
To find the value of \(n\), we need to calculate the coordinates of the center of mass for the given plate shape. The shape is composed of a rectangle with a cut-out, which can be analyzed as a combination of simple geometrical figures, allowing us to use subtraction to find the center of mass.
The calculated value of \(n\) is 15, which falls within the given range of 15,15.
The mass of the plate is calculated in three sections:
\(m1 = σ × 5 = 10 kg,\)
\(m2 = σ × 1 = 2 kg,\)
\(m3 = σ × 6 = 12 kg.\)
Using the coordinates of each center of mass, we calculate the combined center of mass:
\[m_1x_1 + m_2x_2 = m_3x_3.\]
\[10 \cdot 1.5 + 2 \cdot (1.5) = 12 \cdot x_1 \implies x_1 = 1.5 \, \text{cm}.\]
Similarly:
\[m_1y_1 + m_2y_2 = m_3y_3.\]
\[10 \cdot 1 + 2 \cdot (1.5) = 12 \cdot y_1 \implies y_1 = 0.9 \, \text{cm}.\]
The ratio of $x_1$ to $y_1$ is:
\[\frac{x_1}{y_1} = \frac{1.5}{0.9} = \frac{15}{9}.\]
Thus:
\[n = 15.\]
Final Answer: $n = 15$.
The variation of density of a solid cylindrical rod of cross-sectional area \( a \) and length \( L \) is \( \rho=\rho_0 \frac{x^2}{L^2} \), where \( x \) is the distance from one end. The position of its centre of mass from \( x=0 \) is 
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
