I-V graph for a metallic wire at two different temperature T₁ and T₂ is as shown in the figure below.

Which of the two temperatures is lower and why?

1

(a)

Since, slope of 1 > slope of 2
R₁ < R₂
\(\Rightarrow \) Also, we know that resistance is directly proportional to the temperature.
Therefore, T₂ > T_1.

Graphing showing the variation of current versus voltage for material Gaas is shown in the figure. Identify the region of

(i) negative resistance.
(ii) where Ohm's law is obeyed.

1

(a)

(i) DE is the region of negative resistance because the slope of curve in this part is negative.
(ii) BC is the region where Ohm's law is obeyed because in this part, the current varies linearly with the voltage.

In an experiment of meter bridge, the balancing length of the wire is l. What would be its value, if the radius of the meter bridge wire is double? Justify your answer.

1

(a)

The balancing length remains same as per relation.
\(\frac { R }{ S } =\frac { l }{ 100-l } \)
The balancing length is independent of radius bridge wire provided that it is uniform throughout.

The emf of a cell is always greater than its terminal voltage. Why?

1

(a)

The emf of a cell is greater than its terminal voltage because there is some potential drop across the cell due to its small internal resistance.

A 10 V battery of negligible internal resistance is connected across a 200 V battery and a resistance of 38 \(\Omega \) as shown in the figure. Find the value of the current in the circuit.

1

(a)

Since, the positive terminal of the batteries are connected together, so the equivalent emf of the batteries is given by E = 200 - 10 = 190 V
Hence, the current in the circuit is given 
\(I=\frac { E }{ R } =\frac { 190 }{ 38 } =5A\).