A neutron star with magnetic moment of magnit | vedclass

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Question

$(a)$ Given, power radiated $P$ is

$P=\mu_{ 0 }^{x} m^{y} \omega^{z} c^{u}$

Substituting dimensions of different physical quantities involved, w

Vedclass · Accepted Answer

$x=1, y=2, z=4$ and $u=-3$

Vedclass · Answer

$(a)$ Given, power radiated $P$ is

$P=\mu_{ 0 }^{x} m^{y} \omega^{z} c^{u}$

Substituting dimensions of different physical quantities involved, we have $\left[ ML ^{2} T ^{-3}ight]=\left[ MLT ^{-2} A ^{-2}ight]^{x}\left[ L ^{2} A ight]^{y}\left[ T ^{-1}ight]^{z} \left[ LT ^{-1}ight]^{u}$

Equating powers of fundamental quantities, we have

$x=1 \ldots 	ext { (i) }$

$x+2 y+u =2 \ldots(ii)$

$-2 x-z-u =-3 \ldots(iii)$

$-2 x+y =0 \ldots(iv)$

From Eq. $(i)$, putting the value of $x$ in Eq. $(iv)$, we get

$\Rightarrow \quad-2 	imes 1+y=0 \Rightarrow y=2 \quad \ldots(v)$

Now, from E.q. $(i)$ and $(v)$, putting the values of $x$ and $y$ in Eq. $(ii)$, we get

$\Rightarrow \quad 1+2 	imes 2+u=2 \Rightarrow u=-3 \quad \ldots(vi)$

Now, again from Eqs. $(i)$ and $(vi)$, putting the values of $x$ and $u$ in Eq. $(iii)$, we get

$\Rightarrow-2 	imes 1-z+3=-3 \Rightarrow z=4$

So, $x=1, y=2, z=4, u=-3$

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