A 2 m long tube closed at one end is lowered vertically into water until the closed end is flush with the water surface, Calculate h as shown in the following figure: (Barometric pressure =1 atm =10 m of hydrostatic water head, temperature =25° C, density of water =1 g / ml, neglect water vapour pressure)

Question

A 2 m long tube closed at one end is lowered vertically into water until the closed end is flush with the water surface, Calculate h as shown in the following figure: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/40df10844e94cb5877e7e904a5759a50-.png" /> (Barometric pressure =1 atm =10 m of hydrostatic water head, temperature =25° C, density of water =1 g / ml, neglect water vapour pressure) (A) 1.01 m (B) 0.29 m (C) 1.71 m (D) 0.92 m

Tardigrade Admin · Accepted Answer

Let the cross section area be a now as per Boyle's law P 1 V1=P2 V2 (10+h) h × a=10 × 2 × a ⇒ 10 h+h2=20 ⇒ h=1.71 m

$1.01 m$

$0.29 m$

$1.71 m$

$0.92 m$

Let the cross section area be $a$ now as per Boyle's law
$P_{1} V_{1} = P_{2} V_{2}$
$(10 + h) h \times a = 10 \times 2 \times a$
$\Rightarrow 10 h + h^{2} = 20$
$\Rightarrow h = 1.71 m$

1.01m

0.29m

1.71m

0.92m

Let the cross section area be a now as per Boyle's law P1​V1​=P2​V2​ (10+h)h×a=10×2×a ⇒10h+h2=20 ⇒h=1.71m

$1.01 m$

$0.29 m$

$1.71 m$

$0.92 m$

Let the cross section area be $a$ now as per Boyle's law
$P_{1} V_{1} = P_{2} V_{2}$
$(10 + h) h \times a = 10 \times 2 \times a$
$\Rightarrow 10 h + h^{2} = 20$
$\Rightarrow h = 1.71 m$