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As observer is at O So height of water observed by observer

  $\Rightarrow \frac{H}{{\mu }_{\omega }}=\frac{H}{\left(4/3\right)}=\frac{3H}{4}$

given diagram (17.5 - H) is height of observer

$So\frac{3H}{4}=17.5-H$          

$\frac{7H}{4}=17.5$            

7H = 70

H = 10

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The Medicine subjects that are the most important for a successful career in this field are:

• Anatomy

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Length of perpendicular from origin to xcosec a - y sec a = k cot2 a and x sin a + ycos a = k sin 2a are

$p=kcot\alpha sin\alpha cos\alpha =\frac{k}{2}.\frac{cos2\alpha }{sin2\alpha }sin2\alpha =\frac{k}{2}cos2\alpha \Rightarrow \frac{2p}{k}=cos2\alpha$

and $q=ksin2\alpha \Rightarrow \frac{q}{k}=sin2\alpha$

on solving these two we get 4p² + q² = k²

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Thermal resistance of spherical shell = $\frac{r_2-r_1}{4\pi K{r}_1{r}_2}$  

Rate of heat flow = $\frac{\Delta T}{R}=\frac{\left({\theta }_2-{\theta }_1\right)}{\left(\frac{r_2-r_1}{4\pi K{r}_1{r}_2}\right)}=\frac{4\pi K{r}_1{r}_2\left({\theta }_2-{\theta }_1\right)}{\left(r_2-r_1\right)}$

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${\left(t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}_x={\left(t_{mean}\right)}_y$
$\frac{\mathcal{l}n2}{{\lambda }_x}=\frac{1}{{\lambda }_y}$

${\lambda }_x=\left(\mathcal{l}n2\right){\lambda }_y$              

Given N_x = N_y = N₀

So activity A = $\lambda N$

$As{\lambda }_x<{\lambda }_y\Rightarrow A_x<A_y$           

So y will decay faster than x.

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Heisenberg uncertainty principle :

$\begin{array}{l}\Rightarrow \Delta x\Delta p\ge ?\\ \Rightarrow \Delta x\left(m\Delta v\right)\ge ?\end{array}$

$\Rightarrow \Delta x\left(m\sqrt[]{\frac{3RT}{m}}\right)\ge ?$

$\Rightarrow \Delta x\propto \frac{1}{\sqrt[]{m}}$

$f\left(x\right)=\left|x^2-2x-3\right|e^{\left|9x^2-12x+4\right|}$

$\Rightarrow f\left(x\right)=\{\begin{array}{l}\left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},x<-1\\ -\left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},-1\le x<3\\ \left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},x\ge 3\end{array}$

$f\text{'}\left(-1^{-}\right)\ne f\left(-1^{+}\right),$

$f\text{'}\left(3^{-}\right)\ne f\left(3^{+}\right)$

Number of non differential points is 2 at x = -1, 3.

$B_1=\frac{{\mu }_{0}l}{4\pi y}\left[sin{\theta }_1+sin90°\right]-------\left(i\right)$

B due to OX wire

$B_2=\frac{{\mu }_{0}l}{4\pi x}\left[sin{\theta }_2+sin90°\right]$       

As in the diagram direction of B₁ and B₂ are in downward direction

So B = B₁ + B₂

$B=\frac{{\mu }_{0}l}{4\pi y}\left(sin{\theta }_1+1\right)+\frac{{\mu }_{0}l}{4\pi x}\left(sin{\theta }_2+1\right)$

$B=\frac{{\mu }_{0}l}{4\pi xy}\left[\left(x+y\right)+\sqrt[]{x^2+y^2}\right]$  

Amity Online MBA curriculum is divided into semester patterns. The course duration is of two-years, which is divided into four semesters. Each semester offers a new set of subjects. Generic management subjects are taught in the initial semesters. Whereas, more course-specific subjects are taught in the later semesters.

The Medicine syllabus is usually seen as challenging. It is very detailed and complex and requires regular study and memorisation of difficult subjects from your side.
A majority of students find medical education to be both intellectually and physically exhausting, particularly because of the balance between theory, practicals and hospital responsibilities. Having said that, the level of difficulty may differ on the basis of the Medicine course type or college.

Medical courses try to maintain a balance between theory and practical applications. The focus in the early years is on theoretical knowledge of topics like anatomy and biochemistry. As you move forward, clinical rotations, patient interaction and case study learning are also taught as part of the syllabus.
This combination makes sure that you not only understand fundamental ideas but also get critical diagnostic and therapeutic abilities for use in actual medical settings.

$e^{4x}+2e^{3x}-e^x-6=0$

$e^x=t\in \left(0,\infty \right)$

$t^4+2t^3-t-6=0$

Let f (t) = t⁴ + 2t³ – t – 6

f’ (t) = 4t³ + 6t² – 1f

Þf’ (0) = -1, f’ (+ $\infty )$ = + $\infty$

For t > 0

Þ f’ (t) = 0 has only one root.

One solution of f (t) = 0 is possible

Acceleration due to gravity at r distance above the surface = $\frac{GM}{{\left(R+r\right)}^2}$  

Acceleration due to gravity at r distance below the surface = $\frac{GM}{R^3}\left(R-r\right)$

So, ratio = $\frac{\left(R-r\right){\left(R+r\right)}^2}{R^3}=\frac{\left(R-r\right)\left(R^2+r^2+2Rr\right)}{R^3}=1+\frac{r}{R}-\frac{r^2}{R^2}-\frac{r^3}{R^3}$

In give diagram Diode 1 and 2 are in forward bias with R = $30\Omega$ and Diode 3 is reverse bias with R = infinite l₁ current is flowing through 20 $\Omega$  

So $\frac{l_1}{2}and\frac{l_1}{2}$ current will flow through Diode 1 and 2. As resistance is same applying KCl in ABCD Loop $-\frac{l_1}{2}\times 130-\frac{l_1}{2}\times 130-l_1\times 20+200=0$

-100l₁ + 200 = 0

l₁ = 2