JEE MAIN - Physics (2019 - 10th April Evening Slot - No. 21)
In the formula X = 5YZ2
, X and Z have dimensions of capacitance and magnetic field, respectively. What are
the dimensions of Y in SI units?
[M–3L–2T8A4]
[M–2L–2T6A3]
[M–1L–2T4A2]
[M–2L0 T–4A–2]
Explanation
Capacitance (C) = $${{{Q^2}} \over {2E}}$$ = $${{\left[ {{A^2}{T^2}} \right]} \over {\left[ {M{L^2}{T^{ - 2}}} \right]}}$$ = $$\left[ {{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}} \right]$$ X
Magnetic field (B) = $${F \over {IL}}$$ = $${{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ A \right]\left[ L \right]}}$$ = $${\left[ {M{T^{ - 2}}{A^{ - 1}}} \right]}$$ = Z
Given,
X = 5YZ2
$$ \therefore $$ Y = $${X \over {5{Z^2}}}$$
[Y] = $${{\left[ {{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}} \right]} \over {{{\left[ {M{T^{ - 2}}{A^{ - 1}}} \right]}^2}}}$$
$$ \therefore $$ [Y] = [M–3L–2T8A4]
Magnetic field (B) = $${F \over {IL}}$$ = $${{\left[ {ML{T^{ - 2}}} \right]} \over {\left[ A \right]\left[ L \right]}}$$ = $${\left[ {M{T^{ - 2}}{A^{ - 1}}} \right]}$$ = Z
Given,
X = 5YZ2
$$ \therefore $$ Y = $${X \over {5{Z^2}}}$$
[Y] = $${{\left[ {{M^{ - 1}}{L^{ - 2}}{T^4}{A^2}} \right]} \over {{{\left[ {M{T^{ - 2}}{A^{ - 1}}} \right]}^2}}}$$
$$ \therefore $$ [Y] = [M–3L–2T8A4]
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