The acceleration of the system is given by:
\[ a = \frac{m_2 - m_1}{m_1 + m_2} g \]
Given \(a = \frac{g}{\sqrt{2}}\), substitute in the equation:
\[ \frac{g}{\sqrt{2}} = \frac{m_2 - m_1}{m_1 + m_2} g \]
Simplifying:
\[ \frac{1}{\sqrt{2}} = \frac{m_2 - m_1}{m_1 + m_2} \]
Cross-multiplying:
\[ \sqrt{2}(m_2 - m_1) = m_1 + m_2 \]
Rearranging:
\[ m_1(\sqrt{2} + 1) = m_2(\sqrt{2} - 1) \]
The ratio of masses is:
\[ \frac{m_1}{m_2} = \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \]
To solve this problem, we will analyze the motion of the two blocks connected by a string passing over a pulley. The masses of the blocks are \( m_1 \) and \( m_2 \) (where \( m_2 > m_1 \)), and the acceleration of the system is given as \( \frac{g}{\sqrt{2}} \).
Let's begin by applying Newton's second law to both masses:
Here, \( T \) is the tension in the string, and \( a = \frac{g}{\sqrt{2}} \) is the acceleration of the system.
From the two equations, we have:
Equating the two expressions for tension \( T \), we get:
\(m_1g + m_1a = m_2g - m_2a\)
Plug in the value of \( a = \frac{g}{\sqrt{2}} \) into the equation:
\(m_1g + m_1\left(\frac{g}{\sqrt{2}}\right) = m_2g - m_2\left(\frac{g}{\sqrt{2}}\right)\)
Simplify the equation:
\(g(m_1 + \frac{m_1}{\sqrt{2}}) = g(m_2 - \frac{m_2}{\sqrt{2}})\)
Cancel out the common factor \( g \):
\(m_1(1 + \frac{1}{\sqrt{2}}) = m_2(1 - \frac{1}{\sqrt{2}})\)
Rearrange to express the ratio of the masses:
\(\frac{m_1}{m_2} = \frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}}\)
Simplify the expression using the conjugate:
\(\frac{1 - \frac{1}{\sqrt{2}}}{1 + \frac{1}{\sqrt{2}}} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{(\sqrt{2} - 1)(1 - \frac{1}{\sqrt{2}})}{(\sqrt{2}^2 - 1^2)}\)
This simplifies to:
\(\frac{\sqrt{2} - 1}{\sqrt{2} + 1}\)
Thus, the correct option for the ratio \(\frac{m_1}{m_2}\) is:
\(\frac{\sqrt{2} - 1}{\sqrt{2} + 1}\)


If A and B are two events such that \( P(A \cap B) = 0.1 \), and \( P(A|B) \) and \( P(B|A) \) are the roots of the equation \( 12x^2 - 7x + 1 = 0 \), then the value of \(\frac{P(A \cup B)}{P(A \cap B)}\)