", "content": "$${{12} \\over t}{e^{ - 3t}}V$$ "}, {"identifier": "B", "content": "$$6\\left( {1 - {e^{ - t/0.2}}} \\right)V$$ "}, {"identifier": "C", "content": "$$12{e^{ - 5t}}V$$ "}, {"identifier": "D", "content": "$$6{e^{ - 5t}}V$$ "}] | ["C"] | null | Growth in current in $$L{R_2}$$ branch when switch is closed is given by
<br><br>$$i = {E \over {{R_2}}}\left[ {1 - {e^{ - {R_2}t/L}}} \right]$$
<br><br>$$ \Rightarrow {{di} \over {dt}} = {E \over {{R_2}}}.{{{R_2}} \over L}.{e^{ - {R_{2t/L}}}}$$
<br><br>$$ = {E \over L}{e^{{{{R_2}t} \over L}}}$$
<br><br>Hence, potentia... | mcq | aieee-2009 | 8,965 |
99RX6CCpD5rzXB9z | physics | alternating-current | growth-and-decay-of-current | In the circuit shown below, the key $$K$$ is closed at $$t=0.$$ The current through the battery is
<img src="data:image/png;base64,UklGRhYMAABXRUJQVlA4IAoMAADQiACdASqqArcBP4G+1mY2LywnINIpesAwCWlu+EfodSUZNlvx5j/ytpbaA9if7nlzwv+Q0qn0APKQ72H7bvtRBCNG5tO1p2tO1p2tO1p2tO1p2tOvn2A8opWNT7AeUUrGp9gPKKVjU+wHlFNaBRSsan2A8opWNT7A... | [{"identifier": "A", "content": "$${{V{R_1}{R_2}} \\over {\\sqrt {R_1^2 + R_2^2} }}$$ at $$t=0$$ and $${V \\over {{R_2}}}$$ at $$t = \\infty $$ "}, {"identifier": "B", "content": "$${V \\over {{R_2}}}$$ at $$\\,t = 0$$ and $${{V\\left( {{R_1} + {R_2}} \\right)} \\over {{R_1}{R_2}}}$$ at $$t = \\infty $$ "}, {"identifie... | ["B"] | null | At $$t=0,$$ no current will flow through $$L$$ and $${R_1}$$
<br><br>$$\therefore$$ Current through battery $$ = {V \over {{R_2}}}$$
<br><br>At $$t = \infty ,$$
<br><br>effective resistance, $${{\mathop{\rm R}\nolimits} _{eff}} = {{{R_1}{R_2}} \over {{R_1} + {R_2}}}$$
<br><br>$$\therefore$$ Current through battery $$... | mcq | aieee-2010 | 8,966 |
77IfQmGHKpIKU2PC | physics | alternating-current | growth-and-decay-of-current | In an $$LCR$$ circuit as shown below both switches are open initially. Now switch $${S_1}$$ is closed, $${S_2}$$ kept open. ($$q$$ is charge on the capacitor and $$\tau $$ $$=RC$$ is Capacitance time constant). Which of the following statement is correct ?
<img src="data:image/png;base64,UklGRgIKAABXRUJQVlA4IPYJAAAQc... | [{"identifier": "A", "content": "Work done by the battery is half of the energy dissipated in the resistor "}, {"identifier": "B", "content": "$$t = \\,\\tau ,\\,q = CV/2$$ "}, {"identifier": "C", "content": "At $$t = \\,2\\tau ,\\,q = CV\\left( {1 - {e^{ - 2}}} \\right)$$"}, {"identifier": "D", "content": "At $$t = \\... | ["C"] | null | Charge on he capacitor at any time $$t$$ is given by
<br><br>$$q = CV\left( {1 - {e^{t/\tau }}} \right)$$
<br><br>at $$t = 2\tau $$
<br><br>$$q = CV\left( {1 - {e^{ - 2}}} \right)$$ | mcq | jee-main-2013-offline | 8,968 |
z6vlUAAY4ivffbsW | physics | alternating-current | growth-and-decay-of-current | In the circuit shown here, the point $$'C'$$ is kept connected to point $$'A'$$ till the current flowing through the circuit becomes constant. Afterward, suddenly, point $$'C'$$ is disconnected from point $$'A'$$ and connected to point $$'B'$$ at time $$t=0.$$ Ratio of the voltage across resistance and the inductor at... | [{"identifier": "A", "content": "$${e \\over {1 - e}}$$ "}, {"identifier": "B", "content": "$$1$$ "}, {"identifier": "C", "content": "$$-1$$ "}, {"identifier": "D", "content": "$${{1 - e} \\over e}$$ "}] | ["C"] | null | Applying kirchhoffs law of voltage in closed loop
<br><br>$$ - {V_R} - {V_C} = 0 \Rightarrow {{{V_R}} \over {{V_C}}} = - 1$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263603/exam_images/fal6gjqea4kd4lrkini1.webp" loading="lazy" alt="JEE Main 2014 (Offline) Physics... | mcq | jee-main-2014-offline | 8,969 |
T88M1CToFCt9GBN2 | physics | alternating-current | growth-and-decay-of-current | <p>An $$LCR$$ circuit is equivalent to a damped pendulum. In an $$LCR$$ circuit the capacitor is charged to $${Q_0}$$ and then connected to the $$L$$ and $$R$$ as shown below :
</p><p><img src="data:image/png;base64,UklGRpAIAABXRUJQVlA4IIQIAACQpgCdASoAAyMCP4G812a2LiwnoLD4+sAwCWlu4XEUDmNwvxQAjBoAVzu5nfJiNtmQesC1+PWrZkH... | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l91fp8ku/1310e8e1-f91c-4040-93ce-16996684616a/bfcf6ce0-47dd-11ed-9a49-57ec402e0bd4/file-1l91fp8kv.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l91fp8ku/1310e8e1-f91c-4040-93ce-16996684616a/bfc... | ["C"] | null | From $$KVL$$ at any time $$t$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l91fscjm/37fdaa04-6a6c-40c7-bca1-bc62db93c167/164d2620-47de-11ed-9a49-57ec402e0bd4/file-1l91fscjn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l91fscjm/37fdaa04-6a6c-40c7-bca1-bc62db93c167... | mcq | jee-main-2015-offline | 8,970 |
llMccTVLLZ3zXzCWEYLRb | physics | alternating-current | growth-and-decay-of-current | A series <i>LR</i> circuit is connected to a voltage source with
<br/>V(t) = V<sub>0 </sub> sin$$\Omega $$t. After very large time, current <i>I(t)</i> behaves as
<br/>(t<sub>0</sub> >> $${L \over R}$$) : | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l80z49lv/48fcd01f-5490-4374-87e9-68d30fc0aed8/af2adf30-33d0-11ed-bcbd-5353ee2ab2b1/file-1l80z49lw.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l80z49lv/48fcd01f-5490-4374-87e9-68d30fc0aed8/af2... | ["B"] | null | Current in LR circuit,
<br><br>$${\rm I} = {{{V_0}} \over {\sqrt {{R^2} + {w^2}{L^2}} }}\sin \left( {\omega t - {\pi \over 2}} \right)$$
<br><br>it will be a sinusoidal wave. | mcq | jee-main-2016-online-9th-april-morning-slot | 8,972 |
pVjtQedPEhKmcZuPG8DPi | physics | alternating-current | growth-and-decay-of-current | In the circuit shown,
<br/><br/><img src="data:image/png;base64,UklGRoQNAABXRUJQVlA4IHgNAADQwQCdASoAAzoCP4HA12Y2L7inITK5YxAwCWlu/GwZp70HZ1+/rt/mLXG0G7T/37jWoCHwWgDuz5532Uyunof/kNR3OO5x3OO5x3OO5x3OO5x3OO5x3OO5x3OO5x3OO5x3OO5x3OO5x3OO5x3OO5x3OO5x3NtkkDa2L9N+Pxn60Ome0z2mezfI/BGjwyW76wHDKapE8YnjE5EJQT59Axkw1rAGWEOqmt31gOG... | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264190/exam_images/d2moa8ekc47chgdteoet.webp\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2019 (Online) 11th January Morning Slot Physics - Alternating Current ... | ["A"] | null | From time t = 0 to t = t<sub>0</sub>, growth of current takes place and after that decay of current takes place.
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264190/exam_images/d2moa8ekc47chgdteoet.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt... | mcq | jee-main-2019-online-11th-january-morning-slot | 8,973 |
9dHGgEj6QRHN21KURKyMZ | physics | alternating-current | growth-and-decay-of-current | In the figure shown, a circuit contains two identical resistors with resistance R = 5$$\Omega $$ and an inductance with L = 2mH. An ideal battery of 15 V is connected in the circuit. What will be the current through the battery long after the switch is closed ?
<br/><br/><img src="data:image/png;base64,UklGRvQLAABXRUJQ... | [{"identifier": "A", "content": "6 A"}, {"identifier": "B", "content": "7.5 A"}, {"identifier": "C", "content": "3 A"}, {"identifier": "D", "content": "5.5 A"}] | ["A"] | null | Ideal inductor will behave like zero resistance long time after switch is closed
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265905/exam_images/zi3zkykhhf8ireddmwam.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th J... | mcq | jee-main-2019-online-12th-january-morning-slot | 8,974 |
K5I4kHCDK9VBY64KG53rsa0w2w9jwzhfdkz | physics | alternating-current | growth-and-decay-of-current | A coil of self inductance 10 mH and resistance 0.1 $$\Omega $$ is connected through a switch to a battery of internal
resistance 0.9 $$\Omega $$. After the switch is closed, the time taken for the current to attain 80% of the saturation
value is: [take ln 5 = 1.6]
| [{"identifier": "A", "content": "0.324 s"}, {"identifier": "B", "content": "0.002 s"}, {"identifier": "C", "content": "0.103 s"}, {"identifier": "D", "content": "0.016 s"}] | ["D"] | null | L = 10 × 10<sup>–3</sup> H, r<sub>1</sub> = 0.1 $$\Omega $$<br><br>
$$i = \varepsilon \left\{ {1 - {e^{ - 1/2}}} \right\}$$<br><br>
$${i_{saturation}}{\rm{ }} = {\rm{ }}\varepsilon $$<br>
$$80\% {\rm{ }}{i_{saturation}}{\rm{ }} = {\rm{ }}0.8{\rm{ }}\varepsilon $$<br><br>
0.8 = 1 - e<sup>-t/2</sup> ; e<sup>-t/2</sup> = ... | mcq | jee-main-2019-online-10th-april-evening-slot | 8,975 |
AoOLRRzNY46fQyhrJe3rsa0w2w9jx6qjn49 | physics | alternating-current | growth-and-decay-of-current | Consider the LR circuit shown in the figure. If the switch S is closed at t = 0 then the amount of charge that
passes through the battery between t = 0 and t = $${L \over R}$$
is :
<img src="data:image/png;base64,UklGRvQFAABXRUJQVlA4IOgFAABwTQCdASrsAukAP4HA2mY2MKwnILVZAsAwCWlu4XExG/P58n9mGPTaR9h8oIHNyHVVT7L0SvfDG2zPvD... | [{"identifier": "A", "content": "$${{2.7EL} \\over {{R^2}}}$$"}, {"identifier": "B", "content": "$${{EL} \\over {2.7{R^2}}}$$"}, {"identifier": "C", "content": "$${{7.3EL} \\over {{R^2}}}$$"}, {"identifier": "D", "content": "$${{EL} \\over {7.3{R^2}}}$$"}] | ["B"] | null | $$I = {E \over R}\left( {1 - {e^{ - {t \over \tau }}}} \right)$$ where $$\tau $$ = $${L \over R}$$
<br><br>We know $$dq = Idt$$
<br>$$ \Rightarrow $$ $$\int\limits_0^Q {dq} = {E \over R}\int\limits_0^\tau {\left( {1 - {e^{ - {t \over \tau }}}} \right)dt} $$
<br>$$ \Rightarrow $$ Q $$ = {E \over R}\left[ t \right]_0... | mcq | jee-main-2019-online-12th-april-evening-slot | 8,976 |
wMxJsw83Tk4t09PURnjgy2xukfajdg43 | physics | alternating-current | growth-and-decay-of-current | A series L-R circuit is connected to a battery of emf V. If the circuit is switched on at t = 0, then
the time at which the energy stored in the inductor reaches $$\left( {{1 \over n}} \right)$$ times of its maximum value, is : | [{"identifier": "A", "content": "$${L \\over R}\\ln \\left( {{{\\sqrt n } \\over {\\sqrt n + 1}}} \\right)$$"}, {"identifier": "B", "content": "$${L \\over R}\\ln \\left( {{{\\sqrt n } \\over {\\sqrt n - 1}}} \\right)$$"}, {"identifier": "C", "content": "$${L \\over R}\\ln \\left( {{{\\sqrt n + 1} \\over {\\sqrt n ... | ["B"] | null | P.E. in inductor, $$U = {1 \over 2}L{I^2}$$<br><br>$$U \propto {I^2}$$<br><br>$${U \over {{U_0}}} = {\left( {{I \over {{I_0}}}} \right)^2}$$<br><br>$${1 \over n} = {\left( {{I \over {{I_0}}}} \right)^2}$$<br><br>$$I = {{{I_0}} \over {\sqrt n }}$$<br><br>We know, $$I = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)$$<b... | mcq | jee-main-2020-online-4th-september-evening-slot | 8,977 |
caUO97TMDmEHWY7vcf7k9k2k5f63fee | physics | alternating-current | growth-and-decay-of-current | An emf of 20 V is applied at time t = 0 to a circuit containing in series 10 mH inductor and 5 $$\Omega $$
resistor. The ratio of the currents at time t = $$\infty $$ and at t = 40 s is close to : (Take e<sup>2</sup> = 7.389) | [{"identifier": "A", "content": "1.06"}, {"identifier": "B", "content": "0.84"}, {"identifier": "C", "content": "1.15"}, {"identifier": "D", "content": "1.46"}] | ["A"] | null | i = i<sub>0</sub>(1 - $${e^{ - {{Rt} \over L}}}$$)
<br><br>i<sub>$$\infty $$</sub> = i<sub>0</sub>(1 - $${e^{ - \infty }}$$) = i<sub>0</sub>
<br><br>$$ \therefore $$ $${{{i_\infty }} \over {{i_{40s}}}}$$ = $${{{i_0}} \over {{i_0}\left( {1 - {e^{ - {{5 \times 40} \over {10 \times {{10}^{ - 3}}}}}}} \right)}}$$
<br><br> ... | mcq | jee-main-2020-online-7th-january-evening-slot | 8,978 |
cbtaZCxZW8ds4GCWV01kmhoyh37 | physics | alternating-current | growth-and-decay-of-current | An RC circuit as shown in the figure is driven by a AC source generating a square wave. The output wave pattern monitored by CRO would look close to :<br/><br/><img src="data:image/png;base64,UklGRjAFAABXRUJQVlA4ICQFAAAwJgCdASoaAWoAPm0ylUekIqIhLPLYOIANiWlu3WBpKf+i/8x7bP9ZyxeTjUxX4/5n+Zfz74C/ve4x/x282Y0/1n8k8UP+O4pH8ivh... | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267107/exam_images/gutueod6kn6fuof8gk6q.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2021 (Online) 16th March Morning Shift Physics - Alternating Current Que... | ["D"] | null | Assuming AC start with positive voltage, when +ve voltage is across input then the capacitor start charging, trying to reach saturation value, till there is +ve voltage across input, when $$-$$ve voltage of AC appears across input, the capacitor starts discharging till there is $$-$$ve voltage across input and this pro... | mcq | jee-main-2021-online-16th-march-morning-shift | 8,979 |
1kryxzkav | physics | alternating-current | growth-and-decay-of-current | Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then T<sub>1</sub> is connected to T<sub>2</sub>. As the current in R = 6$$\Omega$$ attains a maximum value of steady state level, T<sub>1</sub> is disconnected from T<sub>2</sub> and immediately connected to T<sub>3</sub>. Potentia... | [] | null | 3 | What T<sub>1</sub> and T<sub>2</sub> are connected, then the steady state current in the inductor $$I = {6 \over 6} = 1A$$<br><br>When T<sub>1</sub> and T<sub>3</sub> are connected then current through inductor remains same. So potential difference across 3$$\Omega$$<br><br>V = Ir = 1 $$\times$$ 3 = 3 Volt | integer | jee-main-2021-online-27th-july-morning-shift | 8,981 |
1l56v40l0 | physics | alternating-current | growth-and-decay-of-current | <p>If L, C and R are the self inductance, capacitance and resistance respectively, which of the following does not have the dimension of time?</p> | [{"identifier": "A", "content": "RC"}, {"identifier": "B", "content": "$${L \\over R}$$"}, {"identifier": "C", "content": "$$\\sqrt{LC}$$"}, {"identifier": "D", "content": "$${L \\over C}$$"}] | ["D"] | null | <p>$$U = {1 \over 2}L{i^2} = {1 \over 2}C{V^2}$$</p>
<p>So, $$\left[ {{L \over C}} \right] = {{{V^2}} \over {{i^2}}} = {R^2}$$ is not the dimension of time.</p> | mcq | jee-main-2022-online-27th-june-evening-shift | 8,983 |
1l57q0r56 | physics | alternating-current | growth-and-decay-of-current | <p>The current flowing through an ac circuit is given by</p>
<p>I = 5 sin(120$$\pi$$t)A</p>
<p>How long will the current take to reach the peak value starting from zero?</p> | [{"identifier": "A", "content": "$${1 \\over {60}}$$ s"}, {"identifier": "B", "content": "60 s"}, {"identifier": "C", "content": "$${1 \\over {120}}$$ s"}, {"identifier": "D", "content": "$${1 \\over {240}}$$ s"}] | ["D"] | null | <p>$$\omega = 120\pi $$</p>
<p>$$ \Rightarrow T = {1 \over {60}}\sec $$</p>
<p>The current will take its peak value in $${T \over 4}$$ time</p>
<p>So $$t = {T \over 4}$$</p>
<p>$$ = {1 \over {240}}s$$</p> | mcq | jee-main-2022-online-27th-june-morning-shift | 8,984 |
1l6p56ir0 | physics | alternating-current | growth-and-decay-of-current | <p>A coil of inductance 1 H and resistance $$100 \,\Omega$$ is connected to a battery of 6 V. Determine approximately :</p>
<p>(a) The time elapsed before the current acquires half of its steady - state value.</p>
<p>(b) The energy stored in the magnetic field associated with the coil at an instant 15 ms after the circ... | [{"identifier": "A", "content": "t = 10 ms; U = 2 mJ"}, {"identifier": "B", "content": "t = 10 ms; U = 1 mJ"}, {"identifier": "C", "content": "t = 7 ms; U = 1 mJ"}, {"identifier": "D", "content": "t = 7 ms; U = 2 mJ"}] | ["C"] | null | <p>$$i(t) = {V \over R}(1 - {e^{ - Rt/L}})$$ ...... (1)</p>
<p>$${L \over R} = {1 \over {100}}s \Rightarrow {L \over R} = 10\,ms$$ ...... (2)</p>
<p>$${V \over {2R}} = {V \over R}(1 - {e^{ - Rt/L}})$$</p>
<p>$$ \Rightarrow {e^{ - Rt/L}} = {1 \over 2} \Rightarrow t = {L \over R}\ln 2 = 6.93\,ms$$</p>
<p>$$U = {1 \over 2... | mcq | jee-main-2022-online-29th-july-morning-shift | 8,985 |
1l6rj1imr | physics | alternating-current | growth-and-decay-of-current | <p>A capacitor of capacitance 500 $$\mu$$F is charged completely using a dc supply of 100 V. It is now connected to an inductor of inductance 50 mH to form an LC circuit. The maximum current in LC circuit will be _______ A.</p> | [] | null | 10 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7e9lrut/71df2bd1-d468-4054-adb5-aa2728ee9596/aa5ffe60-2753-11ed-a077-1f1e3989e798/file-1l7e9lruu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7e9lrut/71df2bd1-d468-4054-adb5-aa2728ee9596/aa5ffe60-2753-11ed-a077-1f1e3989e798... | integer | jee-main-2022-online-29th-july-evening-shift | 8,986 |
1lguyaz5y | physics | alternating-current | growth-and-decay-of-current | <p><img src="data:image/png;base64,UklGRtgLAABXRUJQVlA4IMwLAAAwqQCdASoAA5gBP4HA2mS2MS0nIpJJSsAwCWlu4XHEDmNwvV6MtJ/y25s/vPEfEWwovS1P/1OF/sK32FbniLt1H4zwcZ4OM8HGeDjPBxng4zQ6Q+TUZnXkWz5hVD/N9jHDPzI+H5fThowJ0YEvYzs/F2fi7Pxdf5RBTctnizzoA+zG20NbQb0JcWiYLIkBPQGaKljG2q7FZbeIISUr2owjAUZ89Bbd3678UZrXqmGiYntkzVtn4uz8XZzFKXViQsz/c... | [{"identifier": "A", "content": "$$\\mathrm{A}=\\mathrm{X}_{\\mathrm{L}}, \\mathrm{B}=\\mathrm{R}$$"}, {"identifier": "B", "content": "$$\\mathrm{A}=\\mathrm{X}_{L}, \\mathrm{~B}=Z$$"}, {"identifier": "C", "content": "$$\\mathrm{A}=\\mathrm{X}_{\\mathrm{C}}, \\mathrm{B}=\\mathrm{X}_{\\mathrm{L}}$$"}, {"identifier": "D"... | ["C"] | null | $$
\begin{aligned}
& \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{(2 \pi \mathrm{f}) \mathrm{C}} \\\\
& \therefore \mathrm{X}_{\mathrm{C}} \propto \frac{1}{\mathrm{f}} \\\\
& \therefore \text { Curve } \mathrm{A} \\\\
& \mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}=(2 \pi \mathrm{f}) \mathrm{L} \\\\
& \ther... | mcq | jee-main-2023-online-11th-april-morning-shift | 8,989 |
1lh02oh0z | physics | alternating-current | growth-and-decay-of-current | <p>An oscillating LC circuit consists of a $$75 ~\mathrm{mH}$$ inductor and a $$1.2 ~\mu \mathrm{F}$$ capacitor. If the maximum charge to the capacitor is $$2.7 ~\mu \mathrm{C}$$. The maximum current in the circuit will be ___________ $$\mathrm{mA}$$</p> | [] | null | 9 | <p>The maximum current in an LC circuit can be found using the following formula related to simple harmonic motion:</p>
<p>$I_{\text{max}} = \omega Q_{\text{max}}$,</p>
<p>where:</p>
<ul>
<li>$I_{\text{max}}$ is the maximum current,</li>
<li>$\omega$ is the angular frequency, and</li>
<li>$Q_{\text{max}}$ is the maximu... | integer | jee-main-2023-online-8th-april-morning-shift | 8,990 |
rdKd8e6bBqQc6MiC | physics | alternating-current | quality-factor | For an RLC circuit driven with voltage of amplitude vm and frequency $${\omega _0}$$ = $${1 \over {\sqrt {LC} }}$$ the current exhibits resonance. The quality factor, Q is given by : | [{"identifier": "A", "content": "$${{CR} \\over {{\\omega _0}}}$$ "}, {"identifier": "B", "content": "$${{{\\omega _0}L} \\over R}$$ "}, {"identifier": "C", "content": "$${{{\\omega _0}R} \\over L}$$ "}, {"identifier": "D", "content": "$${R \\over {\\left( {{\\omega _0}C} \\right)}}$$ "}] | ["B"] | null | Quality factor (Q) = $${{Angular\,\,{\mathop{\rm Re}\nolimits} sonance} \over {Bandwith}}$$
<br><br>= $${{{1 \over {\sqrt {LC} }}} \over {{R \over L}}}$$
<br><br>= $${{{\omega _0}} \over {{R \over L}}}$$
<br><br>= $${{{\omega _0}L} \over R}$$ | mcq | jee-main-2018-offline | 8,992 |
rqaFZ6jbWJ8liGdw1Djgy2xukfrpjmiy | physics | alternating-current | quality-factor | An AC circuit has R= 100 $$\Omega $$, C = 2 $$\mu $$F and L = 80 mH, connected in series. The quality factor of the
circuit is : | [{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "0.5"}, {"identifier": "D", "content": "400"}] | ["B"] | null | $$Q = {1 \over R}\sqrt {{L \over C}} $$
<br><br>= $${1 \over {100}}\sqrt {{{80 \times {{10}^{ - 3}}} \over {2 \times {{10}^{ - 6}}}}} $$
<br><br>= $${1 \over {100}}\sqrt {40 \times {{10}^3}} $$
<br><br>= $${{200} \over {100}}$$ = 2 | mcq | jee-main-2020-online-6th-september-morning-slot | 8,993 |
gIZPhIbsx762wdakI71klriw8n1 | physics | alternating-current | quality-factor | A resonance circuit having inductance and resistance 2 $$\times$$ 10<sup>$$-$$4</sup> H and 6.28$$\Omega$$ respectively oscillates at 10 MHz frequency. The value of quality factor of this resonator is ___________. [$$\pi$$ = 3.14] | [] | null | 2000 | Given, L = 2 $$\times$$ 10<sup>$$-$$4</sup> H, R = 6.28 $$\Omega$$, f<sub>0</sub> = 10 MHz = 10 $$\times$$ 10<sup>6</sup> Hz<br/><br/>$$\therefore$$ Quality factor $$ = {\omega _0}{L \over R} = 2\pi {f_0}{L \over R}$$<br/><br/>$$ = 2\pi \times 10 \times {10^6} \times {{2 \times {{10}^{ - 4}}} \over {6.28}}$$<br/><br/>... | integer | jee-main-2021-online-24th-february-morning-slot | 8,994 |
1ldyemcnl | physics | alternating-current | quality-factor | <p>In the circuit shown in the figure, the ratio of the quality factor and the band width is ___________ s.</p>
<p><img src="data:image/png;base64,UklGRnALAABXRUJQVlA4IGQLAAAwswCdASoAA9cBP4HA22W2MK2nIZKpGsAwCWlu4XBkSmNwvj6V3N7My351eO/X6skGrw9VMf/3q9JJp8rdm1cdZtXG2l7ilMQL8pNHAyCXz0BQSUwUElMFBJS+jBcWM25x2n4LixmxsjgN4cfIh... | [] | null | 10 | Bandwidth $\Delta \omega=\frac{R}{L}$
<br/><br/>
Quality factor $Q=\frac{1}{R} \sqrt{\frac{L}{C}}$
<br/><br/>
So $\frac{Q}{\Delta \omega}=\frac{\frac{1}{R} \sqrt{\frac{L}{C}}}{\frac{R}{L}}$
<br/><br/>
$$
\begin{aligned}
& =\frac{L^{\frac{3}{2}}}{R^{2} \sqrt{C}} \\\\
& =\frac{3^{\frac{3}{2}}}{10^{2}\left(27 \times 10^{-... | integer | jee-main-2023-online-24th-january-morning-shift | 8,995 |
7j0rzgh4L20GoRNY | physics | atoms-and-nuclei | alpha-particle-scattering-and-rutherford-model-of-atom | An $$\alpha $$-particle of energy $$5$$ $$MeV$$ is scattered through $${180^ \circ }$$ by a fixed uranium nucleus. The distance of closest approach is of the order of | [{"identifier": "A", "content": "$${10^{ - 12}}\\,cm$$ "}, {"identifier": "B", "content": "$${10^{ - 10}}\\,cm$$ "}, {"identifier": "C", "content": "$$1A$$ "}, {"identifier": "D", "content": "$${10^{ - 15a}}\\,cm$$ "}] | ["A"] | null | <b>KEY NOTE :</b>
<br><br>Distance of closest approach
<br><br>$${r_0} = {{Ze\left( {2e} \right)} \over {4\pi {\varepsilon _0}E}}$$
<br><br>Energy, $$E = 5 \times {10^6} \times1.6 \times {10^{ - 19}}J$$
<br><br>$$\therefore$$ $${r_0} = {{9 \times {{10}^9} \times \left( {92 \times 1.6 \times {{10}^{ - 19}}} \right)\le... | mcq | aieee-2004 | 8,997 |
Mz2bMY7kQcZWsWlb | physics | atoms-and-nuclei | alpha-particle-scattering-and-rutherford-model-of-atom | It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its
energy is p<sub>d</sub>; while for its similar collision with carbon nucleus at rest, fractional loss of energy is p<sub>c</sub>. The values of p<sub>d</sub> and p<sub>c</sub> are respectively : | [{"identifier": "A", "content": "(0, 1)"}, {"identifier": "B", "content": "(0.89, 0.28)"}, {"identifier": "C", "content": "(0.28, 0.89)"}, {"identifier": "D", "content": "(0, 0)"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266433/exam_images/sj7otzqwlwlgwdcfoecv.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Offline) Physics - Atoms and Nuclei Question 206 English Explanation 1">
<br><br>Applying conservation o... | mcq | jee-main-2018-offline | 8,999 |
3IHS4G6RTcILZNfZnl7k9k2k5gwb5n8 | physics | atoms-and-nuclei | alpha-particle-scattering-and-rutherford-model-of-atom | The graph which depicts the results of
Rutherford gold foil experiment with
$$\alpha $$-particales is :<br/>
$$\theta $$ : Scattering angle<br/>
Y : Number of scattered $$\alpha $$-particles detected
(Plots are schematic and not to scale) | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265454/exam_images/hkqbvkbayaw5jf770yyg.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 8th January Morning Slot Physics - Atoms and Nuclei Questi... | ["C"] | null | From Ernest Rutherford formula.
Which indicates that number of particles (Y)
that would be deflected by an angle ‘$$\theta $$’ due to
scattering is
<br>$${Y_\theta } = {K \over {{{\sin }^4}{\theta \over 2}}}$$ where K = constant
<br><br>Option (C) is the correct answer. | mcq | jee-main-2020-online-8th-january-morning-slot | 9,000 |
d3Ucn2JdtK89K0HxIijgy2xukf3vdpvr | physics | atoms-and-nuclei | alpha-particle-scattering-and-rutherford-model-of-atom | Hydrogen ion and singly ionized helium atom are accelerated, from rest, through the same potential
difference. The ratio of final speeds of hydrogen and helium ions is close to : | [{"identifier": "A", "content": "2 : 1"}, {"identifier": "B", "content": "1 : 2"}, {"identifier": "C", "content": "5 : 7"}, {"identifier": "D", "content": "10 : 7"}] | ["A"] | null | We know, kinetic energy K = qV
<br><br>also $$K = {{{P^2}} \over {2m}}$$<br><br>$$ \therefore $$ $$qV = {{{P^2}} \over {2m}} = {{{m^2}{v^2}} \over {2m}}$$<br><br>$$V = \sqrt {{{2qv} \over m}} $$<br><br>$$V \propto \sqrt {{q \over m}} $$<br><br>$${{{V_H}} \over {{V_{He}}}} = {{\sqrt {{e \over m}} } \over {\sqrt {{e \ove... | mcq | jee-main-2020-online-3rd-september-evening-slot | 9,001 |
1l54852gc | physics | atoms-and-nuclei | alpha-particle-scattering-and-rutherford-model-of-atom | <p>$$\sqrt {{d_1}} $$ and $$\sqrt {{d_2}} $$ are the impact parameters corresponding to scattering angles 60$$^\circ$$ and 90$$^\circ$$ respectively, when an $$\alpha$$ particle is approaching a gold nucleus. For d<sub>1</sub> = x d<sub>2</sub>, the value of x will be ____________.</p> | [] | null | 3 | <p>Impact parameter $$\propto$$ $$\cot {\theta \over 2}$$</p>
<p>$$ \Rightarrow \sqrt {{{{d_1}} \over {{d_2}}}} = {{\sqrt 3 } \over 1}$$</p>
<p>$$ \Rightarrow {d_1} = 3{d_2}$$</p>
<p>$$ \Rightarrow x = 3$$</p> | integer | jee-main-2022-online-29th-june-morning-shift | 9,002 |
1lgswq2kx | physics | atoms-and-nuclei | alpha-particle-scattering-and-rutherford-model-of-atom | <p>The energy of $$\mathrm{He}^{+}$$ ion in its first excited state is, (The ground state energy for the Hydrogen atom is $$-13.6 ~\mathrm{eV})$$ :</p> | [{"identifier": "A", "content": "$$-13.6 ~\\mathrm{eV}$$"}, {"identifier": "B", "content": "$$-27.2 ~\\mathrm{eV}$$"}, {"identifier": "C", "content": "$$-3.4 ~\\mathrm{eV}$$"}, {"identifier": "D", "content": "$$-54.4 ~\\mathrm{eV}$$"}] | ["A"] | null | The energy levels of a one-electron ion can be described by the formula:
<br/><br/>
$$
E_n = -\frac{Z^2}{n^2} \times E_0
$$
<br/><br/>
where $E_n$ is the energy of the nth level, Z is the atomic number (number of protons), n is the principal quantum number, and $E_0$ is the ground state energy of the hydrogen atom (-13... | mcq | jee-main-2023-online-11th-april-evening-shift | 9,004 |
1lgvtkqin | physics | atoms-and-nuclei | alpha-particle-scattering-and-rutherford-model-of-atom | <p>If 917 $$\mathop A\limits^o $$ be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be ___________ $$\mathop A\limits^o $$.</p> | [] | null | 3668 | <p>The energy difference formula for transitions between energy levels in a hydrogen atom, which is given by </p>
<p>$
\Delta E = -13.6 \, \text{eV} \times \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)
$</p>
<p>where ($n_1$) and ($n_2$) are the initial and final energy levels, respectively. For the Lyman series, the e... | integer | jee-main-2023-online-10th-april-evening-shift | 9,005 |
jaoe38c1lsfm0ryi | physics | atoms-and-nuclei | alpha-particle-scattering-and-rutherford-model-of-atom | <p>Given below are two statements:</p>
<p>Statement I : Most of the mass of the atom and all its positive charge are concentrated in a tiny nucleus and the electrons revolve around it, is Rutherford's model.</p>
<p>Statement II : An atom is a spherical cloud of positive charges with electrons embedded in it, is a speci... | [{"identifier": "A", "content": "Both Statement I and Statement II are true\n"}, {"identifier": "B", "content": "Statement I is true but Statement II is false\n"}, {"identifier": "C", "content": "Statement I is false but Statement II is true\n"}, {"identifier": "D", "content": "Both statement I and statement II are fal... | ["B"] | null | <p>According to Rutherford atomic model, most of mass of atom and all its positive charge is concentrated in tiny nucleus & electron revolve around it.</p>
<p>According to Thomson atomic model, atom is spherical cloud of positive charge with electron embedded in it.</p>
<p>Hence, Statement I is true but statement II fa... | mcq | jee-main-2024-online-29th-january-evening-shift | 9,008 |
lv5gt1qz | physics | atoms-and-nuclei | alpha-particle-scattering-and-rutherford-model-of-atom | <p>In an alpha particle scattering experiment distance of closest approach for the $$\alpha$$ particle is $$4.5 \times 10^{-14} \mathrm{~m}$$. If target nucleus has atomic number 80 , then maximum velocity of $$\alpha$$-particle is __________ $$\times 10^5 \mathrm{~m} / \mathrm{s}$$ approximately.</p>
<p>($$\frac{1}{4 ... | [] | null | 156 | <p>$$\begin{aligned}
& \frac{1}{2} m v_0^2=\frac{1}{4 \pi \epsilon_0} \frac{z(e)}{r} \\
& \frac{1}{2} \times 6.72 \times 10^{-27} v_0^2=9 \times 10^9 \times \frac{80 \times 2 \times 1.6 \times 1.6 \times 10^{-19} \times 10^{-19}}{4.5 \times 10^{-14}} \\
& v_0^2=\frac{2 \times 9 \times 80 \times 1.6 \times 1.6 \times 2}... | integer | jee-main-2024-online-8th-april-morning-shift | 9,009 |
t5PziT3qZsxzNgR8 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | If $$13.6$$ $$eV$$ energy is required to ionize the hydrogen atom, then the energy required to remove an electron from $$n=2$$ is | [{"identifier": "A", "content": "$$10.2$$ $$eV$$ "}, {"identifier": "B", "content": "$$0$$ $$eV$$ "}, {"identifier": "C", "content": "$$3.4$$ $$eV$$ "}, {"identifier": "D", "content": "$$6.8$$ $$eV.$$ "}] | ["C"] | null | <b>KEY CONCEPT : </b>
<br><br>The energy of nth orbit of hydrogen is given by
<br><br>$${E_n} = {{13.6} \over {{n^2}}}eV/$$ atom
<br><br>For $$n=2,$$ $${E_n} = {{ - 13.6} \over 4} = - 3.4eV$$
<br><br>Therefore the energy required to remove electron from
<br><br>$$n = 2$$ is $$+3.4eV.$$ | mcq | aieee-2002 | 9,010 |
aeFwtWYlFHKpp345 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | If the binding energy of the electron in a hydrogen atom is $$13.6eV,$$ the energy required to remove the electron from the first excited state of $$L{i^{ + + }}$$ is | [{"identifier": "A", "content": "$$30.6$$ $$eV$$ "}, {"identifier": "B", "content": "$$13.6$$ $$eV$$ "}, {"identifier": "C", "content": "$$3.4$$ $$eV$$ "}, {"identifier": "D", "content": "$$122.4$$ $$eV$$ "}] | ["A"] | null | $${E_n} = - {{13.6} \over {{n^2}}}{Z^2}eV/$$atom
<br><br>For lithium ion $$Z=3;$$ $$n=2$$ (for first excited state)
<br><br>$${E_n} = - {{13.6} \over {{2^2}}} \times {3^2} = - 30.6eV$$ | mcq | aieee-2003 | 9,013 |
QebZoD6k0EMI6BOq | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?
<img src="data:image/png;base64,UklGRjAHAABXRUJQVlA4ICQHAADQWACdASoBAmYBP4G21mW2LSwnINDZ8sAwCWlu/HyYom9nZ1+/q1/hLVfPH8ueh3+J8K5XxkniQ0ojQQiuPCyqr++3gA8LKqv77eADGGEOfvk... | [{"identifier": "A", "content": "$$iv$$ "}, {"identifier": "B", "content": "$$iii$$"}, {"identifier": "C", "content": "$$ii$$"}, {"identifier": "D", "content": "$$i$$ "}] | ["B"] | null | <b>KEY CONCEPT :</b> $$E = Rhc\left[ {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right]$$
<br><br>$$E$$ will be maximum for the transition for which
<br><br>$$\left[ {{1 \over {{n_1}^2}} - {1 \over {{n_2}^2}}} \right]$$ is maximum. Here $${n_2}$$ is the higher energy level
<br><br>Clearly, $$\left[ {{1 \over {{n_1}^2}... | mcq | aieee-2005 | 9,014 |
YQo15g0qID6qndlx | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | Which of the following transitions in hydrogen atoms emit photons of highest frequency ? | [{"identifier": "A", "content": "$$n = 1$$ to $$n=2$$ "}, {"identifier": "B", "content": "$$n = 2$$ to $$n=6$$ "}, {"identifier": "C", "content": "$$n = 6$$ to $$n=2$$ "}, {"identifier": "D", "content": "$$n = 2$$ to $$n=1$$ "}] | ["D"] | null | We have no find the frequency of emitted photons. For emission of photons the transition must take place from a higher energy level to a lower energy level which are given only in options $$(c)$$ and $$(d)$$.
<br><br>Frequency is given by
<br><br>$$hv = - 13.6\left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)$$
<... | mcq | aieee-2007 | 9,015 |
nA8jwc73aR0R2zBS | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | Suppose an electron is attracted towards the origin by a force $${k \over r}$$ where $$'k'$$ is a constant and $$'r'$$ is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the $${n^{th}}$$ orbital of the electron is found to be $$'{r_n}'$$ and the kinetic energy of the e... | [{"identifier": "A", "content": "$${T_n} \\propto {1 \\over {{n^2}}},{r_n} \\propto {n^2}$$ "}, {"identifier": "B", "content": "$${T_n}$$ independent of $$n,{r_n} \\propto n$$"}, {"identifier": "C", "content": "$${T_n} \\propto {1 \\over n},{r_n} \\propto n$$ "}, {"identifier": "D", "content": "$${T_n} \\propto {1 \\ov... | ["B"] | null | When $$F = {k \over r} = $$ centripetal force, then $${k \over r} = {{m{v^2}} \over r}$$
<br><br>$$ \Rightarrow m{v^2} = $$ constant $$ \Rightarrow $$ kinetic energy is constant
<br><br>$$ \Rightarrow T$$ is independent of $$n.$$ | mcq | aieee-2008 | 9,016 |
s3BeYEZV7vNh9876 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | The transition from the state $$n=4$$ to $$n=3$$ in a hydrogen like atom result in ultra violet radiation. Infrared radiation will be obtained in the transition from : | [{"identifier": "A", "content": "$$3 \\to 2$$ "}, {"identifier": "B", "content": "$$4 \\to 2$$"}, {"identifier": "C", "content": "$$5 \\to 4$$"}, {"identifier": "D", "content": "$$2 \\to 1$$"}] | ["C"] | null | It is given that transition from the state $$n=4$$ to $$n=3$$ in a hydrogen like atom result in ultraviolet radiation. For infrared radiation the energy gap should be less. The only option is $$5$$ $$ \to 4.$$ | mcq | aieee-2009 | 9,017 |
vcMFHQ01FVRjO7lC | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | Energy required for the electron excitation in $$L{i^{ + + }}$$ from the first to the third Bohr orbit is : | [{"identifier": "A", "content": "$$36.3$$ $$eV$$ "}, {"identifier": "B", "content": "$$108.8$$ $$eV$$"}, {"identifier": "C", "content": "$$122.4$$ $$eV$$ "}, {"identifier": "D", "content": "$$12.1$$ $$eV$$ "}] | ["B"] | null | Energy of excitation,
<br><br>$$\Delta E = 13.6\,{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)eV$$
<br><br>$$ \Rightarrow \Delta E = 13.6{\left( 3 \right)^2}\left( {{1 \over {{1^2}}} - {1 \over {{3^2}}}} \right) = 108.8\,eV$$ | mcq | aieee-2011 | 9,018 |
D0BrAVlk651aQ3Pd | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | Hydrogen atom is excited from ground state to another state with principal quantum number equal to $$4.$$ Then the number of spectral lines in the emission spectra will be : | [{"identifier": "A", "content": "$$2$$ "}, {"identifier": "B", "content": "$$3$$ "}, {"identifier": "C", "content": "$$5$$ "}, {"identifier": "D", "content": "$$6$$ "}] | ["D"] | null | The possible number of the spectral lines is given
<br><br>$$ = {{n\left( {n - 1} \right)} \over 2} = {{4\left( {4 - 1} \right)} \over 2} = 6$$
<br><br> | mcq | aieee-2012 | 9,019 |
h4cMuNp7bssJ79B8 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | A diatomic molecule is made of two masses $${m_1}$$ and $${m_2}$$ which are separated by a distance $$r.$$ If we calculate its rotational energy by applying Bohr's rule of angular momentum quantization, its energy will be given by: ($$n$$ is an integer) | [{"identifier": "A", "content": "$${{{{\\left( {{m_1} + {m_2}} \\right)}^2}{n^2}{h^2}} \\over {2m_1^2m_2^2{r^2}}}$$ "}, {"identifier": "B", "content": "$${{{n^2}{h^2}} \\over {2\\left( {{m_1} + {m_2}} \\right){r^2}}}$$ "}, {"identifier": "C", "content": "$${{2{n^2}{h^2}} \\over {\\left( {{m_1} + {m_2}} \\right){r^2}}}$... | ["D"] | null | The energy of the system of two atoms of diatomic
<br><br>molecule $$E = {1 \over 2}I\omega $$
<br><br>where $$I=$$ moment of inertia
<br><br>$$\omega = $$ Angular velocity $$ = {L \over I}.$$
<br><br>$$L=$$ Angular momentum
<br><br>$$I = {1 \over 2}\left( {{m_1}{r_1}^2 + {m^2}{r_2}^2} \right)$$
<br><br>Thus, $$E = {1... | mcq | aieee-2012 | 9,020 |
EMS4NwMsxXTATMl3 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | In a hydrogen like atom electron make transition from an energy level with quantum number $$n$$ to another with quantum number $$\left( {n - 1} \right)$$. If $$n > > 1,$$ the frequency of radiation emitted is proportional to : | [{"identifier": "A", "content": "$${1 \\over n}$$ "}, {"identifier": "B", "content": "$${1 \\over {{n^2}}}$$ "}, {"identifier": "C", "content": "$${1 \\over {{n^{{3 \\over 2}}}}}$$"}, {"identifier": "D", "content": "$${1 \\over {{n^3}}}$$"}] | ["D"] | null | $$\Delta $$E $$ = 13.6{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$$
<br><br>As $$\Delta $$E = h$$\upsilon $$
<br><br>$$ \Rightarrow $$ $$\Delta $$E $$ \propto $$ $$\upsilon $$ $$ \propto $$ $$\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$$
<br><br>Here n<sub>1</sub> = n - 1 and n<sub>2</sub> = ... | mcq | jee-main-2013-offline | 9,021 |
GRStPOuEY8L8EZaW | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | The radiation corresponding to $$3 \to 2$$ transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field $$3 \times {10^{ - 4}}\,T.$$ If the radius of the larger circular path followed by these electrons is $$10.0$$ $$mm$$, the work function of the me... | [{"identifier": "A", "content": "$$1.8$$ $$eV$$ "}, {"identifier": "B", "content": "$$1.1$$ $$eV$$ "}, {"identifier": "C", "content": "$$0.8$$ $$eV$$ "}, {"identifier": "D", "content": "$$1.6$$ $$eV$$ "}] | ["B"] | null | Radius of circular path followed by electron is given by,
<br><br>$$r = {{m\upsilon } \over {qB}} = {{\sqrt {2meV} } \over {eB}} = {1 \over B}\sqrt {{{2m} \over e}V} $$
<br><br>$$ \Rightarrow V = {{{B^2}{r^2}e} \over {2m}} = 0.8V$$
<br><br>For transition between $$3$$ to $$2.$$
<br><br>$$E = 13.6\left( {{1 \over 4} - ... | mcq | jee-main-2014-offline | 9,022 |
X6C549hC3b9UUpwC | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | Hydrogen $$\left( {{}_1{H^1}} \right)$$, Deuterium $$\left( {{}_1{H^2}} \right)$$, singly ionised Helium $${\left( {{}_2H{e^4}} \right)^ + }$$ and doubly ionised lithium $${\left( {{}_3L{i^6}} \right)^{ + + }}$$ all have one electron around the nucleus. Consider an electron transition from $$n=2$$ to $$n=1.$$ If the w... | [{"identifier": "A", "content": "$$4{\\lambda _1} = 2{\\lambda _2} = 2{\\lambda _3} = {\\lambda _4}$$ "}, {"identifier": "B", "content": "$${\\lambda _1} = 2{\\lambda _2} = 2{\\lambda _3} = {\\lambda _4}$$ "}, {"identifier": "C", "content": "$${\\lambda _1} = {\\lambda _2} = 4{\\lambda _3} = 9{\\lambda _4}$$ "}, {"iden... | ["C"] | null | Wave number $${1 \over \lambda } = R{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {{n^2}}}} \right]$$
<br><br>$$ \Rightarrow \lambda \propto {1 \over {{Z^2}}}$$
<br><br>By question $$n=1$$ and $${n_1} = 2$$
<br><br>Then, $${\lambda _1} = {\lambda _2} = 4{\lambda _3} = 9{\lambda _4}$$ | mcq | jee-main-2014-offline | 9,023 |
f1VCtY5QI3pxZNBP | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | As an electron makes a transition from an excited state to the ground state of a hydrogen - like atom/ion : | [{"identifier": "A", "content": "kinetic energy decreases, potential energy increases but total energy remains same "}, {"identifier": "B", "content": "kinetic energy and total energy decrease but potential energy increases "}, {"identifier": "C", "content": "its kinetic energy increases but potential energy and total ... | ["C"] | null | $$U = - K{{z{e^2}} \over r};\,\,T.E = {k \over 2}{{ze^2} \over r}$$
<br><br>$$K.E = {k \over 2}{{z{e^2}} \over r}.$$ Here $$r$$ decreases | mcq | jee-main-2015-offline | 9,024 |
nJ7izufyR29BNU7Sikprv | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | A hydrogen atom makes a transition from n = 2 to n = 1 and emits a photon. This photon strikes a doubly ionized lithium atom (z = 3) in excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for the process is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "5"}] | ["C"] | null | Energy released when hydrogen atom makes transition from n = 2 to n = 1 is,
<br><br>E<sub>1</sub> = 13.6 $$ \times $$ $$\left( {{1 \over {{1^2}}} - {1 \over {{z^2}}}} \right)$$
<br><br>= $${3 \over 4} \times 13.6\,\,\,$$ eV
<br><br>Energy required to remove a electron from n<sup>th</sup> excited state of doubly ioniz... | mcq | jee-main-2016-online-9th-april-morning-slot | 9,025 |
4yXZDD7lTq8o9hRM | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | Some energy levels of a molecule are shown in the figure. The
ratio of the wavelengths r = $${{\lambda _1}}$$/$${{\lambda _2}}$$, is given by: <br/><br/>
<img src="data:image/png;base64,UklGRlQMAABXRUJQVlA4IEgMAAAw5gCdASoAA7gCP4HA3mU2Ma8nIVO5CsAwCWlu/CHZUFsnZ1+/s9/rvW/cisOcTY7J8F4Su83b489gyg///W2QudDTPhmLXdDTPhmLXdDTPh... | [{"identifier": "A", "content": "r = 1/3"}, {"identifier": "B", "content": "r = 4/3"}, {"identifier": "C", "content": "r = 2/3"}, {"identifier": "D", "content": "r = 3/4"}] | ["A"] | null | From energy level diagram, using $$\Delta E = {{hc} \over \lambda }$$
<br><br>For wavelength $${{\lambda _1}}$$, $$\Delta E =$$ –E – (–2E) = $${{hc} \over {{\lambda _1}}}$$<br><br>For wavelength $${{\lambda _2}}$$, $$\Delta E =$$ –E – (–$${{4E} \over 3}$$) = $${{hc} \over {{\lambda _2}}}$$
<br><br>$$ \therefore $$ r = ... | mcq | jee-main-2017-offline | 9,026 |
4Z7EBmzMEEHC9RcXJc6tZ | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | The acceleration of an electron in the first orbit of the hydrogen atom (n = 1) is : | [{"identifier": "A", "content": "$${{{h^2}} \\over {{\\pi ^2}{m^2}{r^3}}}$$"}, {"identifier": "B", "content": "$${{{h^2}} \\over {{8\\pi ^2}{m^2}{r^3}}}$$"}, {"identifier": "C", "content": "$${{{h^2}} \\over {{4\\pi ^2}{m^2}{r^3}}}$$"}, {"identifier": "D", "content": "$${{{h^2}} \\over {{4\\pi }{m^2}{r^3}}}$$"}] | ["C"] | null | <p>The speedy of the particle in the orbit of an atom is $$v = {{{I^2}} \over {2h{\varepsilon _0}}}$$</p>
<p>We have the radius of the first orbit is</p>
<p>$$r = {{{h^2}{\varepsilon _0}} \over {\pi m{e^2}}}$$</p>
<p>$$ \Rightarrow {\varepsilon _0} = {{r\pi m{e^2}} \over {{h^2}}}$$</p>
<p>Therefore, the acceleration of... | mcq | jee-main-2017-online-9th-april-morning-slot | 9,027 |
J0N3KwLGZxT2gBT4GnmmY | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | According to Bohr’s theory, the time averaged magnetic field at the centre (i.e. nucleus) of a hydrogen atom due to the motion of electrons in the n<sup>th</sup> orbit is proportional to : (n = principal quantum number)
| [{"identifier": "A", "content": "$${n^{ - 4}}$$ "}, {"identifier": "B", "content": "$${n^{ - 5}}$$ "}, {"identifier": "C", "content": "n<sup>$$-$$3</sup> "}, {"identifier": "D", "content": "n<sup>$$-$$2</sup>"}] | ["B"] | null | Magnetic field at the center of neucleus of H-atom
<br><br>B = $${{{\mu _0}I} \over {2{r_n}}}$$
<br><br>Radius of n<sup>th</sup> orbital,
<br><br>r<sub>n</sub> = $${{{n^2}{h^2}{\varepsilon _0}} \over {m\pi Z{e^2}}}$$
<br><br>$$\therefore\,\,\,$$ r<sub>n</sub> $$ \propto $$ n<sup>2</sup>
<br><br>velocity of electron ... | mcq | jee-main-2017-online-8th-april-morning-slot | 9,028 |
1LvjgkZaOgSrgOKXTShwX | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | Muon ($$\mu $$<sup>$$-$$</sup>) is a negatively charged (|q| = |e|) particle with a mass m<sub>$$\mu $$</sub> = 200 m<sub>e</sub>, where m<sub>e</sub> is the mass of the electron and e is the electronic charge. If $$\mu $$<sup>$$-$$</sup> is bond to a proton to form a hydrogen like atom, identify the correct statement... | [{"identifier": "A", "content": "(A), (B), (D)"}, {"identifier": "B", "content": "(A), (C), (D)"}, {"identifier": "C", "content": "(B), (D)"}, {"identifier": "D", "content": "(C), (D)"}] | ["B"] | null | <p>(i) Radius of the orbit is given by</p>
<p>$$r = {{{h^2}} \over {4\pi m{e^2}}} \times {{{n^2}} \over Z}$$</p>
<p>Only, m<sub>$$\mu$$</sub> = 200 m<sub>e</sub> rest are same. So, statement (A) is correct: The radius of muonic orbit is 200 times smaller than that of the electron.</p>
<p>(ii) Velocity of particle in an... | mcq | jee-main-2018-online-15th-april-evening-slot | 9,029 |
PavhhsYmKCP6uAqubzoe5 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | The energy required to remove the electron from a singly ionized Helium atom is $$2.2$$ times the energies required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is : | [{"identifier": "A", "content": "$$20$$ $$eV$$ "}, {"identifier": "B", "content": "$$34$$ $$eV$$ "}, {"identifier": "C", "content": "$$79$$ $$eV$$ "}, {"identifier": "D", "content": "$$109$$ $$eV$$ "}] | ["C"] | null | Energy required to remove e<sup>$$-$$</sup> from singly ionized Helium atom
<br><br>E<sub>1</sub> = $${{13.6{z^2}} \over {{n^2}}}$$ = $${{13.6 \times {2^2}} \over {{1^2}}}$$ = 54.4 eV.
<br><br>Let, E<sub>2</sub> = energy required to remove e<sup>$$-$$</sup>from He $$-$$ atom
<br><br>$$\therefore\,\,\,\,$$ According t... | mcq | jee-main-2018-online-15th-april-morning-slot | 9,030 |
Pirx9xZT9W7Gitl2 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | If the series limit frequency of the Lyman series is $${\nu _L}$$, then the series limit frequency of the Pfund series is: | [{"identifier": "A", "content": "$${\\nu _L}/25$$ "}, {"identifier": "B", "content": "$$25{\\nu _L}$$"}, {"identifier": "C", "content": "$$16{\\nu _L}$$ "}, {"identifier": "D", "content": "$${\\nu _L}/16$$ "}] | ["A"] | null | <b><u>Note</u> :</b>
<br><br>(1) In Lyman Series, transition happens in n = 1 state
<br>from n = 2, 3, . . . . . $$ \propto $$
<br><br>(2) In Balmer Series, transition happens in n = 2 state
<br>from n = 3, 4, . . . . . $$ \propto $$
<br><br>(3) In Paschen Series, ... | mcq | jee-main-2018-offline | 9,031 |
keWq8KI8WjGXxYdp | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let
$${\lambda _n}$$, $${\lambda _g}$$ be the de Broglie wavelength of the electron in the n<sup>th</sup> state and the ground state respectively. Let
$${\Lambda _n}$$ be the wavelength of the emitted photon in the tran... | [{"identifier": "A", "content": "$${\\Lambda _n} \\approx A + {B \\over {\\lambda _n^2}}$$"}, {"identifier": "B", "content": "$${\\Lambda _n} \\approx A + B{\\lambda _n}$$ "}, {"identifier": "C", "content": "$$\\Lambda _n^2 \\approx A + B\\lambda _n^2$$ "}, {"identifier": "D", "content": "$$\\Lambda _n^2 \\approx \\lam... | ["A"] | null | We know,
<br><br>Wavelength of emitted photon from n<sub>2</sub> state to n<sub>1</sub> state is
<br><br>$${1 \over \lambda }$$ = RZ<sup>2</sup> $$\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$
<br><br>Here electron comes from n<sup>th</sup> state to ground state (n = 1), <br><br>then the wavele... | mcq | jee-main-2018-offline | 9,032 |
YvwRcQgcrtD5hs3MO63rsa0w2w9jx7eqe54 | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | The electron in a hydrogen atom first jumps from the third excited state to the second excited state and
subsequently to the first excited state. The ratio of the respective wavelengths, $${{{\lambda _1}} \over {{\lambda _2}}}$$, of the photons emitted
in this process is : | [{"identifier": "A", "content": "$${{22} \\over 5}$$"}, {"identifier": "B", "content": "$${7 \\over 5}$$"}, {"identifier": "C", "content": "$${9 \\over 7}$$"}, {"identifier": "D", "content": "$${{20} \\over 7}$$"}] | ["D"] | null | n = 1 (Ground state)
<br>n = 2 (First excitate state)
<br>n = 3 (Second excitate state)
<br>n = 4 (Third excitate state)
<br><br>$${{hc} \over {{\lambda _1}}} = 13.6\left( {{1 \over 9} - {1 \over {16}}} \right)$$
<br><br>$${{hc} \over {{\lambda _2}}} = 13.6\left( {{1 \over 4} - {1 \over 9}} \right)$$
<br><br>$${{{\lamb... | mcq | jee-main-2019-online-12th-april-evening-slot | 9,033 |
rzTPPFCfSfjSKFadMd3rsa0w2w9jx7eb8bl | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | Consider an electron in a hydrogen atom revolving in its second excited state (having radius 4.65 $$\mathop A\limits^o $$). The
de-Broglie wavelength of this electron is : | [{"identifier": "A", "content": "6.6 $$\\mathop A\\limits^o $$"}, {"identifier": "B", "content": "3.5 $$\\mathop A\\limits^o $$"}, {"identifier": "C", "content": "9.7 $$\\mathop A\\limits^o $$"}, {"identifier": "D", "content": "12.9 $$\\mathop A\\limits^o $$"}] | ["C"] | null | For second excited state n = 3
<br><br>$$mvr = {{3h} \over {2\pi }}$$ ........(1)
<br><br>$$mv = {h \over \lambda }$$ .........(2)
<br><br>Dividing (1) by (2), we get
<br><br>$$r = {{3\lambda } \over {2\pi }}$$
<br><br>$$ \Rightarrow $$ $$\lambda = {{2\pi r} \over 3}$$ = $${{2 \times 3.14 \times 4.65} \over 3}$$ = 9.7... | mcq | jee-main-2019-online-12th-april-evening-slot | 9,034 |
16y1MXImK4m6rKpp06JAz | physics | atoms-and-nuclei | bohr's-model-and-hydrogen-spectrum | Radiation coming from transitions
n = 2 to n = 1 of hydrogen atoms fall on He<sup>+</sup>
ions in n = 1 and n = 2 states. The possible
transition of helium ions as they absorb energy
from the radiation is : | [{"identifier": "A", "content": "n = 1 $$ \\to $$ n = 4"}, {"identifier": "B", "content": "n = 2 $$ \\to $$ n = 5"}, {"identifier": "C", "content": "n = 2 $$ \\to $$ n = 4"}, {"identifier": "D", "content": "n = 2 $$ \\to $$ n = 3"}] | ["C"] | null | Energy released for tension n = 2 to n = 1 of hydrogen atom
<br><br>
$$E = 13.6{Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$<br><br>
Z = 1, n<sub>1</sub> = 1, n<sub>2</sub> = 2<br><br>
$$E = 13.6 \times 1 \times \left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)$$<br><br>
$$E = 13.6 \times {3 \over ... | mcq | jee-main-2019-online-8th-april-morning-slot | 9,038 |
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