Model Paper 1 — Question 13

(a) Both the graphs are correct because when the resistors are joined in series, the equivalent resistance is more than when they are joined in parallel.

Therefore,

(i) On V-I graph, the straight line obtained for a series combination will have higher slope than that for a parallel combination because it's slope gives resistance.

(ii) On I-V graph, the straight line obtained for a series combination will have less slope than that for a parallel combination because it's slope gives conductance which is reciprocal of resistance.

(b) Given,

Length of the wire is increased by 25%.

Let original length = L,

New length = L' = L + 0.25L = 1.25L

If length increases by 25%, to keep the volume constant, area must decrease.

So,

Volume = constant

Let original area be A and new area be A'.

⇒ L ⋅ A = L′ ⋅ A′

$\text A'= \dfrac{\text{L⋅A}}{\text L'}= \dfrac{\text{L⋅A}}{1.25\text L}= \dfrac{\text A}{1.25}=0.8\text A$

So new area = 0.8A.

As,

Original Resistance :

$\text R= \text ρ\dfrac{\text L}{\text A}$

New Resistance :

$\text R'= \text ρ\dfrac{\text L'}{\text A'}=\text ρ\dfrac{1.25\text L}{0.8\text A}=\text ρ\dfrac{\text L}{\text A}\times\dfrac{1.25}{0.8}=1.5625R$

% Increase in Resistance = (1.5625 − 1) × 100 % = 56.25 %

Change in Resistivity:

Resistivity (ρ) is a material property and it does not depend on length, area, or shape.

So,

% change in resistivity = 0%

(c) Although the same current flows through the live wire and the bulb filament, only the filament glows because it has very high resistance. This high resistance converts electrical energy into heat and light, making the filament white-hot and causing it to glow. On the other hand, the live wire has very low resistance, so it does not get heated up significantly and therefore does not glow.