COMEDK 2023 Morning Shift | Current Electricity Question 10 | Physics

A galvanometer having a resistance of $$8 \Omega$$ is shunted by a wire of resistance $$2 \Omega$$. If the total current is $$1 \mathrm{~A}$$, the part of it passing through the shunt will be

0.25 A

0.8 A

0.2 A

0.5 A

COMEDK 2023 Morning Shift

MCQ (Single Correct Answer)

-0

Two cells with the same emf $$E$$ and different internal resistances $$r_1$$ and $$r_2$$ are connected in series to an external resistance $$R$$. If the potential difference across the first cell is zero then value of $$R$$.

$$\sqrt{r_1 r_2}$$

$$r_1+r_2$$

$$r_1-r_2$$

$$\frac{r_1+r_2}{2}$$

COMEDK 2023 Evening Shift

MCQ (Single Correct Answer)

-0

The current sensitivity of a galvanometer having 20 divisions is $$10 \mu \mathrm{A} /$$ div. If the resistance of the galvanometer is $$100 \Omega$$ then the value of the resistance to be used to convert this galvanometer in to an voltmeter to read up to $$1 \mathrm{~V}$$ is :

$$4 \times 10^{-6} \Omega$$ in series with the galvanometer.

$$4900 \Omega$$ in parallel with the galvanometer.

$$4 \times 10^{-3} \Omega$$ in series with the galvanometer.

$$4900 \Omega$$ in series with the galvanometer.

COMEDK 2023 Evening Shift

MCQ (Single Correct Answer)

-0

The electric flux from cube of side $$1 \mathrm{~m}$$ is '$$\Phi$$' When the side of the cube is made $$3 \mathrm{~m}$$ and the charge enclosed by the cube is made one third of the original value, then the flux from the bigger cube will be :

$$\frac{\Phi}{3}$$

$$\Phi$$

$$3 \Phi$$

$$9 \Phi$$

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