To solve this problem, we need to understand the relationship between excess pressure inside a soap bubble and its radius, and subsequently its volume.
The formula for excess pressure (\(\Delta P\)) inside a soap bubble is given by:
\(\Delta P = \frac{4T}{R}\)
where \(T\) is the surface tension of the liquid and \(R\) is the radius of the soap bubble.
According to the question, the excess pressure inside the first bubble is thrice that inside the second bubble, so:
\(\Delta P_1 = 3 \Delta P_2\)
By substituting the formula for excess pressure, we get:
\(\frac{4T}{R_1} = 3 \times \frac{4T}{R_2}\)
After simplifying, the above equation becomes:
\(\frac{1}{R_1} = \frac{3}{R_2}\)
This implies:
\(R_2 = 3R_1\)
The volume of a sphere (or bubble) is given by:
\(V = \frac{4}{3} \pi R^3\)
Thus, the volumes of the bubbles can be represented as:
After substituting \(R_2 = 3R_1\) into the equation for \(V_2\), we find:
\(V_2 = \frac{4}{3} \pi (27R_1^3) = 27 \times \frac{4}{3} \pi R_1^3 = 27V_1\)
Therefore, the ratio of the volumes of the first and second bubble is:
\(\frac{V_1}{V_2} = \frac{1}{27}\)
Based on this calculation, the correct answer is 1 : 27.
The excess pressure \( P_{\text{excess}} \) inside a soap bubble is given by the formula:
\[ P_{\text{excess}} = \frac{4T}{r}, \] where \( T \) is the surface tension and \( r \) is the radius of the bubble.
Let the radii of the two bubbles be \( r_1 \) and \( r_2 \), and their excess pressures be \( P_1 \) and \( P_2 \), respectively.
Given:
\[ P_1 = 3P_2. \]
Using the formula for excess pressure:
\[ \frac{4T}{r_1} = 3 \times \frac{4T}{r_2}. \]
Cancelling common terms:
\[ \frac{1}{r_1} = 3 \times \frac{1}{r_2} \Rightarrow r_1 = \frac{r_2}{3}. \]
Since the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \), the ratio of the volumes is:
\[ \frac{V_1}{V_2} = \left( \frac{r_1}{r_2} \right)^3 = \left( \frac{1}{3} \right)^3 = \frac{1}{27}. \]
Thus, the ratio of the volumes is \( 1 : 27 \).
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,


\( x \) is a peptide which is hydrolyzed to 2 amino acids \( y \) and \( z \). \( y \) when reacted with HNO\(_2\) gives lactic acid. \( z \) when heated gives a cyclic structure as below:

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)