Current Electricity Test

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Current Electricity Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

On interchanging the resistances, the balance point of a meter bridge shifts to the left by $$10$$cm. The resistance of their series combination is $$1k\Omega$$. How much was the resistance on the left slot before interchanging the resistances?

A
$$550 \Omega$$

B
$$910 \Omega$$

C
$$990 \Omega$$

D
$$505 \Omega$$

Solution

$$R_1 + R_2 = 1000$$
$$R_2 = 1000 - R_1$$

On balancing condition,

$$R_1(100-l)=(1000 - R_1)l$$ ............ (1)

On interchanging resistance

On balancing condition

$$(1000 - R_1)(110-l) = R_1 (l-10)$$

$$R_1(l-10) = (1000 - R_1)(110-l)$$ .............. (2)

On dividing (1) by (2)

$$\dfrac{100-l}{l-10} = \dfrac{l}{110-l}$$

$$l = 55$$

On putting $$l = 55$$ in equation (1)

we get $$R_1(100 - 55) = (1000 - R_1)55$$

On solving we will get $$R_1 = 550\Omega $$
Question 2
1 / -0

In a Wheatstone's bridge, there resistances $$\mathrm{P},\ O$$ and $$\mathrm{R}$$ connected in the three arms and the fourth arm is formed by two resistances $$\mathrm{S}_{1}$$ and $$\mathrm{S}_{2}$$ connected in parallel. The condition for bridge to be balanced will be

A
$$\displaystyle \dfrac{\mathrm{P}}{\mathrm{O}}=\dfrac{\mathrm{R}}{\mathrm{S}_{1}+\mathrm{S}_{2}}$$

B
$$\displaystyle \dfrac{\mathrm{P}}{\mathrm{O}}=\dfrac{2\mathrm{R}}{\mathrm{S}_{1}+\mathrm{S}_{2}}$$

C
$$\displaystyle \dfrac{\mathrm{P}}{\mathrm{O}}=\dfrac{\mathrm{R}(\mathrm{S}_{1}+\mathrm{S}_{2})}{\mathrm{S}_{1}\mathrm{S}_{2}}$$

D
$$\displaystyle \dfrac{\mathrm{P}}{\mathrm{O}}=\dfrac{\mathrm{R}(\mathrm{S}_{1}+\mathrm{S}_{2})}{2\mathrm{S}_{1}\mathrm{S}_{2}}$$

Solution

Equivalent resistance of the resistors $$S_1$$ and $$S_2$$ connected in parallel $$\dfrac{1}{S} =\dfrac{1}{S_1}+ \dfrac{1}{S_2} $$
$$\implies$$ $$S = \dfrac{S_1S_2}{S_1+S_2}$$
For balanced bridge, $$\dfrac{P}{O} = \dfrac{R}{S}$$ $$\implies \dfrac{P}{O} = \dfrac{R(S_1+S_2)}{S_1S_2}$$
Question 3
1 / -0

The potential difference $$(V_A-V_B)$$ between the points A and B in the given figure is.

A
+ 9 V

B
-3 V

C
+3 V

D
+6 V

Solution

Given : $$I = 2$$ A
Using Kirchhoff's law $$V_A - I(2) - 3 - I(1) - V_B = 0$$
$$\therefore$$ $$V_A - 2\times 2 - 3 - 2\times 1 -V_B = 0$$
OR $$V_A - 9 - V_B = 0$$
$$\implies V_A - V_B = +9$$ V
Question 4
1 / -0

A steady current flow in a metallic conductor of non-uniform cross-section. The quantity/quantities remaining constant along the whole length of the conductor is/are:

A
current, electric field and drift speed

B
drift speed only

C
current and drift speed

D
current only

Solution

When a steady current flows in a metallic conductor of non-uniform cross-section then the drift speed is $$V_d=I/neA$$ and the electric field $$E=I/ \rho A$$
$$\Rightarrow V_d\propto \dfrac{1}{A}$$ and $$E\propto \dfrac{1}{A}$$
$$\Rightarrow$$ Only current remains constant.
In a metallic conductor of non-uniform cross-section, only the current remains constant along the entire length of the conductor.
Question 5
1 / -0

One Volt is equal to -

A
1 Joule

B
1 Newton/Coulomb

C
1 Joule/Coulomb

D
1 Coulomb/Newton

Solution

$$\bullet $$ Volt is the electrical unit of voltage or potential difference $$(V)$$.

$$\bullet $$ $$V$$ is related to Change in Potential Energy $$\Delta U$$ as
$$ V = \cfrac{\Delta U}{q}$$

From above formula, We get $$1$$ Volt potential difference when a change of $$1$$ joule of Potential energy is there in moving a charge of $$1$$ Coulomb.

Therefore, $$1$$ Volt = $$1$$ Joule/Coulomb.
Question 6
1 / -0

Kirchoffs law of junctions is also called the law of conservation of:

A
Energy

B
Charge

C
Momentum

D
Angular momentum

Solution

Kirchoff's second law:- According to the law, at a particular junction the charge entering the junction is equal to charge exiting that junction.
$$ \Rightarrow$$ There is no charge accumulated at junction. This is also called as law of conservation of charge.
Question 7
1 / -0

In comparing e.m.f.s of 2 cells with the help of potentiometer, at the balance point, the current flowing through the wire is taken from:

A
1) Any one of these cells

B
2) both of these cells

C
3) Battery in the main circuit

D
4) From an unknown source

Solution

emf of secondary cell $$ = I \ r \ l$$
where
I - current in the main circuit from battery.
r - resistance per unit length of wire.
l - balancing length.
Question 8
1 / -0

The potential gradients on the potentiometer wire are$$V_{1}$$ and $$V_{2}$$ with an ideal cell and a real cell of same emf in the primary circuit then

A
$$V_{1}=V_{2}$$

B
$$V_{1} > V_{2}$$

C
$$V_{1} < V_{2}$$

D
$$V_{1} \leq V_{2}$$

Solution

In an ideal cell the internal resistance is zero.
$$ \therefore$$ Whole P.D is across potentiometer.
But in an real cell the internal resistance is not zero.
$$ \Rightarrow$$ There will be some potential drop across internal resistance.
$$ \therefore$$ P.D across potentiometer < Emf of cell.
$$ \therefore V_{1} > V_{2}$$.
Question 9
1 / -0

A steady current is passing through a linear conductor of non uniform cross-section. The net quantity of charge crossing any cross section per second is

A
Independent of area of cross-section

B
Directly proportional to the length of the conductor

C
Directly proportional to the area of cross section.

D
Inversely proportional to the length of the conductor

Solution

The quantity of charge passing through any cross-section per second is nothing but current flowing in the conductor.
As given, the current is steady current, means current is uniform throughout the cross-section. Hence it is independent of Area of cross section.
Question 10
1 / -0

Assertion(A) : The e.m.f of the cell in secondary circuit must be less than e.m.f of cell in primary circuit in potentiometer.
Reason (R): Balancing length cannot be more than length of potentiometer wire.

A
Both A & R are true and R is the correct explanation of A

B
Both A & R are true and R is not a correct explanation of A

C
A is true but R is false

D
A is false but R is true

Solution

When the e.m.f of secondary cell is greater than the primary cell, it implies that the balancing length is greater than the available length. Therefore we cant measure the e.m.f of secondary cell.
$$\therefore$$ E. M.F of of secondary cell < E.M.F of primary cell.
Hence assertion and reason both are correct and reason is the correct explanation of the assertion.

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Current Electricity Test - 22

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Current Electricity Test - 22

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Question 1

$$550 \Omega$$

$$910 \Omega$$

$$990 \Omega$$

$$505 \Omega$$

Solution

Question 2

$$\displaystyle \dfrac{\mathrm{P}}{\mathrm{O}}=\dfrac{\mathrm{R}}{\mathrm{S}_{1}+\mathrm{S}_{2}}$$

$$\displaystyle \dfrac{\mathrm{P}}{\mathrm{O}}=\dfrac{2\mathrm{R}}{\mathrm{S}_{1}+\mathrm{S}_{2}}$$

$$\displaystyle \dfrac{\mathrm{P}}{\mathrm{O}}=\dfrac{\mathrm{R}(\mathrm{S}_{1}+\mathrm{S}_{2})}{\mathrm{S}_{1}\mathrm{S}_{2}}$$

$$\displaystyle \dfrac{\mathrm{P}}{\mathrm{O}}=\dfrac{\mathrm{R}(\mathrm{S}_{1}+\mathrm{S}_{2})}{2\mathrm{S}_{1}\mathrm{S}_{2}}$$

Solution

Question 3

+ 9 V

-3 V

+3 V

+6 V

Solution

Question 4

current, electric field and drift speed

drift speed only

current and drift speed

current only

Solution

Question 5

1 Joule

1 Newton/Coulomb

1 Joule/Coulomb

1 Coulomb/Newton

Solution

Question 6

Energy

Charge

Momentum

Angular momentum

Solution

Question 7

1) Any one of these cells

2) both of these cells

3) Battery in the main circuit

4) From an unknown source

Solution

Question 8

$$V_{1}=V_{2}$$

$$V_{1} > V_{2}$$

$$V_{1} < V_{2}$$

$$V_{1} \leq V_{2}$$

Solution

Question 9

Independent of area of cross-section

Directly proportional to the length of the conductor

Directly proportional to the area of cross section.

Inversely proportional to the length of the conductor

Solution

Question 10

Both A & R are true and R is the correct explanation of A

Both A & R are true and R is not a correct explanation of A

A is true but R is false

A is false but R is true

Solution

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