Tag: physics

$f=900 N$
$a=\pi \times (\frac {1.8}{2})^2=\pi \times \frac {(1.8)^2}{4}$
$A=\pi \times (\frac {36}{2})^2=\pi \times \frac {(36)^2}{4}$
Now $\frac {f}{a}=\frac {F}{A}$
$\Rightarrow \frac {900\times 4}{\pi (1.8)^2}=\frac {F\times 4}{\pi \times (36)^2}$
$\therefore F=\frac {900\times 4}{\pi \times 1.8\times 1.8}\times \frac {\pi \times 36\times 36}{4}$
$=36\times 10^4N$.

$\begin{array}{l} \dfrac { f }{ { 10 } } =\dfrac { { 1000 } }{ { 225 } }  \ f=\dfrac { { 10000 } }{ { 225 } }  \ f=44.44N \end{array}$

The hydraulic machine uses Pascal's principle, which states that states that the pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid.
The hydraulic machine is used to multiple force or to transmit force from one location to another. 
i)     hydraulic brakes
ii)    hydraulic press
ii)    hydraulic lift

We know that $P _{1} = P _{2}$

$\Rightarrow \dfrac{F _{1}}{A _{1}} = \dfrac{F _{2}}{A _{2}} $

$\Rightarrow \dfrac{A _{1}}{A _{2}} = \dfrac{F _{1}}{F _{2}}  = \dfrac{1}{30} $

$\Rightarrow  1:30 $

Answer is A.

Pressure is the amount of force acting per unit area. That is, $P=F/A$.
where:
$p$ is the pressure,
$F$ is the normal force,
$A$ is the area of the surface on contact. Let us consider A = $\pi { r }^{ 2 }$.
Therefore, $\dfrac { { F } _{ 1 } }{ { \pi { r } _{ 1 } }^{ 2 } } =\dfrac { { F } _{ 2 } }{ { \pi { r } _{ 2 } }^{ 2 } } $.
In this case, $\dfrac { 1500 }{ { 20 }^{ 2 } } =\dfrac { W }{ { 1 }^{ 2 } } ,\quad W\quad =\quad 3.75\quad kg.$
Hence, weight to be placed on the small piston sufficient to lift a car of mass 1500 kg is 3.75 kg.

To avoid vaporization in the pipe line, the pipe line over the bridge is laid such that it is not more than 6.4m above hydraulic gradient

Weight on the ram  $F  =  mg=1000\times 10 = 10000$

The maximum number of jets generally employed in an impulse turbine without jet interference are 6.

Using   $\dfrac{F _1}{A _1} = \dfrac{F _2}{A _2}$