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Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by $63 \times 10^{-4} \mathrm{~N}$ force due to the weight of the water. The surface tension of water is $7 \times 10^{-2} \mathrm{~N} / \mathrm{m}$. The inner diameter of the capillary tube is nearly $(\pi=22 / 7)$A)$6.3 \times 10^{-1} \mathrm{~m}$B)$3 \times 10^{-2} \mathrm{~m}$C)$7 \times 10^{-2} \mathrm{~m}$D)$9 \times 10^{-2} \mathrm{~m}$
Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by $63 \times 10^{-4} \mathrm{~N}$ force due to the weight of the water. The surface tension of water is $7 \times 10^{-2} \mathrm{~N} / \mathrm{m}$. The inner diameter of the capillary tube is nearly $(\pi=22 / 7)$
A)$6.3 \times 10^{-1} \mathrm{~m}$
B)$3 \times 10^{-2} \mathrm{~m}$
C)$7 \times 10^{-2} \mathrm{~m}$
D)$9 \times 10^{-2} \mathrm{~m}$
Physics
Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by $63 \times 10^{-4} \mathrm{~N}$ force due to the weight of the water. The surface tension of water is $7 \times 10^{-2} \mathrm{~N} / \mathrm{m}$. The inner diameter of the capillary tube is nearly $(\pi=22 / 7)$
A)$6.3 \times 10^{-1} \mathrm{~m}$
B)$3 \times 10^{-2} \mathrm{~m}$
C)$7 \times 10^{-2} \mathrm{~m}$
D)$9 \times 10^{-2} \mathrm{~m}$

Correct answer is B.

Upwards force due to surface tension,
$F_{u}=(2 \pi R) \times T$
where, $2 \pi R \rightarrow$ circumference of tube
$F_{u}=(2 \pi R) \times T$
Because surface tension is force per unit length,
$\begin{array}{l}
\Rightarrow 75 \times 10^{-4}=(2 \pi R) \times 6 \times 10^{-2} \\
(2 R)=\frac{75 \times 10^{-4}}{6 \times 10^{-2}\times \pi}=3 \times 10^{-2} \mathrm{~m}
\end{array}$

i

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