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Q. No.A triangular wire frame, in the form of an equilateral triangle \(PQR\) moves with a uniform velocity into a region where there is a uniform magnetic field \(B\). The edge \(PQ\) is parallel to the boundary of the region and the velocity \(v\) is perpendicular to it. The emf(\(E\)) induced within the frame is plotted as a function of time \(t,\) starting from when the frame enters the magnetic field. \(E\) is given by:
1.
\(Bv^2t\)
2.
\(2Bv^2t\)
3.
\(\dfrac{\sqrt3}{2}Bv^2t\)
4.
\(\dfrac{2}{\sqrt3}Bv^2t\)
Subtopic: Motional emf |
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A wire, bent into the shape of a right angled triangle \(PQR,\) lies with its side \(PR\) parallel to a current carrying wire, and side \(QR\) perpendicular to it. The loop lies in the plane of the wire. EMF induced in the loop when it is moved with constant speed along \(PR\) is \(\varepsilon_1\) and it is \(\varepsilon_2\) when moved along \(QR\) with the same constant speed. Then,
1. \(\varepsilon_1=0,\varepsilon_2\neq0\)
2. \(\varepsilon_1\neq0,\varepsilon_2=0\)
3. \(\varepsilon_1=0,\varepsilon_2=0\)
4. \(\varepsilon_1\neq0,\varepsilon_2\neq0\)
1. \(\varepsilon_1=0,\varepsilon_2\neq0\)
2. \(\varepsilon_1\neq0,\varepsilon_2=0\)
3. \(\varepsilon_1=0,\varepsilon_2=0\)
4. \(\varepsilon_1\neq0,\varepsilon_2\neq0\)
Subtopic: Motional emf |
From NCERT
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In the circuit shown in the adjoining figure, the switch was kept at the position \('1'\) for a long time. The switch \(K\) is suddenly (and smoothly) shifted to position \('2'.\)
The current through the cell, just after the shift, is:
The current through the cell, just after the shift, is:
| 1. | \(\dfrac{V_0}{2R}\) | 2. | \(\dfrac{V_0}{R}\) |
| 3. | \(\dfrac{3V_0}{4R}\) | 4. | zero |
Subtopic: LR circuit |
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A straight horizontal wire of mass \(m\) and length \(l,\) and having a negligible resistance can slide freely on a pair of conducting parallel rails, placed vertically. The rails are connected at the top by a capacitor \(C.\) A uniform magnetic field \(B\) exists in the region, perpendicular to the plane of the rails. The wire:
| 1. | falls with uniform velocity. |
| 2. | accelerates down with acceleration less than \(g\). |
| 3. | accelerates down with acceleration equal to \(g\). |
| 4. | moves down and eventually comes to rest. |
Subtopic: Motional emf |
71%
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An inductor \((L)\) and a resistor \((R)\) are connected in series and a battery is connected, as shown in the figure. Once the current becomes steady, the power in the resistance is \(P_R\) and the energy stored in the inductor is \(U_L.\) The switch is suddenly (and smoothly) toggled to the position \(B\) allowing the inductor to discharge. The time in which the energy stored becomes \(\dfrac12\) its initial value is:
| 1. | \(\dfrac{U_L}{P_R}\) | 2. | \(\dfrac{U_L~\mathrm {ln}2}{P_R}\) |
| 3. | \(\dfrac{2U_L~\mathrm{ln 2}}{P_R}\) | 4. | \(\dfrac{2U_L}{P_R}\) |
Subtopic: LR circuit |
54%
From NCERT
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A circular wire of radius \(R\) is placed in a uniform magnetic field \(B,\) which acts into the plane as shown. The wire is given a half-turn about a diameter. The resistance per unit length of the wire is \(\lambda.\) The total charge flowing through the wire is:
| 1. | \(\dfrac{2BR}{\lambda}\) | 2. | \(\dfrac{BR}{\lambda}\) |
| 3. | \(\dfrac{BR}{2\lambda}\) | 4. | zero |
Subtopic: Motional emf |
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The self-inductance of a long solenoid of cross-section \(A,\) total length \(L\) and total number of turns \(N,\) is (approximately):
| 1. | \(\dfrac{\mu_0A}{L}\cdot N\) | 2. | \(\dfrac{\mu_0A}{L}\cdot N^2\) |
| 3. | \(\dfrac{\mu_0L^3}{A}\cdot N\) | 4. | \(\dfrac{\mu_0L^3}{A}\cdot N^2\) |
Subtopic: Self - Inductance |
81%
From NCERT
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The \(\text{DC}\) time constant of an \(L\)-\(R\) circuit is the same as that of an \(R\)-\(C\) circuit where the inductor in the first circuit was replaced by a capacitor. The value of the resistance \(R\) equals:
| 1. | \(\dfrac{1}{2\pi}\sqrt{\dfrac{L}{C}}\) | 2. | \(\sqrt{\dfrac{L}{C}}\) |
| 3. | \(2\pi\sqrt{\dfrac{L}{C}}\) | 4. | \(2\sqrt{\dfrac{L}{C}}\) |
Subtopic: LR circuit |
58%
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A \(3~\mu\text{F}\) capacitor is charged with \(6~\mu \text{C}\) and connected across a \(1~\text{mH}\) inductance. The rate of change of current is:
1. \(2\) A/s
2. \(2\times10^{-3}\) A/s
3. \(2\times10^{3}\) A/s
4. \(2\times10^{-6}\) A/s
1. \(2\) A/s
2. \(2\times10^{-3}\) A/s
3. \(2\times10^{3}\) A/s
4. \(2\times10^{-6}\) A/s
Subtopic: Self - Inductance |
72%
From NCERT
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A conducting circular wire of radius \(r\) is moving with constant velocity \(v\) towards the right in a uniform magnetic field \(B.\) We consider two points \(X,Y\) such that chord \(XY\) is perpendicular to the velocity \(v\) and is at a distance \(x\) from the centre \((O)\) of the circle. The EMF induced between \(X,Y\) is \(\varepsilon.\) Then, \(\varepsilon\) is proportional to:
1. \(x\)
2. \(\sqrt{r^2-x^2}\)
3. \(r\)
4. \(x\sqrt{r^2-x^2}\)
1. \(x\)
2. \(\sqrt{r^2-x^2}\)
3. \(r\)
4. \(x\sqrt{r^2-x^2}\)
Subtopic: Motional emf |
73%
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