1
GATE ME 2008
MCQ (Single Correct Answer)
+2
-0.6
The gap between a moving circular plate and a stationary surface is being continuously reduced, as the circular plate comes down at a uniform speed $$V$$ towards the stationary bottom surface, as shown in the figure. In the process, the fluid contained between the two plates flows out radially. The fluid is assumed to be incompressible and inviscid. GATE ME 2008 Fluid Mechanics - Fluid Kinematics Question 13 English

The radial component of the fluid acceleration at $$r=R$$ is

A
$${{3{V^2}R} \over {4{h^2}}}$$
B
$${{{V^2}R} \over {4{h^2}}}$$
C
$${{{V^2}R} \over {2{h^2}}}$$
D
$${{{V^2}h} \over {4{R^2}}}$$
2
GATE ME 2007
MCQ (Single Correct Answer)
+2
-0.6
Which combination of the following statements about steady incompressible forced vortex flow is correct?

P: Shear stress is zero at all points in the flow.
Q: Velocity is directly proportional to the radius from the centre of the vortex.
R: Total mechanical energy per unit mass is constant in the entire flow field.
S: Total mechanical energy per unit mass is constant in the entire flow field.

A
$$P$$ and $$Q$$
B
$$R$$ and $$S$$
C
$$P$$ and $$R$$
D
$$P$$ and $$S$$
3
GATE ME 2005
MCQ (Single Correct Answer)
+2
-0.6
A leaf is caught in a whirlpool. At a given instant, the leaf is at a distance of $$120$$ $$m$$ from the centre of the whirlpool. The whirlpool can be described by the following velocitry distribution ;
$${V_r} = - \left( {{{60 \times {{10}^3}} \over {2\pi r}}} \right)m/s$$
and $${V_\theta } = - \left( {{{300 \times {{10}^3}} \over {2\pi r}}} \right)m/s.$$

Where $$r$$ (in meters) is the distance from the centre of the whirlpool . What will be the distance of the leaf from the centre when it has moved through half a revolution?

A
$$48$$ $$m$$
B
$$64$$ $$m$$
C
$$120$$ $$m$$
D
$$142$$ $$m$$
4
GATE ME 2004
MCQ (Single Correct Answer)
+2
-0.6
For a fluid flow through a divergent pipe of length $$L$$ having inlet and outlet radii of $${R_1}$$ and $${R_2}$$ respectively and a constant flow rate of $$Q,$$ assuming the velocity to be axial and uniform at any cross- section , the acceleration at the exit is
A
$${{2Q\left( {{R_1} - {R_2}} \right)} \over {\pi LR_2^3}}$$
B
$${{2{Q^2}\left( {{R_1} - {R_2}} \right)} \over {\pi LR_2^3}}$$
C
$${{2{Q^2}\left( {{R_1} - {R_2}} \right)} \over {{\pi ^2}L{R_2}^5}}$$
D
$${{2{Q^2}\left( {{R_2} - {R_1}} \right)} \over {{\pi ^2}L{R_2}^5}}$$
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