Question 1 |
The steady velocity field in an inviscid fluid of
density 1.5 is given to be \vec{V}=(y^2-x^2)\hat{i}+(2xy)\hat{j}
Neglecting body forces, the pressure gradient at
(x = 1, y = 1) is
10\hat{j} | |
20\hat{i} | |
-6\hat{i}-6\hat{j} | |
-4\hat{i}-4\hat{j} |
Question 1 Explanation:
By Euler's equation of motion,
\begin{aligned} x:-\frac{\partial p}{\partial x}+\rho g_x&=\rho \frac{Du}{Dt}\\ y:-\frac{\partial p}{\partial y}+\rho g_y&=\rho \frac{Dv}{Dt}\\ \end{aligned}
Neglecting body forces (i.e. g_x=g_y=0 )
\begin{aligned} \frac{\partial p}{\partial x}&=-\rho \frac{Du}{Dt}=-\rho \left [ u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y} \right ]\\ &=-\rho [(y^2-x^2)(-2x)+(2xy)(2y)]\\ &=-\rho[-2xy^2+2x^3+4xy^2]\\ &=-\rho(2xy^2+2x^3)\\ &=-1.5 \times (2 \times 1 \times 1^2 +2\times1^3 )\\ &=-6 Pa/m \end{aligned}
Similarly,
\begin{aligned} \frac{\partial p}{\partial y}&=-\rho \frac{Dv}{Dt}\\ &=-\rho \left [ u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y} \right ]\\ &=-\rho [(y^2-x^2)(2y)+(2xy)(2x)]\\ &=-\rho[2y^3-2x^2y+4x^2y]\\ &=-\rho(2y^3+2x^2y)\\ &=-1.5 \times (2 \times 1^3 +2\times1^2 \times 1 )\\ &=-6 Pa/m \end{aligned}
The pressure gradient vector is given by
\triangledown p=\frac{\partial p}{\partial x}\hat{i}+\frac{\partial p}{\partial y}\hat{j}=-6\hat{i}-6\hat{j}
\begin{aligned} x:-\frac{\partial p}{\partial x}+\rho g_x&=\rho \frac{Du}{Dt}\\ y:-\frac{\partial p}{\partial y}+\rho g_y&=\rho \frac{Dv}{Dt}\\ \end{aligned}
Neglecting body forces (i.e. g_x=g_y=0 )
\begin{aligned} \frac{\partial p}{\partial x}&=-\rho \frac{Du}{Dt}=-\rho \left [ u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y} \right ]\\ &=-\rho [(y^2-x^2)(-2x)+(2xy)(2y)]\\ &=-\rho[-2xy^2+2x^3+4xy^2]\\ &=-\rho(2xy^2+2x^3)\\ &=-1.5 \times (2 \times 1 \times 1^2 +2\times1^3 )\\ &=-6 Pa/m \end{aligned}
Similarly,
\begin{aligned} \frac{\partial p}{\partial y}&=-\rho \frac{Dv}{Dt}\\ &=-\rho \left [ u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y} \right ]\\ &=-\rho [(y^2-x^2)(2y)+(2xy)(2x)]\\ &=-\rho[2y^3-2x^2y+4x^2y]\\ &=-\rho(2y^3+2x^2y)\\ &=-1.5 \times (2 \times 1^3 +2\times1^2 \times 1 )\\ &=-6 Pa/m \end{aligned}
The pressure gradient vector is given by
\triangledown p=\frac{\partial p}{\partial x}\hat{i}+\frac{\partial p}{\partial y}\hat{j}=-6\hat{i}-6\hat{j}
Question 2 |
The velocity field in a fluid is given to be
\vec{V}=4(xy)\hat{i}+2(x^2-y^2)\hat{j}
Which of the following statement(s) is/are correct?
MSQ
\vec{V}=4(xy)\hat{i}+2(x^2-y^2)\hat{j}
Which of the following statement(s) is/are correct?
MSQ
The velocity field is one-dimensional. | |
The flow is incompressible | |
The flow is irrotational | |
The acceleration experienced by a fluid particle is zero at (x = 0, y = 0). |
Question 2 Explanation:
For given flow,
u=4xy, v = 2(x^2- y^2)
As velocity field is function of two space variables, flow is two dimensional.
\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=4y-4y
Therefore, flow is incompressible.
\omega _z=\frac{1}{2}\left ( \frac{\partial v}{\partial x} -\frac{\partial u}{\partial y}\right )=\frac{1}{2}(4x-4x)=0
Therefore, flow is irrotational.
\begin{aligned} a_x &=u\frac{\partial u}{\partial x} +v\frac{\partial u}{\partial y}\\ &= 4xy(4y)+2(x^2-y^2)(4x)\\ &= 16xy^2+8x^3-8xy^2 \\ &=16 \times 0 \times 0^2+8 \times 0^3-8 \times 0 \times 0^2 \\ &= 0 \end{aligned}
\begin{aligned} a_y &=u\frac{\partial v}{\partial x} +v\frac{\partial v}{\partial y}\\ &= 4xy(4x)+2(x^2-y^2)(-4y)\\ &= 16x^2y-8x^2y+8y^3 \\ &=16 \times 0^2 \times 0-8 \times 0^2 \times 0+8 \times 0^3 \\ &= 0\\ |\vec{a}|&=\sqrt{a_x^2+a_y^2}=0 \end{aligned}
u=4xy, v = 2(x^2- y^2)
As velocity field is function of two space variables, flow is two dimensional.
\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=4y-4y
Therefore, flow is incompressible.
\omega _z=\frac{1}{2}\left ( \frac{\partial v}{\partial x} -\frac{\partial u}{\partial y}\right )=\frac{1}{2}(4x-4x)=0
Therefore, flow is irrotational.
\begin{aligned} a_x &=u\frac{\partial u}{\partial x} +v\frac{\partial u}{\partial y}\\ &= 4xy(4y)+2(x^2-y^2)(4x)\\ &= 16xy^2+8x^3-8xy^2 \\ &=16 \times 0 \times 0^2+8 \times 0^3-8 \times 0 \times 0^2 \\ &= 0 \end{aligned}
\begin{aligned} a_y &=u\frac{\partial v}{\partial x} +v\frac{\partial v}{\partial y}\\ &= 4xy(4x)+2(x^2-y^2)(-4y)\\ &= 16x^2y-8x^2y+8y^3 \\ &=16 \times 0^2 \times 0-8 \times 0^2 \times 0+8 \times 0^3 \\ &= 0\\ |\vec{a}|&=\sqrt{a_x^2+a_y^2}=0 \end{aligned}
Question 3 |
A steady two-dimensional flow field is specified by
the stream function
\psi =kx^3y
where x and y are in meter and the constant k = 1 \; m^{-2}s^{-1}. The magnitude of acceleration at a point (x,y) = (1 m, 1 m) is ________ m/s^2 (round off to 2 decimal places).
\psi =kx^3y
where x and y are in meter and the constant k = 1 \; m^{-2}s^{-1}. The magnitude of acceleration at a point (x,y) = (1 m, 1 m) is ________ m/s^2 (round off to 2 decimal places).
2.42 | |
1.25 | |
3.62 | |
4.24 |
Question 3 Explanation:
Given,
Stream function,
\begin{aligned} \psi &=kx^3y; \; k=1m^{-2}s^{-1}\\ u&=-\frac{\partial \psi }{\partial y}\Rightarrow u=-x^3\\ v&=-\frac{\partial \psi }{\partial x}\Rightarrow v=3x^2y\\ \vec{V}&=-x^3\hat{i}+3x^2y\hat{j}\\ a_x&=\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\\ a_x&=-x^3(-3x^2)\Rightarrow a_x=3x^5\\ At\; &(1,1), \; a_x=3m/sec^2\\ a_y&=\frac{\partial v}{\partial t}+u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}\\ a_y&=-x^3(6xy)+3(x^2y)(3x^2)\\ &=-6x^4y+9x^4y=3x^4y\\ \vec{a}&=3x^5\hat{i}+3x^4y\hat{j}\\ At& \;\;(1,1)\\ \vec{a}&=3\hat{i}+3\hat{j}\\ \Rightarrow |a|=\sqrt{3^2+3^2}&=4.24m/s^2 \end{aligned}
Stream function,
\begin{aligned} \psi &=kx^3y; \; k=1m^{-2}s^{-1}\\ u&=-\frac{\partial \psi }{\partial y}\Rightarrow u=-x^3\\ v&=-\frac{\partial \psi }{\partial x}\Rightarrow v=3x^2y\\ \vec{V}&=-x^3\hat{i}+3x^2y\hat{j}\\ a_x&=\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}+w\frac{\partial u}{\partial z}\\ a_x&=-x^3(-3x^2)\Rightarrow a_x=3x^5\\ At\; &(1,1), \; a_x=3m/sec^2\\ a_y&=\frac{\partial v}{\partial t}+u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}+w\frac{\partial v}{\partial z}\\ a_y&=-x^3(6xy)+3(x^2y)(3x^2)\\ &=-6x^4y+9x^4y=3x^4y\\ \vec{a}&=3x^5\hat{i}+3x^4y\hat{j}\\ At& \;\;(1,1)\\ \vec{a}&=3\hat{i}+3\hat{j}\\ \Rightarrow |a|=\sqrt{3^2+3^2}&=4.24m/s^2 \end{aligned}
Question 4 |
A two dimensional flow has velocities in x and y directions given by u=2xyt and v=-y^2 t, where t denotes time. The equation for streamline passing through x=1, \; y=1 is
x^2y=1 | |
xy^2=1 | |
x^2y^2=1 | |
x/y^2=1 |
Question 4 Explanation:
\begin{aligned} u &=2 x y t \\ v &=-y^{2} t \\ \frac{d x}{u} &=\frac{d y}{v}=\frac{d z}{w} \\ \frac{d x}{2 x y t} &=\frac{d y}{-y^{2} t} \\ -y d x &=2 x d y \\ \ln x y^{2} &=c \\ x y^{2} &=1 \end{aligned}
Question 5 |
For a two-dimensional, incompressible flow having velocity components u and v in the x and y directions, respectively, the expression
\frac{\partial (u^2)}{\partial x}+\frac{\partial (uv)}{\partial y}
can be simplified to
\frac{\partial (u^2)}{\partial x}+\frac{\partial (uv)}{\partial y}
can be simplified to
u\frac{\partial u}{\partial x}+u\frac{\partial v}{\partial y} | |
2u\frac{\partial u}{\partial x}+u\frac{\partial v}{\partial y} | |
2u\frac{\partial u}{\partial x}+v\frac{\partial v}{\partial y} | |
u\frac{\partial u}{\partial x}+v\frac{\partial v}{\partial y} |
Question 5 Explanation:
\frac{\partial\left(u^{2}\right)}{\partial x}+\frac{\partial(u v)}{\partial y}
By differentiating:
\Rightarrow 2 u\left[\frac{\partial u}{\partial x}\right]+u \frac{\partial v}{\partial y}+v \frac{\partial u}{\partial y}
\Rightarrow u \frac{\partial u}{\partial x}+\left[u \frac{\partial u}{\partial x}+u \frac{\partial v}{\partial y}\right]+v \frac{\partial u}{\partial y}
According to continity eq. : \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0
So, u\frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}
By differentiating:
\Rightarrow 2 u\left[\frac{\partial u}{\partial x}\right]+u \frac{\partial v}{\partial y}+v \frac{\partial u}{\partial y}
\Rightarrow u \frac{\partial u}{\partial x}+\left[u \frac{\partial u}{\partial x}+u \frac{\partial v}{\partial y}\right]+v \frac{\partial u}{\partial y}
According to continity eq. : \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0
So, u\frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}
Question 6 |
Consider a flow through a nozzle, as shown in the figure below:
The air flow is steady, incompressible and inviscid. The density of air is 1.23 kg/m^3. The pressure difference (p_1 - p_{atm}) is _______ kPa (round off to the nearest integer)
The air flow is steady, incompressible and inviscid. The density of air is 1.23 kg/m^3. The pressure difference (p_1 - p_{atm}) is _______ kPa (round off to the nearest integer)
1522.12 | |
1.52 | |
321.32 | |
3.21 |
Question 6 Explanation:
\begin{aligned} A_{1} V_{1} &=A_{2} V_{2} \\ 0.2 \times V_{1} &=0.02 \times 50 \\ V_{1} &=\frac{1}{10} \times 50=5 \mathrm{m} / \mathrm{s} \end{aligned}
Applying BE
\begin{aligned} \frac{P_{1}}{w}+\frac{V_{1}^{2}}{2 g}+z_{1} &=\frac{P_{2}}{w}+\frac{V_{2}^{2}}{2 g}+z_{2} \; \; \; (\because z_{1}=z_{2})\\ \frac{P_{1}-P_{2}}{\rho_{\text {air }} g} &=\frac{V_{2}^{2}-V_{1}^{2}}{2 g} \\ P_{1}-P_{2} &=\left(\frac{50^{2}-5^{2}}{2}\right) \times 1.23\\ &=1522.125 \mathrm{Pa}=1.52 \mathrm{kPa} \end{aligned}
Question 7 |
The velocity field of an incompressible flow in a Cartesian system is represented by
\vec{V}=2(x^2-y^2)\hat{i}+v\hat{j}+3\hat{k}
Which one of the following expressions for v is valid?
\vec{V}=2(x^2-y^2)\hat{i}+v\hat{j}+3\hat{k}
Which one of the following expressions for v is valid?
-4xz+6xy | |
-4xy-4xz | |
4xz-6xy | |
4xy+4xz |
Question 7 Explanation:
\vec{V}=\left(2 x^{2}-2 y^{2}\right) \hat{i}+v \hat{j}+3 \hat{k}
For Incompressible flow
\begin{aligned} \frac{\partial u}{\partial x}+\frac{\partial V}{\partial y}+\frac{\partial W}{\partial z} &=0 \\ 4 x+\frac{\partial V}{\partial y} &=0 \\ \frac{\partial V}{\partial y} &=-4 x \\ v &=-4 x y+f(x, z) \end{aligned}
f(x,z) is an arbitary function of x and z
Hence the most suitable answer is option (B)
For Incompressible flow
\begin{aligned} \frac{\partial u}{\partial x}+\frac{\partial V}{\partial y}+\frac{\partial W}{\partial z} &=0 \\ 4 x+\frac{\partial V}{\partial y} &=0 \\ \frac{\partial V}{\partial y} &=-4 x \\ v &=-4 x y+f(x, z) \end{aligned}
f(x,z) is an arbitary function of x and z
Hence the most suitable answer is option (B)
Question 8 |
For a steady flow, the velocity field is \overrightarrow{\textrm{v}}=(-x^{2}+3y)\hat{i}+(2xy)\hat{j} , The magnitude of the acceleration of a particle at (1, -1) is
2 | |
1 | |
2\sqrt{5} | |
0 |
Question 8 Explanation:
Given velocity field,
\vec{V}=\left(-x^{2}+3 y\right) \hat{i}+(2 x y) \hat{j}
where \quad u=-x^{2}+3 y \text { and } v=2 x y
The acceleration components along x and y -axis.
\begin{aligned} a_{x}&=u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}\\ \text{and}\quad a_{y} &=u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y} \\ &=\left(-x^{2}+3 y\right) \times(-2 x)+2 x y \times 3 \\ &=2 x^{3}-6 x y+6 x y=2 x^{3}\\ \text{and}\quad a_{y} &=\left(-x^{2}+3 y\right) \times 2 y+2 x y \times 2 x \\ &=-2 y x^{2}+6 y^{2}+4 x^{2} y=2 y x^{2}+6 y^{2} \end{aligned}
At point (1,-1)
\begin{aligned} a_{x}&=2\\ \text{and}\quad a_{y} &=2 \times(-1) \times 1+6 \times(-1)^{2} \\ &=-2+6=4 \\ \vec{a} &=a_{x} \hat{i}+a_{y} \hat{j}=2 i+4 \hat{j} \end{aligned}
Resultant acceleration:
\begin{aligned} a &=\sqrt{4+16}=\sqrt{20} \\ &=\sqrt{4 \times 5}=2 \sqrt{5} \end{aligned}
\vec{V}=\left(-x^{2}+3 y\right) \hat{i}+(2 x y) \hat{j}
where \quad u=-x^{2}+3 y \text { and } v=2 x y
The acceleration components along x and y -axis.
\begin{aligned} a_{x}&=u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}\\ \text{and}\quad a_{y} &=u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y} \\ &=\left(-x^{2}+3 y\right) \times(-2 x)+2 x y \times 3 \\ &=2 x^{3}-6 x y+6 x y=2 x^{3}\\ \text{and}\quad a_{y} &=\left(-x^{2}+3 y\right) \times 2 y+2 x y \times 2 x \\ &=-2 y x^{2}+6 y^{2}+4 x^{2} y=2 y x^{2}+6 y^{2} \end{aligned}
At point (1,-1)
\begin{aligned} a_{x}&=2\\ \text{and}\quad a_{y} &=2 \times(-1) \times 1+6 \times(-1)^{2} \\ &=-2+6=4 \\ \vec{a} &=a_{x} \hat{i}+a_{y} \hat{j}=2 i+4 \hat{j} \end{aligned}
Resultant acceleration:
\begin{aligned} a &=\sqrt{4+16}=\sqrt{20} \\ &=\sqrt{4 \times 5}=2 \sqrt{5} \end{aligned}
Question 9 |
Consider the two-dimensional velocity field given by \overrightarrow{\mathrm{v}}=(5+a_{1}x+b_{1}y)\hat{i}\: + \: (4+a_{2}x+b_{2}y)\hat{j} ,where a_1,b_1,a_2 text{ and }b_2 are constants. Which one of the following conditions needs to be satisfied for the flow to be incompressible?
a_{1}+\, b_{1}=0 | |
a_{1}+\, b_{2}=0 | |
a_{2}+\, b_{2}=0 | |
a_{2}+\, b_{1}=0 |
Question 9 Explanation:
Given that the flow is incompressible.
\therefore \quad \operatorname{div} \bar{V}=0
\frac{\partial}{\partial x}\left(5+a_{1} x+b_{1} y\right)+\frac{\partial}{\partial y}\left(8+a_{2} x+b_{2} y\right)=0
a_{1}+b_{2}=0
\therefore \quad \operatorname{div} \bar{V}=0
\frac{\partial}{\partial x}\left(5+a_{1} x+b_{1} y\right)+\frac{\partial}{\partial y}\left(8+a_{2} x+b_{2} y\right)=0
a_{1}+b_{2}=0
Question 10 |
For a two-dimensional flow, the velocity field is \vec{u}=\frac{x}{x^{2}+y^{2}}\hat{i}+\frac{y}{x^{2}+y^{2}}\hat{j}, where \hat{i} and \hat{j} are the basis vectors in the x-y Cartesian coordinate system. Identify the CORRECT statements from below.
(1)The flow is incompressible
(2)The flow is unsteady.
(3) y-component of acceleration, a_{y}=\frac{-y}{(x^{2}+y^{2})^{2}}
(4) x-component of acceleration,a_{x}=\frac{-(x+y)}{(x^{2}+y^{2})^{2}}
(1)The flow is incompressible
(2)The flow is unsteady.
(3) y-component of acceleration, a_{y}=\frac{-y}{(x^{2}+y^{2})^{2}}
(4) x-component of acceleration,a_{x}=\frac{-(x+y)}{(x^{2}+y^{2})^{2}}
(2) and (3) | |
(1) and (3) | |
(1) and (2) | |
(3) and (4) |
Question 10 Explanation:
For 2 \mathrm{D} -flow velocity field is:
\begin{aligned} \vec{V}&=\frac{x}{x^{2}+y^{2}} \hat{i}+\frac{y}{x^{2}+y^{2}} \hat{j}\\ so \quad u&=\frac{x}{x^{2}+y^{2}} and v=\frac{y}{x^{2}+y^{2}} \\ a_{x} &=u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y} \\ &=\frac{x}{x^{2}+y^{2}} \frac{\partial}{\partial x}\left[\frac{x}{x^{2}+y^{2}}\right] \\ &+\frac{y}{x^{2}+y^{2}} \frac{\partial}{\partial y}\left[\frac{x}{x^{2}+y^{2}}\right] \\ &= \frac{x}{x^{2}+y^{2}}\left[\frac{x(-2 x)}{\left(x^{2}+y^{2}\right)^{2}}+\frac{1}{x^{2}+y^{2}}\right]\\ &+\frac{y}{x^{2}+y^{2}} \left[\frac{x(2 y)}{\left(x^{2}+y^{2}\right)^{2}}\right]\\ &=\frac{-2 x^{3}}{\left(x^{2}+y^{2}\right)^{3}}+\frac{x}{\left(x^{2}+y^{2}\right)^{2}}\\ &+ \frac{2xy^{2}}{\left(x^{2}+y^{2}\right)^{3}}\\ &=\frac{-2 x^{3}+x\left(x^{2}+y^{2}\right)-2 x y^{2}}{\left(x^{2}+y^{2}\right)^{3}} \\ &=\frac{-2 x^{3}+x^{3}+x y^{2}-2 x y^{2}}{\left(x^{2}+y^{2}\right)^{3}}\\ &=\frac{-x^{3}-x y^{2}}{\left(x^{2}+y^{2}\right)^{3}} \\ &=\frac{-x\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)^{3}} \end{aligned}
\begin{aligned} a_{x} &=\frac{x}{\left(x^{2}+y^{2}\right)^{2}} \\ a_{y} &=u \frac{\partial v}{\partial x}+u \frac{\partial v}{\partial y} \\ &=\frac{x}{x^{2}+y^{2}} \frac{\partial}{\partial x}\left[\frac{y}{x^{2}+y^{2}}\right] \\& +\frac{y}{x^{2}+y^{2}} \frac{\partial}{\partial y}\left[\frac{y}{x^{2}+y^{2}}\right] \\ &=\frac{x}{x^{2}+y^{2}} \frac{(-2 x y)}{\left(x^{2}+y^{2}\right)^{2}}\\& +\frac{y}{x^{2}+y^{2}} \times\left[\frac{-2 y^{2}}{\left(x^{2}+y^{2}\right)^{2}}+\frac{1}{x^{2}+y^{2}}\right]\\ &=\frac{-2 x^{2} y}{\left(x^{2}+y^{2}\right)^{3}}+\frac{2 y^{3}}{\left(x^{2}+y^{2}\right)^{3}}\\&+\frac{y}{\left(x^{2}+y^{2}\right)^{2}} \\ &=\frac{-2 x^{2} y-2 y^{3}+y\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)^{3}} \\ &=\frac{-x^{2} y-y^{3}}{\left(x^{2}+y^{2}\right)^{3}}=\frac{y\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)^{3}} \\ a_{y}&=\frac{-y}{\left(x^{2}+y^{2}\right)^{2}} \\ \end{aligned}
\begin{aligned} \vec{V}&=\frac{x}{x^{2}+y^{2}} \hat{i}+\frac{y}{x^{2}+y^{2}} \hat{j}\\ so \quad u&=\frac{x}{x^{2}+y^{2}} and v=\frac{y}{x^{2}+y^{2}} \\ a_{x} &=u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y} \\ &=\frac{x}{x^{2}+y^{2}} \frac{\partial}{\partial x}\left[\frac{x}{x^{2}+y^{2}}\right] \\ &+\frac{y}{x^{2}+y^{2}} \frac{\partial}{\partial y}\left[\frac{x}{x^{2}+y^{2}}\right] \\ &= \frac{x}{x^{2}+y^{2}}\left[\frac{x(-2 x)}{\left(x^{2}+y^{2}\right)^{2}}+\frac{1}{x^{2}+y^{2}}\right]\\ &+\frac{y}{x^{2}+y^{2}} \left[\frac{x(2 y)}{\left(x^{2}+y^{2}\right)^{2}}\right]\\ &=\frac{-2 x^{3}}{\left(x^{2}+y^{2}\right)^{3}}+\frac{x}{\left(x^{2}+y^{2}\right)^{2}}\\ &+ \frac{2xy^{2}}{\left(x^{2}+y^{2}\right)^{3}}\\ &=\frac{-2 x^{3}+x\left(x^{2}+y^{2}\right)-2 x y^{2}}{\left(x^{2}+y^{2}\right)^{3}} \\ &=\frac{-2 x^{3}+x^{3}+x y^{2}-2 x y^{2}}{\left(x^{2}+y^{2}\right)^{3}}\\ &=\frac{-x^{3}-x y^{2}}{\left(x^{2}+y^{2}\right)^{3}} \\ &=\frac{-x\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)^{3}} \end{aligned}
\begin{aligned} a_{x} &=\frac{x}{\left(x^{2}+y^{2}\right)^{2}} \\ a_{y} &=u \frac{\partial v}{\partial x}+u \frac{\partial v}{\partial y} \\ &=\frac{x}{x^{2}+y^{2}} \frac{\partial}{\partial x}\left[\frac{y}{x^{2}+y^{2}}\right] \\& +\frac{y}{x^{2}+y^{2}} \frac{\partial}{\partial y}\left[\frac{y}{x^{2}+y^{2}}\right] \\ &=\frac{x}{x^{2}+y^{2}} \frac{(-2 x y)}{\left(x^{2}+y^{2}\right)^{2}}\\& +\frac{y}{x^{2}+y^{2}} \times\left[\frac{-2 y^{2}}{\left(x^{2}+y^{2}\right)^{2}}+\frac{1}{x^{2}+y^{2}}\right]\\ &=\frac{-2 x^{2} y}{\left(x^{2}+y^{2}\right)^{3}}+\frac{2 y^{3}}{\left(x^{2}+y^{2}\right)^{3}}\\&+\frac{y}{\left(x^{2}+y^{2}\right)^{2}} \\ &=\frac{-2 x^{2} y-2 y^{3}+y\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)^{3}} \\ &=\frac{-x^{2} y-y^{3}}{\left(x^{2}+y^{2}\right)^{3}}=\frac{y\left(x^{2}+y^{2}\right)}{\left(x^{2}+y^{2}\right)^{3}} \\ a_{y}&=\frac{-y}{\left(x^{2}+y^{2}\right)^{2}} \\ \end{aligned}
There are 10 questions to complete.
In question 18, the vorticity vector is to be found, but its written ‘velocity vector’. Please rectify that.
Thank you Spandan Samanta,
We have updated the answer.
In question 23, the question and the answer given are completely different, please check that.
Thank you Spandan Samanta,
We have updated the answer as option A.
still not solved..kindly read the question and explaination first
Question no 26 wrong answer is given. Please corrrect it..
please correct the ques. no. 29