Question

two masses (m_1) and (m_2) are attached to a mass - less spring balance as shown in figure 1. the spring balance is in a mass - less, friction - less system. the reading of the spring balance is (a) ((m_1 - m_2)g) (b) (\frac{m_1m_2g}{m_1 + m_2}) (c) (\frac{2m_1m_2g}{m_1 + m_2}) (d) (\frac{m_1 + m_2}{m_1m_2}g)

Explanation:

Step1: Analyze forces on masses

Let the acceleration of the system be \(a\). For mass \(m_1\), \(m_1g - T_1=m_1a\). For mass \(m_2\), \(T_2 - m_2g=m_2a\). Also, \(T_1 = T_2\) (since it's the same string).

Step2: Solve for acceleration

From \(m_1g - T_1=m_1a\) and \(T_2 - m_2g=m_2a\) with \(T_1 = T_2\), we get \(m_1g - m_1a=m_2g + m_2a\). Rearranging gives \(a=\frac{(m_1 - m_2)g}{m_1 + m_2}\).

Step3: Find tension in string

Substitute \(a\) into \(m_1g - T_1=m_1a\). \(T_1=m_1g - m_1\times\frac{(m_1 - m_2)g}{m_1 + m_2}=\frac{m_1(m_1 + m_2)g-m_1(m_1 - m_2)g}{m_1 + m_2}=\frac{2m_1m_2g}{m_1 + m_2}\).
The reading of the spring - balance is equal to the tension in the string.

Answer:

\(\frac{2m_1m_2g}{m_1 + m_2}\) (assuming the options are related to the tension value in the string which is the reading of the spring - balance)