Current Electricity Test

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Current Electricity Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

An ideal battery of emf $$2V$$ and a series resistance $$R$$ are connected in the primary circuit of a potentiometer of length $$1$$ m and resistance $$\Omega $$. The value of $$R$$ to give a potential difference of $$5$$ mV across the 10cm of potentiometer wire is:

A
$$180 \Omega $$

B
$$190 \Omega $$

C
$$195 \Omega $$

D
$$200 \Omega $$

Solution

Let,

$$L = 1 m = 100 cm$$

$$l =10cm$$

Voltage drop across potentiometer wire is

$$V = 5\times10^{-3}\dfrac{100}{10} = 0.05 V$$

Current will be

$$I = \dfrac{2}{R + R^{'}} = \dfrac{2}{R + 5}$$

We know that $$V = IR^{'}$$

where, $$R^{'}$$ is resistance of potentiometer wire

$$0.05 = \dfrac{2}{R + 5}5$$

$$R+5 = 200$$

$$R = 195 \Omega$$
Question 2
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The length of a potentiometer wire is $$l$$. A cell of emf $$E$$ is balanced at a length $$\dfrac{l}{5}$$ from the positive end of the wire. If the length of the wire is increased by $$\dfrac{l}{2}$$, the distance of the balance point with the same cell is

A
$$\dfrac{2}{15}l$$

B
$$\dfrac{3}{15}l$$

C
$$\dfrac{3}{10}l$$

D
$$\dfrac{4}{10}l$$

Solution

Here,
The e.m.f. of the cell is
$$E = \dfrac {kl}{5}$$, where $$k$$ is potential gradient.
Also $$k = \dfrac {E_{0}}{l}$$, where $$E_{0}$$ is the e.m.f. of the battery in potentiometer circuit.
$$\therefore E = \dfrac {E_{0}}{5}$$
When the length of potentiometer wire is increased by $$\dfrac {l}{2}$$ the potential gradient becomes
$$k = \dfrac {E_{0}}{\left(l+\dfrac {l}{2}\right)}$$

$$k = \dfrac {2E_{0}}{3l}$$
If $$l_{n}$$ is new balancing length then e.m.f. of cell is
$$E = kl_{n} = \left(\dfrac {2E_{0}}{3l}\right)l_{n}$$

$$\therefore E = \dfrac {E_{0}}{5} = \left(\dfrac {2E_{0}}{3l}\right)l_{n}$$

$$\therefore l_{n} = \dfrac{3l}{10}$$
Question 3
1 / -0

Resistance of a resistor at temperature $$t^{0}C$$ is $$R_{t}=R_{0}(1+\alpha t+\beta t^{2})$$. Here $$R_{0}$$ is the resistance at $$0^{0}C$$. The temperature coefficient of resistance at temperature $$t^{0}C$$ is

A
$$\dfrac{(1+\alpha t+\beta t^{2})}{\alpha +2\beta t}$$

B
$$\alpha +2\beta t$$

C
$$\dfrac{\alpha +2\beta t}{(1+\alpha t+\beta t^{2})}$$

D
$$\dfrac{\alpha +2\beta t}{2(1+\alpha t+\beta t^{2})}$$

Solution

Resistance of a resistor at temperature $$t^{0}C$$ is

$$R_{t}=R_{0}(1+\alpha t+\beta t^{2})$$.

The temperature coefficient of resistance is given by:

Temperature coefficient of resistance$$= \dfrac{1}{R_t} \dfrac{dR}{dt} $$.

Temperature coefficient of resistance $$=\dfrac{R_0}{R_0(1+\alpha t+\beta t^{2})} \dfrac {d}{dt}(1+\alpha t+\beta t^{2})$$

Temperature coefficient of resistance$$=\dfrac{\alpha +2\beta t}{(1+\alpha t+\beta t^{2})}$$.
Question 4
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A potential difference of $$2 V$$ exists across a potentiometer wire of $$2 m$$ length. When the potential difference across a $$2\Omega $$ resistance of a second circuit is measured by the potentiometer wire, it amounts to $$5 mm$$ balancing length. The current in the second circuit is

A
2.5 mA

B
3 mA

C
4.5 mA

D
1 mA

Solution

P.G $$ = \dfrac{\Delta V}{l} = \dfrac{2}{2} = 1 \dfrac{v}{m}$$

$$ \Rightarrow \dfrac{\Delta V}{l} = 1$$

$$ \Rightarrow \Delta V across \ resistance = \dfrac{1 \times 5}{1000} = 0.005 V$$

$$ \Rightarrow I = \dfrac{\Delta V}{r} = 0.0025 = 2.5mA$$
Question 5
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Two unknown resistrance X and Y are connected to left and right gaps of a meter bridge and the balancing point is obtained at 80 cm from left. When a 10$$\Omega $$ resistance is connected in paralle to x, the balance point is 50 cm from left. The values of X and Y respectively are

A
40$$\Omega $$ , 9 $$\Omega $$

B
30$$\Omega $$ , 7.5$$\Omega $$

C
20$$\Omega $$ , 6$$\Omega $$

D
10$$\Omega $$ , 3$$\Omega $$

Solution

Case 1
$$ \Rightarrow \dfrac{X}{Y} = \dfrac{80}{20} = 4$$
$$ X = 4Y$$

Case 2

$$ \dfrac{\dfrac{10X}{10+X}}{Y} = \dfrac{50}{50} = 1$$
$$ \Rightarrow \dfrac{10}{10+X} \times 4 = 1$$

$$ 10 + X = 40$$

$$ X = 30 \Omega$$

$$ Y = \dfrac{30}{4} = 7.5 \Omega$$
Question 6
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The resistance of a 240 V , 200 W electric bulb when hot is 10 times the resistance when cold.The resistance at room temperature and the temperature coefficient of the filament are
(given working temperature of the filament is 2000$$^{0}$$C)

A
$$28.8\Omega,4.5\times 10^{-3}/^{0}C $$

B
$$14.4\Omega,4.5\times 10^{-3}/^{0}C $$

C
$$28.8\Omega,3.5\times 10^{-3}/^{0}C $$

D
$$14.4\Omega,3.5\times 10^{-3}/^{0}C $$

Solution

$$R_2 = \dfrac{V^2}{P}$$
$$ R _2 = \dfrac{240 \times 240}{200} = 288 \Omega$$
$$R_1 = \dfrac{R_2}{10} = 28.8 \Omega$$

$$ R_2 = R_{1}(1+\alpha \Delta T)$$

$$ R_{2} = 10 R_{0}$$

$$ 9 = \alpha \Delta T$$
As $$2000^{0} C > > > $$ Room temp,
we can take
$$ \Delta T = 2000^{0}C$$
$$ \alpha = \dfrac{9}{2000}$$
$$ \therefore \alpha= 4.5 \times 10^{-3}$$
Question 7
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Aluminium$$\left ( \alpha =4\times 10^{-3}K^{-1} \right )$$ resistance of 60$$\Omega $$ and carbon $$\left ( \alpha =0.5\times 10^{-3}K^{-1} \right )$$ resistance 40$$\Omega $$ are connected in parallel. The combination is heated. The effective resistance is

A
Greater than 24$$\Omega $$

B
Less than 24$$\Omega $$

C
Greater than 40$$\Omega $$

D
Greater than 100$$\Omega $$

Solution

$$ R_{eff} = \dfrac{R_{A} R_{C}}{R_{A}+R_{C}}$$

Before heating $$ = \dfrac{60 \times 40}{100} = 24 \Omega$$

$$ \dfrac{\Delta R_{eff}}{R_{eff}} = \dfrac{-\Delta R_{A}}{R_{A}}- \dfrac{\Delta R_{B}}{R_{B}}$$

$$ \alpha_{eff} \Delta T = \Delta T(-\alpha_{A}-\alpha_{B})$$

$$ \Rightarrow \alpha_{eff} = -(\alpha_{A}+\alpha_{B})$$
Question 8
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Consider a thin square sheet of side $$L$$ and thickness $$t$$, made of a material of resistivity $$\rho $$. The resistance between two opposite faces, shown by the shaded areas in the figure is____

A
directly proportional to L

B
directly proportional to t

C
independent of L

D
independent of t

Solution

As $$R= \rho \dfrac{l}{a}$$

Therefore $$R=\rho \dfrac{l}{l\times t}=\dfrac{\rho}{t}$$
Question 9
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If a wire is stretched to make it $$0.1\%$$ longer, its resistance will

A
increase by $$0.05\%$$

B
increase by $$0.2\%$$

C
decrease by $$0.2\%$$

D
decrease by $$0.05\%$$.

Solution

$$\displaystyle \mathrm{R}=\mathrm{\rho}\frac{\ell}{\mathrm{A}}=\frac{\mathrm{\rho}\ell}{A} \times \dfrac {l}{l}=\dfrac{\rho l^2}{Vol^m}$$

$$ \mathrm{R}\propto\ell^{2}$$

$$ \displaystyle \frac{\Delta \mathrm{R}}{\mathrm{R}}=2\frac{\Delta\ell}{\ell}$$

$$ \displaystyle \frac{\Delta \mathrm{R}}{\mathrm{R}}=2\times 0.1 \%$$

$$ \displaystyle \frac{\Delta \mathrm{R}}{\mathrm{R}}=0.2 \%$$

Option B
Question 10
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Consider a long conductor, the middle of which is earthed. If the potential difference across the two ends of the conductor is $$220\ V$$, then what is the potential at the ends and the middle point?

A
$$220\ V$$ all over the conductor

B
$$110\ V$$ and $$-110\ V$$ at ends and $$0\ V$$ at the mid point

C
$$0\ V$$ at ends and $$220\ V$$ at midpoint

D
$$-220\ V$$ all over the conductor

Solution

Let the potential at the ends be $$V_1\,and\, V_2$$.
Since the middle point is earthed, its potential needs to be zero.
Also,
$$ V_1 - V_2 = 220 $$
$$ \Rightarrow V_1 = 110\ V \,and\, V_2 = -110\ V $$

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Re-Attempt Weekly Quiz Competition

Current Electricity Test - 39

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Current Electricity Test - 39

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Question 1

$$180 \Omega $$

$$190 \Omega $$

$$195 \Omega $$

$$200 \Omega $$

Solution

Question 2

$$\dfrac{2}{15}l$$

$$\dfrac{3}{15}l$$

$$\dfrac{3}{10}l$$

$$\dfrac{4}{10}l$$

Solution

Question 3

$$\dfrac{(1+\alpha t+\beta t^{2})}{\alpha +2\beta t}$$

$$\alpha +2\beta t$$

$$\dfrac{\alpha +2\beta t}{(1+\alpha t+\beta t^{2})}$$

$$\dfrac{\alpha +2\beta t}{2(1+\alpha t+\beta t^{2})}$$

Solution

Question 4

2.5 mA

3 mA

4.5 mA

1 mA

Solution

Question 5

40$$\Omega $$ , 9 $$\Omega $$

30$$\Omega $$ , 7.5$$\Omega $$

20$$\Omega $$ , 6$$\Omega $$

10$$\Omega $$ , 3$$\Omega $$

Solution

Question 6

$$28.8\Omega,4.5\times 10^{-3}/^{0}C $$

$$14.4\Omega,4.5\times 10^{-3}/^{0}C $$

$$28.8\Omega,3.5\times 10^{-3}/^{0}C $$

$$14.4\Omega,3.5\times 10^{-3}/^{0}C $$

Solution

Question 7

Greater than 24$$\Omega $$

Less than 24$$\Omega $$

Greater than 40$$\Omega $$

Greater than 100$$\Omega $$

Solution

Question 8

directly proportional to L

directly proportional to t

independent of L

independent of t

Solution

Question 9

increase by $$0.05\%$$

increase by $$0.2\%$$

decrease by $$0.2\%$$

decrease by $$0.05\%$$.

Solution

Question 10

$$220\ V$$ all over the conductor

$$110\ V$$ and $$-110\ V$$ at ends and $$0\ V$$ at the mid point

$$0\ V$$ at ends and $$220\ V$$ at midpoint

$$-220\ V$$ all over the conductor

Solution

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