JEE MAIN - Physics (2020 - 8th January Evening Slot - No. 13)
Two liquids of densities $${\rho _1}$$ an $${\rho _2}$$ ($${\rho _2}$$ = 2$${\rho _1}$$) are
filled up behind a square wall of side 10 m as
shown in figure. Each liquid has a height of
5 m. The ratio of the forces due to these liquids
exerted on upper part MN to that at the lower part
NO is (Assume that the liquids are not mixing)
_8th_January_Evening_Slot_en_13_1.png)
1/3
1/2
1/4
2/3
Explanation
Force F1 on MN = $${{\rho gh} \over 2} \times A$$
Force F2 on NO = $$\left( {\rho gh + {{2\rho gh} \over 2}} \right) \times A$$
$${{{F_1}} \over {{F_2}}} = {1 \over 4}$$
Force F2 on NO = $$\left( {\rho gh + {{2\rho gh} \over 2}} \right) \times A$$
$${{{F_1}} \over {{F_2}}} = {1 \over 4}$$
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