Question

The area of one end of a U-tube is 0.01 m2and that of the other end is 1 m2.when a force was applied on the liquid at the first end ,the force experienced at the other end was 25000 N. What was the force applied on the liquid at the first end?​

Answer 1

Answer:

Given:

The area of one end of a U- tube is 0.01 metre square and that of the other end is 1 m². When a force was applied on the liquid at the 1st end, the force experienced at the other end is 20000 N.

To find:

Force applied at the 1st end.

Calculation:

Pascal's Law states that the pressure exerted by fluid will be same at all points at a specified depth.

Hence,

\sf{\therefore \: P1 = P2}∴P1=P2

\sf{ = > \: \dfrac{f1}{a1} = \dfrac{f2}{a2} }=>

a1

f1

=

a2

f2

\sf{ = > \: \dfrac{f1}{0.01} = \dfrac{20000}{1} }=>

0.01

f1

=

1

20000

\sf{ = > \: \dfrac{f1}{ {10}^{ - 2} } = \dfrac{20000}{1} }=>

10

−2

f1

=

1

20000

\sf{ = > \: f1= 20000 \times {10}^{ - 2} }=>f1=20000×10

−2

\sf{ = > \: f1= 200 \: N }=>f1=200N

So, the force exerted at the first end is 200 N.

Answer 2

Answer:

$\huge\fcolorbox{red}{lime}{Answer \ \: : }$

As we know that pressure head remains constant at a particular horizontal level.

[tex] \bold{⇒ \: A1 = 0.01 {m}^{2} } \\\bold{⇒ \: A2

[tex] \bold{⇒ \: A1 = 0.01 {m}^{2} } \\\bold{⇒ \: A2 = 1 {m}^{2} } \\ \bold{⇒ \: F1

[tex] \bold{⇒ \: A1 = 0.01 {m}^{2} } \\\bold{⇒ \: A2 = 1 {m}^{2} } \\ \bold{⇒ \: F1 = ?} \\ \bold{⇒ \: P1=P2

[tex] \bold{⇒ \: A1 = 0.01 {m}^{2} } \\\bold{⇒ \: A2 = 1 {m}^{2} } \\ \bold{⇒ \: F1 = ?} \\ \bold{⇒ \: P1=P2} \\ \bold{⇒ \: F2=20000N} \\ \bold{ ⇒ \: \frac{F1

[tex] \bold{⇒ \: A1 = 0.01 {m}^{2} } \\\bold{⇒ \: A2 = 1 {m}^{2} } \\ \bold{⇒ \: F1 = ?} \\ \bold{⇒ \: P1=P2} \\ \bold{⇒ \: F2=20000N} \\ \bold{ ⇒ \: \frac{F1}{F2} = \frac{A1}{A2} } \\ \bold{⇒ \: F1 =F2 \times \frac{A1}{A2} = 20000 \times \frac{0.01}{1} = 200N}