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A neutron star with magnetic moment of magnitude $\mathrm{m}$ is spinning with angular velocity $\omega$ about its magnetic axis. The electromagnetic power P radiated by it is given by $\mu_{0}^{\mathrm{x}} \mathrm{m}^{\mathrm{y}} \omega^{\mathrm{z}} \mathrm{c}^{\mathrm{u}}$ where $\mu_{0}$ and $\mathrm{c}$ are the permeability and speed of light in free space, respectively. Then
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The correct answer is:
$x=1, y=2, z=4$ and $u=-3$
$\mathrm{P}=\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-3}\right]$
$\mu_{0}=\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2} \mathrm{I}^{-2}\right]$
$\mathrm{m}=\left[\mathrm{I} \mathrm{L}^{2}\right]$
$\omega=\left[\mathrm{T}^{-1}\right]$
$\mathrm{C}=\left[\mathrm{L} \mathrm{T}^{-1}\right]$
$\therefore\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-3}\right]=\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2} \mathrm{I}^{-2}\right]^{\mathrm{x}}\left[\mathrm{I} \mathrm{L}^{2}\right]^{\mathrm{y}}\left[\mathrm{T}^{-1}\right]^{\mathrm{z}}\left[\mathrm{LT}^{-1}\right]^{\mathrm{l}}$
$\mathrm{x}=1, \mathrm{y}=2, \mathrm{z}=4, \mathrm{u}=-3$
$\mu_{0}=\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2} \mathrm{I}^{-2}\right]$
$\mathrm{m}=\left[\mathrm{I} \mathrm{L}^{2}\right]$
$\omega=\left[\mathrm{T}^{-1}\right]$
$\mathrm{C}=\left[\mathrm{L} \mathrm{T}^{-1}\right]$
$\therefore\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-3}\right]=\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2} \mathrm{I}^{-2}\right]^{\mathrm{x}}\left[\mathrm{I} \mathrm{L}^{2}\right]^{\mathrm{y}}\left[\mathrm{T}^{-1}\right]^{\mathrm{z}}\left[\mathrm{LT}^{-1}\right]^{\mathrm{l}}$
$\mathrm{x}=1, \mathrm{y}=2, \mathrm{z}=4, \mathrm{u}=-3$
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