Question

21.
A car of mass 1000 kg is supported on piston of tube B for repairs. It is connected to tube A which is of 1/5
diameter of tube B. Tubes are full of oil. Force F applied on piston of A to support the car is
(a) 200 kg wt
(b) 200 newton
(c)40 kg wt
(d) 40 newton​

Answer 1

Given:

• Mass of car  = 1000 kg

• Diameter of tube $A = \frac{1}{5} th$  of tube $B$

To Find:

• Force exerted at A

Solution:

According to Pascal's Law the pressure exerted at one point is equally distributed throughout the liquid. So,

Pressure at point A is equal to pressure at point B

                         $P_A = P_B$

 now  

                     $P_B = \frac{force}{area}$

                     $P_B = \frac{weight \ of \ car }{area }$

                           $= \frac{mg}{(\frac{ \pi d_B^2}{4} )}$

so,

                  $P_B = \frac{1000 \times 9.81}{(\frac{ \pi d_B^2}{4} )}$

                  $P_B = \frac{9810}{(\frac{ \pi d_B^2}{4} )}$    

so,

                 $P_A = \frac{9810}{(\frac{ \pi d_B^2}{4} )}$

             $\frac{F_A}{(\frac{ \pi d_A^2}{4} )} = \frac{9810}{(\frac{ \pi d_B^2}{4} )}$

or

             $F_A = 9810 \times \frac{ d_A^2}{d_B^2}$

Now     $d_A = \frac{d_B}{5}$

force exerted at A

              $F_A = \frac{9810}{25}$

or

        $F_A = 40N = 40 kg.wt$

Thus, the force exerted at A is $40N$

Answer 2

Answer:

40kg wt i sthe answer of your question.

hope it helps