Question

In the arrangement shown above, a block of mass 2700 kg is in equilibrium on applying a force F. The value of force F ? ($d_{liquid}=0.75gc{m}^{-3}is$)

• A. $147N$
• B. $1755.18N$
• C. $153N$
• D. $918N$

Answer 1

The correct option is B $1755.18N$

Total pressure acting on side of
$A_1=\frac{F}{A_1}+h\rho g$
$=\frac{F}{6\times {10}^{-4}}+(2\times 0.75\times 9.8\times {10}^3)pa$
pressure acting on side$A_2=\frac{2700\times 9.8}{90\times {10}^{-4}}pa$

According to Pascal's law
Pressure acting on side of $A_1=$ pressure acting on side $A_2$
$\frac{F}{6\times {10}^{-4}}+(2\times 0.75\times 9.8\times {10}^3)$$=\frac{2700\times 9.8}{90\times {10}^{-4}}$

$\frac{F}{6\times {10}^{-4}}=294\times {10}^4-14.7\times {10}^3$

$F=6\times {10}^{-4}(292.53\times {10}^4)$
$F=1755.18N$