A square plate of 10 cm side moves parallel to another
plate with a relative velocity of 10 cm s, both plates
immersed in water. If the viscous force is 2 x 10°N,
calculate the perpendicular distance between the
plates.
Ans. 0.05 cm.
Answers
The perpendicular distance between the plates is 0.05 cm.
Explanation:
=> Here, It is given that
A square plate of 10 cm side moves parallel to another plate.
Thus, A = l*b = 10cm*10 cm = 100 cm²
Velocity of plate, α_v = 10 cm/ s
Viscous force, F = 200 dyne
Viscosity of water, n = 0.01 poise
=> The perpendicular distance d(x) between the plates:
F = n * A * d(v)/ d(x)
d(x) = n * A * d(v) / F
= 0.01 * 100 * 10 / 200
= 1 / 20
= 0.05 cm
Thus, the perpendicular distance between the plates is 0.05 cm.
Learn more:
Q:1 A horizontal jet of water coming out of a pipe of area of cross - section 20cm^2 hits a vertical wall with a velocity of 10m/s and rebounds with the same speed. The force exerted by water on the wall is?
Click here: https://brainly.in/question/206950
Answer:
A square plate of 10 cm side moves parallel to another plate.
Thus, A = l×b = 10cm*10 cm = 100 cm²
Velocity of plate, α_v = 10 cm/ s
Viscous force, F = 200 dyne
Viscosity of water, n = 0.01 poise
= The perpendicular distance d(x) between the plates:
F = n × A × d(v)/ d(x)
d(x) = n × A × d(v) / F
= 0.01 × 100 × 10 / 200
= 1 / 20
= 0.05 cm Ans...
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Explanation: