JEE MAIN - Physics (2020 - 3rd September Morning Slot - No. 3)
When a long glass capillary tube of radius
0.015 cm is dipped in a liquid, the liquid rises
to a height of 15 cm within it. If the contact angle
between the liquid and glass to close to 0o, the
surface tension of the liquid, in milliNewton m–1,
is [$$\rho $$(liquid) = 900 kgm–3, g = 10 ms–2]
(Give answer in closest integer) _____.
(Give answer in closest integer) _____.
Answer
101
Explanation
Capillary rise
h = $${{2T\cos \theta } \over {\rho gr}}$$
$$ \Rightarrow $$ T = $${{\rho grh} \over {2\cos \theta }}$$
= $${{\left( {900} \right)\left( {10} \right)\left( {15 \times {{10}^{ - 5}}} \right)\left( {15 \times {{10}^{ - 2}}} \right)} \over 2}$$
= 1012.5 $$ \times $$ 10–4
= 101.25 × 10–3 = 101.25 mN/m
$$ \simeq $$ 101.00 mN/m
h = $${{2T\cos \theta } \over {\rho gr}}$$
$$ \Rightarrow $$ T = $${{\rho grh} \over {2\cos \theta }}$$
= $${{\left( {900} \right)\left( {10} \right)\left( {15 \times {{10}^{ - 5}}} \right)\left( {15 \times {{10}^{ - 2}}} \right)} \over 2}$$
= 1012.5 $$ \times $$ 10–4
= 101.25 × 10–3 = 101.25 mN/m
$$ \simeq $$ 101.00 mN/m
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