Properties of Matter 4 Question 9
11. If $r=5 \times 10^{-4} \mathrm{~m}, \rho=10^3 \mathrm{~kg} \mathrm{~m}^{-3}, g=10 \mathrm{~ms}^{-2}$, $T=0.11 \mathrm{Nm}^{-1}$, the radius of the drop when it detaches from the dropper is approximately
(a) $1.4 \times 10^{-3} \mathrm{~m}$
(b) $3.3 \times 10^{-3} \mathrm{~m}$
(c) $2.0 \times 10^{-3} \mathrm{~m}$
(d) $4.1 \times 10^{-3} \mathrm{~m}$
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Answer:
Correct Answer: 11. (a)
Solution:
- $\frac{2 \pi r^{2} T}{R}=m g=\frac{4}{3} \pi R^{3} \cdot \rho \cdot g$
$\therefore \quad R^{4}=\frac{3 r^{2} T}{2 \rho g}=\frac{3 \times\left(5 \times 10^{-4}\right)^{2}(0.11)}{2 \times 10^{3} \times 10}$
$$ =4.125 \times 10^{-12} m^{4} $$
$\therefore R=1.425 \times 10^{-3} m \approx 1.4 \times 10^{-3} m$
$\therefore$ Correct option is (a).
