A nichrome wire X with length (l) & cross-sectional area (A) is connected to a 10 V source and another nichrome wire Y with length (2l) & cross-sectional area (A/2), is connected to a 20 V source.

(a) Compare the resistances of wires X and Y. [Given that the resistivity of nichrome is (ρ).]

(b) Compare the electrical power consumed by each wire.

(c) Compare the masses of these wires. (Given that the density of nichrome is d.)

(d) State True or False : Wire X and wire Y both show the same rise in temperature in the same time.

Given,

• Length of nichrome wire X (l_X) = l
• Cross-sectional area of nichrome wire X (A_X) = A
• Potential difference across nichrome wire X (V_X) = 10 V
• Length of nichrome wire Y (l_Y) = 2l
• Cross-sectional area of nichrome wire Y (A_Y) = A/2
• Potential difference across nichrome wire Y (V_Y) = 20 V
• resistivity of nichrome = ρ

(a) Let 'R_X' and 'R_Y' be the resistance of nichrome wire X and Y respectively.

Now,

$\text R_\text X = \text ρ \dfrac {\text l}{\text A}$

And

$\text R$\text Y = \text ρ \dfrac {2\text l}{\dfrac {\text A}{2}} =\text ρ \dfrac {4\text l}{\text A} = 4\text R\text X

As,

$\dfrac{\text R$\text Y}{\text R\text X} = 4

So, R_X : R_Y = 1 : 4

(b) $\text P$\text X = \dfrac{{\text V\text X}^2}{\text R\text X} = \dfrac{10^2}{\text R\text X} = \dfrac{100}{\text R_\text X}

And

$\text P$\text Y = \dfrac{{\text V\text Y}^2}{\text R\text Y} = \dfrac{20^2}{4\text R\text X} = \dfrac{400}{4\text R\text X} =\dfrac{100}{\text R\text X} = \text P_\text X

So, P_X : P_Y = 1 : 1

(c) Given,

Mass density of nichrome = d

Then,

Mass of wire X (m_X) = d x volume = d x Al

And,

Mass of wire Y (m_Y) = d x volume = d x 2l x $\dfrac{\text A}{2}$ = d x Al = m_X

So, m_X : m_Y = 1 : 1

(d) True.

Reason —

Let, heat gained by the wire X be 'Q_X' and change in temperature be 'ΔT_X'. Similarly, heat gained by the wire Y be 'Q_Y' and change in temperature be 'ΔT_Y'.

Again, let specific heat capacity of nichrome be 'c' and same time be 't'.

Then,

Q_X = P_X x t = m_Xc x ΔT_X

and

Q_Y = P_Y x t = m_Yc x ΔT_Y

As, P_X = P_Y then Q_X = Q_Y for same time t.

⇒ m_Xc x ΔT_X = m_Yc x ΔT_Y

⇒ m_X x ΔT_X = m_Y x ΔT_Y

So,

ΔT_X : ΔT_Y = m_X : m_Y = 1 : 1

Hence, wire X and wire Y both show the same rise in temperature in the same time.