Physics And Measurement Question 203

Question: Force F is given in terms of time $ t $ and distance $ x $ by F =A sin C $ t $ + B cos $ Dx $ . Then the dimensions of $ A/B $ and $ C/D $ are

Options:

A) $ [M _{{}}^{0}L _{{}}^{0}T _{{}}^{0}],[M _{{}}^{0}L _{{}}^{0}T _{{}}^{-1}] $

B) $ [M _{{}}^{{}}L _{{}}^{{}}T _{{}}^{-2}],[M _{{}}^{0}L _{{}}^{-1}T _{{}}^{0}] $

C) $ [M _{{}}^{0}L _{{}}^{0}T _{{}}^{0}],[M _{{}}^{0}L _{{}}^{{}}T _{{}}^{-1}] $

D) $ [M _{{}}^{0}L _{{}}^{1}T _{{}}^{-1}],[M _{{}}^{0}L _{{}}^{0}T _{{}}^{0}] $

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Answer:

Correct Answer: C

Solution:

[c] $ \frac{A}{B}=\frac{Force}{Force}=[M^{0}L^{0}T^{0}] $ Ct = angle

$ \Rightarrow $ $ C=\frac{Angle}{Time}=\frac{1}{T}={{T}^{-1}} $ Dx = angle

$ \Rightarrow $ $ D=\frac{Angle}{Dis\tan ce}=\frac{1}{L}={{L}^{-1}} $

$ \therefore \frac{C}{D}=\frac{{{T}^{-1}}}{{{L}^{-1}}}=[M^{0}L{{T}^{-1}}] $



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