F = A cos Bx + C sin Dt
Then, the dimensions of $\frac { D } { B }$ are :-
- A $\mathrm { M } ^ { \circ } \mathrm { L } ^ { \circ } \mathrm { T } ^ { \circ }$
- B $\mathrm { M } ^ { \circ } \mathrm { L } ^ { \circ } \mathrm { T } ^ { - 1 }$
- C $\mathrm { M } ^ { \circ } \mathrm { L } ^ { - 1 } \mathrm {~T} ^ { \circ }$
- D $\mathrm { M } ^ { \circ } \mathrm { LT } ^ { - 1 }$ PPM043
Show answer & explanation
Correct answer: D
$\begin{array}{l}\mathrm{Bx}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~L}^{0} \Rightarrow \mathrm{BL}^{1}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \Rightarrow \mathrm{~B}=\mathrm{M}^{0} \mathrm{~L}^{-1} \mathrm{~T}^{0} \\ \mathrm{Dt}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \Rightarrow \mathrm{DT}^{1}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \Rightarrow \mathrm{D}=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1} \\ \frac{\mathrm{D}}{\mathrm{B}}=\frac{\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}}{\mathrm{M}^{0} \mathrm{~L}^{-1} \mathrm{~T}^{0}}=\mathrm{L}^{1} \mathrm{~T}^{-1}\end{array}$