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

An uncharged capacitor with a solid dielectric is connected to a similar air capacitor charged to a potential of ${\mathrm{V}}_0$. If the common potential after sharing of charges becomes $\mathrm{V}$, then the dielectric constant of the dielectric must be –

1. $\frac{{\mathrm{V}}_0}{\mathrm{V}}$                               

2. $\frac{\mathrm{V}}{{\mathrm{V}}_0}$

3. $\frac{{\mathrm{V}}_0-\mathrm{V}}{\mathrm{V}}$                         

4. $\frac{{\mathrm{V}}_0-\mathrm{V}}{{\mathrm{V}}_0}$

Two parallel conducting plates $5\mathrm{mm}$ apart are held horizontally one above the other. The upper plate is maintained at a positive potential of $15<mspace\; width="1em"></mspace>\mathrm{kV}$ while the lower plate is earthed. If a small oil drop of relative density $0.92$ and of radius $5<mspace\; width="1em"></mspace>\mu \mathrm{m}$ remains stationary between the plates, then the charge on the drop will be.

1.  $10\mathrm{e}$                     

2.  $8\mathrm{e}$ 

3.  $5\mathrm{e}$                       

4.  $3\mathrm{e}$

An infinite number of charges (each of magnitude $1<mspace\; width="1em"></mspace>\mu \mathrm{c}$) are placed along the $\mathrm{X}$-axis at x = 1, 2, 4, 8....metre. If the charges are alternately of opposite sign, then the potential at the point x = 0 due to these charges will be.

1.  $6\times {10}^3<mspace\; width="1em"></mspace>\mathrm{V}$                       

2.  $9\times {10}^3<mspace\; width="1em"></mspace>\mathrm{V}$

3.  $1.8\times {10}^4<mspace\; width="1em"></mspace>\mathrm{V}$                   

4.  $1.2\times {10}^4<mspace\; width="1em"></mspace>\mathrm{V}$

Figure shows a ball having a charge \(q\) fixed at a point $\mathrm{A}$. Two identical balls having charges \(+q\) and \(–q\) and mass \(‘m’\) each are attached to the ends of a light rod of length $\mathrm{}$\(2 a\). The rod is free to rotate about a fixed axis perpendicular to the plane of the paper and passing through the mid-point of the rod. The system is released from the situation as shown in the figure. The angular velocity of the rod when the rod becomes horizontal will be:

A proton and an $\mathrm{\alpha }$-particle are at a distance r from each other. After letting them free if they move to infinity, the kinetic energy of the proton will be -

1.  $8{\mathrm{Ke}}^2/5\mathrm{r}$                           

2.  $2{\mathrm{Ke}}^2/5\mathrm{r}$

3.  $8{\mathrm{Ke}}^2/\mathrm{r}$                             

4.  ${\mathrm{Ke}}^2/5\mathrm{r}$

${10}^3$ small water drops, each of radius r and each carrying charge q, combine to form one bigger drop. The potential of a bigger drop as compared to that of a smaller drop will be.

1.  ${10}^5<mspace\; width="1em"></mspace>\mathrm{V}$                     

2.  ${10}^3<mspace\; width="1em"></mspace>\mathrm{V}$

3.  $10<mspace\; width="1em"></mspace>\mathrm{V}$                       

4.  ${10}^2<mspace\; width="1em"></mspace>\mathrm{V}$

The plates of a parallel plate capacitor have an area of $90<mspace\; width="1em"></mspace>{\mathrm{cm}}^2$ each and are separated by $2.5<mspace\; width="1em"></mspace>\mathrm{mm}$. The capacitor is charged by a 400 volt supply. How much electrostatic energy is stored by the capacitor?

1.  $2.55\times {10}^{-6}<mspace\; width="1em"></mspace>\mathrm{J}$                         

2.  $1.55\times {10}^{-6}<mspace\; width="1em"></mspace>\mathrm{J}$

3.  $8.15\times {10}^{-6}<mspace\; width="1em"></mspace>\mathrm{J}$                         

4.  $5.5\times {10}^{-6}<mspace\; width="1em"></mspace>\mathrm{J}$

An alpha particle with kinetic energy $10<mspace\; width="1em"></mspace>\mathrm{MeV}$ is heading towards a stationary nucleus of atomic number $50$. Calculate the distance of closest approach.

1.  $1.4\times {10}^{-15}<mspace\; width="1em"></mspace>\mathrm{m}$                         

2.  $4\times {10}^{-15}<mspace\; width="1em"></mspace>\mathrm{m}$

3.  $14.4\times {10}^{-15}<mspace\; width="1em"></mspace>\mathrm{m}$                       

4.  $8\times {10}^{-15}<mspace\; width="1em"></mspace>\mathrm{m}$