Mercury has an angle of contact equal to 140 degree with sodium soda lime glass. A narrow tube of radius 1.000 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 N/m . Density of mercury = 13.6 × 10³ kg/m cube
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Hi
Here is your answer,
Given , angle of contact (θ) = 140°
Radius of tube (r) = 1 mm = 10⁻³ m
Surface tension (S) = 0.465 N/m
Density of mercury (ρ) = 13.6 × 10³ kg/m³
Height of liquid rise or fall due to surface tension (h)
= 2S cos θ/rρg = 2 × 0.465 × cos 140°/1 × 10⁻³ × 13.6 × 10³ × 9.8
= 2 × 0.465 × (- 0.7660) / 10⁻³ × 13.6 × 10³ × 9.8
= - 5.34 mm
Hence, the mercury level will depressed by 5.34 mm.
Hope it helps you!
Here is your answer,
Given , angle of contact (θ) = 140°
Radius of tube (r) = 1 mm = 10⁻³ m
Surface tension (S) = 0.465 N/m
Density of mercury (ρ) = 13.6 × 10³ kg/m³
Height of liquid rise or fall due to surface tension (h)
= 2S cos θ/rρg = 2 × 0.465 × cos 140°/1 × 10⁻³ × 13.6 × 10³ × 9.8
= 2 × 0.465 × (- 0.7660) / 10⁻³ × 13.6 × 10³ × 9.8
= - 5.34 mm
Hence, the mercury level will depressed by 5.34 mm.
Hope it helps you!
Answered by
0
Answer:
1) According, to the third law of motion by newton, every action has an equal and opposite reaction.
2) so, if the bullet exerts a momentum when going forward, same amount of momentum will push the gun backward
3) Hence, the momentum of both would be equal
10 g * v = 1000 g * 5 m/s
v = 500 m/s
Explanation:
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