An 1:50 model of an ogee spillway crest records an acceleration of $1.5 m / s e c^{2}$ at a certain location. The homologous value of acceleration in the prototype is
1.5

Open in App

Solution

The correct option is A 1.5
$(F_{r})_{m} = (F_{r})_{p}$
$\Rightarrow \frac{V_{m}}{\sqrt{g L_{m}}} = \frac{V_{p}}{\sqrt{g L_{p}}}$
$\Rightarrow \frac{V_{m}}{V_{p}} = \sqrt{\frac{L_{m}}{L_{p}}} = \sqrt{\frac{1}{50}}$
$\frac{a_{m}}{a_{p}} = \frac{V_{m} / T_{m}}{V_{p} / T_{P}} = \frac{V_{m}}{V_{p}} . \frac{T_{p}}{T_{m}}$
$\Rightarrow \frac{1.5}{a_{p}} = \sqrt{\frac{1}{50}} . \frac{T_{p}}{T_{m}}$ ....(i)

Also, $\frac{V_{m}}{\sqrt{g L_{m}}} = \frac{V_{p}}{\sqrt{g L_{p}}}$
$\Rightarrow \frac{L_{m}}{T_{m} \sqrt{g L_{m}}} = \frac{L_{p}}{T_{p} \sqrt{g L_{p}}}$
$\Rightarrow \frac{T_{m}}{T_{p}} = \sqrt{\frac{L_{m}}{L_{p}}} = \sqrt{\frac{1}{50}}$

$∴$ From equation (i),

$\frac{1.5}{a_{p}} = \sqrt{\frac{1}{50}} \times \sqrt{50} = 1$
$∴ a_{p} = 1.5 m / s^{2}$

Suggest Corrections

Similar questions

m^{3} / s e c .

m^{3} / s e c .

0.3 m^{3} / s e c .

m^{3} / s e c

1000 m^{3} / s

1

40

2

m / s e c

m^{3} / s .

Join BYJU'S Learning Program