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Force F is given as F = X cos(Pt) + Y sin(Qs). If t and s are time and distance then find the dimension of \(\frac{P}{X} and \frac{Q}{Y}.\)
1. M-1L-1T1, M-1L-2T2
2. M2LT, MLT
3. ML2T, MLT
4. MLT2, MLT

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Correct Answer - Option 1 : M-1L-1T1, M-1L-2T2

CONCEPT:

Dimensions:

  • Dimensions of a physical quantity are the powers to which the fundamental units are raised to obtain one unit of that quantity.

EXPLANATION:

​Given F = X cos(Pt) + Y sin(Qs), t = time and s = distance

The dimension of F, t and s is given as,

⇒ [F] = [M1L1T-2]

⇒ [t] = [M0L0T1]

⇒ [s] = [M0L1T0]

  • Since the trigonometric function is a dimensionless quantity. So, Pt and Qs is also a dimensionless quantity.

So,

⇒ [P][t] = [M0L0T0]

⇒ [P][M0L0T1] = [M0L0T0]

⇒ [P] = [M0L0T-1]     -----(1)

⇒ [Q][s] = [M0L0T0]

⇒ [Q][M0L1T0] = [M0L0T0]

⇒ [Q] = [M0L-1T0]     -----(2)

  • The dimension of the F, X and Y must be equal.

⇒ [F] = [X] = [Y] = [M1L1T-2]     -----(3)

By equation 1 and equation 3,

\(\Rightarrow \frac{\left [ P \right ]}{\left [ X \right ]}=\frac{\left [ M^{0}L^{0}T^{-1} \right ]}{\left [ M^{1}L^{1}T^{-2} \right ]}\)

\(\Rightarrow \frac{\left [ P \right ]}{\left [ X \right ]}=\left [ M^{-1}L^{-1}T^{1} \right ]\)

By equation 1 and equation 3,

\(\Rightarrow \frac{\left [ Q \right ]}{\left [ Y \right ]}=\frac{\left [ M^{0}L^{-1}T^{0} \right ]}{\left [ M^{1}L^{1}T^{-2} \right ]}\)

\(\Rightarrow \frac{\left [ P \right ]}{\left [ X \right ]}=\left [ M^{-1}L^{-2}T^{2} \right ]\)

  • Hence, option 1 is correct.

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