- A$12$
- B$18$
- ✓$15$
- D$20$
$=5 \times 10^{9} Hz$
Relative permittivity, $\in_{ r }=2$
and Relative permeability, $\mu_{ r }=2$
Since speed of light in a medium is given by,
$v =\frac{1}{\sqrt{\mu \in}}=\frac{1}{\sqrt{\mu_{ s } \mu_{0} \cdot \in_{ s } \in_{0}}}$
$v =\frac{1}{\sqrt{\mu_{ s } \in_{ x }}} \frac{1}{\sqrt{\mu_{0} \in_{0}}}=\frac{ C }{\sqrt{\mu_{ s } \in_{ r }}}$
Where $C$ is speed of light is vacuum.
$\therefore v =\frac{3 \times 10^{8}}{\sqrt{4}}=\frac{30 \times 10^{7}}{2} m / s$
$=15 \times 10^{7} m / s$
$\therefore$ Ans. is $15$
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(Round off to the nearest integer) $\left[{h}=4.14 \times 10^{-15} \,{eVs}, {c}=3 \times 10^{8}\, {ms}^{-1}\right.]$
Reason $(R):$ Electrostatic force of repulsion between two protons reduces net nuclear forces between them.