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A galvanometer has a resistance of 25 ohm and a maximum of 0.01 A current can be passed through it. In order to change it into an ammeter of range 10 A, the shunt resistance required is

5/999 ohm

10/999 ohm

20/999 ohm

25/999 ohm

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Explanation

$i_{g} = i \frac{S}{G + S}$ ⇒ $0 .01 = 10 \frac{S}{25 + S}$

$\Rightarrow 1000 S = 25 + S$ ⇒ $S = \frac{25}{999} Ω$ .

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Two resistances of 400 Ω and 800 Ω are connected in series with a 6-volt battery of negligible internal resistance. A voltmeter of resistance 10,000 Ω is used to measure the potential difference across 400 Ω. The error in the measurement of potential difference in volts approximately is :

0.01

0.02

0.03

0.05

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The length of a wire of a potentiometer is 100 cm, and the emf of its standard cell is E volt. It is employed to measure the e.m.f of a battery whose internal resistance is 0.5 Ω. If the balance point is obtained at l = 30 cm from the positive end, the e.m.f. of the battery is :

$\frac{30 E}{100}$

$\frac{30 E}{100 .5}$

$\frac{30 E}{(100 - 0 .5)}$

$\frac{30 (E - 0 .5 i)}{100}$ , where i is the current in the potentiometer

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Explanation

From the principle of potentiometer $V \propto l$

⇒ $\frac{V}{E} = \frac{l}{L}$ ; where V = emf of battery, E = emf of standard cell, L = Length of potentiometer wire

$V = \frac{E l}{L} = \frac{30 E}{100}$ .

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The current flowing in a coil of resistance 90 Ω is to be reduced by 90%. What value of resistance should be connected in parallel with it

9 Ω

90 Ω

1000 Ω

10 Ω

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Explanation

$i_{g} = \frac{i}{10}$ ⇒ Required shunt $S = \frac{G}{(n - 1)} = \frac{90}{(10 - 1)} = 10 Ω$

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A galvanometer of 50 ohm resistance has 25 divisions. A current of 4 × 10^–4 ampere gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 volts, it should be connected with a resistance of :

2500 Ω as a shunt

2450 Ω as a shunt

2550 Ω in series

2450 Ω in series

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Explanation

Full deflection current $i_{g} = 25 \times 4 \times 10^{- 4} = 100 \times 10^{- 4} A$

Using $R = \frac{V}{I_{g}} - G = \frac{25}{100 \times 10^{- 4}} - 50 = 2450 Ω$ in series.

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In a metre bridge experiment, the null point is obtained at 20 cm from one end of the wire when resistance X is balanced against another resistance Y. If X < Y, then where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y

50 cm

80 cm

40 cm

70 cm

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Explanation

In balancing condition, $\frac{R_{1}}{R_{2}} = \frac{l_{1}}{l_{2}} = \frac{l_{1}}{100 - l_{1}}$

⇒ $\frac{X}{Y} = \frac{20}{80} = \frac{1}{4}$ .....(i)

and $\frac{4 X}{Y} = \frac{l}{100 - l}$ .....(ii)

From equation (i) and (ii) :

⇒ $\frac{4}{4} = \frac{l}{100 - l}$ ⇒ $l = 50 c m$

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In a potentiometer experiment, the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of 2 Ω, the balancing length becomes 120 cm. The internal resistance of the cell is :

4 Ω

2 Ω

1 Ω

0.5 Ω

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Explanation

$r = R (\frac{l_{1}}{l_{2}} - 1) = 2 (\frac{240}{120} - 1) = 2 Ω$

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Potentiometer wire of length 1 m is connected in series with 490 Ω resistance and 2V battery. If 0.2 mV/cm is the potential gradient, then the resistance of the potentiometer wire is :

4.9 Ω

7.9 Ω

5.9 Ω

6.9 Ω

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Explanation

Potential gradient $x = \frac{e}{(R + R_{h} + r)} . \frac{R}{L}$

⇒ ⇒ R = 4.9 Ω.

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Two uniform wires A and B are of the same metal and have equal masses. The radius of wire A is twice that of wire B. The total resistance of A and B when connected in parallel is :

4 Ω when the resistance of wire A is 4.25 Ω

5 Ω when the resistance of wire A is 4.25 Ω

4 Ω when the resistance of wire B is 4.25 Ω

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Explanation

$\frac{R_{A}}{R_{B}} = {(\frac{r_{B}}{r_{A}})}^{4}$ ⇒ $\frac{R_{A}}{R_{B}} = {(\frac{1}{2})}^{4} = \frac{1}{16}$ ⇒ $R_{B} = 16 R_{A}$

When R_A and R_B are connected in parallel then equivalent resistance $R_{e q} = \frac{R_{A} R_{B}}{(R_{A} + R_{B})} = \frac{16}{17} R_{A}$

If $R_{A} = 4 .25 Ω$ then $R_{e q} = 4 Ω$ i.e. option is correct.

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You are given several identical resistances each of value R = 10 Ω and each capable of carrying maximum current of 1 ampere. It is required to make a suitable combination of these resistances to produce a resistance of 5 Ω which can carry a current of 4 amperes. The minimum number of resistances of the type R that will be required for this job

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Explanation

Suppose n resistors are used for the required job. Suppose equivalent resistance of the combination is R' and according to energy conservation, it's current rating is $i' .$

Energy consumed by the combination = n × (Energy consumed by each resistance)

⇒ $i'^{2} R' = n \times i^{2} R$

⇒ $n = {(\frac{i'}{i})}^{2} \times (\frac{R'}{R}) = {(\frac{4}{1})}^{2} \times (\frac{5}{10}) = 8$

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Yes. You can attempt every Physics question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.