Q152 / 160 · AP EAMCET 2019 (20 Apr Shift 2) · Physics
A metal rod is subjected to cycles of magnetisation at the rate of $42 \mathrm{~Hz}$. Density of the metal is $6 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ and its specific heat capacity is $0.1 \times 10^{-3} \mathrm{cal} \mathrm{kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$. If the area of its $B-H$ loop corresponds to energy density of $10^{-2} \mathrm{Jm}^{-3}$, then the rise in its temperature in one minute is
- A$5^{\circ} \mathrm{C}$
- Correct answerB$10^{\circ} \mathrm{C}$
- C$15^{\circ} \mathrm{C}$
- D$20^{\circ} \mathrm{C}$
Solution
Energy of area of $B-H$ loop,
$\begin{array}{lll} <br />\Delta Q=m s(\Delta \theta) \\<br />\Rightarrow 10^{-2} \times 42 \times 60=6 \times 10^3 \times 0.1 \times 10^{-3} \times 4.2 \times \Delta \theta \\<br />\Rightarrow \Delta \theta=\frac{10^{-2} \times 42 \times 60}{6 \times 10^3 \times 0.1 \times 10^{-3} \times 4.2} \\<br />\Rightarrow \Delta \theta=10^{\circ} \mathrm{C}<br />\end{array}$
$\begin{array}{lll} <br />\Delta Q=m s(\Delta \theta) \\<br />\Rightarrow 10^{-2} \times 42 \times 60=6 \times 10^3 \times 0.1 \times 10^{-3} \times 4.2 \times \Delta \theta \\<br />\Rightarrow \Delta \theta=\frac{10^{-2} \times 42 \times 60}{6 \times 10^3 \times 0.1 \times 10^{-3} \times 4.2} \\<br />\Rightarrow \Delta \theta=10^{\circ} \mathrm{C}<br />\end{array}$
Hence, the correct answer is option: 'B'.