When a proton at rest is accelerated by a potential difference \(V\), its speed is found to be \(v\). The speed of an \(\alpha\text{-particle}\) when accelerated by the same potential difference from rest will be: 

1.	\(v\)	2.	\(v \over \sqrt{2}\)
3.	\(v \sqrt{2}\)	4.	\(2v\)

In the circuit shown in the figure, the energy stored in \(6~\mu\text{F}\) capacitor will be:
      

The figure shows some of the equipotential surfaces. Magnitude and direction of the electric field is given by

1. 200 V/m, making an angle ${120}^0$ with the x-axis

2. 100 V/m, pointing towards the negative x-axis

3. 200 V/m, making an angle $-{60}^0$ with the x-axis

4. 100 V/m, making an angle ${30}^0$ with the x-axis

An air capacitor of capacity $C=10\mu F$ is connected to a constant voltage battery of 12 V. Now the space between the plates is filled with a liquid of dielectric constant 5. The charge that flows now from battery to the capacitor is

1. 120 $\mu C$

2. 699 $\mu C$

3. 480 $\mu C$

4. 24 $\mu C$

A and B are two concentric metallic shells. If A is positively charged and B is earthed, then electric

1.  Field at common centre is non-zero

2.  Field outside B is nonzero

3.  Potential outside B is positive

4.  Potential at common centre is positive

$\mathrm{The} \mathrm{electric} \mathrm{potential} \mathrm{at} \mathrm{a} \mathrm{point} \mathrm{at} \mathrm{distance} \sqrt{3}\mathrm{R} \mathrm{from} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{disc}$
$\mathrm{of} \mathrm{radius} \mathrm{R} \mathrm{lying} \mathrm{in} \mathrm{the} \mathrm{axis} \mathrm{of} \mathrm{the} \mathrm{disc} \mathrm{whose} \mathrm{surface} \mathrm{charge} \mathrm{density} \mathrm{is} \mathrm{\sigma }$
$\mathrm{will} \mathrm{be} \mathrm{given} \mathrm{by}:$
$1. \frac{\sigma }{2{\epsilon }_0}\left[2-\sqrt{3}\right]R 2. \frac{\sigma }{2{\epsilon }_0}\left[2+\sqrt{3}\right]R$
$3.\frac{\sigma }{2{\epsilon }_0}\left[\sqrt{3}-\sqrt{2}\right]R 4. \frac{\sigma }{2{\epsilon }_0}\left[\sqrt{3}+\sqrt{2}\right]R$

An elementary particle of mass m and charge e is projected with velocity v at a much more massive particle of charge Ze, where Z>0. What is the closest possible approach of the incident particle ?

(1) $\frac{Z{e}^2}{2\pi {\epsilon }_{0}m{v}^2}$​

(2) $\frac{Ze}{4\pi {\epsilon }_{0}m{v}^2}$

(3) $\frac{Z{e}^2}{8\pi {\epsilon }_{0}m{v}^2}$​

(4) $\frac{Ze}{8\pi {\epsilon }_{0}m{v}^2}$

Two metallic spheres of radii 2cm and 3cm are given charges 6mC and 4mC respectively. The final charge on the smaller sphere will be if they are connected by a conducting wire

(1) 4mC

(2) 6mC

(3) 5mC

(4) 10mC

The equivalent capacitance between A and B is as the given figure:

1. \(16 \pi \epsilon_ 0 r\)
2. \(4 \pi \epsilon_ 0 r\)
3. \(8 \pi \epsilon_ 0 r\)
4. None of these

Capacitors $C_1=10\mu F and C_2=30\mu F$ are connected in series across a source of emf 20KV. The potential difference across $C_1$ will be

(1) 5 KV

(2) 15 KV

(3) 10 KV

(4) 20 KV