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  • Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A) : Time period of os

Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R).

Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if
density of the liquid is \rho and radius of the drop is \mathrm{r}, then \mathrm{T}=\mathrm{K} \sqrt{\rho \mathrm{r}^{3} / \mathrm{S}^{3/2}} is
dimensionally correct, where \mathrm{K} is dimensionless.
Reason (R): Using dimensional analysis we get R.H.S. having different dimension than that of time period.
In the light of above statements, choose the correct answer from the options given below.
 

Option: 1

Both (A) and (R) are true and (R) is the correct explanation of (A)
 


Option: 2

Both (A) and (R) are true but (R) is not the correct explanation of (A)
 


Option: 3

(A) is true but (R) is false
 


Option: 4

(A) is false but (R) is true


Answers (1)

best_answer

\mathrm{T =K \sqrt{\frac{\rho r^3}{S^{3 / 2}}} }

\mathrm{{[S] } =\left[\frac{F}{\ell}\right]=\left[M^1 L^0 T^{-2}\right] }

\mathrm{{[S] } =\left[M^{1} L^{-3} T^0\right] }

\mathrm{{[r] } =\left[L^{\prime}\right] }

\mathrm{{\left[\sqrt{\frac{\rho r^3}{S^{3 / 2}}}\right] } =\left[\rho^{1 / 2} r^{3 / 2} S^{-3 / 4}\right] }
                  \mathrm{=\left [ M^{1/2}L^{-3/2} \right ]\left [ L^{3/2} \right ]\times \left [ M^{-3/4}T^{-3/2} \right ]}

                   \mathrm{= \left [ M^{-1/4}T^{-3/2} \right ]}

\mathrm{\left [ \sqrt{\frac{\rho r^{3}}{S^{3/2}}} \right ]\neq \left [ T \right ]}

Using dimensional analysis we get RHS having different dimension than that of time period

\therefore \mathrm{\left ( A \right )} is false but \mathrm{\left ( R \right )} is true

Hence 4 is correct option.

Posted by

jitender.kumar

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