A Nichrome wire X with length ( l ) and cross-sectional area ( A ) is connected to a 10 V source and another Nichrome wire Y with length (2 l) and…

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,

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

And

$\text R_\text Y = \text ρ \frac {2\text l}{\frac {\text A}{2}} =\text ρ \frac {4\text l}{\text A} = 4\text R_\text X$

As,

$\frac{\text R_\text Y}{\text R_\text X} = 4$

So, R_X : R_Y = 1 : 4

(b) $\text P_\text X = \frac{{\text V_\text X}^2}{\text R_\text X} = \frac{10^2}{\text R_\text X} = \frac{100}{\text R_\text X}$

And

$\text P_\text Y = \frac{{\text V_\text Y}^2}{\text R_\text Y} = \frac{20^2}{4\text R_\text X} = \frac{400}{4\text R_\text X} =\frac{100}{\text R_\text X} = \text P_\text X$

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 $\frac{\text A}{2}$ = 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_Y : m_X = 1 : 1

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