Question

A barometer kept in an elevator reads $76$ cm when it is ar rest if the elevator goes up with increasing speed.

Answer 1

REF. Image

When the elevator is at rest,

$f_{Hg}g\left(\frac{76}{100}\right)=P_0\Rightarrow \frac{76}{100}=\frac{P_0}{f_{Hg}\left(g\right)}$

(ii) When the elevator is speeding

(let the acceleration be a). Then,

$f_{Hg}(g+a)h^1=P_0$

$h^1=\frac{P_0}{(g+a)f_{Hg}}$

As $a>0,$

$h^1<\frac{P_0}{\left(g\right)f_{Hg}}\Rightarrow h^1<h\Rightarrow h^1<\frac{76}{w_0}m$

So, the new reading will be

less than 76 cm