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All questions of Exam Archive (PYQ) for NEET Exam
A fluid is in streamline flow across a horizontal pipe of variable area of cross section. For this which of the following statements is correct? [NEET Kar. 2013]- a)The velocity is minimum at the narrowest part of the pipe and the pressure is minimum at the widest part of the pipe
- b)The velocity is maximum at the narrowest part of the pipe and pressure is maximum at the widest part of the pipe
- c)Velocity and pressure both are maximum at the narrowest part of the pipe
- d)Velocity and pressure both are maximum at the widest part of the pipe
Correct answer is option 'B'. Can you explain this answer?
A fluid is in streamline flow across a horizontal pipe of variable area of cross section. For this which of the following statements is correct? [NEET Kar. 2013]
a)
The velocity is minimum at the narrowest part of the pipe and the pressure is minimum at the widest part of the pipe
b)
The velocity is maximum at the narrowest part of the pipe and pressure is maximum at the widest part of the pipe
c)
Velocity and pressure both are maximum at the narrowest part of the pipe
d)
Velocity and pressure both are maximum at the widest part of the pipe
 | Srishti Sen answered |
According to Bernoulli’s theorem,
= constant and Av = constantIf A is minimum, v is maximum, P is minimum.
The pressure experienced by a swimmer 20 m below the water surface in a lake is appropriately: (Given density of water = 103 kgm−3, g = 10 ms−2 and 1 atm = 105 Pa) [2024]- a)1 atm
- b)2 atm
- c)3 atm m
- d)4 atm
Correct answer is option 'C'. Can you explain this answer?
The pressure experienced by a swimmer 20 m below the water surface in a lake is appropriately: (Given density of water = 103 kgm−3, g = 10 ms−2 and 1 atm = 105 Pa) [2024]
a)
1 atm
b)
2 atm
c)
3 atm m
d)
4 atm
 | Infinity Academy answered |
To calculate the pressure experienced by the swimmer at a depth of 20 m below the water surface, we need to use the formula for the pressure at a depth in a fluid:
Where:
P = P0 + ρgh
Where:
- P is the total pressure at depth,
- P0 is the atmospheric pressure at the surface (1 atm),
- ρ is the density of the water,
- g is the acceleration due to gravity,
- h is the depth below the surface.
Given:
- ρ = 103 kg/m3 (density of water),
- g = 10 m/s2 (acceleration due to gravity),
- h = 20 m (depth),
- P₀ = 1 atm = 105 Pa (atmospheric pressure).
Now, we calculate the pressure at the depth:
First, calculate the hydrostatic pressure due to the water column:
Total pressure at the depth is the sum of atmospheric pressure and the hydrostatic pressure:
Now, convert the pressure to atmospheres:
Thus, the correct answer is: (c) 3 atm.
First, calculate the hydrostatic pressure due to the water column:
ρgh = (103 kg/m3) × (10 m/s2) × (20 m) = 2 × 105 Pa
Total pressure at the depth is the sum of atmospheric pressure and the hydrostatic pressure:
P = 1 atm + 2 × 105 Pa
Since 1 atm = 105 Pa, we have:
P = 105 Pa + 2 × 105 Pa = 3 × 105 Pa
Now, convert the pressure to atmospheres:
P = 3 × 105 Pa / 105 Pa = 3 atm
Thus, the correct answer is: (c) 3 atm.
The wetability of a surface by a liquid depends primarily on [NEET 2013]- a)surface tension
- b)density
- c)angle of contact between the surface and the liquid
- d)viscosity
Correct answer is option 'C'. Can you explain this answer?
The wetability of a surface by a liquid depends primarily on [NEET 2013]
a)
surface tension
b)
density
c)
angle of contact between the surface and the liquid
d)
viscosity
| | Srishti Sen answered |
Wetability of a surface by a liquid primarily depends on angle of contact between the surface and liquid.
If angle of contact is acute liquids wet the solid and vice-versa.
If angle of contact is acute liquids wet the solid and vice-versa.
A wire of length L and radius r (r << L) and density ρ is kept floating on the surface of a liquid. The maximum radius of the wire for which it may not sink is: (the surface tension of liquid is T) [2024]- a)√(T/(ρg))
- b)√(2T/(ρg))
- c)√(2Tρ/π)
- d)√(2T/(πρg))
Correct answer is option 'D'. Can you explain this answer?
A wire of length L and radius r (r << L) and density ρ is kept floating on the surface of a liquid. The maximum radius of the wire for which it may not sink is: (the surface tension of liquid is T) [2024]
a)
√(T/(ρg))
b)
√(2T/(ρg))
c)
√(2Tρ/π)
d)
√(2T/(πρg))
 | Dr. Mohit Rajpoot answered |
Answer: (D) √(2T/(πρg))
For a long thin wire resting on the liquid surface, the maximum upward force provided by surface tension occurs from the two contact edges (one on each side). The total upward force due to surface tension is F_surface = 2T L.
The weight of the wire is its mass times g. Mass = density × volume = ρ × π r2 L, so the weight is W = ρ π r2 L g.
At the limiting (maximum) radius the upward surface-tension force just balances the weight:
2T L = ρ π r2 L g
Cancel L and solve for r:
r2 = 2T / (π ρ g)
r = √(2T/(π ρ g)), which corresponds to option (D).
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