# Mechanical Operations Questions and Answers – Velocity Patterns and Velocity Gradient

This set of Mechanical Operations Multiple Choice Questions & Answers (MCQs) focuses on “Velocity Patterns and Velocity Gradient”.

1. Which one of the following is the correct unit of velocity gradient?
a) ms-1
b) m3s-1
c) s-1
d) m

Explanation: ms-1 is the unit of velocity, which can be defined as the rate of change of position. ‘m3s-1’ is the unit of throughput, or simply rate of change of volume. ‘m’ is the unit of distance, the route taken by an object. It is also the unit of displacement, the shortest route taken by an object.

2. Velocity gradient has units similar to which of the following quantities?
a) Acceleration
b) Distance
c) Frequency
d) Work

Explanation: The units of velocity gradient are s-1. Its units are same as that of frequency. Frequency is the representation of the number of occurrences of an event, per unit time. Distance has the units, meter. Acceleration is the rate of change of distance, and has the units of ms-2. Work is the product of displacement and force, and has the unit of kgm2s2.

3. Which of the following derivative represents velocity gradient ? (Here, ‘v’ is velocity, ‘t’ is time and ‘x’ is position)
a) dv/dt
b) dv/dx
c) d2v/dx2
d) d2v/dt2

Explanation: Velocity gradient can be defined as the variation in velocity between the adjacent layers. It has the units of s-1. Velocity ‘v’ has the units ms-1 and position x has the units m. In order to get the units of velocity gradient, it is evident that one needs to divide V by x, in other words, velocity gradient is equal to V/x. It can also be obtained when v is differentiated with respect to x, therefore, velocity gradient is also equal to dv/dx.

4. The velocity field is given as v = (xy, xyz, z2xy). What is the velocity gradient at (x, y, z) = (2, 3, 4) ?
a) 6 s-1
b) 12 s-1
c) 28 s-1
d) 0 s-1

Explanation: The formula to calculate velocity gradient is: [dU/dx dU/dy dU/dz dV/dx dV/dy dV/dz dW/dx dW/dy dW/dz]
Here U = xy ; V = xyz ; Z = z2xy
When we differentiate U,V and W, with respect to x, y and z, each, we get:
y x 0 y*z x*z x*y z*z*y z*z*x 2*z*x*y
On substituting (x, y, z) = (2, 3, 4), the determinant becomes: [3 2 0 12 8 6 48 32 48]
On solving the determinant, we get [3 2 0 12 8 6 48 32 48] = 0

5. The acceleration of a body is given by, a = t3 + 3t2 + 2t + 5. Determine the velocity gradient between t1 = 2 seconds and t2 = 5 seconds, at distance x = 5 meters.
a) 305.25 s-1
b) 305 s
c) 60 s-1
d) 61.05 s-1

Explanation: Acceleration ‘a’ can be written as a = dv/dt
dv/dt = t3 + 3t2 + 2t + 5
Take dt to the other side and integrate. Limits on the right hand side are t=2 and t=5
We get,
v = 305.25 ms-1
At x = 5, velocity gradient = (305.25 ms-1) / (5 m)
= 61.05 s-1
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6. What are the dimensions of velocity gradient?
a) T-1
b) Dimensionless
c) L-1
d) M-1

Explanation: Velocity gradient is the ratio of the velocity difference between adjacent layers and the distance between the layers.
Dimensions of velocity = $$\frac {L}{T}$$
Dimensions of distance = L
$$\frac {Dimesnions \, of \, velocity}{Dimensions \, of \, distance} = \frac {1}{T}$$

Sanfoundry Global Education & Learning Series – Mechanical Operations.