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1.Units, Dimensions and Measurement
medium
The dimension of $P = \frac{{{B^2}{l^2}}}{m}$ is
where $B=$ magnetic field, $l=$ length, $m =$ mass
A$ML{T^{ - 3}}$
B$M{L^2}{T^{ - 4}}I^{-2}$
C${M^2}{L^2}{T^{ - 4}}I$
D$ML{T^{ - 2}}{I^{ - 2}}$
Solution
$F = BIL$
$\therefore \,\,{\rm{Dimension}}\,{\rm{of}}\,{\rm{[}}B{\rm{]}} = \frac{{{\rm{[}}F{\rm{]}}}}{{{\rm{[}}I{\rm{]}}\,{\rm{[}}L{\rm{]}}}} = \frac{{[ML{T^{ – 2}}]}}{{[I]\,[L]}}$=$[M{T^{ – 2}}{I^{ – 1}}]$
$[P] = \frac{{{B^2}{l^2}}}{m} = \frac{{{{[M{T^{ – 2}}{I^{ – 1}}]}^2} \times [{L^2}]}}{{[M]}}$ $ = [M{L^2}{T^{ – 4}}{I^{ – 2}}]$
$\therefore \,\,{\rm{Dimension}}\,{\rm{of}}\,{\rm{[}}B{\rm{]}} = \frac{{{\rm{[}}F{\rm{]}}}}{{{\rm{[}}I{\rm{]}}\,{\rm{[}}L{\rm{]}}}} = \frac{{[ML{T^{ – 2}}]}}{{[I]\,[L]}}$=$[M{T^{ – 2}}{I^{ – 1}}]$
$[P] = \frac{{{B^2}{l^2}}}{m} = \frac{{{{[M{T^{ – 2}}{I^{ – 1}}]}^2} \times [{L^2}]}}{{[M]}}$ $ = [M{L^2}{T^{ – 4}}{I^{ – 2}}]$
Standard 11
Physics
