Q. No.The equivalent capacitance of the combination shown in the figure is:
1. \(\dfrac{C}{2}\)
2. \(\dfrac{3C}{2}\)
3. \(3C\)
4. \(2C\)
1. \(\dfrac{C}{2}\)
2. \(\dfrac{3C}{2}\)
3. \(3C\)
4. \(2C\)
| 1. | \(\sqrt{\dfrac{R_1}{R_2}}\) | 2. | \(\dfrac{R^2_1}{R^2_2}\) |
| 3. | \(\dfrac{R_1}{R_2}\) | 4. | \(\dfrac{R_2}{R_1}\) |
Twenty seven drops of same size are charged at \(220~\text{V}\) each. They combine to form a bigger drop. Calculate the potential of the bigger drop:
1. \(1520~\text{V}\)
2. \(1980~\text{V}\)
3. \(660~\text{V}\)
4. \(1320~\text{V}\)
A parallel plate capacitor with cross-sectional area \(A\) and separation \(d\) has air between the plates. An insulating slab of the same area but the thickness of \(\dfrac{d}{2}\) is inserted between the plates as shown in the figure, having a dielectric constant, \(K=4.\) The ratio of the new capacitance to its original capacitance will be:
| 1. | \(2:1\) | 2. | \(8:5\) |
| 3. | \(6:5\) | 4. | \(4:1\) |
| 1. |  |
2. |  |
| 3. |  |
4. |  |
| 1. | dependent on the material property of the sphere |
| 2. | more on the bigger sphere |
| 3. | more on the smaller sphere |
| 4. | equal on both the spheres |
| 1. | \(180^\circ\) | 2. | \(0^\circ\) |
| 3. | \(45^\circ\) | 4. | \(90^\circ\) |
1. \(1.5\times 10^{-6}~\text{J}\)
2. \(4.5\times 10^{-6}~\text{J}\)
3. \(3.25\times 10^{-6}~\text{J}\)
4. \(2.25\times 10^{-6}~\text{J}\)
A hollow metal sphere of radius \(R\) is given \(+Q\) charges to its outer surface. The electric potential at a distance \(\dfrac{R}{3}\) from the centre of the sphere will be:
| 1. | \(\dfrac{1}{4\pi \varepsilon_0}\dfrac{Q}{9R}\) | 2. | \(\dfrac{3}{4\pi \varepsilon_0}\dfrac{Q}{R}\) |
| 3. | \(\dfrac{1}{4\pi \varepsilon_0}\dfrac{Q}{3R}\) | 4. | \(\dfrac{1}{4\pi \varepsilon_0}\dfrac{Q}{R}\) |
| 1. | \(0.9~\mu\text{F}\) | 2. | \(0.09~\mu\text{F}\) |
| 3. | \(0.1~\mu\text{F}\) | 4. | \(0.01~\mu\text{F}\) |
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