AP EAPCET 2022 - 4th July Evening Shift | Current Electricity Question 26 | Physics

The electrons take $$40 \times 10^3$$ s to dirift from one end of a metal wire of length 2 m to its other end. The area of cross-section of the wire is $$4 \mathrm{~mm}^2$$ and it is carrying a current of 1.6 A. The number density of free electrons in the metal wire is

$$8 \times 10^{28} \mathrm{~m}^{-3}$$

$$6 \times 10^{28} \mathrm{~m}^{-3}$$

$$4 \times 10^{28} \mathrm{~m}^{-3}$$

$$5 \times 10^{28} \mathrm{~m}^{-3}$$

AP EAPCET 2022 - 4th July Evening Shift

MCQ (Single Correct Answer)

-0

The current 'I' in the circuit shown in the figure is

AP EAPCET 2022 - 4th July Evening Shift Physics - Current Electricity Question 26 English

$$\varepsilon/R$$

$$-\varepsilon/R$$

$$2\varepsilon/R$$

$$-2\varepsilon/R$$

AP EAPCET 2022 - 4th July Morning Shift

MCQ (Single Correct Answer)

-0

Current density in a cylindrical wire of radius $$R$$ varies with radial distance as $$\beta\left(r+r_0\right)^2$$. The current through the section of the wire shown in the figure is

AP EAPCET 2022 - 4th July Morning Shift Physics - Current Electricity Question 28 English

$$\pi \beta\left[\frac{R^4}{12}+\frac{r_0^2 R^2}{6}+\frac{2 r_0 R^3}{9}\right]$$

$$\pi \beta\left[\frac{R^4}{6}+\frac{r_0^2 R^2}{12}+\frac{r_0 R^3}{9}\right]$$

$$\pi \beta\left[\frac{R^4}{12}+\frac{r_0^2 R^2}{12}+\frac{r_0 R^3}{9}\right]$$

$$\pi \beta\left[\frac{R^4}{8}+\frac{r_0^2 R^2}{6}+\frac{r_0 R^3}{12}\right]$$

AP EAPCET 2022 - 4th July Morning Shift

MCQ (Single Correct Answer)

-0

A cell can supply currents of 1 A and 0.5 A via resistances of $$2.5 \Omega$$ and $$10 \Omega$$, respectively. The internal resistance of the cell is

$$2 \Omega$$

$$3 \Omega$$

$$4 \Omega$$

$$5 \Omega$$

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