For the system shown in the figure, the cylinder on the left at L has a mass of 600kg and a cross sectional area of 800 cm2. The piston on the right, at S has cross sectional area 25cm2 and negligible weight. If the apparatus is filled with oil ρ=0.75g/cm3. Find the force F (in N) required to hold the system in equilibrium. Take g=10m/s2

In equilibrium, 600×10800×10−4=F25×10−4+hρg⇒F25×10−4=608×104−8×0.75×103×10⇒F25×10−4=1.5×104;⇒F=37.5N

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For the system shown in the figure, the cylinder on the left at L has a mass of 600kg and a cross sectional area of $800 c m^{2}$ . The piston on the right, at S has cross sectional area 25cm² and negligible weight. If the apparatus is filled with oil $(ρ = 0.75 g / {cm}^{3})$ . Find the force F (in N) required to hold the system in equilibrium. Take $g = 10 m / s^{2}$

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answer is 37.5.

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Detailed Solution

In equilibrium, $\frac{600 \times 10}{800 \times 10^{- 4}} = \frac{F}{25 \times 10^{- 4}} + h ρ g$

$\Rightarrow \frac{F}{25 \times 10^{- 4}} = \frac{60}{8} \times 10^{4} - 8 \times (0.75 \times 10^{3}) \times 10$

$\Rightarrow \frac{F}{25 \times 10^{- 4}} = 1.5 \times 10^{4}; \Rightarrow F = 37.5 N$

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