SATHEE: Properties of Matter 3 Question 5

Properties of Matter 3 Question 5

11. If $r = 5 \times 10^{- 4} m, ρ = 10^{3} k g m^{- 3}, g = 10 m s^{- 2}$ , $T = 0.11 N m^{- 1}$ , the radius of the drop when it detaches from the dropper is approximately

(a) $1.4 \times 10^{- 3} m$

(b) $3.3 \times 10^{- 3} m$

(d) $4.1 \times 10^{- 3} m$

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Answer:

Correct Answer: 11. $2 m$

Solution:

From equation of continuity $(A v =$ constant $)$

$\frac{π}{4} (8)^{2} (0.25) = \frac{π}{4} (2)^{2} (v)$

Here, $v$ is the velocity of water with which water comes out of the syringe (Horizontally).

Solving Eq. (i), we get $v = 4 m / s$

The path of water after leaving the syringe will be a parabola. Substituting proper values in equation of trajectory.

$y = x \tan θ - \frac{g x^{2}}{2 u^{2} \cos^{2} θ}$

According to question, we have,

$- 1.25 = R \tan 0^{\circ} - \frac{(10) (R^{2})}{(2) (4)^{2} \cos^{2} 0^{\circ}}$

$(R =$ horizontal range $)$

Solving this equation, we get $R = 2 m$

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Properties of Matter 3 Question 5

11. If $r = 5 \times 10^{- 4} m, ρ = 10^{3} k g m^{- 3}, g = 10 m s^{- 2}$ , $T = 0.11 N m^{- 1}$ , the radius of the drop when it detaches from the dropper is approximately

Answer:

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Properties of Matter 3 Question 5

11. If r=5×10−4m,ρ=103kgm−3,g=10ms−2, T=0.11Nm−1, the radius of the drop when it detaches from the dropper is approximately

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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11. If $r = 5 \times 10^{- 4} m, ρ = 10^{3} k g m^{- 3}, g = 10 m s^{- 2}$ , $T = 0.11 N m^{- 1}$ , the radius of the drop when it detaches from the dropper is approximately