Question

(1) 1:2
(3) 4 : 1
(2) 2:1
(4) 1:1

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Answer 1

Given:

↬ In Container-1

Height of liquid= h

Pressure exerted by liquid at bottom of container= P₁

↬ In Container-2

Height of liquid= h/2

Pressure exerted by liquid at bottom of container= P₂

To Find:

Ratio of P₁ and P₂

Solution:

We know that,

• Pressure P exerted by liquid at the bottom of container is given by

$\pink{\boxed{\bf{P=h\rho g}}}$

where,

h is the height of container up to which liquid is filled  

ρ is the density of liquid

g is acceleration due to gravity

$\rule{190}{1}$

Let the density of given liquid be 'ρ'

So,

$\longrightarrow\rm{P_{1}=h\rho g}$

Also,

$\longrightarrow\rm{P_{2}=\dfrac{h}{2}\times\rho g}$

$\longrightarrow\rm{P_{2}=\dfrac{h\rho g}{2}}$

Now,

$\longrightarrow\rm{\dfrac{P_{1}}{P_{2}}=\dfrac{h\rho g}{\dfrac{h\rho g}{2}}}$

$\longrightarrow\rm{\dfrac{P_{1}}{P_{2}}=\dfrac{2\cancel{h\rho g}}{\cancel{h\rho g}}}$

$\longrightarrow\rm{\dfrac{P_{1}}{P_{2}}=\dfrac{2}{1}}$

So,

$\longrightarrow\rm{P_{1}:P_{2}=2:1}$

Hence, the required ratio is 2:1.

Answer 2

Given :

• In container-1, liquid is filled till height h
• In container-2 , liquid is filled till height h/2

To find :

• Ratio of pressure of the liquids in both the containers (P1:P2)

Solution :

As we know that,

$\implies \sf{P \: = \: h \rho g}$

Where,

• P is pressure
• h is height up-to which liquid is filled
• ρ is density of liquid
• g is acceleration due to gravity

For Container 1:

$\implies \sf{P \: = \: h \rho g} \\ \\ \implies \sf{P_1 \: = \: h \rho g} \: \: \: \: \: \: ...(1)$

____________________

For Container 2:

$\implies \sf{P_2 \: = \: \dfrac{h}{2} \rho g} \\ \\ \implies \sf{P_2 \: = \: \dfrac{h \rho g}{2}} \: \: \: \: \: \: ... (2)$

_______________________________

Divide equation (1) by (2)

$\implies \sf{\dfrac{P_1}{P_2} \: = \: \dfrac{\dfrac{h \rho g }{h \rho g}}{2}} \\ \\ \implies \sf{\dfrac{P_1}{P_2} \: = \: \dfrac{2 h \rho g}{h \rho g}} \\ \\ \implies \sf{Ratio \: = \: \dfrac{2}{1}} \\ \\ \implies \sf{Ratio \: = \: 2:1}$