Solveeit Logo
Login

Question

Question: If r= \(5 \times {10}^{-4} m\), \(\rho= {10}^{3}kg{m}^{-3}\), g= \(10{m}/{s}^{2}\), T=\( 0.11N{m}^{-...

If r= 5×10−4m5 \times {10}^{-4} m, ρ=103kgm−3\rho= {10}^{3}kg{m}^{-3}, g= 10m/s210{m}/{s}^{2}, T=0.11Nm−1 0.11N{m}^{-1}, the radius of the drop when it detaches from the dropper is approximately
A. 1.4×10−3m1.4 \times {10}^{-3} m
B. 3.3×10−3m3.3 \times {10}^{-3} m
C. 2.0×10−3m2.0 \times {10}^{-3} m
D. 4.1×10−3m4.1 \times {10}^{-3} m

Explanation

Solution

To solve this problem, first use the expression for net vertical force due to surface time. Substitute the given values in that expression. Then, use the expression for weight of the liquid in the drop. Substitute the expression for volume of the liquid in the expression for weight of the liquid. Then substitute the values and get the weight of the liquid. Before detaching, the weight of the liquid was equal to the net vertical force due to surface tension. So, equate both the expressions and calculate the radius of the drop when it detaches from the drop per.

Complete answer:
Given: r= 5×10−4m5 \times {10}^{-4} m
ρ=103kgm−3\rho= {10}^{3}kg{m}^{-3}
g= 10m/s210{m}/{s}^{2}
T=0.11Nm−1 0.11N{m}^{-1}
Let R be the radius of drop when it detaches from the dropper
The net vertical force due to surface tension is given by,
F=T×2πr2RF= T \times \dfrac {2 \pi {r}^{2}}{R}
Now, substituting values in above equation we get,
F=0.11×2π×5×10−42×1RF= 0.11 \times 2\pi \times {5 \times {10}^{-4}} ^{2} \times \dfrac {1}{R}
⇒F=5.5×10−8R\Rightarrow F= \dfrac {5.5 \times {10}^{-8}}{R}
Let V be the volume of the liquid in the drop
Weight of the liquid in the drop is given by,
W=ρgVW= \rho g V ...(2)
Volume of the liquid in the drop is given by,
V=43πR3V = \dfrac {4}{3} \pi {R}^{3}
Substituting this expression in the equation. (2) we get,
W=ρg(43πR3)W= \rho g (\dfrac {4}{3} \pi {R}^{3})
Substituting the values in above equation we get,
W=103×10×43πR3W= {10}^{3} \times 10 \times \dfrac {4}{3} \pi {R}^{3}
⇒W=4.19×103×R3\Rightarrow W= 4.19 \times {10}^{3} \times {R}^{3}
Before detaching, the net vertical force due to surface tension is equal to the weight of the liquid in the drop.
⇒F=W\Rightarrow F= W
Substituting the values we get,
5.5×10−8R=4.19×103×R3\dfrac {5.5 \times {10}^{-8}}{R}= 4.19 \times {10}^{3} \times {R}^{3}
R4=4.135×10−12{R}^{4}= 4.135 \times {10}^{-12}
Taking out the square root on both the sides we get,
R=1.43×10−3R= 1.43 \times {10}^{-3}
Hence, the radius of the drop when it detached from the dropper is 1.4×10−3m1.4 \times {10}^{-3} m.

So, the correct answer is “Option A”.

Note:
Students should remember that the shape of a drop of water is always spherical. So, use the expression for volume of a sphere while calculating the volume of liquid. This shape is due to the surface tension. The surface tension binds the force between the molecules of water in such a way that these water drops remain in a spherical shape. Surface tension depends on the nature of the liquid, temperature, etc.