Current Electricity Test

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Current Electricity Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

A metallic conductor at 10$$^o$$C connected in the left gap of Meter Bridge gives balancing length 40 cm. When the conductor is at 60$$^o$$C, the balance point shifts by 5 cm. The temperature coefficient of resistance of the material of the wire

A
(1 / 210) /$$^o$$C

B
(1 / 220) /$$^o$$C

C
(1 / 200) /$$^o$$C

D
(1 / 201) /$$^o$$C

Solution

$$ R_{1} - Initial \ R (10^{0} C)$$

$$ R_{2} - Final \ R (20^{0} C)$$

$$ \dfrac{R_{1}}{Q} = \dfrac{40}{60} = \dfrac{2}{3}$$

$$ \dfrac{R_{2}}{Q} = \dfrac{45}{55}$$

$$ \dfrac{R_{1}}{R_{2}} = \dfrac{2 \times 11}{3 \times 9}$$

$$ \dfrac{R_{1}}{R_{2}} = \dfrac{22}{27}$$

$$ \dfrac{1}{1+ \alpha \Delta T} = \dfrac{22}{27} \Rightarrow \dfrac{22}{27} = 1+ \alpha \Delta T$$

$$ \Rightarrow \dfrac{5}{22 \times 50} = \alpha \Rightarrow \alpha = \dfrac{1}{220}$$
Question 2
1 / -0

A cell of emf e$$_{1}$$ in the secondary circuit gives null deflection for 1.5m length of potentiometer of wire length 10m. If another cell of emf e$$_{2}$$ is connected in series with e$$_{1}$$then null deflection was obtained for 2.5 m length. Then e$$_{1}$$: e$$_{2}$$ is:

A
3 : 5

B
5 : 3

C
3 : 2

D
2 : 3

Solution

Formula $$\dfrac { e_{ 1 } }{ e_{ 2 } } =\dfrac { l_{ 1 } }{ l_{ 2 } } $$

$$\Rightarrow \dfrac { e_{ 1 } }{ e_{ 1 }+e_{ 2 } } =\dfrac { 1.5 }{ 2.5 }$$ ( here $$e_{ 2 }=e_{ 1 }+e_{ 2 }$$ )

$$\dfrac { e_{ 1 } }{ e_{ 1 }+e_{ 2 } } =\dfrac{3}{5}$$

$$5e_{ 1 }=3e_{ 1 }+3e_{ 2 }$$

$$\Rightarrow 2e_{ 1 }=3e_{ 2 }$$

$$\Rightarrow \dfrac { e_{ 1 } }{ e_{ 2 } } =\dfrac{3}{2}$$
Question 3
1 / -0

The resultant resistance of two resistors when connected in series is 48 ohm. The ratio of their resistances is 3 : 1. The value of each resistance is

A
20$$\Omega $$ ,28$$\Omega $$

B
32$$\Omega $$ ,16$$\Omega $$

C
36$$\Omega $$, 12$$\Omega $$

D
24$$\Omega $$, 24$$\Omega $$

Solution

$$ R_{1} + R_{2} = 48 $$ ohm

$$ \frac{R_{1}}{R_{2}} = \frac{3}{1}$$

$$ \Rightarrow R_{1} = 3 R_{2}$$

$$ \Rightarrow 4R_{2} = 48 \Rightarrow R_{2} = 12 $$ ohm

$$ R_{1} = 36 $$ ohm
Question 4
1 / -0

If the electric current in a lamp decreases by $$5\ \%$$, then the power output decreases by:

A
$$20\ \%$$

B
$$10\ \%$$

C
$$5\ \%$$

D
$$2.5\ \%$$

Solution

Hint: Power consumed by resistor is $$P = {i^2}R$$ ,where $$i$$ is the current flowing through the resistor and $$R$$ is the total resistance.

Correct option: B

Explanation for correct answer:

Step 1: Find the decrease in power of the lamp.

As, power of the lamp, $$P = {i^2}R$$

$$ \Rightarrow \dfrac{{dP}}{P} = 2\dfrac{{di}}{i}$$

$$ \Rightarrow \dfrac{{dP}}{P} = 2 \times 5\% $$

$$ \Rightarrow \dfrac{{\vartriangle P}}{P} = 10\% $$

So, the decrease in the power of the lamp is $$10\% $$ .
Question 5
1 / -0

A series battery of six lead accumulators, each of emf 2.0V and internal resistance 0.50$$\Omega $$ is charged by a 100 V dc supply. The series resistance should be used in the charging circuit in order to limit the current to 8.0A is

A
4$$\Omega $$

B
6$$\Omega $$

C
8$$\Omega $$

D
10$$\Omega $$

Solution

$$ \Rightarrow (100-12) = I[3+R]$$

$$ \dfrac{88}{3+R} = 8 $$

$$ \Rightarrow R = 8\Omega$$
Question 6
1 / -0

If in a Wheatstone bridge the battery and Galvanometer are interchanged, the condition for balance

A
is disturbed

B
is not disturbed

C
depends on the internal resistance of the bridge

D
depends on the values of the resistances in the bridge

Solution

Case 1

$$ \dfrac{R_{1}}{R_{2}} = \dfrac{R_{3}}{R_{4}}$$

Case 3
Simplified

$$ \dfrac{R_{1}}{R_{3}} = \dfrac{R_{2}}{R_{4}}$$

$$ \dfrac{R_{1}}{R_{2}} = \dfrac{R_{3}}{R_{4}}$$

Condition hasn't changed.
Question 7
1 / -0

Two equal resistances are connected in the gaps of a meter bridge. If the resistance in the left gap is increased by $$10\%$$, the balancing point shift :

A
$$10\%$$ to right

B
$$10\%$$ to left

C
$$9.6\%$$ to right

D
$$4.8\%$$ to right
Solution
Hint: In a meter bridge, $$\dfrac{{{R_{left}}}}{{{R_{right}}}} = \dfrac{l}{{100 - l}}$$ , where $${R_{left}}$$ and $${R_{right}}$$ are the resistances on the left and the right gap of the meter bridge respectively and $$l$$ is the balancing length.

Correct option: D

Explanation for correct answer:

Step 1: Find the shift in balancing point.
- Initially, $$\dfrac{R}{R} = \dfrac{l}{{100 - l}}$$
$$ \Rightarrow l = 50cm$$
- After the increase of resistance in left gap by $$10\% $$,
$$\dfrac{{1.1R}}{R} = \dfrac{l}{{100 - l}}$$

$$ \Rightarrow l = 52.38cm$$
- The shift in balancing point, $$\dfrac{{\Delta l}}{l} \times 100 = \dfrac{{2.38}}{{50}} \times 100 =4.76\approx 4.8\% $$
So, the shift in the balancing point is $$4.8\% $$ .
Question 8
1 / -0

In a potentiometer experiment, the balancing length of potentiometer of a cell of e.m.f 1.5V in the secondary is 440 cm. A resistance 5$$\Omega $$ is connected between the terminals of cell, the balancing length is 400 cm.Then
a) internal resistance of the cell is 0.5$$\Omega $$
b) terminal voltage of the cell is 15/11V
c) Potential gradient of the potentiometer wire is $$\frac{1.5}{440}V/cm$$
d) potential difference across the potoentiometer wire of length 10m is nearly 3.4V

A
a,b are correct

B
a,b and c are correct

C
a,b and d are correct

D
a,b,c and d are correct

Solution

$$ r= R(\dfrac{l_{1}}{l_{2}} - 1)$$
$$ = 5(\dfrac{440}{400} -1) = 0.5\Omega$$

$$ \Rightarrow 1.5 = I(5.5)$$
$$ I = \dfrac{1.5}{5.5}$$

$$ \Rightarrow Terminal Voltage = E-IR$$
$$ = 1.5-\dfrac{1.5}{5.5} \times 0.5$$
$$ = 1.5(\dfrac{5.5-0.5}{5.5}) = \dfrac{15}{11} V$$

P.G $$ = \dfrac{emf balanced}{length required}$$
$$ = \dfrac{1.5}{440} \dfrac{V}{cm}$$

$$ \Rightarrow \dfrac{\Delta V}{l} = \dfrac{1.5}{440}$$

$$ \Delta V = \dfrac{1.5}{440} \times 1000$$

$$ = 3.4V$$
$$ \therefore a,b,c,d $$
are true
Question 9
1 / -0

The V - I graph for a conductor at temperature T$$_{1}$$ and T$$_{2}$$ are shown in fig.
(T$$_{2}$$ - T$$_{1}$$) is proportional to

A
$$Cos 2\theta $$

B
$$Sin 2\theta $$

C
$$Cot 2\theta $$

D
$$Tan 2\theta $$

Solution

We know that for conductors
$$resistance \propto temperature$$
i.e $$R_{1} \propto T_{1} \Rightarrow \tan\theta \propto T_{1} \Rightarrow \tan\theta = k T_{1}$$
and $$R_{2} \propto T_{2} \Rightarrow \tan(90^{\circ} -\theta) \propto T_{2} \Rightarrow \cot\theta = k T_{2}$$
where k is constant
$$k(T_{2} - T_{1}) = (\cot\theta - \tan\theta)$$

$$k(T_{2} - T_{1}) = (\dfrac{\cos\theta}{\sin\theta} - \dfrac{\sin\theta}{\cos\theta})$$

$$k(T_{2} - T_{1}) = (\dfrac{\cos^{2}\theta - \sin^{2}\theta}{\sin\theta\cos\theta})$$

$$k(T_{2} - T_{1}) = \dfrac{\cos2\theta}{\sin\theta\cos\theta}$$

$$k(T_{2} - T_{1}) = 2\cot2\theta$$

$$\Rightarrow T_{2} - T_{1} \propto \cot2\theta$$
Question 10
1 / -0

The temperature coefficient resistivity of a material is $$0.0004/K$$. When the temperature of the material is increased by $$50^{o}C$$, its resistivity increases by $$2\times 10^{-8}\ ohm-meter$$.The initial resistivity of the material of the resistance

A
$$50\times 10^{-8}$$

B
$$90\times 10^{-8}$$

C
$$100\times 10^{-8}$$

D
$$200\times 10^{-8}$$

Solution

$$(\rho-\rho_{0}) = \rho_{0} \propto \Delta T$$ (using standard result)

$$ \Rightarrow 2 \times10^{-8} = \rho_{0} \times4 \times 10^{-4} \times50$$
$$\Rightarrow 2\times 10^{-8} = \rho_{0}\times2 \times 10^{-2}$$

$$\Rightarrow \rho_{0} = 100\times10^{-8} ohm-meter$$

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Current Electricity Test - 38

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Current Electricity Test - 38

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Question 1

(1 / 210) /$$^o$$C

(1 / 220) /$$^o$$C

(1 / 200) /$$^o$$C

(1 / 201) /$$^o$$C

Solution

Question 2

3 : 5

5 : 3

3 : 2

2 : 3

Solution

Question 3

20$$\Omega $$ ,28$$\Omega $$

32$$\Omega $$ ,16$$\Omega $$

36$$\Omega $$, 12$$\Omega $$

24$$\Omega $$, 24$$\Omega $$

Solution

Question 4

$$20\ \%$$

$$10\ \%$$

$$5\ \%$$

$$2.5\ \%$$

Solution

Question 5

4$$\Omega $$

6$$\Omega $$

8$$\Omega $$

10$$\Omega $$

Solution

Question 6

is disturbed

is not disturbed

depends on the internal resistance of the bridge

depends on the values of the resistances in the bridge

Solution

Question 7

$$10\%$$ to right

$$10\%$$ to left

$$9.6\%$$ to right

$$4.8\%$$ to right

Solution

Question 8

a,b are correct

a,b and c are correct

a,b and d are correct

a,b,c and d are correct

Solution

Question 9

$$Cos 2\theta $$

$$Sin 2\theta $$

$$Cot 2\theta $$

$$Tan 2\theta $$

Solution

Question 10

$$50\times 10^{-8}$$

$$90\times 10^{-8}$$

$$100\times 10^{-8}$$

$$200\times 10^{-8}$$

Solution

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