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Physics NEET Questions — Free MCQ Practice

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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):

A.

Resistance nearly 6 k Ω in series

B.

6 k Ω in parallel

C.

500 Ω in series

D.

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 :

A.

19 G

B.

20 G

C.

$\frac{G}{20}$

D.

$\frac{G}{19}$

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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 :

A.

$V^{'} = 2 V$

B.

$V^{'} > 2 V$

C.

$V^{'} > > 2 V$

D.

$V' < 2 V$

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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 :

A.

Low resistance in parallel

B.

High resistance in parallel

C.

High resistance in series

D.

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 :

A.

$\frac{1}{40}$

B.

$\frac{1}{4}$

C.

$\frac{1}{140}$

D.

$\frac{1}{10}$

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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)

A.

9

B.

6

C.

3

D.

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 :

A.

0.09 Ω

B.

0.03 Ω

C.

0.3 Ω

D.

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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The dimension of the magnetic field intensity B is:

A.

${MLT}^{- 2} A^{- 1}$

B.

${MT}^{- 2} A^{- 1}$

C.

${ML}^{2} {TA}^{- 2}$

D.

$M^{2} {LT}^{- 2} A^{- 1}$

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Explanation

(b) F = Bil $\Rightarrow [B] = \frac{[F]}{[i] [L]} = \frac{{MLT}^{- 2}}{AL} = {MT}^{- 2} A^{- 1}$

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An electron moves with a constant speed v along a circle of radius r. Its magnetic moment will be (e is the electron's charge)

A.

evr

B.

$\frac{1}{2}$ evr

C.

${πr}^{2}$ ev

D.

2 $π$ rev

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Explanation

(b) $M = i ({πr}^{2}) = \frac{e v}{2 πr} \times {πr}^{2} \Rightarrow M = \frac{1}{2} evr$

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The field normal to the plane of a wire of n turns and radius r which carries a current i is measured on the axis of the coil at a small distance h from the centre of the coil. This is smaller than the field at the centre by the fraction

A.

$\frac{3}{2} \frac{h^{2}}{r^{2}}$

B.

$\frac{2}{3} \frac{h^{2}}{r^{2}}$

C.

$\frac{3}{2} \frac{r^{2}}{h^{2}}$

D.

$\frac{2}{3} \frac{r^{2}}{h^{2}}$

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Explanation

(a) Field at the centre $B_{1} = \frac{μ_{0}}{4 π} \times \frac{2 πin}{r} = \frac{μ_{0}}{2} \cdot \frac{ni}{r}$

Field at a distance h from the centre

$B_{2} = \frac{μ_{0}}{4 π} \cdot \frac{2 {πnir}^{2}}{(r^{2} + h^{2})^{3 / 2}} = \frac{μ_{0}}{2} \cdot \frac{{nir}^{2}}{r^{3} {(1 + \frac{h^{2}}{r^{2}})}^{3 / 2}} = B_{1} {(1 + \frac{h^{2}}{r^{2}})}^{- 3 / 2} = B_{1} (1 - \frac{3}{2} \cdot \frac{h^{2}}{r^{2}})$

(By binomial theorem)

Hence $B_{2}$ is less than $B_{1}$ by a fraction $= \frac{3}{2} \frac{h^{2}}{r^{2}}$

Physics MCQs for NEET — Practice Questions with Answers

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