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Q. No.
A capacitance is formed by connecting two metallic balls of radius \(r\) by a conducting wire, and two oppositely charged identical metallic hemispheres \((A,B)\) slightly larger than the balls. The separation between the hemispheres and the respective balls is \(d.\) The capacitance between \(A,B\) is:

 
1. \(\dfrac{4\pi\varepsilon_0r^2}{d}\) 2. \(\dfrac{2\pi\varepsilon_0r^2}{d}\)
3. \(\dfrac{\pi\varepsilon_0r^2}{d}\) 4. \(\dfrac{\pi\varepsilon_0r^2}{2d}\)
Subtopic:  Capacitance |
Level 4: Below 35%

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The arrangement shown in the figure is set up with capacitors initially uncharged, and the circuit is completed. A potential difference is imposed across \(AB\) so that the charge on the upper capacitor is doubled without changing its sign.

          
Then, \(V_{A}-V_{B}=\)
1. \(E_0\)
2. \(2E_0\)
3. \(-E_0\)
4. zero
Subtopic:  Capacitance |
Level 3: 35%-60%

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A \(12 ~\mu \text{F}\) capacitor is charged by means of a \(6~\text{V}\) battery and the charged capacitor and the battery are connected in series so that their combined potential difference is twice as much. When a second unknown capacitor (initially uncharged) is connected across this combination, the first capacitor is observed to lose half of its initial charge. The capacitance of the unknown capacitor is:
1. \(4 ~\mu \text{F}\) 2. \(6~ \mu \text{F}\)
3. \(24 ~\mu \text{F}\) 4. \(36 ~\mu \text{F}\)
 
Subtopic:  Capacitance |
Level 4: Below 35%

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Two concentric metallic spheres, surface areas \(A_1,A_2\) and separation \(d\), have a capacitance \(C_0.\) If a parallel plate capacitor is built with the same separation \(d,\) and has the same capacitance \(C_0\) then its plate area will be:
1. \(\dfrac{A_1+A_2}{2}\) 2. \(\sqrt{A_1A_2}\)
3. \(\dfrac{2A_1A_2}{A_1+A_2}\) 4. \(\dfrac{A_1^2A_2^2}{A_1+A_2}\)
Subtopic:  Capacitance |
Level 3: 35%-60%

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Two large plane metallic plates of surface area \(A\) are placed parallel to each other, and they are charged oppositely with equal charges: \(+Q,-Q.\) The separation between the two plates is \(d,\) which is very small. A small positive charge \(q~(q\ll Q)\) is placed exactly midway between the two plates. The force on \(q\) due to the remaining charges \((\text{i.e.,}+Q,-Q)\) is: \(\Bigg(k={\large\frac{1}{4\pi\varepsilon_0}}\Bigg)\)
                            
1. \(k{\Large\frac{2qQ}{\big(d^2/4\big)}}\) 2. \({\Large\frac{qQ}{\varepsilon_0A}}\)
3. \({\Large\frac{2qQ}{\varepsilon_0A}}\) 4. zero
Subtopic:  Capacitance |
Level 3: 35%-60%

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A parallel plate capacitor is charged by giving it a total charge \(Q,\) resulting in an electric field \(E\) between its plates. A small charge \(q\) is placed between the plates of the capacitor. The force on \(q\) is \(F_1\) and the force between the two plates of the capacitor is \(F_2.\) Then:
1. \(F_1=qE,~F_2=QE\) 2. \(F_1=2qE,~F_2=QE\)
3. \(F_1=2qE,~F_2={\Large\frac{QE}{2}}\) 4. \(F_1=qE,~F_2={\Large\frac{QE}{2}}\)
Subtopic:  Capacitance |
Level 3: 35%-60%

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Three metallic spheres of radii \(r_1,~ r_2,~ r_3\) are connected by very long conducting wires to form an equilateral triangle. The capacitance of the system is:

        
1. \(4 \pi \varepsilon_{0}\left(r_{1}+r_{2}+r_{3}\right)\)
2. \(4 \pi \varepsilon_{0} \dfrac{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}}{r_{1}+r_{2}+r_{3}}\)
3. \(4 \pi \varepsilon_{0}\left(\dfrac{1}{r_{1}}+\dfrac{1}{r_{2}}+\dfrac{1}{r_{3}}\right)^{-1}\)
4. \(4 \pi \varepsilon_{0} \sqrt{r_{1}^{2}+r_{2}^{2}+r_{3}^{2}}\)
Subtopic:  Capacitance |
 52%
Level 3: 35%-60%

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Four uncharged capacitors and resistances are connected as shown and a potential difference is applied between \(P\) and \(Q\).
It is observed that \(V_A=V_B\) after a sufficiently long time.
Then:
1. \(\dfrac{C_1}{C_2}=\dfrac{C_3}{C_4}\) 2. \(\dfrac{C_1}{C_2}=\dfrac{R_3~C_3}{R_4~C_4}\)
3. \(\dfrac{C_1}{C_2}=\dfrac{R_4}{R_3}\) 4. \(\dfrac{C_1}{C_2}=\dfrac{R_4~C_3}{R_3~C_4}\)
Subtopic:  Capacitance |
 51%
Level 3: 35%-60%

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Consider a spherical conductor of radius \(r\), centred at the point \(O\). A point charge \(q(q>0)\) is placed outside the sphere, at a distance '\(d\)' from its centre \((O)\) \((d>r)\) and the sphere is earthed.
                                  
Given below are two statements: 
Assertion (A): The electric flux due to the external charge \(q\) and the induced charges on the sphere through the spherical surface \(S\) shown dotted in the diagram, is \(\left(-\dfrac{q}{\varepsilon_0}\right)\).
Reason (R): Negative charges are induced on the surface of the conducting sphere due to the positive charge \(q\) in the vicinity, and the potential of the conducting sphere is zero. 
 
1. Both (A) and (R) are True and (R) is the correct explanation of (A).
2. Both (A) and (R) are True but (R) is not the correct explanation of (A).
3. (A) is True but (R) is False.
4. (A) is False but (R) is True.
Subtopic:  Capacitance |
 52%
Level 3: 35%-60%

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The capacitance of a sphere of radius \(a,\) surface area \(A\) is:
1. \(\dfrac{4\pi\varepsilon_0A}{a}\)
2. \(\dfrac{2\pi\varepsilon_0A}{a}\)
3. \(\dfrac{\varepsilon_0A}{a}\)
4. \(\dfrac{\varepsilon_0A}{2\pi a}\)
Subtopic:  Capacitance |
 55%
Level 3: 35%-60%

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