Properties Of Matter Ques 75
- If $r=5 \times 10^{-4} m, \rho=10^{3} kg m^{-3}, \quad g=10 ms^{-2}$, $T=0.11 Nm^{-1}$, the radius of the drop when it detaches from the dropper is approximately
(a) $1.4 \times 10^{-3} m$
(b) $3.3 \times 10^{-3} m$
(c) $2.0 \times 10^{-3} m$
(d) $4.1 \times 10^{-3} m$
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Answer:
Correct Answer: 75.$2 m$
Solution:
- From equation of continuity $(A v=$ constant $)$
$$ \frac{\pi}{4}(8)^{2}(0.25)=\frac{\pi}{4}(2)^{2}(v) $$
Here, $v$ is the velocity of water with which water comes out of the syringe (Horizontally).
Solving Eq. (i), we get $\quad v=4\ \text{m}/\text{s}$
The path of water after leaving the syringe will be a parabola. Substituting proper values in equation of trajectory.
$$ y=x \tan \theta-\frac{g x^{2}}{2 u^{2} \cos ^{2} \theta} $$
According to the question, we have,
$$ -1.25=R \tan 0^{\circ}-\frac{(10)\left(R^{2}\right)}{(2)(4)^{2} \cos ^{2} 0^{\circ}} $$
$(R=$ horizontal range $)$
Solving this equation, we get $R=2 m$
BETA
