The temp at which the resistance of a conductor becomes 30% more than that of its resistance at 47°C will be:
(given the value of temperature coefficient of resistance of the conductor is $2×10^{-4}K^{-1}$)

1547 K

The correct answer is Option (3) → 1547 K

Resistance (R) is,

$R_T=R_0(1+αΔT)$

$⇒1.3R_0=R_0(1+αΔT)$

$⇒ΔT=\frac{1.3-1}{α}$

$⇒ΔT=\frac{0.3}{2×10^{-4}}=1500K$

$∴T=47+1500=1547K$