Question 20: Properties of Matter

Question

The surface tension of a soap bubble is 0.03 N/m. The work done in increasing the diameter of bubble from 2 cm to 6 cm is $\alpha \pi 	imes 10^{-4}$ J. The value of $\alpha$ is _________. (Take $\pi = 3.14$)

Accepted Answer

Young's Modulus is given by :
$$ \mathrm{Y}=\frac{	ext { Stress }}{	ext { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \mathrm{~L} / \mathrm{L}}=\frac{\mathrm{FL}}{\mathrm{~A} \Delta \mathrm{~L}} $$
Where:
F is the applied force (load)
L is the original length.
A is the area of cross-section
$\quad \Delta \mathrm{L}$ is the change in length (stretching)
According to question, both wires are stretched by the same magnitude ( $\Delta \mathrm{L}_{\mathrm{A}}=\Delta \mathrm{L}_{\mathrm{B}}$ ) under the same given load $\left(\mathrm{F}_{\mathrm{A}}=\mathrm{F}_{\mathrm{B}}ight)$.
Therefore, for both wires:
$$ Y \propto \frac{L}{A} $$
Let $Y_A$ and $Y_B$ be the Young's moduli for wires $A$ and $B$ :
$$ \frac{Y_A}{Y_B}=\frac{\left(\frac{L_A}{A_A}ight)}{\left(\frac{L_B}{A_B}ight)}=\frac{L_A}{L_B} 	imes \frac{A_B}{A_A} $$
The given values :
$\mathrm{L}_{\mathrm{A}}=6.0 \mathrm{~cm}, \mathrm{~L}_{\mathrm{B}}=5.4 \mathrm{~cm}$
$\mathrm{A}_{\mathrm{A}}=3.0 	imes 10^{-5} \mathrm{~m}^2, \mathrm{~A}_{\mathrm{B}}=4.5 	imes 10^{-5} \mathrm{~m}^2$
$$ \begin{gathered} \frac{Y_A}{Y_B}=\frac{6.0}{5.4} 	imes \frac{4.5 	imes 10^{-5}}{3.0 	imes 10^{-5}} \ \frac{Y_A}{Y_B}=\frac{5}{3}=\frac{x}{3} \Rightarrow x=5 \end{gathered} $$
So, the value of x is 5 . Hence, the correct option is (D).

Answer

0.86

Answer

0.64

Answer

1.92

Answer

7.68

Properties of Matter

The surface tension of a soap bubble is 0.03 N/m. The work done in increasing the diameter of bubble from 2 cm to 6 cm is \alpha \pi \times 10^{-4} J. The value of \alpha is _________. (Take \pi = 3.14)

Choose the correct answer: