
A narrow tube of radius $1mm$ made of glass is dipped mercury. By what amount does the mercury dipped down in the relative to the mercury surface outside. Surface tension of mercury is $0.465N{m^{ - 1}}$, angle of contact $140^\circ $ and density of mercury $13.6 \times {10^3}kg{m^{ - 3}}$.
Answer
491.1k+ views
Hint: We have to see the values which are given to us now using the surface tension formula which is related to the angle of contact and the dip height of the narrow tube. Then after some rearrangement and calculation we can find the value of the height.
Complete step by step answer:
As per the problem we have a narrow tube of radius $1mm$ made of glass is dipped mercury having surface tension of mercury is $0.465N{m^{ - 1}}$, angle of contact $140^\circ $ and density of mercury $13.6 \times {10^3}kg{m^{ - 3}}$.
We know,
Surface tension formula for a narrow tube is given as,
$s = \dfrac{{h\rho gr}}{{2\cos \theta }}$
Where,
Surface tension of mercury is $0.465N{m^{ - 1}}$.
Angle of contact, $\theta $ is $140^\circ $.
Density of mercury, $\rho $ is $13.6 \times {10^3}kg{m^{ - 3}}$.
Acceleration due to gravity, g is $9.8m{s^{ - 2}}$.
Radius of the narrow tube, r is $1mm = {10^{ - 3}}m$.
Now putting the respective values in the above surface tension formula we will get,
$0.465N{m^{ - 1}} = \dfrac{{h\left( {13.6 \times {{10}^3}kg{m^{ - 3}}} \right)\left( {9.8m{s^{ - 2}}} \right)\left( {{{10}^{ - 3}}m} \right)}}{{2\cos 140^\circ }}$
Rearranging the above equation we will get,
$\dfrac{{0.465N{m^{ - 1}} \times 2\cos 140^\circ }}{{\left( {13.6 \times {{10}^3}kg{m^{ - 3}}} \right)\left( {9.8m{s^{ - 2}}} \right)\left( {{{10}^{ - 3}}m} \right)}} = h$
$ \Rightarrow h = \dfrac{{0.465 \times \left( { - 1.532} \right)}}{{\left( {13.6} \right)\left( {9.8} \right)}}m = \dfrac{{ - 0.712}}{{133.28}}m$
Now on simplifying the above equation we will get,
$h = - 0.00534m$
Note: Surface tension is the tendency of liquid surfaces or we can say the property of the surface of liquid that allows it to resist the external force. Remember that this phenomena can be observed in the spherical shape of small water drops of any kind of liquid. Examples of this phenomenon are soap bubbles, others are insects walking on water and floating of needles on the surface of water.
Complete step by step answer:
As per the problem we have a narrow tube of radius $1mm$ made of glass is dipped mercury having surface tension of mercury is $0.465N{m^{ - 1}}$, angle of contact $140^\circ $ and density of mercury $13.6 \times {10^3}kg{m^{ - 3}}$.
We know,
Surface tension formula for a narrow tube is given as,
$s = \dfrac{{h\rho gr}}{{2\cos \theta }}$
Where,
Surface tension of mercury is $0.465N{m^{ - 1}}$.
Angle of contact, $\theta $ is $140^\circ $.
Density of mercury, $\rho $ is $13.6 \times {10^3}kg{m^{ - 3}}$.
Acceleration due to gravity, g is $9.8m{s^{ - 2}}$.
Radius of the narrow tube, r is $1mm = {10^{ - 3}}m$.
Now putting the respective values in the above surface tension formula we will get,
$0.465N{m^{ - 1}} = \dfrac{{h\left( {13.6 \times {{10}^3}kg{m^{ - 3}}} \right)\left( {9.8m{s^{ - 2}}} \right)\left( {{{10}^{ - 3}}m} \right)}}{{2\cos 140^\circ }}$
Rearranging the above equation we will get,
$\dfrac{{0.465N{m^{ - 1}} \times 2\cos 140^\circ }}{{\left( {13.6 \times {{10}^3}kg{m^{ - 3}}} \right)\left( {9.8m{s^{ - 2}}} \right)\left( {{{10}^{ - 3}}m} \right)}} = h$
$ \Rightarrow h = \dfrac{{0.465 \times \left( { - 1.532} \right)}}{{\left( {13.6} \right)\left( {9.8} \right)}}m = \dfrac{{ - 0.712}}{{133.28}}m$
Now on simplifying the above equation we will get,
$h = - 0.00534m$
Note: Surface tension is the tendency of liquid surfaces or we can say the property of the surface of liquid that allows it to resist the external force. Remember that this phenomena can be observed in the spherical shape of small water drops of any kind of liquid. Examples of this phenomenon are soap bubbles, others are insects walking on water and floating of needles on the surface of water.
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