Q. No.A thin uniform rod of mass \(M\) and length \(L\) is suspended from one of its ends, '\(A\)', so that it can rotate freely about it. A charge '\(q\)' is fixed to its lower end \(B\). A uniform horizontal Electric field is switched on and the rod rotates about \(A\), finally coming to equilibrium – making an angle of \(45^{\circ}\) with the vertical. If the acceleration due to gravity is '\(g\)', then,
1. \(qE =Mg\)
2. \(2qE =Mg\)
3. \(qE =2Mg\)
4. \(\sqrt{2}qE =Mg\)
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| 1. | \(\dfrac{1}{r^2}\) | 2. | \(\dfrac{1}{r}\) |
| 3. | \(r\) | 4. | \(\dfrac{1}{r^3}\) |
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Then,
| 1. | \(q,Q\) are of the same sign and \(|q|=|Q|\) |
| 2. | \(q,Q\) are of opposite signs and \(|q|=|Q|\) |
| 3. | \(q,Q\) are of the same sign and \(|q|<|Q|\) |
| 4. | \(q,Q\) are of opposite signs and \(|q|>|Q|\) |
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The electric field within the medium due to the charge \(Q\) at some point \(P\) is \(\vec E_{Q}\). The Electric field at the same point \(P\) due to induced charge within the medium is \(\vec E_{m}\). Then,
| 1. | \(|\vec E_m|=\left|\dfrac{\vec E_Q}{K}\right|,\) and the two fields are in opposite directions. |
| 2. | \(|\vec E_Q|=\left|\dfrac{\vec E_m}{K}\right|,\) and the two fields are in the same direction. |
| 3. | \(|\vec E_Q+\vec E_m|=\left|\dfrac{\vec E_Q}{K}\right|,\) and the two fields are in opposite directions. |
| 4. | \(|\vec E_Q+\vec E_m|=\left|\dfrac{\vec E_m}{K}\right|,\) and the two fields are in the same direction. |
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Now, identical point charges (same magnitude \(q\) each), are placed at the four corners of a square of side \(a.\) When one of the charges is removed, the electric field at the center becomes \(E_s.\) Then,
| 1. | \(\dfrac{E_s}{2}=\dfrac{E_C}{3}\) | 2. | \(\dfrac{E_s}{3}=\dfrac{E_C}{2}\) |
| 3. | \(\dfrac{E_s}{\sqrt2}=\dfrac{E_C}{\sqrt3}\) | 4. | \(\dfrac{E_s}{\sqrt3}=\dfrac{E_C}{\sqrt2}\) |
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Then:
| 1. | \(E_1 < 0, E_2 > 0, E_3 < 0\) |
| 2. | \(E_1 < 0, E_2 = 0, E_3 < 0\) |
| 3. | \(E_1 = 0, E_2 > 0, E_3 = 0\) |
| 4. | \(E_1 < 0, E_2 > 0, E_3 > 0\) |
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1. \(v^2=u^2+2gR\)
2. \(v^2=u^2+2gR\left(\frac{qE}{mg}\right)\)
3. \(v^2=u^2+gR\)
4. \(v^2=u^2+gR\left(\frac{qE}{mg}\right)\)
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Note: \(A,B,C\) are conductors. Other charges may be present in the vicinity.
| 1. | I, III | 2. | II |
| 3. | I, II, III | 4. | none of I, II, III |
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The mid-point of side \(AB\) is \(P\) while the circumcenter of \(ABC\) is \(O\). Let the electric field at \(P\) be \(E_p\) and that at \(O\) be \(E_O.\)
Then, \(E_O:E_P=\)
| 1. | \(\dfrac{2}{9}\) | 2. | \(\dfrac{4}{9}\) |
| 3. | \(\dfrac{9}{2}\) | 4. | \(\dfrac{9}{4}\) |
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A pair of field lines are drawn, connecting the charges \(q_1, q_2\) in addition to the straight field line connecting them. From the above diagram,
| 1. | \(q_1>0, q_2<0~\text{and}~|q_1|>|q_2|\) |
| 2. | \(q_1<0, q_2>0~\text{and}~|q_1|>|q_2|\) |
| 3. | \(q_1>0, q_2<0~\text{and}~|q_1|<|q_2|\) |
| 4. | \(q_1<0, q_2>0~\text{and}~|q_1|<|q_2|\) |
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