When an air bubble of radius r rises from the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the depth of the lake be $h$. The atmospheric pressure $P_0$ is equivalent to a water column of height $H = 10 \ m$. At the bottom of the lake

Vedclass · Accepted Answer

$9.53$

Vedclass · Answer

(C) Let the depth of the lake be $h$. The atmospheric pressure $P_0$ is equivalent to a water column of height $H = 10 \ m$. At the bottom of the lake, the pressure $P_1 = P_0 + h ho g = (H + h) ho g$. At the surface, the pressure $P_2 = P_0 = H ho g$. Since the temperature is constant, we use Boyle's Law: $P_1 V_1 = P_2 V_2$. Given $V_1 = \frac{4}{3} \pi r^3$ and $V_2 = \frac{4}{3} \pi (\frac{5r}{4})^3$. Substituting the values: $(H + h) ho g 	imes \frac{4}{3} \pi r^3 = H ho g 	imes \frac{4}{3} \pi (\frac{5r}{4})^3$. $(10 + h) = 10 	imes (\frac{5}{4})^3$. $10 + h = 10 	imes \frac{125}{64} = \frac{1250}{64} = 19.53125$. $h = 19.53125 - 10 = 9.53125 \ m \approx 9.53 \ m$.

Similar Questions

$A$ student has to answer $10$ out of $13$ questions in an examination,choosing at least $5$ questions from the first $6$ questions. The number of choices available to the student is:

The length $x$ of a rectangle is decreasing at the rate of $3 \text{ cm/min}$ and the width $y$ is increasing at the rate of $2 \text{ cm/min}$. When $x = 10 \text{ cm}$ and $y = 6 \text{ cm}$,find the rate of change of the perimeter.

The equation of the circle whose diameter is the common chord of the circles $x^2+y^2+2x+3y+2=0$ and $x^2+y^2+2x-3y-4=0$ is

The square of the difference of the slopes of the lines represented by the equation $x^2(\sec^2 \theta - \sin^2 \theta) - 2xy \tan \theta + y^2 \sin^2 \theta = 0$ is:

What are the metal ions present in carnallite?

Vedclass Test Series

Exam Paper Generator

Online Exam Module