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JEE Main 28-Jan-2025 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 28-Jan-2025 Paper - Shift 14 Marks
Question
Given below is the plot of the molar conductivity vs $\sqrt{\text { concentration }}$ for KCl in aqueous solution.
Image
If, for the higher concentration of KCl solution, the resistance of the conductivity cell is $100 \Omega$, then the resistance of the same cell with the dilute solution is ' $x$ ' $\Omega$.
The value of $x$ is ___________ (Nearest integer)
✓

Answer

150
$\mathrm{R}=\rho \frac{\ell}{\mathrm{A}}$
$
\begin{gathered}
\kappa=\mathrm{G}_{\mathrm{l}} \mathrm{G}^{*} \mathrm{G}=\frac{1}{\mathrm{R}} ; \kappa=\frac{1}{\rho} \\
\mathrm{G}^{*}=\frac{\ell}{\mathrm{A}} \\
\mathrm{R}=\text { Resistance } \\
\rho=\text { Resistivity } \\
\frac{\ell}{\mathrm{A}}=\text { cell constant }\left(\mathrm{G}^{*}\right) \\
\frac{\kappa_{\mathrm{c}}}{\kappa_{\mathrm{d}}}=\frac{\mathrm{R}_{\mathrm{d}}}{\mathrm{R}_{\mathrm{c}}} ; \lambda_{\mathrm{m}}=\frac{\kappa \times 1000}{\mathrm{C}} \\
\frac{\kappa_{\mathrm{c}}}{\kappa_{\mathrm{d}}}=\frac{\left(\lambda_{\mathrm{m}} \cdot \mathrm{C}\right)}{\left(\lambda_{\mathrm{m}} \cdot \mathrm{C}\right)_{\mathrm{d}}}=\frac{\mathrm{R}_{\mathrm{d}}}{\mathrm{R}_{\mathrm{c}}} \quad \begin{array}{l}
\mathrm{c}=\text { concentrated sol. } \\
\mathrm{d}=\text { diluted solution }
\end{array} \\
\frac{100 .(0.15)^{2}}{150 .(0.1)^{2}}=\frac{\mathrm{R}_{\mathrm{d}}}{100} \quad \\
\mathrm{R}_{\mathrm{d}}=150 \Omega
\end{gathered}
$

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