$$\begin{gathered} \mathrm{A} \mathrm{battery} \mathrm{of} \mathrm{emf} \mathrm{of} 4\mathrm{V} \mathrm{gives} \mathrm{a} \mathrm{current} \mathrm{of} 0.2\mathrm{A} \mathrm{when} \mathrm{connected} \mathrm{across} \mathrm{a} \mathrm{wire} \mathrm{of} \mathrm{resistance} 18\mathrm{\Omega }. \\ T\mathrm{he} \mathrm{internal} \mathrm{resistance} of the battery is \end{gathered}$$

1.  $1\mathrm{\Omega }$

2.  $2\mathrm{\Omega }$

3.  $4\mathrm{\Omega }$

4.  $20\mathrm{\Omega }$

If a resistance coil is made by joining in parallel two resistances each of 20$\Omega$. An emf of 2V is applied across this coil for 100 seconds. The heat produced in the coil is

(1) 20 J

(2) 10 J

(3) 40 J

(4) 80 J

Which of the following graph correctly represents the variation of mobility $\left(\mathrm{\mu }\right)$ of electrons with applied electric field (E) in a metallic conductor?

1.		2.
3.		4.

The drift velocity of the electrons in a current-carrying  metallic conductor is of the order of 

1.  ${10}^4 \mathrm{m}/\mathrm{s}$

2.  ${10}^8 \mathrm{m}/\mathrm{s}$

3.  ${10}^0\mathrm{m}/\mathrm{s}$

4.  ${10}^{-4}\mathrm{m}/\mathrm{s}$

If I be the current limit of a fuse wire of length l and radius r, then select the appropriate relation

(1) $I \alpha l$

(2) $I \alpha r^2$

(3) $I \alpha l^0$

(4) $I \alpha r^0$

The sensitivity of a potentiometer can be increased by

1.  increasing the potential gradient

2.  decreasing the length of potentiometer wire

3.  decreasing the current in the potentiometer wire

4.  all of these

Three copper wires have their lengths in the ratio 5:3:1 and their masses are in the ratio 1:3:5. Their electrical resistance will be in the ratio of :

1. 5:3:1

2. 1:3:5

3. 125:15:1

4. 1:15:125

The current I as shown in the circuit will be:

1.  10 A

2.  $\frac{20}{3}$ $\mathrm{A}$ $$

3.  $\frac{2}{3}$ $\mathrm{A}$ $$ $$

4.  $\frac{5}{3}$ $\mathrm{A}$

The current in a wire varies with time according to the relation i= (3+2t) A. The amount of charge passing a cross section of the wire in the time interval t=0 to t=4.0 sec would be: (where t is time in seconds)

When the length and area of cross-section both are doubled, then its resistance :

(1) Will become half

(2) Will be doubled

(3) Will remain the same

(4) Will become four times