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### For an electromagnetic wave traveling in free space ....

For an electromagnetic wave traveling in free space, the relation between average energy densities due to electric $(U_e)$ and magnetic $(U_m)$ fields is:

(A) $U_e > U_m$
(B) $U_e = U_m$
(C) $U_e \neq U_m$
(D) $U_e < U_m$

Solution

We have, ${U_e} = \frac{1}{2}{ \in _0}{E^2}$ and ${U_m} = \frac{1}{2}\frac{{{B^2}}}{{{\mu _0}}}$

$\therefore \frac{{{U_e}}}{{{U_m}}} = {\mu _0}{ \in _0} \times \frac{{{E^2}}}{{{B^2}}}$

Using $\frac{E}{B} = c = \frac{1}{{\sqrt {{\mu _0}{ \in _0}} }}$,

$\frac{{{U_e}}}{{{U_m}}} = \frac{1}{{{c^2}}} \times {c^2} = 1$

### $f(x)=x^6+2x^4+x^3+2x+3 $$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$$n=?$
Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R$. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$