Question

two pistons of hydraulic press have diameter 24 cm and 2cm. Calculate the force exerted by the longer piston when 40 kg wt is placed on smaller piston​

Answer 1

Explanation:

☢️PLEASE MARK AS BRAINLIEST☢️

Answer 2

Answer:

5760kg wt

Explanation:

Given :

• Diameter of hydraulic press = 24 cm and 2cm

Radius = 24/2 = 12 cm

Radius = 2/2 = 1 cm

To Calculate :

• Force exerted by the longer piston when 40 kg wt is placed on smaller piston.

Formula Going To Be applied :

• $\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \: area \: of \: circle = \: \pi {r}^{2}$
• Will be applying Pascal's law ,

P1 = P2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \: area \: of \: circle = \pi \times 1 {} {cm}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \: \pi {cm}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \: area \: of \: circle = \pi( {12)}^{2} c {m}^{2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \: area \: of \: circle = \pi144 {cm}^{2}$

We know , F1 = 40 kg wt.

We have to find F2.

Now, we will apply Pascal's law , which says that

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \: \dfrac{F 1}{A1} = \dfrac{F2}{ A2}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf \: \dfrac{40}{\pi} = \dfrac{F2}{12} \\ \\ cross \: multiplication$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies \: F2 = \dfrac{144\pi \times 40}{\pi}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \implies \sf \: 5760 \: kg \: wt$