Current Electricity Test

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Current Electricity Test - 74

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Weekly Quiz Competition

Question 1
1 / -0

Statement I: In a meter-bridge experiment, the null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance.
Statement II: Resistance of a metal increases with the increase in temperature.

A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I.

B
Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I.

C
Statement I is true, Statement II is false.

D
Statement I is false, Statement II is true.

Solution
Question 2
1 / -0

Express which of the following steups can be used to verify Ohm's law?

A

B

C

D

Solution

In ohm's law, we check $$V = IR$$ where I is the current flowing through a resister and V is the potential difference across that resistor. Only option (a) fits the criteria. Remember that ammeter is connected in series with resistance and voltmeter in parallel with the resistance.
Question 3
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To verify Ohm's law, a student is provided with a test resistor $$R_T$$ a high resistance $$R_1$$, a small resistance $$R_2,$$ two identical galvanometer $$G_1$$ and $$G_2$$ and a variable voltage source V. The correct circuit to carry out the experiment is

A

B

C

D

Solution
Question 4
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The temperature coefficient of resistance of a wire is $$0.00125$$ per $$^{0}C$$. At $$300 K$$, its resistance is $$1 \ ohm$$. This resistance of the wire will be $$2 \ ohm$$ at

A
$$1154 K$$

B
$$1100 K$$

C
$$1400 K$$

D
$$1127 K$$

Solution

Given : $$\alpha = 0.00125/^oC$$
Using $$R_T = R_{273} (1 + \alpha (T- 273) )$$
For $$T = 300$$, $$R_{300} = 1\Omega$$
$$\therefore$$   $$R_{300} = R_{273} (1 + \alpha (300- 273) )$$
OR   $$1 = R_{273} (1 + 0.00125 (300- 273) )$$ $$\implies R_{273} = 0.967\Omega$$

Let at temperature be $$T$$ when $$R_T = 2\Omega$$
$$\therefore$$     $$R_T = R_{273} (1 + \alpha (T- 273) )$$
OR    $$2 = 0.967(1 + 0.00125 (T- 273) )$$
OR $$T- 273 = 854.6$$
$$\implies$$ $$T = 273 + 854.6 = 1127.6 K$$
Question 5
1 / -0

The equivalent resistance between $$A$$ and $$B$$ ( of the circuit as shown ) is

A
$$4.5\Omega$$

B
$$12\Omega$$

C
$$5.4\Omega$$

D
$$20\Omega$$

Solution

Clearly from the figure,
Resistors $$7\Omega$$ and $$3\Omega$$ are in parallel; $$6\Omega$$
and $$4\Omega $$ are in parallel and both in series.

So $$R_{eq}=\dfrac{7\times 3}{7+3}+\dfrac{4\times 6}{4+6}=4.5\Omega$$
Question 6
1 / -0

Two cells of emf approximately 5V and 10V are to be accurately compared using a potentiometer of length 400cm.

A
The battery that runs the potentiometer should have voltage of 8V.

B
The battery of potentiometer can have a voltage of 15V and R adjusted so that the
potential drop across the wire slightly exceeds 10V.

C
The first portion of 50 cm of the wire itself should have a potential drop of 10V.

D
The potentiometer is usually used for comparing resistances and not voltages.

Solution

Here, emf of primary cells are 5V and 10V. So the potential drop across potentiometer wire must be slightly more than that larger emf 10 V. So the battery should be of 15V and about 4V potential is dropped by using rheostat or resistances. So verifies answer(b).
Question 7
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Which is a wrong statement

A
The Wheatstone bridge is most sensitive when all the four resistance are of the same order

B
In a balanced Wheatstone bridge, interchanging the position of galvanometer and cell affects the balance of the bridge

C
Kirchhoff's first law (for currents meeting at a junction in a electric circuit) expresses the conversion of change

D
The rheostat can be used a a potential divider

Solution

if the galvanometer and cell are interchanged, the position of the balance point remains unchanged.
In balance Wheatstone bridge, the arms of galvanometer and cell can be interchanged without affecting the balance of the bridge. Hence B staement is incorrect
Question 8
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Kirchhoff $$I$$ law and $$II$$ law of current, proves the

A
Conservation of charge and energy

B
Conservation of current and energy

C
Conservation of mass and charge

D
None of thee

Solution

Kirchhoff's current law (1st Law) states that current flowing into a node (or a junction) must be equal to current flowing out of it. This is a consequence of charge conservation.
Kirchhoff's voltage law (2nd Law) states that the sum of all voltages around any closed loop in a circuit must equal zero. This is a consequence of charge conservation and also conservation of energy.
Question 9
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The conductors of resistance $$1 \Omega, 2 \Omega$$ and $$3\Omega$$ are in series in a circuit. What is the equivalent resistance in the circuit ?

A
Less than $$1\Omega$$

B
Less than $$3\Omega$$

C
More than $$1 \Omega$$

D
More than $$3 \Omega$$

Solution

Given:
$$R_1=1 \Omega$$
$$R_2=2 \Omega$$
$$R_3=3 \Omega$$
All are in series
So $$R_{eq}=R_1+R_2+R_3=1+2+3=6 \Omega$$
Question 10
1 / -0

Two bulbs, one of 50 watt and another of 25 watt are connected in series to the mains. The ratio of the currents through them is

A
$$2 : 1$$

B
$$1 : 2$$

C
$$1 : 1$$

D
Without voltage, cannot be calculated

Solution

IN series combination, same amount of current flows through elements
The bulbs are in series, hence they will have the same current through them.