In a hydraulic lift, used at a service station the radius of the large and small pistons are in the ratio of 20:1. What weight placed on the small piston will be sufficient to lift a car of mass 1500 Kg?

Open in App

Solution

Step 1: Given data
We are given the ratio of the radius of the large and the small piston. If $r_{1}$ is the radius of the large piston and $r_{2}$ is the radius of the small piston, the ratio is given as,
$\frac{r_{1}}{r_{2}} = \frac{20}{1}$
Weight of car, $W_{2} = 1500 k g$
Step 2: Formula used
According to Pascal's law, pressure applied to a liquid at one point is distributed evenly across the liquid in all directions.
$P_{1} = P_{2} P = \frac{F}{A} [w h e r e P = p r e s s u r e, F = F o r c e, A = a r e a]$
Step 3: Calculation
We can determine the force required to lift the car's weight using the relationship between pressures for small and large pistons.
The pressure applied on the larger piston is equal to the pressure applied on the smaller piston.
$P_{1} = P_{2} [P_{1} = p r e s s u r e a p p l i e d o n l a r g e r p i s t o n, P_{2} = p r e s s u r e a p p l i e d o n s m a l l e r p i s t o n] \frac{W_{1}}{π {r_{1}}^{2}} = \frac{W_{2}}{π {r_{2}}^{2}} [A r e a = π r^{2}] W_{2} = \frac{{r_{2}}^{2}}{{r_{1}}^{2}} W_{1} [W_{1} = w e i g h t o f l a r g e r p i s t o n, W_{2} = w e i g h t o f s m a l l e r p i s t o n]$
Now we will use the values that are already given. By doing so we obtain the force required to be
$W_{2} = {(\frac{1}{20})}^{2} \times 1500 = 3.75 k g$
Hence, the weight to be placed on the small piston sufficient to lift a car of mass of 1500 kg is 3.75 kg.

Suggest Corrections

Similar questions

0.20 m^{2}

1.20 \times 10^{4} N

0.90 m^{2}

2 c m^{2}

80 c m^{2}

3500 k g

500 c m^{2}

g = 9.8 m / s^{2} .

Join BYJU'S Learning Program