Question

when the system shown in the adjoining figure is in equilibrium and the areas of cross section of small piston and bigger PISTON are a and 10a. Mass on smaller and bigger piston are m and M ..... please see the pic and answer step by step.....

Answer 1

System is in equilibrium.
Therefore,
Pressure due to smaller piston = Pressure due to larger piston
mg / a = Mg / (10a) ……[∵ Pressure = Force / Area]
M = 10 m

(c) is the correct option.

Answer 2

Mass on bigger area M=10 m. Option C is correct.

Step-by-step explanation:

• Given data

        Small piston area = a  $\mathbf{(m^{2})}$

        Mass on small piston =m (kg)

        Bigger piston area = 10 a   $\mathbf{(m^{2})}$

        Mass on bigger piston = M (kg)

        Value of gravity acceleration = g  $\mathbf{(\frac{m}{s^{2}})}$

• So

        Weight or force on small piston = mg (N)

        Weight or force on bigger piston = Mg (N)

• We know that

        $\mathbf{\textrm{Pressure (P)}=\frac{Force}{Area}}$

        So

        $\mathbf{\textrm{Pressure on small piston}\ (P_{s})=\frac{mg}{a}}$    ...1)

        $\mathbf{\textrm{Pressure on bigger piston}\ (P_{b})=\frac{Mg}{10a}}$     ...2)

• Now from law of Pascal

        Pressure on bigger area = Pressure on small area

        $\mathbf{P_{b}=P_{s}}$

        $\mathbf{\frac{Mg}{10a}=\frac{mg}{a}}$

        So

        $\mathbf{M=\frac{mg}{a}\times \frac{10a}{g}}$

        M =10 m