Current Electricity Test

Result

Current Electricity Test - 54

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A lamp is marked 60 W, 220 V. If it operates at 200 V, the rate of consumption of energy will _____

A
decrease

B
increase

C
remain unchanged

D
first increase then decrease

Solution

Given power of bulb =60W
Voltage=220V
Power $$=\dfrac{V^2}{R}$$
$$60=\dfrac{220\times220}{R}$$
$$R=(\dfrac{220\times220}{60})\Omega$$
$$R=(\dfrac{220\times22}{6})\Omega$$ (1)
When voltage = 200V, resistance will be the same
$$P=\dfrac{200\times200}{R}$$
From 1,
$$=\dfrac { 200\times 200\times 6 }{ 220\times 22 } $$

$$\dfrac { 200\times 20\times 6 }{ 22\times 22 } $$

$$=49.5 W$$

We see that power gets decreased.
Power means rate of consumption of energy, so rate of consumption of energy decreases.
Option A is correct.
Question 2
1 / -0

Equivalent resistance of the above combination will be:

A
$$\displaystyle 5\Omega $$

B
$$\displaystyle 5/4\Omega $$

C
$$\displaystyle 10\Omega $$

D
$$\displaystyle 20\Omega $$

Solution

Equivalent resistance of resistors connected in series
$$R_{eq} = R_1+R_2+R_3+R_4$$

$$\therefore$$ $$R_{eq} = 5+5+5+5 =20\Omega$$
Question 3
1 / -0

Find the effective resistance, when $$1\Omega$$, $$10\Omega$$ and $$4\Omega$$ resistances are connected in series.

A
$$25\Omega$$

B
$$15\Omega$$

C
$$30\Omega$$

D
$$\frac {20}{27}\Omega$$

Solution

In a series combination of resistors, the equivalent resistance is given by-

$$R_{eq}=R_1+R_2+R_3+...$$

Here-

$$R_{eq}=1+10+4$$

$$\implies R_{eq}=15\Omega$$

Answer-(B)
Question 4
1 / -0

Wire of a certain material is stretched slowly be dep10. Its new resistance and specific resistance becomes respectively.

A
Both remain same

B
1.1 times in both

C
1.2 times , 1.1 times

D
1.21 times, same

Solution

Resistance of the wire $$R = \dfrac{\rho l^2}{V}$$
Length of the wire after it is stretched $$l' = 1.1 l$$
$$\therefore$$ New resistance $$R' = \dfrac{\rho (1.1 l)^2}{V} = 1.21 R$$
Specific resistance is an intrinsic property of a material and it does not depend on the dimension of the conductor. Hence specific resistance remains the same even when the wire is stretched.
Question 5
1 / -0

Two wires of resistances 10 $$\Omega$$ and 5 $$\Omega$$ are connected in series. The effective resistances is _______ $$\Omega$$

A
15

B
20

C
30

D
40

Solution

we know that in the series combination of two resistors, the equivalent resistance is given by-

$$R_{eq}=R_1+R_2$$

Here-

$$R_{eq}=10+5$$

$$\Rightarrow R_{eq}=15\Omega$$
Question 6
1 / -0

A source of electromotive force (emf) is a:

A
Cell

B
Battery

C
Generator

D
All of these

Solution

A source of EMF can be cell, battery or generator.
Question 7
1 / -0

Find the effective resistance when $$\displaystyle 1\Omega ,2\Omega $$ and $$\displaystyle 3\Omega $$ resistances are connected in series

A
$$\displaystyle 2\Omega $$

B
$$\displaystyle 1\Omega $$

C
$$\displaystyle 6\Omega $$

D
$$\displaystyle 5\Omega $$

Solution

Effective resistance of resistors connected in series $$R_{s} = R_1+R_2+R_3$$
$$\implies$$ $$R_s = 1+2+3 = 6\Omega$$
Question 8
1 / -0

A wire of radius r has resistance R. If it is stretched to a radius $$\displaystyle\frac{r}{2}$$, its resistance will be

A
$$16\,\text{R}$$

B
$$2\,\text{R}$$

C
$$4\,\text{R}$$

D
$$0$$

Solution

Let length & cross sectional area of wire be $$\ell$$ & a respectively. $$\rho$$ be the specific resistance, then
$$R=\rho\displaystyle\frac{\ell}{a}$$
If radius becomes half, area becomes $$\displaystyle\frac{1}{4}th$$ or cross sectional area after the stretch $$=\displaystyle\frac{a}{4}$$. Let its length increases to $$\ell'$$. Since volume remains the same in the process,
$$\ell a=\ell'\times\displaystyle\frac{a}{4}\;\Rightarrow\;\ell'=4\;\ell$$
Let R' be the resistance of stretched wire,
$$R'=\rho\displaystyle\frac{4\ell}{a/4}=16\times\rho\displaystyle\frac{\ell}{a}=16R$$
Question 9
1 / -0

The current in the given circuit is

A
$$0.3\;\text{amp}$$

B
$$0.4\;\text{amp}$$

C
$$0.1\;\text{amp}$$

D
$$0.2\;\text{amp}$$

Solution

Two batteries are joined with opposite polarity so, total e.m.f.=$$5-2=3\;V$$
Total resistance = $$10+20=30\Omega$$
Current = $$\displaystyle\frac{3}{30}=0.1\;\text{A}$$
Question 10
1 / -0

A cell of emf $$5$$ V can supply a total energy of $$9000$$ J, then the total charge that can be obtained from the cell would be _____ C

A
$$180$$

B
$$18000$$

C
$$1800$$

D
$$18$$

Solution

EMF = 5V
Energy = 9000J
Power = Voltage $$\times$$ current
P = VI ($$\varepsilon $$ = energy )(t = time )(t = time)(it = q)
$$\varepsilon =VIt\\ \varepsilon =Vq\\ \dfrac { 9000 }{ 5 } = 1800C\\ q=1800c$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Current Electricity Test - 54

Result

Current Electricity Test - 54

Score

-

Rank

-

SHARING IS CARING

Question 1

decrease

increase

remain unchanged

first increase then decrease

Solution

Question 2

$$\displaystyle 5\Omega $$

$$\displaystyle 5/4\Omega $$

$$\displaystyle 10\Omega $$

$$\displaystyle 20\Omega $$

Solution

Question 3

$$25\Omega$$

$$15\Omega$$

$$30\Omega$$

$$\frac {20}{27}\Omega$$

Solution

Question 4

Both remain same

1.1 times in both

1.2 times , 1.1 times

1.21 times, same

Solution

Question 5

15

20

30

40

Solution

Question 6

Cell

Battery

Generator

All of these

Solution

Question 7

$$\displaystyle 2\Omega $$

$$\displaystyle 1\Omega $$

$$\displaystyle 6\Omega $$

$$\displaystyle 5\Omega $$

Solution

Question 8

$$16\,\text{R}$$

$$2\,\text{R}$$

$$4\,\text{R}$$

$$0$$

Solution

Question 9

$$0.3\;\text{amp}$$

$$0.4\;\text{amp}$$

$$0.1\;\text{amp}$$

$$0.2\;\text{amp}$$

Solution

Question 10

$$180$$

$$18000$$

$$1800$$

$$18$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.