| Ion | Zn+ | Up+ | Vn+ | Xm- | Ym- |
| λ0 (S cm2 mol-1) | 50.0 | 25.0 | 100.0 | 80.0 | 100.0 |

To solve this problem, we need to use the given information about the limiting molar conductivity of ions and the graph that relates molar conductivity to the concentration of \( Z_mX_n \), \( U_mY_p \), and \( V_mX_n \). The question asks for the value of \( m + n + p \), where these are the stoichiometric coefficients of the ions involved in the electrolyte reactions.
1. Analyzing the Data Given:
The limiting molar conductivities (\( \lambda^0 \)) of the ions are provided for \( U_mY_p \) (250 S cm2 mol-1) and \( V_mX_n \) (440 S cm2 mol-1), as well as the limiting conductivities of individual ions:
2. Using the Graph of Molar Conductivity:
The plot of the molar conductivity (\( \Lambda \)) of \( Z_mX_n \) vs \( c^{1/2} \) shows a linear relationship, where the slope can provide insights into the relationship between the ionic concentrations and the molar conductivity. The slope gives the effective concentration-related contribution to the overall molar conductivity.
3. Solving for \( m + n + p \):
By using the given molar conductivities and interpreting the graph, we can relate the stoichiometric coefficients of the ions involved in the compound \( Z_mX_n \), \( U_mY_p \), and \( V_mX_n \) through the observed conductivities. By applying the principles of ionic conductivities and the fact that these are strong electrolytes, we calculate that \( m + n + p = 7 \).
Final Answer:
The value of \( m + n + p \) is 7.
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 