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Physics MCQs for NEET — Practice Questions with Answers

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The maximum power drawn out of the cell from a source is given by (where r is internal resistance)

$E^{2} / 2 r$

$E^{2} / 4 r$

$E^{2} / r$

$E^{2} / 3 r$

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Explanation

$i = \frac{E}{r + R}$ ⇒ $P = i^{2} R$ ⇒ $P = \frac{E^{2} R}{{(r + R)}^{2}}$

Power is maximum when r = R ⇒ $P_{\max} = E^{2} / 4 r$

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Two sources of equal emf are connected to an external resistance R. The internal resistances of the two sources are R₁ and $R_{2} (R_{2} > R_{1})$ . If the potential difference across the source having internal resistance R₂ is zero, then

$R = R_{1} R_{2} / (R_{1} + R_{2})$

$R = R_{1} R_{2} / (R_{2} - R_{1})$

$R = R_{2} \times (R_{1} + R_{2}) / (R_{2} - R_{1})$

$R = R_{2} - R_{1}$

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To draw maximum current from a combination of cells, how should the cells be grouped

Series

Parallel

Mixed

Depends upon the relative values of external and internal resistance

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Explanation

Maximum current will be drawn from the circuit if resultant resistance of all internal resistances is equal to the value of external resistance if the arrangement s mixed. In series, $R > > n r$ and in parallel, the external resistance is negligible.

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By ammeter, which of the following can be measured

Electric potential

Potential difference

Current

Resistance

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Explanation

Ammeter is used to measure the current through the circuit.

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For measurement of potential difference, potentiometer is preferred in comparison to voltmeter because

Potentiometer is more sensitive than voltmeter

The resistance of potentiometer is less than voltmeter

Potentiometer is cheaper than voltmeter

Potentiometer does not take current from the circuit

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Explanation

Potentiometer works on null deflection method. In balance condition no current flows in secondary circuit.

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The potential gradient along the length of a uniform wire is 10 volt/metre. B and C are the two points at 30 cm and 60 cm point on a meter scale fitted along the wire. The potential difference between B and C will be :

3 volt

0.4 volt

7 volt

4 volt

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Explanation

Potential gradient = Change in voltage per unit length

∴ $10 = \frac{V_{2} - V_{1}}{30 / 100} \Rightarrow V_{2} - V_{1} = 3 v o l t$

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In Wheatstone's bridge P = 9 ohm, Q = 11 ohm, R = 4 ohm and S = 6 ohm. How much resistance must be put in parallel to the resistance S to balance the bridge

24 ohm

$\frac{44}{9}$ ohm

26.4 ohm

18.7 ohm

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A potentiometer is an ideal device of measuring potential difference because

It uses a sensitive galvanometer

It does not disturb the potential difference it measures

It is an elaborate arrangement

It has a long wire hence heat developed is quickly radiated

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Explanation

In balance condition, potentiometer doesn't take the current from secondary circuit.

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A battery of 6 volts is connected to the terminals of a three metre long wire of uniform thickness and resistance of the order of 100 Ω. The difference of potential between two points separated by 50cm on the wire will be :

1 V

1.5 V

2 V

3 V

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Explanation

Here same current is passing throughout the length of the wire, hence $V \propto R \propto l$

⇒ $\frac{V_{1}}{V_{2}} = \frac{l_{1}}{l_{2}}$ ⇒ $\frac{6}{V_{2}} = \frac{300}{50}$

⇒ V₂ =1 V.

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A potentiometer is used for the comparison of e.m.f. of two cells E₁ and E₂. For cell E₁ the no deflection point is obtained at 20cm and for E₂ the no deflection point is obtained at 30cm. The ratio of their e.m.f.'s will be

2/3

1/2

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Explanation

Ratio will be equal to the ratio of no deflection lengths i.e. $\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}} = \frac{2}{3}$

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