Related Questions

The ratio of average fluid velocity to the maximum velocity in case of laminar flow of a Newtonion fluid in a circular pipe is
A 0.5
B 1
C 2
D 0.66
An ideal plastic substance indicates no deformation, when stressed upto yield stress, but behaves like a Newtonion fluid beyond yield stress. Which of the following is an ideal plastic?
A Sewage sludge
B Rubber latex
C Blood
D Sugar solution
For turbulent flow of newtonion fluid in a circular cross-section pipe, the ratio of maximum to average fluid velocity is
A 0.5
B 1
C 0.66
The velocity profile exhibited by laminar flow of Newtonion fluids is such that the velocity distribution w.r.t. radius of the circular pipe is a/an __________ with the apex at the centre line of the pipe.
A Hyperbola
B Parabola
C Semi-circle
D Semi-ellipse
For laminar flow of Newtonion fluid in a circular pipe, the velocity distribution is a function of the distance 'd' measured from the centre line of the pipe, and it follows a __________ relationship.
A Logarithmic
B Parabolic
C Hyperbolic
D Linear
The design stress, which is more than the damaging stress, is the least unit stress that will render a member unfit for service before the end of its normal life. The design stress factor or factor of safety indicates the margin between design stress and the __________ stress.
A Working
B Damaging
C Allowable
D None of these
According to maximum shear stress failure criterion, yielding in material occurs, when the maximum shear stress is equal to __________ times the yield stress.
A 0.5
B 2
C $$\sqrt 2 $$
D $$\sqrt {\frac{3}{2}} $$
The strain energy stored in a body due to shear stress, is (where $$\tau $$ = Shear stress, C = Shear modulus and V = Volume of the body)
A $$\frac{{\tau {\text{V}}}}{{2{\text{C}}}}$$
B $$\frac{{2{\text{C}}}}{{\tau {\text{V}}}}$$
C $$\frac{{{\tau ^2}{\text{V}}}}{{2{\text{C}}}}$$
D $$\frac{{2{\text{C}}}}{{{\tau ^2}{\text{V}}}}$$
When a body is subjected to biaxial stress i.e. direct stresses ($${\sigma _{\text{x}}}$$) and ($${\sigma _{\text{y}}}$$) in two mutually perpendicular planes accompanied by a simple shear stress ($${\tau _{{\text{xy}}}}$$ ), then maximum shear stress is
A $$\frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$
B $$\frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$
C $$\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + \tau _{{\text{xy}}}^2} $$
D $$\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + \tau _{{\text{xy}}}^2} $$
When a body is subjected to a direct tensile stress ($${\sigma _{\text{x}}}$$) in one plane accompanied by a simple shear stress ($${\tau _{{\text{xy}}}}$$ ), the maximum shear stress is
A $$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2}\sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
B $$\frac{{{\sigma _{\text{x}}}}}{2} - \frac{1}{2}\sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
C $$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2}\sqrt {\sigma _{\text{x}}^2 - 4\tau _{{\text{xy}}}^2} $$
D $$\frac{1}{2}\sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$

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