
To find the ratio of masses \( m_1 \) and \( m_2 \), we can use the concepts of mechanics involving pulleys and blocks.
Given:
Consider the forces acting on each mass:
Setting these two expressions for tension \( T \) equal gives:
\(m_1g - m_1a = m_2g + m_2a\)
Substituting \( a = \frac{g}{8} \):
\(m_1g - m_1\left(\frac{g}{8}\right) = m_2g + m_2\left(\frac{g}{8}\right)\)
Simplifying further:
\(m_1g\left(1 - \frac{1}{8}\right) = m_2g\left(1 + \frac{1}{8}\right)\)
\(\frac{7m_1g}{8} = \frac{9m_2g}{8}\)
Cancelling \( g \) and simplifying gives:
\(7m_1 = 9m_2\)
Thus, the ratio of the masses is:
\(\frac{m_1}{m_2} = \frac{9}{7}\)
Therefore, the correct answer is \(\frac{9}{7}\).
The acceleration \( a \) of the system is given by:
\[ a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{g}{8}. \]This implies:
\[ 8m_1 - 8m_2 = m_1 + m_2. \]Rearrange terms:
\[ 7m_1 = 9m_2. \]Thus, the ratio of \( m_1 \) to \( m_2 \) is:
\[ \frac{m_1}{m_2} = \frac{9}{7}. \]Therefore, the answer is:
\[ \frac{9}{7}. \]
A particle of mass \(m\) falls from rest through a resistive medium having resistive force \(F=-kv\), where \(v\) is the velocity of the particle and \(k\) is a constant. Which of the following graphs represents velocity \(v\) versus time \(t\)? 
The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1 \(\Omega\) is connected across these points is ______ J. 
The given circuit works as: 
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}