At room temperature \(\left(27^{\circ} \mathrm{C}\right)\), the resistance of a heating element is \(50 \Omega\). The temperature coefficient of the material is \(2.4 \times 10^{-4}{ }^{\circ} \mathrm{C}^1\). The temperature of the element, when its resistance is \(62 \Omega\) is _______ \({ }^{\circ} \mathrm{C}\).

Explanation:
The variation of electrical resistance with temperature is given by the formula:
\( R = R_0(1 + \alpha \Delta T) \)

Where:
\( R = \) resistance at final temperature \( T = 62\ \Omega \)
\( R_0 = \) resistance at initial temperature \( T_0 = 50\ \Omega \)
\( \alpha = \) temperature coefficient of resistance \( = 2.4 \times 10^{-4}\text{ }^{\circ}\text{C}^{-1} \)
\( \Delta T = \) change in temperature \( = T - T_0 \)

Rearranging the terms to solve for \( \Delta T \):
\( \frac{R}{R_0} = 1 + \alpha \Delta T \)
\( \frac{R}{R_0} - 1 = \alpha \Delta T \)
\( \frac{R - R_0}{R_0} = \alpha \Delta T \)

Substituting the given values into the equation:
\( \frac{62 - 50}{50} = (2.4 \times 10^{-4}) \times \Delta T \)
\( \frac{12}{50} = (2.4 \times 10^{-4}) \times \Delta T \)
\( 0.24 = (2.4 \times 10^{-4}) \times \Delta T \)
\( \Delta T = \frac{0.24}{2.4 \times 10^{-4}} \)
\( \Delta T = 1000^{\circ}\text{C} \)

Since \( \Delta T = T - T_0 \):
\( T - 27^{\circ}\text{C} = 1000^{\circ}\text{C} \)
\( T = 1000 + 27 = 1027^{\circ}\text{C} \)

Therefore, option B is the correct choice.