You are traveling in a car during a thunderstorm. In order to protect yourself from lightning, you would prefer to

(1) Touch the nearest electric pole

(2) Take shelter under a tree

(3) Get out and crouch on the ground

(4) Remain in the car

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

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

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

In the circuit shown in figure, energy stored in 6 $\mathrm{\mu F}$ capacitor will be:
 

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 $\mu C$

Two identical metal plates are given charges $Q_1 and Q_2\left(<Q_1\right)$ respectively. If they are now brought close to form a parallel plate capacitor with capacitance $C$, the potential difference between them is :

1. $\frac{Q_1+Q_2}{2C}$

2. $\frac{Q_1+Q_2}{C}$

3. $\frac{Q_1-Q_2}{C}$

4. $\frac{Q_1-Q_2}{2C}$

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

The electric potential at a point at distance 'r' from a short dipole is proportional to

(1) $r^2$

(2) ${\mathrm{r}}^{-1}$

(3) $r^{-2}$

(4) $r^1$

The capacitance of a parallel plate capacitor is C. If a dielectric slab of thickness equal to one-fourth of the plate separation and dielectric constant K is inserted between the plates, then new capacitance become

1. $\frac{KC}{2\left(K+1\right)}$

2. $\frac{2KC}{K+1}$

3. $\frac{5KC}{4K+1}$

4. $\frac{4KC}{3K+1}$