Current Electricity Test

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Current Electricity Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

A potentiometer wire 10 m long has a resistance of 40$$\Omega$$. It is connected in series with a resistance box and a 2V storage cell. If the potential gradient along the wire is 0.1 mV/cm , the resistance in the box is

A
760$$\Omega $$

B
260$$\Omega$$

C
1060$$\Omega$$

D
960$$\Omega$$

Solution

PG$$= 0.1 \dfrac {mV}{cm}$$
$$\Rightarrow \dfrac {\triangle V}{l} = \dfrac {0.1 \times 10^{-3}}{1 \times 10^{-2}}$$
$$\triangle V = l \times 0.01 = 0.1$$
$$\Rightarrow \dfrac {0.1}{2 - 0.1} = \dfrac {40}{R}$$
$$\Rightarrow R = 760 \Omega$$
Question 2
1 / -0

In a potentiometer of ten wires, the balance point is obtained on the sixth wire. To shift the balance point to eighth wire, we should

A
increase resistance in the primary circuit.

B
decrease resistance in the primary circuit.

C
decrease resistance in series with the cell whose e.m.f. has to be measured.

D
increase resistance in series with the cell whose e.m.f. has to be measure.

Solution

By increasing resistance in the primary circuit the P.G of the wire decreases thereby it requires more length to measure the emf of secondary cell.
$$ \therefore $$ By increasing the resistance in series circuit we can move balancing point from sixth to eight wire.
Question 3
1 / -0

A wire of resistance $$20\Omega $$ is bent in the form of a square. The resistance between the ends of diagonal is:

A
$$10\Omega $$

B
$$5\Omega $$

C
$$20\Omega $$

D
$$15\Omega $$

Solution

Since the wire is bend in form of the square and we know that sides of square are equal so each side will have same resistance i.e 5 ohm. Now if we make diagonal two faces in each side of the diagonal will have two sides and resistance will be 10 ohm.
$$R=20 \Omega$$
So, $$\dfrac{R}{4}=5 \Omega$$
$$\Rightarrow$$ $$ R_{1} = 10 \Omega$$; $$R_{2} = 10\Omega$$
So, resistance between the ends of a diagonal = $$ R_{D} = \dfrac{10 \times10}{2 \times10}\Omega = 5\Omega$$
Question 4
1 / -0

Assertion : The equivalent resistance between the points X and Y in the figure is 10.
Reason : According to Wheatstone bridge, points A and C have the same potential.

A
Both (A) and (R) are true and (R) is the correct explanation of A

B
Both (A) and (R) are true but (R) is not the correct explanation of A

C
(A) is true but (R) is false

D
(A) is false but (R) is true

Solution

Simplified circuit:
$${ R }_{xy }=\dfrac {20 \times 20}{2 \times 20} =10\Omega $$
According to Wheatstone condition, A and C are of same potential.
But solving for $${ R }_{xy }$$ and Wheatstone condition are not related.
Question 5
1 / -0

The resistance of a bulb filament is 100$$\Omega $$ at a temperature of 100$$^{0}$$C. If its temperature coefficient of resistance be 0.005 per $$^{0}$$C , its resistance will become 200$$\Omega $$ at a temperature of :

A
300$$^{0}$$C

B
400$$^{0}$$C

C
500$$^{0}$$C

D
200$$^{0}$$C

Solution

Given $$ { R }_{ 0} = 100\ \Omega\ at\ { 100}^{ 0 }c.$$
$$ \alpha =0.005 ; R=200\ \Omega\ at\ T=?$$
$$ \Rightarrow (R-{ R }_{ 0 })={ R }_{ 0 }\alpha (T-{ T }_{ 0 })$$
$$ \Rightarrow 100=100\times \dfrac { 5 }{ 1000 } (T-{ T }_{ 0 })$$
$$ \Rightarrow T=200+{ T }_{ 0 }={ 300 }^{ 0 }c$$
Question 6
1 / -0

Three resistances each of $$3\Omega $$ are connected as shown in figure. The resultant resistance between A and F is:

A
$$9\Omega $$

B
$$2\Omega $$

C
$$4\Omega $$

D
$$1\Omega $$

Solution

To draw the circuit bring point B to D and Point C to point E. From the diagram, inner two resistances are in parallel, and its equivalent is also parallel to the third resistance.
$$\therefore R_{AF} = \dfrac{\dfrac{R}{2} \times R}{\dfrac{3R}{2}}$$
$$ = \dfrac{R}{3}$$
$$\Rightarrow R_{AF} = 1 \Omega$$
Question 7
1 / -0

The resistance of a wire at temperature $$30 ^{\circ}\ C$$ is found to be $$10\ \Omega$$ . Now to increase the resistance by $$10\%,$$ the temperature of the wire must be $$($$ The temperature coefficient of resistance of the material of the wire is $$0.002\; {^{\circ}C}^{-1}$$ and reference temperature is $$0 ^{\circ}\ C)$$.

A
$$35^{\circ}$$C

B
$$36^{\circ}C$$

C
$$33^{\circ}C$$

D
$$83^{\circ}C$$

Solution

$$\displaystyle \frac{R_2}{R_1}=\dfrac{(1+\alpha t_2)}{(1+\alpha t_1)}$$
Here $$R_1=10 \Omega, R_2=10+10\times \frac{1}{10}=11\ \Omega, \alpha=0.002 /^oC, t_1=30 ^oC$$
thus, $$\displaystyle t_2=\frac{1}{\alpha}\left[\frac{R_2}{R_1}(1+\alpha t_1)-1\right]= \frac{1}{0.002}\left[\frac{11}{10}(1+0.002\times 30)-1\right]=83 ^oC$$
Question 8
1 / -0

While connecting $$6$$ cells in a battery in series, in a tape recorder, by mistake one cell is connected with reverse polarity. If the effective resistance of load is $$24$$ ohm and internal resistance of each cell is one ohm and emf $$1.5\ V$$, the current delivered by the battery is

A
$$0.1\ A$$

B
$$0.2\ A$$

C
$$0.3\ A$$

D
$$0.4\ A$$

Solution

Applying KVL to the loop

$$ 4E = I[R+6r]$$

$$ 4 \times 1.5 = I[30]$$

$$ I =\dfrac{6}{30}$$

$$ I = \dfrac{1}{5} = 0.2 A$$
Question 9
1 / -0

A wire has a diameter of $$0.2 mm$$ and a length of $$50 cm$$. The specific resistance of it's material is $$40\times 10^{-6}$$$$ohm cm$$. The current through it, when a potential difference of $$2 V$$ is applied across it, is

A
$$3.14 A$$

B
$$31.4 A$$

C
$$0.314 A$$

D
$$0.0314 A$$

Solution

Given: $$l = 50$$ cm, $$\rho = 40 \times 10^{-6} \Omega$$cm, $$r = 0.1$$ mm $$= 0.01$$ cm
The resistance R is given by,
$$R = \dfrac{\rho l}{A} = \dfrac{40 \times 10^{-6} \times 50}{\pi \times (0.01)^2} \approx 6.37 \Omega$$
Now the current through it when a potential difference of 2V is applied across it is:
$$i=\dfrac{V}{R} = \dfrac{2}{6.37} =0.314$$ A.
Question 10
1 / -0

A carbon filament has resistance of 120$$\Omega $$ at $$0^{0}C$$. The resistance of a copper filament connected in series with carbon so that the combination has same resistance at all temperatures must be
$$\alpha $$ of carbon $$= -7\times 10^{-4}/^{0}C$$; $$\alpha $$ of copper $$=4\times 10^{-3}/^{0}C$$

A
120$$\Omega $$

B
21$$\Omega $$

C
60$$\Omega $$

D
210$$\Omega $$

Solution

$$ R_{ca} = R^{0}_{ca}(1+\alpha_{ca} \Delta T)$$

$$ R_{cu} = R^{0}_{cu}(1+\alpha_{cu} \Delta T)$$

$$ R_{ca} + R_{cu} = R^{0}_{ca} + R^{0}_{cu} + \Delta T(R^{0}_{ca}\alpha_{ca} + R^{0}_{cu}\alpha_{cu})$$

$$ \Rightarrow R_{ca}+R_{cu}$$ is independent of temperature

$$ \Rightarrow R^{0}_{ca}\alpha_{ca} + R^{0}_{cu}\alpha_{cu} = 0$$

$$ 120 \times 7\times 10^{-4} = R^{0}_{cu} \times 4 \times 10^{-3}$$

$$ R^{0}_{cu} = 21 \Omega$$

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Re-Attempt Weekly Quiz Competition

Current Electricity Test - 35

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Current Electricity Test - 35

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SHARING IS CARING

Question 1

760$$\Omega $$

260$$\Omega$$

1060$$\Omega$$

960$$\Omega$$

Solution

Question 2

increase resistance in the primary circuit.

decrease resistance in the primary circuit.

decrease resistance in series with the cell whose e.m.f. has to be measured.

increase resistance in series with the cell whose e.m.f. has to be measure.

Solution

Question 3

$$10\Omega $$

$$5\Omega $$

$$20\Omega $$

$$15\Omega $$

Solution

Question 4

Both (A) and (R) are true and (R) is the correct explanation of A

Both (A) and (R) are true but (R) is not the correct explanation of A

(A) is true but (R) is false

(A) is false but (R) is true

Solution

Question 5

300$$^{0}$$C

400$$^{0}$$C

500$$^{0}$$C

200$$^{0}$$C

Solution

Question 6

$$9\Omega $$

$$2\Omega $$

$$4\Omega $$

$$1\Omega $$

Solution

Question 7

$$35^{\circ}$$C

$$36^{\circ}C$$

$$33^{\circ}C$$

$$83^{\circ}C$$

Solution

Question 8

$$0.1\ A$$

$$0.2\ A$$

$$0.3\ A$$

$$0.4\ A$$

Solution

Question 9

$$3.14 A$$

$$31.4 A$$

$$0.314 A$$

$$0.0314 A$$

Solution

Question 10

120$$\Omega $$

21$$\Omega $$

60$$\Omega $$

210$$\Omega $$

Solution

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