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

Four identical particles each of mass m and charge q are kept at the four corners of a square of length L. The final velocity of these particles after setting them free will be.

1. ${\left[\frac{{\mathrm{Kq}}^2}{\mathrm{mL}}\left(5.4\right)\right]}^{1/2}$                           

2. ${\left[\frac{{\mathrm{Kq}}^2}{\mathrm{mL}}\left(1.35\right)\right]}^{1/2}$

3. ${\left[\frac{{\mathrm{Kq}}^2}{\mathrm{mL}}\left(2.7\right)\right]}^{1/2}$                           

4. Zero

A parallel plate capacitor is charged to a certain potential difference. A slab of thickness 3 mm is inserted between the plates and it becomes necessary to increase the distance between the plates by 2.4 mm to maintain the same potential difference. The dielectric constant of the slab is– 

1.  3                   

2.  5 

3.  2.5               

4.  2

A charge $\mathrm{Q}$ is distributed over two concentric hollow spheres of radii r and $\mathrm{R}<mspace\; width="1em"></mspace>\left(\mathrm{R}<mspace\; width="1em"></mspace>><mspace\; width="1em"></mspace>\mathrm{r}\right)$ such that the surface densities are equal. The potential at the common centre is $\frac{1}{4{\mathrm{\pi \epsilon }}_0}$ times –

1.  $\mathrm{Q}<mspace\; width="1em"></mspace>\left(\frac{\mathrm{r}+\mathrm{R}}{{\mathrm{r}}^2+{\mathrm{R}}^2}\right)$                       

2.  $\frac{\mathrm{Q}}{2}<mspace\; width="1em"></mspace>\left(\frac{\mathrm{r}+\mathrm{R}}{{\mathrm{r}}^2+{\mathrm{R}}^2}\right)$

3.  $2\mathrm{Q}<mspace\; width="1em"></mspace>\left(\frac{\mathrm{r}+\mathrm{R}}{{\mathrm{r}}^2+{\mathrm{R}}^2}\right)$                     

4.  Zero

A charge $+\mathrm{Q}$ is uniformly distributed over a thin ring of radius $\mathrm{R}$, velocity of an electron at the moment it passes through the centre $\mathrm{O}$ of the ring, if the electron was initially at rest at a point $\mathrm{A}$ which is very far away from the centre and on the axis of the ring is

1. $\sqrt{\frac{2\mathrm{kQe}}{\mathrm{mR}}}$                     

2. $\sqrt{\frac{\mathrm{kQe}}{\mathrm{m}}}$

3. $\sqrt{\frac{\mathrm{kme}}{\mathrm{QR}}}$                       

4. $\sqrt{\frac{\mathrm{kQe}}{\mathrm{mR}}}$

The electric potential V as a function of distance x (in metre) is given by: $\mathrm{V}=\left(5{\mathrm{x}}^2+10\mathrm{x}-9\right)<mspace\; width="1em"></mspace>\mathrm{V}.$ The value of the electric field of x = 1m would be -

1.  $20<mspace\; width="1em"></mspace>\mathrm{V}/\mathrm{m}$                 

2.  $6<mspace\; width="1em"></mspace>\mathrm{V}/\mathrm{m}$

3.  $11<mspace\; width="1em"></mspace>\mathrm{V}/\mathrm{m}$                 

4.  $-23<mspace\; width="1em"></mspace>\mathrm{V}/\mathrm{m}$