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1.Units, Dimensions and Measurement
medium
The force is given in terms of time $t$ and displacement $x$ by the equation
${F}={A} \cos {Bx}+{C} \sin {Dt}$
The dimensional formula of $\frac{{AD}}{{B}}$ is -
A$\left[{ML}^{2} {T}^{-3}\right]$
B$\left[{M}^{2} L^{2} {T}^{-3}\right]$
C$\left[{M}^{1} {L}^{1} {T}^{-2}\right]$
D$\left[{M}^{0} {LT}^{-1}\right]$
(JEE MAIN-2021)
Solution
$\theta$ of cos or sin should be dimensionless.
${[ A ]=\left[ MLT ^{-2}\right] }$
${[ B ]=\left[ L ^{-1}\right] }$
${[ D ]=\left[ T ^{-1}\right] }$
${\left[\frac{ AD }{ B }\right]=\frac{\left[ MLT ^{-2}\right]\left[ T ^{-1}\right]}{\left[ L ^{-1}\right]} }$
${\left[\frac{ AD }{ B }\right]=\left[ ML ^{2} T ^{-3}\right] }$
${[ A ]=\left[ MLT ^{-2}\right] }$
${[ B ]=\left[ L ^{-1}\right] }$
${[ D ]=\left[ T ^{-1}\right] }$
${\left[\frac{ AD }{ B }\right]=\frac{\left[ MLT ^{-2}\right]\left[ T ^{-1}\right]}{\left[ L ^{-1}\right]} }$
${\left[\frac{ AD }{ B }\right]=\left[ ML ^{2} T ^{-3}\right] }$
Standard 11
Physics
