In a hydraulic jack as shown, mass of the car $W=800 \,kg , A_1=10 \,cm ^2, A_2=10 \,m ^2$. The minimum force $F$ required to lift the car is $........... N$
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Pressure in a liquid is divided equally so we can say pressure at both the pistons should be same
$\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2} \left\{\begin{array}{l}\text { Where, } \\ F_1=F \\ A_1=10 \,cm ^2 \\ A_2=10 \,m ^2=10 \times 10^4 \,cm ^2 \\ F_2=8000 \,N \end{array}\right.$
Substituting values,
$\frac{F}{10}=\frac{8000}{10 \times 10^4}$
$\Rightarrow F=0.8 \,N$ $\left\{\begin{array}{l}\text { Take } \\ g=10 \,m / s ^2\end{array}\right\}$
$\Rightarrow \frac{F_1}{A_1}=\frac{F_2}{A_2} \left\{\begin{array}{l}\text { Where, } \\ F_1=F \\ A_1=10 \,cm ^2 \\ A_2=10 \,m ^2=10 \times 10^4 \,cm ^2 \\ F_2=8000 \,N \end{array}\right.$
Substituting values,
$\frac{F}{10}=\frac{8000}{10 \times 10^4}$
$\Rightarrow F=0.8 \,N$ $\left\{\begin{array}{l}\text { Take } \\ g=10 \,m / s ^2\end{array}\right\}$
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