Question

In a motor garage, a hydraulic lift is used to lift heavy automobiles such as trucks. The cross-sectional areas of the piston carrying the load and the piston where force is applied are in the ratio 5 : 3. The mass of a truck is 15000 kg. The magnitude of the force applied on the piston to just lift the truck is (Take g$=9.8m{s}^{-2}$)

• A. $8.82\times {10}^4$ N
• B. $8.82\times {10}^3$ N
• C. $8.82\times {10}^2$ N
• D. 882 N

Answer 1

The correct option is A

$8.82\times {10}^4$ N

Let $F_1$ = Force applied on the piston to just lift the truck

$F_2$ = Force applied by the truck

$A_1$ = Area of the piston where force is applied

$A_2$ = Area of the piston carrying the load

Mass of truck, $m=15000$ kg and g$=9.8m{s}^{-2}$.

Now, $F_2=m\times g=15000\times 9.8N$

Now according to Pascal law,

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$,

$F_1$ = $F_2\times \left(\frac{A_1}{A_2}\right)$

It is given $\frac{A_1}{A_2}=\frac{3}{5}$

Therefore, $F_1=15000\times 9.8\times \left(\frac{3}{5}\right)=88200=8.82\times {10}^4$ N.