Terminal Velocity
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A small sphere of radius r falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to
- r3
- r2
- r4
- r5
A balloon of mass M is descending at a constant acceleration a. When a mass m is released from the balloon, it starts rising with the same acceleration a. Assuming that its volume does not change, what is the value of m
- 20
- 10
- 15
- 5
- mr
- m
- mr
- (mr)1/2
- 1:8
- 2:1
- 1:32
- 1:2
- 2
- 1
- 1/2
- 4
Neglect density of air compared to that of oil. (Viscosity of air =18×10−55 N.s/m2, g=10 m/s2, density of oil =900 kg/m3
- 2.5×10−6 m
- 2×10−6 m
- 3×10−6 m
- 4×10−6 m
- Size of body
- Velocity with which it moves
- Viscosity of fluid
- All of the above
- r1/2
- r
- r3/2
- r5
- 9:4
- 27:8
- 3:2
- 4:9
- 2v
- 22/3v
- √2v
- v√2
- directly proportional to R but inversely proportional to v
- directly proportional to both radius R and velocity v
- inversely proportional to R but directly proportional to velocity v
- inversely proportional to both radius R and velocity v
- Upthrust + Weight = Viscous drag
- Weight + Viscous drag = Upthrust
- Viscous drag + Upthrust = Weight
- Viscous drag + Upthrust > Weight
In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0×10–5 m and density 1.2×103kgm–3? Take the viscosity of air at the temperature of the experiment to be 1.8×10–5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.
- 43 × 10–6°C–1
- 25 × 10–6°C–1
- 51 × 10–6°C–1
- 17 × 10–6°C–1
A solid sphere moves at a terminal velocity of 20 ms−1 in air at a place where g=9.8 ms−2. The sphere is taken in a gravity free hall having air at the same pressure and pushed downwards at a speed of 20 ms−1. Then, identify the correct statement(s):
[Assume density of air to be very low.]
- Intial acceleration of sphere will be 9.8 ms−2 downwards.
- Initial acceleration of sphere will be 9.8 ms−2 upwards.
- The magnitude of acceleration of the sphere will decrease as time passes.
- Sphere will eventually stop.
- Less than g
- Greater than g
- Zero
- g
A body rolling on ice with vel. 8m/s comes to rest after traveling 4 m.Compute the co efficient of friction
Neglect the density of air as compared to that of oil. (Take viscosity of air =3.6×10−5 N-s/m2, g=10 m/s2, density of oil ρo=900 kg/m3)
- 2×10−6 m
- 2.5×10−6 m
- 4×10−6 m
- 3×10−6 m
The coefficient of viscosity of air is 1.9×10−5 Ns/m2 and its density is 1.2 kg/m3. Density of water is 1000 kg/m3. Take g=10 m/s2.
- 15 cm/s
- 2.5 cm/s
- 19 cm/s
- 13 cm/s
- 0.4 m/s
- 0.3 m/s
- 0.25 m/s
- 0.1 m/s
Reason: Acceleration due to gravity is g6 on the surface of moon.
- If both assertion and reason are correct and the reason is the correct explanation of the assertion.
- If both assertion and reason are correct and the reason is not a correct explanation of the assertion.
- If the assertion is correct but reason is incorrect.
- If the assertion is incorrect but reason is correct.
A drop of water of radius is falling in air. If the coefficient of viscosity of air is , the terminal velocity of the drop will be: (The density of water and )
- Terminal velocity will be 20 ms−1.
- Terminal velocity will be less than 20 ms−1.
- Terminal velocity will be more than 20 ms−1.
- There will be no terminal velocity.
- 29r2(2σ−ρ)gη
- r2(σ−2ρ)g9η
- 29r2(σ−4ρ)gη
- 29r2(σ−3ρ)gη