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Let the junction temperature be $T$. In steady state, the heat current entering the junction equals the heat current leaving it:
$H_{cu} = H_{br} + H_{st}$
$\frac{K_c A(100 - T)}{L_c} = \frac{K_b A(T - 0)}{L_b} + \frac{K_s A(T - 0)}{L_s}$
Substituting the values:
$\frac{0.92 \times 4 \times (100 - T)}{46} = \frac{0.26 \times 4 \times T}{13} + \frac{0.12 \times 4 \times T}{12}$
$0.02(100 - T) = 0.02T + 0.01T \implies 2 - 0.02T = 0.03T \implies T = 40^\circ\text{C}$
Rate of heat flow in copper rod $H_{cu} = 0.02 \times 4 \times (100 - 40) = 0.08 \times 60 = 4.8 \text{ cal/s}$.
Let $\omega$ be a complex number with $2\omega+1=z$ where $z=\sqrt{-3}$. If $\begin{vmatrix}1&1&1\\1&-\omega^2-1&\omega^2\\1&\omega^2&\omega^7\end{vmatrix}=3k$, find $k$.
Since $z=\sqrt{-3}=i\sqrt{3}$ and $2\omega+1=i\sqrt{3}$, so $\omega=\frac{-1+i\sqrt{3}}{2}=e^{2\pi i/3}$ (a primitive cube root of unity).
Properties: $\omega^3=1$, $1+\omega+\omega^2=0$, so $-\omega^2-1=\omega$ and $\omega^7=\omega$.
Matrix becomes: $\begin{vmatrix}1&1&1\\1&\omega&\omega^2\\1&\omega^2&\omega\end{vmatrix}$
This determinant $= 3(\omega-\omega^2)\cdot... $ After evaluation $= 3i\sqrt{3} = 3z$... so $k=z$.
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