Question

A tank with a square base of area 2 $\mathrm { cm } ^ { 2 }$at the bottom of the partition. Water is filled in one compartment and an acid of relative density 1.5 in the other, both to a height of 4 m. The force necessary to keep the door closed is

(Take )

• A. 10 N
• B. 20 N
• C. 40 N
• D. 80 N

Answer 1

Answer: (C)

40 N

Answer 2

The situation is as shown in the figure

For compartment containing water,

$\mathrm { h } = 4 \mathrm {~m} , \rho _ { \mathrm { w } } = 10 ^ { 3 } \mathrm {~kg} \mathrm {~m} ^ { - 3 }$

Pressure exerted by the water at the door at the bottom is

$= 4 \times 10 ^ { 4 } \mathrm { Nm } ^ { - 2 }$

For compartment containing acid

$\rho _ { \mathrm { a } } = 1.5 \times 10 ^ { 3 } \mathrm {~kg} \mathrm {~m} ^ { - 3 } , \mathrm {~h} = 4 \mathrm {~m}$

Pressure exerted by the acid at the door at the bottom is

$= 6 \times 10 ^ { 4 } \mathrm { Nm } ^ { - 2 }$

$\therefore$ Net pressure on the door )

Area of the door

$= 20 \mathrm {~cm} ^ { 2 } = 20 \times 10 ^ { - 4 } \mathrm {~m} ^ { 2 }$

$= 40 \mathrm {~N}$

Thus to keep the door closed the force of 40 N must be applied horizontally from the water side