Question 1 |

An explosion at time t=0 releases energy E at the origin in a space filled with
a gas of density \rho . Subsequently, a hemispherical blast wave propagates
radially outwards as shown in the figure.

Let R denote the radius of the front of the hemispherical blast wave. The radius R follows the relationship R =kt^aE^b\rho ^c, where k is a dimensionless constant.

The value of exponent a is _____. (Rounded off to one decimal place)

Let R denote the radius of the front of the hemispherical blast wave. The radius R follows the relationship R =kt^aE^b\rho ^c, where k is a dimensionless constant.

The value of exponent a is _____. (Rounded off to one decimal place)

0.1 | |

0.4 | |

0.6 | |

0.8 |

Question 1 Explanation:

Radius, R=f(t, E, \rho)

Where, \mathrm{t}= time, \mathrm{E}= Release energy \& \rho= density

Dimension, \mathrm{R}=[\mathrm{L}]

\begin{aligned} & t=[\mathrm{T}] \\ & E=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right] \\ & \rho=\left[\mathrm{ML}^{-3}\right] \end{aligned}

Fundamental dimensions are \mathrm{M}, \mathrm{L}, \mathrm{T} (basic)

Total no. of variables =4

No. of \pi. terms =4-3=1

So, R=t^{a} E^{b} \rho^{c}

[\mathrm{L}]=[\mathrm{T}]^{\mathrm{a}}\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]^{\mathrm{b}}\left[\mathrm{ML}^{-3}\right]^{\mathrm{c}}

Comparing both side indices of M, L, T

a-2b = 0, 2b-3c = 1, b+c = 0 \Rightarrow b-c

a = 2 b, 2(-c) - 3c = 1

-5c = 1, c =(-1)/5 \Rightarrow b=1/5

So, a=2 \times \frac{1}{5} = \frac{2}{5}

Relation, R=kt^{2/5}E^{1/5}\rho ^{-1/5}=\frac{kt^{2/5}E^{1/5}}{\rho ^{1/5}}

Where, a=\frac{2}{5}=0.40

Question 2 |

A solid block of 2.0 kg mass slides steadily at a velocity V along a vertical wall as shown in the figure below. A thin oil film of thickness h = 0.15 mm provides lubrication between the block and the wall. The surface area of the face of the block in contact with the oil film is 0.04 m2. The velocity distribution within the oil film gap is linear as shown in the figure. Take dynamic viscosity of oil as 7\times 10^{-3} Pa-s and acceleration due to gravity as 10 m/s^{2}. Neglect weight of the oil. The terminal velocity V (in m/s) of the block is _________ (correct to one decimal place).

8.2 | |

8.8 | |

10.8 | |

11.2 |

Question 2 Explanation:

Terminal velocity is a constant velocity i.e. the net acceleration is zero.

\begin{aligned} \text{So,} \qquad \Sigma \mathrm{F}_{\mathrm{net}}&=m a \\ \mathrm{mg}-\tau \mathrm{A} &=0 \\ \tau \mathrm{A} &=\mathrm{mg} \\ \mu \frac{V}{h} A &=\mathrm{mg} \\ \mathrm{T} \times 10^{-3} \times \frac{V}{0.15 \times 10^{-3}} \times 0.04&=2 \times 10 =10.714 \mathrm{m} / \mathrm{s} \end{aligned}

\begin{aligned} \text{So,} \qquad \Sigma \mathrm{F}_{\mathrm{net}}&=m a \\ \mathrm{mg}-\tau \mathrm{A} &=0 \\ \tau \mathrm{A} &=\mathrm{mg} \\ \mu \frac{V}{h} A &=\mathrm{mg} \\ \mathrm{T} \times 10^{-3} \times \frac{V}{0.15 \times 10^{-3}} \times 0.04&=2 \times 10 =10.714 \mathrm{m} / \mathrm{s} \end{aligned}

Question 3 |

In a Lagrangian system, the position of a fluid particle in a flow is described as x=x_{o}e^{-kt} and y=y_{o}e^{kt} where t is the time while x_{o} ,x_{o}, and k are constants. The flow is

unsteady and one-dimensional | |

steady and one-dimensional | |

stead and one-dimensional | |

unsteady and two-dimensional |

Question 3 Explanation:

x direction scalar of velocity field,

\begin{array}{l} u=\frac{d x}{d t} \\ u=-k x \cdot e^{-k t} \end{array}

y direction scalar of velocity field

\begin{aligned} v&=\frac{d y}{d t}\\ v &=k y_{0} e^{k t} \\ \vec{V} &=u \hat{i}+v \hat{j} \\ \vec{v} &=-k x_{0} e^{-k t} \hat{i}+k y_{0} e^{k t} \hat{j} \end{aligned}

u & v are non zero scalar t \geq 0 so it is 2 D flow.

2D possible flow field

\begin{aligned} \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}&=0 \\ \frac{\partial}{\partial x}\left(-k x_{o} e^{-k t}\right)+\frac{\partial}{\partial y}\left(k y_{0} e^{k t}\right)&=0\\ 0+0&=0 \\ \text{ continuity satisfied.} & \\ \frac{\partial u}{\partial t}&=+k^{2} x_{0} e^{-k t} \\ \frac{\partial v}{\partial t}&=k^{2} y_{0} e^{k t} \\ \frac{\partial u}{\partial t} &\neq 0 \\ \frac{\partial v}{\partial t} &\neq 0 \end{aligned}

So, flow is steady.

\begin{array}{l} u=\frac{d x}{d t} \\ u=-k x \cdot e^{-k t} \end{array}

y direction scalar of velocity field

\begin{aligned} v&=\frac{d y}{d t}\\ v &=k y_{0} e^{k t} \\ \vec{V} &=u \hat{i}+v \hat{j} \\ \vec{v} &=-k x_{0} e^{-k t} \hat{i}+k y_{0} e^{k t} \hat{j} \end{aligned}

u & v are non zero scalar t \geq 0 so it is 2 D flow.

2D possible flow field

\begin{aligned} \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}&=0 \\ \frac{\partial}{\partial x}\left(-k x_{o} e^{-k t}\right)+\frac{\partial}{\partial y}\left(k y_{0} e^{k t}\right)&=0\\ 0+0&=0 \\ \text{ continuity satisfied.} & \\ \frac{\partial u}{\partial t}&=+k^{2} x_{0} e^{-k t} \\ \frac{\partial v}{\partial t}&=k^{2} y_{0} e^{k t} \\ \frac{\partial u}{\partial t} &\neq 0 \\ \frac{\partial v}{\partial t} &\neq 0 \end{aligned}

So, flow is steady.

Question 4 |

Assuming constant temperature condition and air to be an ideal gas, the variation in atmospheric
pressure with height calculated from fluid statics is

linear | |

exponential | |

quadratic | |

cubic |

Question 4 Explanation:

Air pressure falls with altitude exponentially

Question 5 |

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: Vorticity is zero at all points in the flow.

R: Velocity is directly proportional to the radius from the centre of the vortex.

S: Total mechanical energy per unit mass is constant in the entire flow field.

P: Shear stress is zero at all points in the flow.

Q: Vorticity is zero at all points in the flow.

R: Velocity is directly proportional to the radius from the centre of the vortex.

S: Total mechanical energy per unit mass is constant in the entire flow field.

P and Q | |

R and S | |

P and R | |

P and S |

Question 5 Explanation:

For forced Vortex flow the relation is given by,

V=r\omega

From above equation, it is shown easily that velocity is directly proportional to the radius from the centre of the vortex (Radius of fluid particle from the axis of rotation)

And also for forced vortex flow,

\begin{aligned} \frac{1}{2} \rho \omega ^2(r_2^2-r_1^2)-\rho g(z_2-z_1)&=0 \\ \Delta K.E. -\Delta P.E.&=0 \\ \Rightarrow \;\; \Delta K.E.&= \Delta P.E. \end{aligned}

Now total mechanical energy per unit mass is constant in the entire flow field.

V=r\omega

From above equation, it is shown easily that velocity is directly proportional to the radius from the centre of the vortex (Radius of fluid particle from the axis of rotation)

And also for forced vortex flow,

\begin{aligned} \frac{1}{2} \rho \omega ^2(r_2^2-r_1^2)-\rho g(z_2-z_1)&=0 \\ \Delta K.E. -\Delta P.E.&=0 \\ \Rightarrow \;\; \Delta K.E.&= \Delta P.E. \end{aligned}

Now total mechanical energy per unit mass is constant in the entire flow field.

There are 5 questions to complete.