SSC JE Mechanical Engineering 2017 Set 1 Paper

${\mathit{P}}_1$ = pressure at section 1
${\mathit{P}}_2$ = pressure at section 2
ρ = density of water
ρm = density of mercury
x = distance
h = distance
S = Density of mercury Consider Section 1 as datum line
$\left(\frac{{\mathit{P}}_1}{\mathit{r}}\right)$ + x + h = $\left(\frac{{\mathit{P}}_2}{\mathit{r}}\right)$ + x + Sh + 0.8
$\frac{{\mathit{P}}_1}{\mathit{r}}-\frac{{\mathit{P}}_2}{\mathit{r}}-0.8$ = (S – 1)h
= (13.6 – 1)h = 12.6 h
By continuity
$\mathit{Q}=\left(\frac{\mathit{\pi }}{4}\right)\times {\left({\mathit{D}}_1\right)}^2\times {\mathit{v}}_1$
$\mathit{Q}=\left(\frac{\mathit{\pi }}{4}\right)\times {\left({\mathit{D}}_2\right)}^2\times {\mathit{v}}_2$
${\mathit{D}}_1=30\mathit{c}\mathit{m}$, ${\mathit{D}}_2=15\mathit{c}\mathit{m}$
Hence
${\mathit{V}}_1=1.6977$ m/sec
${\mathit{V}}_2=6.79$ m/sec
$\left(\frac{{\mathit{P}}_1}{\mathit{r}}\right)+\left(\frac{{\mathit{v}}_1^2}{2\mathit{g}}\right)+{\mathit{z}}_1$
$=\left(\frac{{\mathit{P}}_2}{\mathit{r}}\right)+\left(\frac{{\mathit{v}}_2^2}{2\mathit{g}}\right)+{\mathit{z}}_2$
= 6.792 – 1.69772 2 × 9.81 = 2.2034
$=\left(\frac{{\mathit{P}}_1}{\mathit{r}}\right)-\left(\frac{{\mathit{P}}_2}{\mathit{r}}\right)-0.8$ = 12.6 h = 2.2034
h = 0.175 m
h = 17.5 cm