QJEEM24Ph1622: JEE Main 2024 Physics - Class 11 JEEM24Ph1622 Solution: Three identical spheres each of mass are placed at the corners…

The center of mass (COM) of a system of particles is calculated as:

$X_{COM} = \frac{\sum_{i} m_i x_i}{\sum_{i} m_i}$

$Y_{COM} = \frac{\sum_{i} m_i y_i}{\sum_{i} m_i}$ $$</p> <p>where <span>$m_i$</span> is the mass of the <span>$i$</span>-th particle, and <span>$(x_i, y_i)$</span> are its coordinates.</p> <strong>2. Coordinate Setup:</strong> <p>Let's place the origin at the right angle of the triangle and align the sides along the x and y axes:</p> <p></p><ul> <li>Sphere 1: <span>$(4, 0)$</span></li><br> <li>Sphere 2: <span>$(0, 4)$</span></li><br> <li>Sphere 3: <span>$(0, 0)$</span></li> </ul><p></p> <strong>3. Calculations:</strong> <p>Since all spheres have mass <span>$2M$</span>, we can simplify the COM calculations:</p> <p><mjx-container class="MathJax CtxtMenu_Attached_0" jax="CHTML" display="true" tabindex="0" ctxtmenu_counter="0" style="font-size: 130.9$\left(\frac{4}{3}, \frac{4}{3}\right)$</span>. Its magnitude is:</p> <p><mjx-container class="MathJax CtxtMenu_Attached_0" jax="CHTML" display="true" tabindex="0" ctxtmenu_counter="2" style="font-size: 130.9$\frac{4\sqrt{2}}{x}$</span>. Comparing this to our result, we find that <span>$x = 3$</span>.</p> <strong>Answer:</strong> <p>The value of <span>$x$ is 3.