Given below are two statements: One is labeled as Assertion (A) and other is labeler as Reason (R). Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is and radius of the drop is r, then T = K ρ r 3 S 3 2 is dimensionally correct, where 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.

T=kρr3S3/2 [T] [ML−3]1/2 [L3]1/2 [MLT−2L−1]−34 [T] = [M1/2L−32L32M−34T32] = [M14T3/2]

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

Assertion (A) : Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is and radius of the drop is r, then

$T = K \sqrt[]{\frac{ρ r^{3}}{S^{\frac{3}{2}}}}$ is dimensionally correct, where 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

8 Views|Posted 6 months ago

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6 months ago

Correct Option - 4

Detailed Solution:

$T = k \sqrt[]{\frac{ρ r^{3}}{S^{3 / 2}}}$

$[T]$ ${[M L^{- 3}]}^{1 / 2} {[L^{3}]}^{1 / 2} {[M L T^{- 2} L^{- 1}]}^{\frac{- 3}{4}}$

$[T]$ $= [M^{1 / 2} L^{- \frac{3}{2}} L^{\frac{3}{2}} M^{- \frac{3}{4}} T^{\frac{3}{2}}]$

$= [M^{\frac{1}{4}} T^{3 / 2}]$

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