Question

A thin glass plate of area A and mass m is hinged along one of sides.the speed with which the air should be blown parrell to the upper surface to hold the plate horizontal is Y√mg/18₱A . if ₱ is density of air. then value Y is ?
where Y is constant.​

Answer 1

Given:

A thin glass plate of area A and mass m is hinged along with one of the sides.

The speed with which the air should be blown parallel to the upper surface to hold the plate horizontal is Y =$\sqrt{\frac{mg}{18pA} }$

P is the density of air.

To find: The value of Y?

Explanation:

by applying bernoulli's theorem for above the plate and below the plate, we have,        

                          $p_{1} = p_{2} +\frac{1}{2} pr^{2}$

                          $p_{1}-p_{2} =\frac{1}{2} pr^{2}$

the upward force is,

                         $(p_{1} - p_{2})A = \frac{1}{2} pr^{2}A$

∴                                $\frac{1}{2} pv^{2}A = mg$

                                      $v = \sqrt{\frac{2mg}{pA} }$

                                         $= 6\sqrt{\frac{mg}{18pA} }$

                                          $y$ $= 6$

Hence the value of y is 6.

Answer 2

Answer:

The value of Y is 6.

Explanation:

According to the statement,

let the below plate...(i)

And the above plate...(ii)

we will apply Bernoulli's theorem for it..

Here, p₁ and p₂ are pressure. So,

p₁ = p₂+1/2ρv²

p₁-p₂=1/2рv²

As we know that P=F/A. So, PA=F

So, the upward force is:

(p₁-p₂)A= 1/2рv²A

hence, 1/2р2рv²A= mg

We get, v=√2mg/рA

Multiplying 18 inside the root,

v= 6√mg/18рA

Hence, the value of constant Y is 6.