physics : class11 Excess Pressure in a Liquid Jet : Surface Tension : Properties of Bulk Matter MCQ Questions and Answers
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Excess Pressure in a Liquid Jet
MCQ questions and answers.
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MCQ Questions
1. A cylindrical liquid jet of radius 0.50 mm is formed in air. If the surface tension of the liquid is 0.072 N m−1, what is the excess pressure inside the jet over the surrounding air?
A.
72 Pa
B.
144 Pa
C.
288 Pa
D.
36 Pa
Correct Answer:
144 Pa
Explanation:
For a cylindrical liquid jet, the excess pressure is ΔP = T/r. Thus, ΔP = 0.072/(0.50 × 10−3) = 144 Pa.
2. A liquid jet has an excess pressure of 360 Pa when its surface tension is 0.060 N m−1. What is the radius of the jet?
A.
0.083 mm
B.
0.167 mm
C.
0.333 mm
D.
1.67 mm
Correct Answer:
0.167 mm
Explanation:
For a liquid jet, ΔP = T/r, so r = T/ΔP = 0.060/360 = 1.67 × 10−4 m = 0.167 mm.
3. Two cylindrical jets of the same liquid have radii r and 3r. What is the ratio of their excess pressures, ΔP(r) : ΔP(3r)?
A.
1 : 3
B.
3 : 1
C.
1 : 9
D.
9 : 1
Correct Answer:
3 : 1
Explanation:
Since the excess pressure in a liquid jet is ΔP = T/r, pressure is inversely proportional to radius. Therefore, ΔP(r)/ΔP(3r) = 3.
4. A liquid jet of radius 0.20 mm has an excess pressure of 500 Pa. If its radius is reduced to 0.125 mm without changing the liquid, what will be its new excess pressure?
A.
312.5 Pa
B.
625 Pa
C.
800 Pa
D.
1000 Pa
Correct Answer:
800 Pa
Explanation:
For the same liquid, ΔP1r1 = ΔP2r2. Hence, ΔP2 = 500 × (0.20/0.125) = 800 Pa.
5. Which force-balance relation correctly gives the excess pressure inside a cylindrical liquid jet of radius r and length L, having surface tension T?
A.
ΔP(πrL) = 2TL
B.
ΔP(2rL) = 2TL
C.
ΔP(2πrL) = 2TL
D.
ΔP(2rL) = πTL
Correct Answer:
ΔP(2rL) = 2TL
Explanation:
On taking a longitudinal half of the jet, the pressure force is ΔP × 2rL, while surface tension acts along the two cut edges with total force 2TL. Therefore, ΔP(2rL) = 2TL and ΔP = T/r.
6. The diameter of a cylindrical liquid jet is 1.0 mm and its surface tension is 0.050 N m−1. What is the excess pressure inside the jet?
A.
25 Pa
B.
50 Pa
C.
100 Pa
D.
200 Pa
Correct Answer:
100 Pa
Explanation:
Since r = d/2, the jet relation can be written as ΔP = 2T/d. Thus, ΔP = 2(0.050)/(1.0 × 10−3) = 100 Pa.
7. Two liquid jets have the same radius. The surface tension of liquid X is twice that of liquid Y. What is the ratio of their excess pressures, ΔPX : ΔPY?
A.
1 : 2
B.
2 : 1
C.
1 : 4
D.
4 : 1
Correct Answer:
2 : 1
Explanation:
For jets of equal radius, ΔP = T/r shows that excess pressure is directly proportional to surface tension. Hence, ΔPX : ΔPY = 2 : 1.
8. A liquid jet of radius 0.25 mm is in contact with air at a pressure of 1.00 × 10^5 Pa. If the liquid surface tension is 0.050 N m−1, what is the absolute pressure inside the jet?
A.
1.001 × 10^5 Pa
B.
1.002 × 10^5 Pa
C.
1.004 × 10^5 Pa
D.
1.020 × 10^5 Pa
Correct Answer:
1.004 × 10^5 Pa
Explanation:
The excess pressure is ΔP = T/r = 0.050/(0.25 × 10−3) = 200 Pa. Therefore, the absolute internal pressure is 100000 + 200 = 100200 Pa, which is 1.002 × 10^5 Pa.
9. A cylindrical liquid jet has radius r and surface tension T. If both its radius and surface tension are doubled, how does its excess pressure change?
A.
It becomes four times as large
B.
It becomes twice as large
C.
It remains unchanged
D.
It becomes half as large
Correct Answer:
It remains unchanged
Explanation:
The excess pressure is ΔP = T/r. Replacing T by 2T and r by 2r gives ΔP′ = 2T/(2r) = T/r, so it remains unchanged.
10. A liquid jet of radius 0.40 mm has an excess pressure of 150 Pa. What surface tension would be required for another jet of radius 0.25 mm to have the same excess pressure?
A.
0.025 N m−1
B.
0.0375 N m−1
C.
0.060 N m−1
D.
0.096 N m−1
Correct Answer:
0.0375 N m−1
Explanation:
For the second jet, T = ΔPr = 150 × 0.25 × 10−3 = 0.0375 N m−1. This follows directly from ΔP = T/r.
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