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When water rises in a capillary tube of radius `r` to height h, then its potential energy `U_(1)` if capillary tube of radius 2r is dipped in same water then potential energy of water is `U_(2)` then `U_(1):U_(2)` will be
A. `1:1`
B. `1:2`
C. `2:1`
D. `1:4`

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Correct Answer - A
`(u_(1))/(u_(2)) = (m_(1)gh_(1))/(m_(2)gh_(2)) = (r_(1)n_(1))/(r_(2)n_(2))=(r_(1)r_(2))/(n_(2)r_(1))`
` (u_(1))/(u_(2))=1/1`

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