A) 71.23 dyne/cm
B) 72.13 dyne/cm
C) 31.27 dyne/cm
D) 21.37 dyne/cm
Correct Answer: B
Solution :
The ring is in contact with water along its inner and outer circumference so, when pulled out the total force on it due to surface tension will be |
\[F=T\left( 2p{{r}_{1}}+2p{{r}_{2}} \right)\] |
\[\Rightarrow T=\frac{mg}{2\pi \left( {{r}_{1}}+{{r}_{2}} \right)}\left[ \because F=mg \right]\] |
i.e. \[T=\frac{3.97\times 980}{3.14\times \left( 8.5+8.7 \right)}\] |
\[=72.13dyne/cm\] |
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