NEET प्रश्न (MCQ) हिंदी में — उत्तर और विस्तृत हल सहित

1.  कार्य = स्थितिज ऊर्जा में वृद्धि

$$\begin{gathered} = {\mathrm{Kq}}_1{\mathrm{q}}_2\left(\frac{1}{{\mathrm{r}}_2}-\frac{1}{{\mathrm{r}}_1}\right) \\ = 9\times {10}^9\times 8\times {10}^{-6}\times 12\times {10}^{-6} \left(\frac{1}{6\times {10}^{-2}}-\frac{1}{10\times {10}^{-2}}\right) \\ = 9\times 8\times 12 \left(\frac{1}{6}-\frac{1}{10}\right) \times {10}^{-1} =5.8 \mathrm{J} \end{gathered}$$

$$\begin{gathered} 1. {\mathrm{q}}_{\mathrm{r}} + {\mathrm{q}}_{\mathrm{R}} =\mathrm{Q} ........\left(1\right) \\ \frac{{\mathrm{q}}_{\mathrm{r}}}{{\mathrm{q}}_{\mathrm{R}}}=\frac{4{\mathrm{\pi r}}^2\mathrm{\sigma }}{4{\mathrm{\pi R}}^2\mathrm{\sigma }}=\frac{{\mathrm{r}}^2}{{\mathrm{R}}^2} \\ \Rightarrow \frac{{\mathrm{q}}_{\mathrm{r}}}{{\mathrm{q}}_{\mathrm{R}}}=\frac{{\mathrm{r}}^2}{{\mathrm{R}}^2} ........\left(2\right) \\ (1)\; और\; (2)\; से \\ {\mathrm{q}}_{\mathrm{r}}=\frac{{\mathrm{Qr}}^2}{{\mathrm{R}}^2+{\mathrm{r}}^2} और {\mathrm{q}}_{\mathrm{R}}=\frac{{\mathrm{QR}}^2}{{\mathrm{R}}^2+{\mathrm{r}}^2} \\ अतः, \mathrm{V}=\frac{{\mathrm{q}}_{\mathrm{r}}}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}+\frac{{\mathrm{q}}_{\mathrm{R}}}{4{\mathrm{\pi \epsilon }}_0\mathrm{R}} \\ \Rightarrow \mathrm{V}=\frac{\mathrm{Q}}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\mathrm{r}+\mathrm{R}}{{\mathrm{r}}^2 + {\mathrm{R}}^2}\right) \end{gathered}$$

$$\begin{gathered} 1. \overrightarrow{\mathrm{E}}=-\frac{\mathrm{dV}}{\mathrm{dx}}\hat{\mathrm{i}} \\ \Rightarrow \overrightarrow{\mathrm{E}}=\left(-10\mathrm{x}-10\right) \hat{\mathrm{i}} \\ \mathrm{At} \mathrm{x}=12 \mathrm{m}, \overrightarrow{\mathrm{E}}=-20 \hat{\mathrm{i}} \\ \Rightarrow \left|\overrightarrow{\mathrm{E}}\right|=20 {\mathrm{Vm}}^{-1} \end{gathered}$$

$$\begin{gathered} 3. \overrightarrow{\mathrm{E}}=\frac{100}{{\mathrm{x}}^2} \hat{\mathrm{x}} \\ {\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}=-\underset{10}{\overset{20}{\int }}\overrightarrow{\mathrm{E}}.\mathrm{d}\overrightarrow{\mathrm{a}}=-\underset{10}{\overset{20}{\int }}\frac{100}{{\mathrm{x}}^2}\mathrm{dx}=+100{\left[\frac{1}{\mathrm{x}}\right]}_{10}^{20} \\ = 100\left[\frac{1}{20}-\frac{1}{10}\right] = -5\mathrm{V} \\ \therefore \left|{\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}\right| = 5\mathrm{V} \end{gathered}$$

$$\begin{gathered} 1. {\mathrm{U}}_{\infty }=0, {\mathrm{U}}_0\left(\mathrm{केंद्र}\right)=-\frac{\mathrm{KQe}}{\mathrm{R}} \\ \mathrm{हानि} \mathrm{में} \mathrm{स्थितिज}.\mathrm{ऊर्जा}. =\frac{\mathrm{KQe}}{\mathrm{R}}, \mathrm{लाभ} \mathrm{में} \mathrm{गतिज}.\mathrm{ऊर्जा}. =\frac{1}{2}{\mathrm{mV}}^2 =\frac{\mathrm{KQe}}{\mathrm{R}} \\ \mathrm{V}=\sqrt{\frac{2\mathrm{KQe}}{\mathrm{Rm}}} \end{gathered}$$