$f=f_{0} \sqrt{\frac{c+v}{c+v}}$ ; $V=$ relative speed of approach
$f_{0}=10\, \mathrm{GHz}$
$\therefore f=10 \sqrt{\frac{c+\frac{c}{2}}{c-\frac{c}{2}}}=10 \sqrt{3}=17.3\, \mathrm{GHz}$
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$1.$ In the graphs below, the resistance $\mathrm{R}$ of a superconductor is shown as a function of its temperature $\mathrm{T}$ for two different magnetic fields $\mathrm{B}_1$ (solid line) and $\mathrm{B}_2$ (dashed line). If $\mathrm{B}_2$ is larger than $\mathrm{B}_1$ which of the following graphs shows the correct variation of $\mathrm{R}$ with $\mathrm{T}$ in these fields?
MCQ $Image$
$2.$ A superconductor has $T_C(0)=100 \mathrm{~K}$. When a magnetic field of 7.5 Tesla is applied, its $\mathrm{T}_{\mathrm{c}}$ decreases to $75 \mathrm{~K}$. For this material one can definitely say that when
$(A)$ $\mathrm{B}=5$ Tesla, $\mathrm{T}_{\mathrm{c}}(\mathrm{B})=80 \mathrm{~K}$
$(B)$ $\mathrm{B}=5$ Tesla, $75 \mathrm{~K}<\mathrm{T}_{\mathrm{c}}(\mathrm{B})<100 \mathrm{~K}$
$(C)$ $\mathrm{B}=10 \mathrm{Tesla}, 75 \mathrm{~K}<\mathrm{T}_{\mathrm{c}}<100 \mathrm{~K}$
$(D)$ $\mathrm{B}=10$ Tesla, $\mathrm{T}_{\mathrm{c}}=70 \mathrm{~K}$
Give the answer question $1, 2$
