Electricity Test

Result

Electricity Test - 74

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Weekly Quiz Competition

Question 1
1 / -0

1 kWh is equal to

A
$$3.6 \times 10^6 MJ$$

B
$$3.6 \times 10^5 MJ$$

C
$$3.6 \times 10^2 MJ$$

D
$$3.6 MJ$$

Solution

1 kilowatt hour is the energy produced by 1 kilowatt power source in 1 hour.

$$1kWh=1kW\times 1hour=1000\times 3600 W.s$$

$$\implies 1kWh=3.6\times 10^6J$$

$$\implies 1kWh=3.6MJ$$

Answer-(D)
Question 2
1 / -0

A circular conductor is made of a uniform wire of resistance 2 x $$10^{-3}$$ ohm/metre and the diameter of this circular conductor is 2 metres. Then the resistance measured between the ends of the diameter is (in ohms)

A
$$\pi \times10^{-3}$$

B
$$2 \pi \times10^{-3}$$

C
$$4 \pi \times10^{-3}$$

D
$$4 \times10^{-3}$$

Solution

The resistance per unit length is given to be $$ R' = 2 \times 10^{-3} $$
The length can be calculated as $$ L = \pi \times D $$, where D is given diameter of 2 m.
Hence $$ L = 2 \pi $$

Therefore,

Resistance $$ = 2 \times 10^{-3} \times 2\pi = 4 \pi \times 10^{-3} Ohms. $$

Now when the resistance is measured between ends of any diameter, it can
be seen as parallel connection of two resistances R/2 each (R being the total
resistance of the entire loop). So the resistance between two end points of a diameter will be R/4 i.e. $$\pi \times 10^{-3} Ohms. $$
Question 3
1 / -0

If the filament of bulb D breaks then does bulb C give more light, less light, the same light or no light?

A
more light

B
less light

C
same light

D
no light

Solution

It is assumed that every bulb is identical and has same resistance($$R$$).
So C and D are in parallel.And A , B and combination of (C and D) all these three are in series.
Initially resistance here was $$R/2$$ (C and D both have $$R$$ in parallel).So resistances in series are $$R$$ ,$$R$$ and $$R/2$$. Since resistance across third element is less than first two elements so less portion of cell voltage will be shared here.

Now bulb D breaks.So combination of C and D will reduced to only C.Now only C is there so net resistance here is $$R$$ .So now A ($$R$$ resistance) , B ($$R$$ resistance) and C ($$R$$ resistance) are in series and have equal resistances so voltage across C will be $$1/3$$ of cell voltage.

In other words after breaking of D voltage across C has increased from before hence it will glow more brightly than before.
Question 4
1 / -0

Do the arrows in the diagram indicate the direction of electron flow or conventional flow of electricity?

A
Conventional flow of current

B
Flow of Electron

C
Flow of positive charge

D
None of the above

Solution

We know that conventional current flows from positive terminal of battery or cell to negative terminal of cell.Since in the figure the current is shown from positive terminal of cell (represented by bigger line) hence it is conventional current shown in the figure.
Direction of electron flow is opposite to conventional current direction.
Question 5
1 / -0

An electric lamp whose resistance of 10 ohm, and a conductor of 2 ohm resistance are connected in series with a 6 V battery. The total current through the circuit and the potential difference across the electric lamp are :

A
3.6 A, 6 V

B
0.5 A, 5 V

C
2.0 A, 0.2 V

D
0.3 A, 3 V

Solution

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Question 6
1 / -0

In the network shown, each resistance is equal to R. The equivalent
resistance between diagonally opposite corners is:

A
R

B
$$\frac{R}{3}$$

C
$$\frac{2R}{3}$$

D
$$\frac{4R}{3}$$

Solution
Question 7
1 / -0

The equivalent resistance of eight equal resistances in series is $$48$$ ohms. What would be the equivalent resistance if they are connected in parallel?

A
$$\displaystyle \frac {3}{4} \Omega$$

B
$$\displaystyle \frac {4}{3} \Omega$$

C
$$4 \Omega$$

D
$$3 \Omega$$

Solution

Let the resistance of each resistor is $$R$$.

In series we know, equivalent resistance is $$R_s=nR=8R=48\Omega$$
$$\implies R=6\Omega$$

In paralle we know, equivalent resistance is $$R_p=\dfrac{R}{n}$$
$$\implies R_p=\dfrac{6}{8}=\dfrac{3}{4}\Omega$$

Answer-(A)
Question 8
1 / -0

Arrange the order of power dissipated in the given circuits, if the same current enters at point $$A$$ in all the circuits and resistance of each resistor is $$R$$.

A
$$(ii) > (iii) > (iv) > (i)$$

B
$$(iii) > (ii) > (iv) > (i)$$

C
$$(iv) > (iii) > (ii) > (i)$$

D
$$(i) > (ii) > (iii) > (iv)$$

Solution
Question 9
1 / -0

When current is passed through a conductor, the heat produced in it depends upon
I. The material of the conductor
II. The current flowing through the conductor.
III. The time for which the current flows.

A
I and II only

B
I and III only

C
II and III only

D
All I, II and III

Solution
Question 10
1 / -0

Two heater coils separately take $$10$$ minute and $$5$$ minute to boil certain amount of water. If both the coils are connected in series, the time taken will be

A
$$15$$ min

B
$$7.5$$ min

C
$$3.33$$ min

D
$$2.5$$ min

Solution