Question

The force F is given in terms of time t and displacement x by the equation$F{\text{ }} = {\text{ }}A{\text{ }}cosBx{\text{ }} + {\text{ }}C{\text{ }}sinDt$. The dimensional formula of D/B is______

A. ${\text{[}}{{\text{M}}^{\text{0}}}{{\text{L}}^{\text{0}}}{{\text{T}}^{\text{0}}}{\text{]}}$
B. ${\text{[}}{{\text{M}}^{\text{0}}}{{\text{L}}^{\text{0}}}{{\text{T}}^{ - 1}}{\text{]}}$
C. ${\text{[}}{{\text{M}}^{\text{0}}}{{\text{L}}^{ - 1}}{{\text{T}}^{\text{0}}}{\text{]}}$
D. ${\text{[}}{{\text{M}}^{\text{0}}}{{\text{L}}^1}{{\text{T}}^{ - 1}}{\text{]}}$

Answer 1

Dimensions play a key role in physics. Various quantities have various dimensions. Dimensions of the main fundamental quantities are M for mass while L for length T for time K for temperature while A for the current. Dimensional analysis is useful in solving various problems and checking for the validity of the options. We use one of the properties of dimensions here to solve this question.

Complete answer:
There are some rules which we use in case of problems involving dimension usage. Like in the case of any formula, dimensions of the left hand part of the equation must be the same as dimensions of the right hand part of the equation. We can add or subtract quantities of the same dimensions. Logarithm function doesn’t have any dimensions and all trigonometric functions don’t have any dimensions
and exponential functions also don’t have any dimensions.
If we have a product of two physical quantities and the dimension of one quantity is inverse of the dimension of the other quantity then that product will be dimensionless.
We use all these rules in dimensions to solve this.
The equation given is $F{\text{ }} = {\text{ }}A{\text{ }}cosBx{\text{ }} + {\text{ }}C{\text{ }}sinDt$
Sin function and cos function should be dimensionless.
So $Bx$ must be dimensionless and $Dt$ must be dimensionless.
Dimension of $x$ is L so dimension of $B$must be ${{\text{L}}^{{\text{ - 1}}}}$ and dimension of t is T so the dimension of D must be ${{\text{T}}^{{\text{ - 1}}}}$
Hence have
$\dfrac{{\text{D}}}{{\text{B}}}{\text{ = }}\dfrac{{{{\text{T}}^{{\text{ - 1}}}}}}{{{{\text{L}}^{{\text{ - 1}}}}}}{\text{ = [}}{{\text{M}}^{\text{0}}}{{\text{L}}^{\text{1}}}{{\text{T}}^{{\text{ - 1}}}}{\text{]}}$

So, the correct answer is “Option D”.

Note:
There are some quantities which have units while don't have dimensions. Few instances for that are plane angle which has unit or radian but no dimensions and solid angle which has a unit of steradian and no dimension and angular displacement also has unit of radian but no dimension.