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A galvanometer has 30 divisions and a sensitivity 16 $μ A / div.$ It can be converted into a voltmeter to read 3 V by connecting (approximately):

Resistance nearly 6 k Ω in series

6 k Ω in parallel

500 Ω in series

It cannot be converted

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In a circuit, 5 percent of total current passes through a galvanometer. If the resistance of the galvanometer is G then the value of the shunt is :

19 G

20 G

$\frac{G}{20}$

$\frac{G}{19}$

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Explanation

$\frac{i_{g}}{i} = \frac{S}{G + S}$

⇒ $\frac{5}{100} = \frac{S}{G + S}$

⇒ $S = \frac{G}{19}$

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A voltmeter has a range 0-V with a series resistance R. With a series resistance 2R, the range is 0-V'. The correct relation between V and V' is :

$V^{'} = 2 V$

$V^{'} > 2 V$

$V^{'} > > 2 V$

$V' < 2 V$

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Explanation

For conversion of galvanometer (of resistances) into voltmeter, a resistance R is connected in series.

∴ $i_{g} = \frac{V_{}}{R + G}$ and $i_{g} = \frac{{V^{'}}_{}}{2 R + G}$

⇒ $\frac{V}{R + G} = \frac{{V_{}}^{'}}{2 R + G}$ ⇒ $\frac{V^{'}}{V_{}} = \frac{2 R + G}{R + G} = \frac{2 (R + G) - G}{(R + G)}$

$= 2 - \frac{G}{(R + G)}$ ⇒ ${V^{'_{}}}_{} = 2 V - \frac{V_{} G}{(R + G)}$ ⇒ ${V^{'}}_{} < 2 V_{}$

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If an ammeter is to be used in place of a voltmeter then we must connect with the ammeter a :

Low resistance in parallel

High resistance in parallel

High resistance in series

Low resistance in series

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Explanation

If ammeter is used in place of voltmeter (i.e. in parallel) it may damage due to large current in circuit. Hence to control this large amount of current a high resistance must be connected in series.

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A galvanometer of resistance 36 Ω is changed into an ammeter by using a shunt of 4 Ω. The fraction f₀ of total current passing through the galvanometer is :

$\frac{1}{40}$

$\frac{1}{4}$

$\frac{1}{140}$

$\frac{1}{10}$

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Explanation

$\frac{i_{g}}{i} = \frac{S}{G + S} = \frac{4}{36 + 4} = \frac{4}{40} = \frac{1}{10}$

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A galvanometer, having a resistance of 50 Ω gives a full scale deflection for a current of 0.05 A. The length in meter of a resistance wire of area of cross-section 2.97× 10^–2 cm² that can be used to convert the galvanometer into an ammeter which can read a maximum of 5 A current is (Specific resistance of the wire = 5 × 10^–7Ωm)

1.5

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Explanation

$\frac{i}{i_{g}} = 1 + \frac{G}{S} \Rightarrow \frac{5}{0 .05} = 1 + \frac{50}{S}$

⇒ $S = \frac{50}{99} = \frac{ρ \times l}{A}$ ⇒ $l = \frac{50}{99} \times \frac{2 .97 \times 10^{- 2} \times 10^{- 4}}{5 \times 10^{- 7}} = 3 m$ .

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An ammeter reads up to 1 ampere. Its internal resistance is 0.81 ohm. To increase the range to 10 A the value of the required shunt is :

0.09 Ω

0.03 Ω

0.3 Ω

0.9 Ω

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Explanation

$\frac{i}{i_{g}} = 1 + \frac{G}{S}$ ⇒ $\frac{10}{1} = 1 + \frac{0 .81}{S} \Rightarrow S = 0 .09 Ω$ .

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A moving coil galvanometer of resistance 100Ω is used as an ammeter using a resistance 0.1Ω. The maximum deflection current in the galvanometer is 100 mA. Find the minimum current in the circuit so that the ammeter shows maximum deflection :

100.1 mA

1000.1 mA

10.01 mA

1.01 mA

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Explanation

$I_{G} \times G = (I - I_{G}) S$

⇒ I = $(1 + \frac{G}{S}) I_{G}$

⇒ I = 100.1 mA

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A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms needed to be connected in series with the coil will be

99995

9995

10³

10⁵

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Explanation

Voltage sensitivity $= \frac{Current sensitivity}{Resistance of galvanometer G}$

⇒ $G = \frac{10}{2} = 5 Ω$ .

Here $i_{g} =$ Full scale deflection current $= \frac{150}{10} = 15 m A$ .

V = Voltage to be measured = 150 × 1 = 150 V.

Hence $R = \frac{V}{i_{g}} - G = \frac{150}{15 \times 10^{- 3}} - 5 = 9995 Ω$ .

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The electric charge in uniform motion produces :

An electric field only

A magnetic field only

Both electric and magnetic field

Neither electric nor magnetic field

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Explanation

A movable charge produces electric field and magnetic field both.

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

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

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