Current Electricity Test

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Current Electricity Test - 68

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Weekly Quiz Competition

Question 1
1 / -0

In the network atom in figure, the equivalent resistance between the point A and B is :-

A
$$3 \ \Omega $$

B
$$2 \ \Omega $$

C
$$4 \ \Omega $$

D
$$1 \ \Omega $$

Solution

Between points A and B , $$R_3$$ is parallel to $$R_4$$ and $$ R_5$$ whereas $$R_4$$ and $$R_5$$ are in series.

We know that equivalent resistance in series is given as,
$$R_{eq}=R_1+R_2$$

Equivalent resistance in parallel is given as,
$$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1}+\dfrac{1}{R_2}$$

Since $$R_4$$ and $$R_5$$ are in series, $$R_{45} = 2+4 = 6 \ \Omega$$

Since, $$R_{45}$$ and $$R_3$$ are in parallel,
$$\dfrac{1}{R_{eq}}=\dfrac{1}{3}+\dfrac{1}{6}=\dfrac{2+1}{6}=\dfrac{3}{6}$$

$$R_{eq}=2 \ \Omega $$
Question 2
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If two resistances $${R_1} = \left( {10.0 \pm 0.1} \right)\Omega \;and\;{R_2} = \left( {20.0 \pm 0.1} \right)\Omega \;are\;connected$$ in series, then the maximum percentage error in its equivalent resistance will be

A
1/5%

B
2/3%

C
2%

D
4%

Solution

Given,
$$R_1=(10\pm0.1)\Omega$$
$$R_2=(20\pm0.1)\Omega$$
In series combination,
$$R_{eq}=R_1+R_2$$
$$R_{eq}=(10\pm0.1)+(20\pm0.1)=10+20+(\pm0.1)+(0.1)$$
$$R_{eq}=(30\pm0.2)\Omega$$
The maximum percentage error
$$\dfrac{0.2}{30}\times 100=\dfrac{2}{3}$$%
The correct option is B.
Question 3
1 / -0

If the emf of the battery in primary circuit of a potetimeter is doubled the value of potential gradient will:-

A
Remain Uncharged

B
Be halved

C
Be doubled

D
None of these

Solution

$$\begin{array}{l} P.G=\left( { \frac { E }{ { R+{ R_{ w } } } } } \right) \frac { { { R_{ w } } } }{ { { l_{ w } } } } \\ Then, \\ If\, E\, is\, double\, then\, potential\, gradient\, is\, also\, double. \\ Hence,\, option\, \, C\, is\, the\, correct\, answer. \end{array}$$
Question 4
1 / -0

Find the value of I ?

A
-1.4 A

B
-2.1 A

C
2 A

D
-4.2 A

Solution

$$\begin{array}{l} { i_{ 1 } }+{ i_{ 2 } }+I=0 \\ \Rightarrow \dfrac { { 10-2 } }{ 5 } +\dfrac { { 5-2 } }{ 6 } +I=0 \\ \Rightarrow \dfrac { 8 }{ 5 } +\dfrac { 3 }{ 6 } +I=0 \\ \therefore I=-2.1A \end{array}$$
$$\therefore$$ Option $$B$$ is correct.
Question 5
1 / -0

When resistor are connected in series what remains same.?

A
Current

B
Voltage

C
Resistance

D
power

Solution

From the figure we can see
$${ V }_{ 1 }=I{ R }_{ 1 }$$
$${ V }_{ 2 }=I{ R }_{ 2 }$$
$${ V }_{ 3 }=I{ R }_{ 3 }$$
Hence, current is constant when resistor are connected in series.
Question 6
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A long straight wire of radius a carries a steady current $$I.$$ The current is uniformly distributed over its cross-section. The ratio o the magnetic fields $$B$$ and $$B',$$ at radial distances $$\dfrac{a}{2}$$ and $$2a$$ respectively, from the axis of the wire is

A
$$\dfrac{1}{4}$$

B
$$\dfrac{1}{2}$$

C
$$1$$

D
$$4$$

Solution

Consider two amperian loops of radius $$a/2$$ and $$2a$$ as shown in the diagram$$.$$
Applying ampere's circuital law for these loops$$,$$ we get
$$\oint {B.dL = } {\mu _0}{I_{enclosed}}$$
For the smaller loop$$,$$
$$ \Rightarrow B \times 2\pi \dfrac{a}{2} = {\mu _0} \times \dfrac{1}{{\pi {a^2}}} \times {\left( {\dfrac{a}{2}} \right)^2}$$
$$ = {\mu _0}I \times \dfrac{1}{4} = \dfrac{{{\mu _0}I}}{4}$$
$$ \Rightarrow {B_1} = \dfrac{{{\mu _0}I}}{{4\pi a}}$$
$$B' \times 2\pi \left( {2a} \right) = {\mu _0}I$$
$$\dfrac{B}{{B'}} = \dfrac{{{\mu _0}I}}{{4\pi a}} \times \dfrac{{4\pi a}}{{{\mu _0}I}} = 1$$
Hence,
option $$(C)$$ is correct answer.
Question 7
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The magnetic field of $$2\times { 10 }^{ 2 } T$$ acts at right angles to a coil of area $$100 cm^{2}$$ with $$50$$ turns. The average emf induced in the coil is $$0.1 V$$, When it is removed from the field in time $$L$$.The value of $$t$$ is

A
$$0.1 s$$

B
$$1 s$$

C
$$0.01 s$$

D
$$20 s$$

Solution
Question 8
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If $$i=t^2 0 < t < T$$ then r.m.s value of current is

A
$$\dfrac {T^2} {\sqrt2}$$

B
$$\dfrac {T^2} {2}$$

C
$$\dfrac {T^2} {\sqrt5}$$

D
none of these

Solution

$$\begin{array}{l} i={ t^{ 2 } } \\ \left\langle { r.m.s } \right\rangle =\left\langle { { l^{ 2 } } } \right\rangle =\sqrt { \dfrac { { { { \left( { { t^{ 2 } } } \right) }^{ 2 } }dt } }{ { dt } } } \\ =\dfrac { { { t^{ 5 } } } }{ { 4t } } ={ \left[ { \int _{ 0 }^{ 1 }{ \dfrac { { { t^{ 4 } } } }{ 4 } } } \right] ^{ \dfrac { 1 }{ 2 } } } \\ \left\langle { r.m.s } \right\rangle ={ \left[ { \dfrac { { { t^{ 4 } } } }{ 4 } } \right] ^{ \dfrac { 1 }{ 2 } } }=\dfrac { { { T^{ 2 } } } }{ 2 } \\ \therefore { l_{ r.m.s } }=\dfrac { { { T^{ 2 } } } }{ 2 } \end{array}$$
Hence, option $$B$$ is the correct answer.
Question 9
1 / -0

Equivalent resistance between $$A$$ and $$B$$ is

A
$$3 \Omega$$

B
$$6 \Omega$$

C
$$1.5 \Omega$$

D
$$4.5 \Omega$$

Solution

The correct option is C.
Question 10
1 / -0

The equivalent resistance of fig shown between 'a' and 'b' is

A
$$\dfrac {2r} {8}$$

B
$$\dfrac {5r} {8}$$

C
$$\dfrac {r} {8}$$

D
$$\dfrac {3r} {8}$$

Solution

$$\therefore$$ Option $$B$$ is correct.