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.  $\frac{3}{\mathrm{KV}}$

Maximum charge stored on a metal sphere of radius $15$ cm may be $7.5<mspace\; width="1em"></mspace>\mu \mathrm{C}$. The potential energy of the sphere in this case is :

1.  $9.67<mspace\; width="1em"></mspace>\mathrm{J}$                   

2.  $0.25<mspace\; width="1em"></mspace>\mathrm{J}$ 

3.  $3.25<mspace\; width="1em"></mspace>\mathrm{J}$                    

4.  $1.69<mspace\; width="1em"></mspace>\mathrm{J}$