In the figure, two conducting concentric spherical shells are shown.

                           

If the electric potential at the centre is 20V and the electric potential at the surface of outer shell is 5V, then the potential at the surface of inner shell is –

1.  5V                   

2.  15V 

3.  20V                 

3.  cannot be determined as radii are not given

The circuit was in the shown state for a long time. Now if the switch S is closed then the charge that flows through the switch S, will be –

1.  $\frac{400}{3}<mspace\; width="1em"></mspace>\mathrm{\mu C}$                               

2.  $100<mspace\; width="1em"></mspace>\mathrm{\mu C}$

3.  $\frac{100}{3}<mspace\; width="1em"></mspace>\mathrm{\mu C}$                               

4.  $50<mspace\; width="1em"></mspace>\mathrm{\mu C}$

The potential at a certain point in an electric field is 200 V. The work done in carrying an electron upto that point will be.

1.  $3.2\times {10}^{-17}<mspace\; width="1em"></mspace>\mathrm{J}$                      

2.  $-3.2\times {10}^{-17}<mspace\; width="1em"></mspace>\mathrm{J}$

3.  $200<mspace\; width="1em"></mspace>\mathrm{J}$                                     

4.  $-200<mspace\; width="1em"></mspace>\mathrm{J}$

Two charged conducting spheres of radii ${\mathrm{r}}_1$ and ${\mathrm{r}}_2$ are at the same potential. The ratio of their surface charge densities will be -

1.  ${\mathrm{r}}_1/{\mathrm{r}}_2$                               

2.  ${\mathrm{r}}_2/{\mathrm{r}}_1$

3.  ${{\mathrm{r}}_1}^2/{{\mathrm{r}}_2}^2$                           

4.  ${{\mathrm{r}}_2}^2/{{\mathrm{r}}_1}^2$

At the mid point of a line joining an electron and a proton, the values of E and V will be.

1. $\mathrm{E}=0,<mspace\; width="1em"></mspace>\mathrm{V}\ne 0$                   

2. $\mathrm{E}\ne 0,<mspace\; width="1em"></mspace>\mathrm{V}=0$

3. $\mathrm{E}\ne 0,<mspace\; width="1em"></mspace>\mathrm{V}\ne 0$                   

4. $\mathrm{E}=0,<mspace\; width="1em"></mspace>\mathrm{V}=0$

A charge of $10\mu \mathrm{C}$ is kept at the origin of $\mathrm{X}–\mathrm{Y}$ coordinate system. The potential difference in volts between two points (a, 0) and $\left(\mathrm{a}/\sqrt{2}<mspace\; width="1em"></mspace>,<mspace\; width="1em"></mspace>\mathrm{a}/\sqrt{2}\right)$ will be.

1.  Zero                             

2.  $9\times {10}^4$

3.  $\frac{9\times {10}^4}{\mathrm{a}}$                           

4.  $\frac{9\times {10}^4}{\sqrt{2}}$

Find equivalent capacitance between $\mathrm{X}$ and $\mathrm{Y}$ if each capacitor is $4<mspace\; width="1em"></mspace>\mathrm{\mu F}$.

1.  $4<mspace\; width="1em"></mspace>\mathrm{\mu F}$                                     

2.  $2<mspace\; width="1em"></mspace>\mathrm{\mu F}$

3.  $12<mspace\; width="1em"></mspace>\mathrm{\mu F}$                                   

4.  $1<mspace\; width="1em"></mspace>\mathrm{\mu F}$

Two point charge of $8<mspace\; width="1em"></mspace>\mathrm{\mu C}$ and $12<mspace\; width="1em"></mspace>\mathrm{\mu C}$ are kept in air at a distance of 10 cm from each other. The work required to change the distance between them to 6 cm will be.

1.  $5.8<mspace\; width="1em"></mspace>\mathrm{J}$                       

2.  $4.8<mspace\; width="1em"></mspace>\mathrm{J}$

3.  $3.8<mspace\; width="1em"></mspace>\mathrm{J}$                       

4.  $2.8<mspace\; width="1em"></mspace>\mathrm{J}$

Two parallel plate capacitors of capacitances C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the region between the plates of the capacitor C is completely filled with a material of dielectric constant K. The potential difference across the capacitors now becomes –

1.  $\frac{3\mathrm{V}}{\mathrm{K}+2}$                             

2.  $\mathrm{KV}$

3.  $\frac{\mathrm{V}}{\mathrm{K}}$                                 

4.  $\frac{3}{\mathrm{KV}}$