Current Electricity Test

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Current Electricity Test - 63

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Question 1
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A potentiometer wire of length $$L$$ and resistance $$10\Omega$$ is connected in series with a battery of e.m.f. $$2.5\ V$$ and a resistance in its primary circuit. The null point corresponding to a cell of e.m.f. $$1\ V$$ is obtained at a distance $$L/2$$. If the resistance in the primary circuit is doubled then the position of new null point will be

A
$$0.4\ L$$

B
$$0.5\ L$$

C
$$0.6\ L$$

D
$$0.8\ L$$

Solution

Initial current in the circuit $$= \dfrac {2.5V}{R + 10\Omega}$$
Potential difference across $$\dfrac {L}{2}$$ or $$5\Omega = \dfrac {2.5}{R + 10}\times 5 = 1V$$
because a cell of $$1\ volt$$ balances.
From this $$R = 2.5\Omega$$
$$\therefore$$ When $$R$$ is doubled,
$$i = \dfrac {2.5}{5 + 10} A$$ If this balances at a length $$x, \dfrac {2.5}{15} \times R' = 1\ V$$ where $$R'$$ is the resistance corresponding to the length $$x$$.
$$\Rightarrow R' = \dfrac {15\times 1}{5/2} = 6\Omega$$ i.e., $$x = \dfrac {L\times 6}{10} = 0.6\ L$$.
Question 2
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The total electrical resistance between the points $$A$$ and $$B$$ of the circuit shown is:

A
$$9.23\Omega $$

B
$$15\Omega $$

C
$$30\Omega $$

D
$$100\Omega $$

Solution

$$\cfrac { 1 }{ Req. } =\cfrac { 1 }{ 30\Omega } +\cfrac { 1 }{ 20\Omega } +\cfrac { 1 }{ 40\Omega } $$
$$Req.=\cfrac { 120 }{ 13 } \Omega =9.23\Omega $$
Question 3
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If specific resistance of a potentiometer wire is $$10^{-7}\Omega m$$ and current flowing through it is $$0.1\ amp$$., cross-sectional area of wire is $$10^{-6}m^{2}$$ then potential gradient will be

A
$$10^{-2}volt/m$$

B
$$10^{-4}volt/ m$$

C
$$10^{-6}volt/ m$$

D
$$10^{-8}volt/ m$$

Solution

Resistance of the potentiometer wire $$R = \rho\dfrac{l}{A}$$
Potential difference across the wire $$V = Il$$
So. potential gradient $$\dfrac {V}{l} = \dfrac {IR}{l} = \dfrac {I\rho l}{Al} $$
Or $$\dfrac{V}{l} = \dfrac{I\rho }{A}= \dfrac {0.1\times 10^{-7}}{10^{-6}} = 0.01 = 10^{-2}V/m$$.
Question 4
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A potentiometer wire of length $$1m$$ has a resistance of $$100\Omega $$. It is connected in series with a resistance and a battery of emf $$2V$$ of negligible resistance. A source of emf $$10mV$$ is balanced against a length of $$40cm$$ of the potentiometer wire. What is the value of the external resistance.

A
$$890\Omega $$

B
$$7900\Omega $$

C
$$680\Omega $$

D
$$740\Omega $$

Solution

Given: A potentiometer wire of length $$1m$$ has a resistance of $$100\Omega$$. It is connected in series with a resistance and a battery of emf $$2V$$ of negligible resistance. A source of emf $$10mV$$ is balanced against a length of $$40cm$$ of the potentiometer wire.
To find the value of the external resistance
Solution:
If $$J$$ is current through the potentiometer wire then
$$J=\dfrac E{R+10}=\dfrac 2{R+10}$$
As the source of e.m.f $$E'=10mV=10\times10^{-3}V$$ is balanced by a length of $$40cm$$ of the potentiometer wire, it follows that $$10\times10^{-3}=J\times $$ resistance of $$40cm$$ of the potentiometer wire.
Now resistance of $$40cm$$ of the potentiometer wire
$$=\dfrac {10}{100}\times 40 =4\Omega$$
So,
$$10\times10^{-3}=\dfrac 2{R+10}\times 4 \\\implies R=790\Omega$$
is the value of the external resistance.
Question 5
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Two coils have a combined resistance of $$12 \Omega$$, when combined series and $$5/3 \Omega$$, when connected in parallel. Their respective resistances are:

A
$$6$$ and $$6$$ ohms

B
$$10$$ and $$2$$ ohms

C
$$5$$ and $$7$$ ohms

D
$$4$$ and $$8$$ ohms

Solution

Let us assume the two resistances to be $$R_1$$ and $$R_2$$.
Let resistances in series and parallel be $$R_s$$ and $$R_p$$ respectively.
Given:
$$R_{s}=12 \Omega=R_1+R_2$$
So,
$$R_1+R_2=12$$ . . . . (i)
$$R_2=12-R_1$$ . . . . (ii)

And,
$$R_{p}=\dfrac{5}{3}\Omega \Rightarrow \dfrac{1}{R_{p}}=\dfrac{3}{5}$$
So,
$$\dfrac{3}{5}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$$

$$\Rightarrow \dfrac{3}{5}=\dfrac{R_1+R_2}{R_1R_2}$$ . . . (iii)

Using $$eq. (i)$$ and $$eq. (ii)$$ in $$eq. (iii)$$
$$\dfrac{3}{5}=\dfrac{12}{R_1(12-R_1)}$$

$$\Rightarrow 12R_1-R_1^2=20$$
$$\Rightarrow R_1^2-12R_1+20=0$$

This is a quadratic equation in $$R_1$$. Let's solve it by factorisation.
$$R_1^2-10R_1-2R_1+20=0$$
$$R_1(R_1-10)-2(R_1-10)=0$$
$$(R_1-10)(R_1-2)=0$$
$$\Rightarrow R_1-10=0$$ or $$R_1-2=0$$
$$\Rightarrow R_1=10$$ or $$R_1=2$$

Putting $$R_1=10$$ in $$eq.(i)$$, $$R_2=2$$
or
Putting $$R_1=2$$ in $$eq.(i)$$, $$R_2=10$$

So option B is correct.
Question 6
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Ravi connected three bulbs with the cells and a switch as shown. When switch is moved to ON position.

A
The bulb X will glow first.

B
The bulb Y will glow first.

C
The bulbs Z and X will glow first.

D
All the bulbs will glow simultaneously.

Solution

All bulbs will glow simultaneously and instantly without any delay. It is because the current is setup in the circuit the moment the switch is closed to provide it a closed path to flow. The three bulbs are in series, so equal current flows through all of them.
Question 7
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Which of the following circuit diagram represents the given torch?

A

B

C

D

Solution

In torch, three batteries are connected in series with the bulb and switch.
Hence Option D is correct
Question 8
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Which of the following I-V graph represents ohmic conductors?

A

B

C

D

Solution

Ohm's law $$I=\dfrac{V}{R}$$ is an equation of straight line. Hence I-V characteristics for ohmic conductors is also straight line and its slope gives, $$m=\dfrac{1}{resistance}$$ of the conductor.
Option A
Question 9
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In parallel combination of n cells, we obtain:

A
More voltage

B
More current

C
Less voltage

D
Less current

Solution

In parallel combination of n cells the net resistance of cells becomes $$\cfrac{1}{n}$$ times. Resistance reduces and current increases.
Question 10
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What is the value of unknown resistance R, of galvanometer shows null deflection in the given meter bridge set up shown in the fig?

A
$$97.50\Omega$$

B
$$105\Omega$$

C
$$220\Omega$$

D
$$110\Omega$$

Solution

Given that, the galvanometer shows null deflection. Hence the given combination is Wheatstone bridge.
Let $$k$$ be the resistance per unit length of the meter bridge wire.
Applying Wheatstone bridge principle we get:-

$$\dfrac{20k}{55}=\dfrac{80k}{R}$$

$$\implies R= \dfrac{80\times 55}{20} =220\ \Omega$$