A cylinder containing water, stands on a table of height H. A small hole is punched in the side of cylinder at its base. The stream of water strikes the ground at a horizontal distance R from the table. Then the depth of water in the cylinder is

Question

A cylinder containing water, stands on a table of height H. A small hole is punched in the side of cylinder at its base. The stream of water strikes the ground at a horizontal distance R from the table. Then the depth of water in the cylinder is (A) H (B) R (C) √R H (D) R2 / 4H

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/54455b6059ec9eaf7373e2fb67213684-.png" /> Let depth of water in cylinder be x So velocity (v) of efflux =√2 g x Time taken (t) by water to travel vertical distance of H =√(2 H/g) ⇒ Range =v × t R=√2 g x × √(2 H/g) Solving this we get (R2/4 H)= x

Q. A cylinder containing water, stands on a table of height $H$ . A small hole is punched in the side of cylinder at its base. The stream of water strikes the ground at a horizontal distance $R$ from the table. Then the depth of water in the cylinder is

$H$

$R$

$R H$

$R^{2} /4 H$

Let depth of water in cylinder be $x$
So velocity $(v)$ of efflux $= 2 gx$
Time taken $(t)$ by water to travel vertical distance of $H$
$= \frac{2 H}{g}$
$\Rightarrow$ Range $= v \times t$
$R = 2 gx \times \frac{2 H}{g}$
Solving this we get
$\frac{R ^{2}}{4 H} = x$

Q. A cylinder containing water, stands on a table of height H. A small hole is punched in the side of cylinder at its base. The stream of water strikes the ground at a horizontal distance R from the table. Then the depth of water in the cylinder is

H

R

RH​

R2/4H

Let depth of water in cylinder be x So velocity (v) of efflux =2gx​ Time taken (t) by water to travel vertical distance of H =g2H​​ ⇒ Range =v×t R=2gx​×g2H​​ Solving this we get 4HR2​=x

Q. A cylinder containing water, stands on a table of height $H$ . A small hole is punched in the side of cylinder at its base. The stream of water strikes the ground at a horizontal distance $R$ from the table. Then the depth of water in the cylinder is

$H$

$R$

$R H$

$R^{2} /4 H$

Let depth of water in cylinder be $x$
So velocity $(v)$ of efflux $= 2 gx$
Time taken $(t)$ by water to travel vertical distance of $H$
$= \frac{2 H}{g}$
$\Rightarrow$ Range $= v \times t$
$R = 2 gx \times \frac{2 H}{g}$
Solving this we get
$\frac{R ^{2}}{4 H} = x$