[3 Mark Questions] - Science STD 10 Questions - Vidyadip

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[3 Mark Questions]

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Question 13 Marks

(i) State the law that explains the heating effect of current with respect to the measurable properties in an electrical circuit.
(ii) List the factors on which the resistance of a conductor depends.

Answer

i. Joules law of heating states that the heat dissipated across a resistor is directly proportional to
(a) the square of the current flowing through it
(b) The resistance of the conductor
(c) duration of flow of current.
$
H =1^2 R t \text { (alternative answer). }
$ ii. Resistance of a conductor depends on
(a) the length of the conductor
(b) the area of the cross section
(c) nature of material
(d) temperature of the conductor.

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Question 23 Marks

For the circuit shown in the diagram given below:

Calculate:

The total effective resistance of the circuit.
The total current drawn from the battery.
The value of current through each resistor.

Answer

$\frac{1}{R} = \frac{1}{R_1} +\frac{1}{R_2} + \frac{1}{R_3}$

$ = \frac{1}{12} + \frac{1}{8} + \frac{1}{30} = \frac{31}{180}$

$\therefore R = \frac{180}{31} \Omega \text{ or 5.8 } \Omega$

$I = \frac{V}{R} = \frac{\text{6 V}}{\text{5.8 } \Omega} = 1.03 \text{ A}$
$I_{1} = \frac{V}{R_1} = \frac{\text{6 V}}{\text{12 } \Omega} = 0.5 \text{ A}$

$I_{2} = \frac{V}{R_2} = \frac{\text{6 V}}{\text{18 } \Omega} = 0.33 \text{ A}$

$I_{3} = \frac{V}{R_3} = \frac{\text{6 V}}{\text{30 } \Omega} = 0.2 \text{ A}$

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Question 33 Marks

Define the term ‘volt’.
State the relation between work, charge and potential difference for an electric circuit.

Calculate the potential difference between the two terminals of a battery if 100 joules of work is required to transfer 20 coulombs of charge from one terminal of the battery to the other.

Answer

One volt is the potential difference between two points in a current carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to another.
Work = charge x potential difference $(\text{OR }\text{W = Q} \times \text{V})$

$V = \frac{W}{Q}$

$V = \frac{\text{100 J}}{\text{20 C}}$

$\therefore V = \text{5 volt}$

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Question 43 Marks

Distinguish between the terms ‘overloading’ and ‘short-circuiting’ as used in domestic circuits.
Why are the coils of electric toasters made of an alloy rather than a pure metal?

Answer

Overloading occurs when an electric circuit draws more current than the permitted value and short circuiting occurs when the ends of a circuit are connected by a conductor of very low resistance as compared to that of circuit.
The coils of electric toaster made of an alloy rather than pure metal because:

Resistivity of alloys is generally higher than its constituent metals.
It has high melting point.
It does not oxidize i.e., burn.

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Question 53 Marks

Two lamps, one rated 60 W at 220 V and the other 40 W at 220 V, are connected in parallel to the electric supply at 220 V.

Draw a circuit diagram to show the connections.
Calculate the current drawn from the electric supply.
Calculate the total energy consumed by the two lamps together when they operate for one hour.

Answer

$\text{I} = \frac{\text{P}}{\text{V}}$
$\text{I}_{1} = \frac{\text{60 W}}{\text{220 V}} = \frac{3}{11}\text{A}$
$\text{I}_{2} = \frac{\text{40 W}}{\text{220 V}} = \frac{2}{11}\text{A}$
$\text{I} = \text{I}_{1} + \text{I}_{2} = \frac{3}{11} + \frac{2}{11} = \frac{5}{11} \text{A} = 0.45 \text{ A}$

$\text{E} = \text{P} \times \text{t}$
$= \text{(40 W + 60 W)} \times 1\text{ h} = \text{100 Wh or 0.1 kWh}$

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Question 63 Marks

Two resistors, with resistances $\text{5 } \Omega \text{ and } \text{10 } \Omega$ respectively are to be connected to a battery of emf 6 V so as to obtain:
(i) Minimum current flowing. (ii) Maximum current flowing.

How will you connect the resistances in each case?
Calculate the strength of the total current in the circuit in the two cases.

Answer

In series

In parallel arrangement

For series arrangement:

$\text{R = R}_{1} + \text{R}_{2} = \text{5 }\Omega + \text{10 }\Omega = \text{15 }\Omega$

$\text{I} = \frac{\text{V}}{\text{R}} = \frac{\text{6V}}{15\Omega} = 0.4\text{A}$

For parallel arrangement:

$\frac{1}{\text{R}} = \frac{1}{\text{R}_{1}} + \frac{1}{\text{R}_{2}} = \frac{1}{5} + \frac{1}{10} = \frac{3}{10}$

$\therefore \text{R} = \frac{10}{3}\Omega$

$\text{I} = \frac{\text{V}}{\text{R}} = \frac{6\text{V}}{\frac{10}{3}\Omega} = 1.8 \text{ A}$

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Question 73 Marks

For the circuit shown in the diagram given below:

Calculate:

The value of current through each resistor.
The total current in the circuit.
The total effective resistance of the circuit.

Answer

$\text{I} = \frac{\text{V}}{\text{R}}$

$\text{I}_{1} = \frac{\text{6V}}{5\Omega} = 1.2\text{A}$

$\text{I}_{2} = \frac{6\text{V}}{10\Omega} = 0.6\text{A}$

$\text{I}_{3} = \frac{6\text{V}}{30\Omega} = 0.2\text{A}$

$\text{I} = \text{I}_{1} + \text{I}_{2} + \text{I}_{3} = 1.2\text{A} + 0.6\text{A} + 0.2\text{A} = \text{2A}$

$\frac{1}{\text{R}} = \frac{1}{\text{R}_{1}} + \frac{1}{\text{R}_{2}}+\frac{1}{\text{R}_{3}}$

$=\frac{1}{5} + \frac{1}{10} + \frac{1}{10}+\frac{1}{30} = \frac{10}{30} = \frac{1}{3}$

$\therefore \text{R} = 3\Omega$

Alternate Answer

$\text{V} = \text{I R}$

$\therefore \text{R} = \frac{\text{V}}{\text{I}} = \frac{6\text{V}}{\text{2A}} = 3\Omega$

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Question 83 Marks

Write Joule’s law of heating.
Two lamps, one rated 100W; 220V, and the other 60W; 220V, are connected in parallel to electric mains supply. Find the current drawn by two bulbs from the line, if the supply voltage is 220V.

Answer

According to Joule’s law of heating, heat produced in a wire is directly proportional to,

Square of current $(1^2 )$
Resistance of wire (R)
Time (t), for which current is passed

Thus,

$H =1^2 \times R \times t \ldots$ (Joule's law of heating)

Current drawn from the first lamp is given by,

$\text{P}_1=\text{V}\times\text{I}_1$

$\therefore\ \text{I}_1=\frac{\text{P}_1}{\text{V}}=\frac{100}{220}=0.45\text{A}$

Current drawn from the second lamp is given by,

$\text{P}_2=\text{V}\times\text{I}_2$

$\therefore\ \text{I}_2=\frac{\text{P}_2}{\text{V}}=\frac{60}{220}=0.27\text{A}$

Thus the total current drawn by two lamps from the line,

which are connected in parallel to each other for supply voltage 220V is,

$\text{I}=\text{I}_1+\text{I}_2=0.45+0.27=0.72\text{A}$

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Question 93 Marks

Show how would you join three resistors, each of resistance $9\Omega$ so that the equivalent resistance of the combination is:

$13.5\Omega$
$6\Omega$

Answer

Consider the resistors $R_1, R_2$ and $R_3$ each of $9 \Omega$ are connected in the circuit.

When one resistor is connected in series with the other two resistors which are connected in parallel to each other, the equivalent resistance in the circuit is,

$\frac{1}{\text{R}_\text{p}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}=\frac{1}{9}+\frac{1}{9}=\frac{2}{9}=0.22$

$\Rightarrow\ \text{R}_\text{p}=4.5\Omega$

$\text{R}_\text{s}=\text{R}_\text{p}+\text{R}_3=4.5\Omega+9\Omega=13.5\Omega$

$\therefore\ \text{R}_{\text{eq}}=13.5\Omega$

Thus, by connecting the resistors in this combination, the equivalent resistance in the circuit, $\text{R}_{\text{eq}}=13.5\Omega.$

When one resistor is connected in parallel with the other two resistors which are connected in series with each other, the equivalent resistance in the circuit is,

$\text{R}_{\text{s}}=\text{R}_1+\text{R}_2=9\Omega+9\Omega=18\Omega$

$\frac{1}{\text{R}_\text{p}}=\frac{1}{\text{R}_\text{s}}+\frac{1}{\text{R}_3}=\frac{1}{18}+\frac{1}{9}=0.166$

$\Rightarrow\ \text{R}_\text{p}=6\Omega$

$\therefore\ \text{R}_{\text{eq}}=6\Omega$

Thus, by connecting the resistors in this combination, the equivalent resistance in the circuit, $\text{R}_{\text{eq}}=6\Omega.$

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Question 103 Marks

List the factors on which the resistance of a conductor in the shape of a wire depends.
Why are metals good conductors of electricity whereas glass is a bad conductor of electricity? Give reason.
Why are alloys commonly used in electrical heating devices? Give reason.

Answer

Factors affecting the resistance of a conductor:

Length of the conductor.
Area of cross-section of the conductor.
Nature of material of the conductor.
Temperature of the conductor.

Metals have low resistivity and glass has high resistivity. So, all the metals are good conductors, while glass is a bad conductor of electricity.
The heating elements of electrical appliances are made of alloys because,

The resistivity of an alloy is much higher than the metal.
An alloy does not undergo oxidation or burn easily even if heated up to higher temperature.

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Question 113 Marks

Given below are the steps for the extraction of copper from its ore. Write the chemical equation of the reactions involved in each case.

Roasting of copper (I) sulphide.
Reduction of copper (I) oxide from copper (I) sulphide.
Electrolytic refining.

Answer

$2\text{Cu}_2\text{S}+3\text{O}_2\xrightarrow{\ \ \ \text{Heat}\ \ \ \ }2\text{Cu}_2\text{O}+2\text{SO}_2$
$2\text{Cu}_2\text{O}+\text{Cu}_2\text{S}\xrightarrow{\ \ \ \text{Heat}\ \ \ \ }6\text{Cu}+\text{SO}_2$
$\text{At anode}\xrightarrow{\ \ \ }\text{Cu}\xrightarrow{\ \ \ }\text{Cu}^{2+}+2\bar{\text{e}}$

$\text{At Cathode}\xrightarrow{\ \ \ }\text{Cu}^{2+}+\text{2e}\xrightarrow{\ \ \ }\text{Cu}$

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Question 123 Marks

The near point of the eye of a person is 50cm. Find the nature and power of the corrective lens required by the person to enable him to see clearly the objects placed at 25cm. from the eye.

Answer

Near point of Hypermetropia eye = 50cm. Book placed at, u = -25cm. Convex lens/ converging lens It will form a virtual image of the abject at near point of defective eye. V = -50cm Lens formula,$\frac{1}{\text{f}}=\frac{1}{\text{v}}-\frac{1}{\text{u}}$
$\frac{1}{\text{f}}=\frac{1}{(-50)}-\frac{1}{(-25)}=\frac1{50}$
$\text{f}=50\text{cm or }0.50\text{m}$
$\text{p}=\frac{1}{\text{f}}=\frac{1}{0.5}=+2\text{D}.$

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Question 133 Marks

Write the mathematical expression for Joule's law of heating.
Compute the heat generated while transferring 96000 coulomb of charge in two hours through a potential difference of 40V.

Answer

a. The mathematical expression of the Joules Law of heating is:
$H=I^2 Rt$
Here, H is a heating effect, I is the current flowing through the device, and t is the time taken
b. Given,
Amount of charge transferred $=96000 C$
Time taken $=2 hrs=2 \times 60 \times 60 sec$
$=7200 sec$
Potential difference $=40 V$
Heat generated $=V \times i \times t$
And we know that;
$i=\frac{ Q }{t}$
So, $H=V Q$
$=40 \times 96000$
$=3.84 \times 10^6 J$

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Question 143 Marks

State the relation correlating the electric current flowing in a conductor and the voltage applied across it. Also draw a graph to show this relationship.
Find the resistance of a conductor if the electric current flowing through it is 0.35A when the potential difference across it is 1.4V.

Answer

$\text{V}\alpha\text{ I}\text{ or }\frac{\text{V}}{\text{I}}=\text{Constant}$

$\text{Or}\text{ V}=1\text{R}$

given I = 0.35A,

V = 1.4V

$\text{R}=\frac{\text{V}}{\text{I}}$

$=\frac{1.4}{0.35}$

$=4.'\Omega$

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Question 153 Marks

A V-1 graph for a nichrome wire is given below. What do you infer from this graph? Draw a labelled circuit diagram to obtain such a graph.

Answer

The graph between V & I is a straight line. So we can conclude that Current flowing through the wire is directly proportional to the potential difference V across it. The resistance of the wire can be calculated as,$\text{R}=\frac{\text{V}}{\text{I}}$
$=\frac{0.4}{0.1}=4\Omega$
We can also conclude that nichrome wire has a constant value of the resistance as $4\Omega.$ Nichrome wire follows ohm's law thus it behaves as an ohmic conductor. The circuit diagram corresponding to the above graph is as shown below.

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Question 163 Marks

State three factors on which the heat produced by an electric current depends. How does it depend on these factors?

Answer

Heat produced by an electric current depends on the following factors:

Heat produced is directly proportional to square of current.
Heat produced is directly proportional to resistance.
Heat produced is directly proportional to the time for which current flows.

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Question 173 Marks

Find the equivalent resistance across the two ends A and B of the following circuit.

Answer

$R_1$ and $R_2$ are in parallel So,$\frac{1}{\text{R}^1}=\frac{1}{\text{R}_2}+\frac{1}{\text{R}_2}$
$\frac{1}{\text{R}^1}=\frac{1}{2}+\frac{1}{2}$
$\frac{1}{\text{R}_1}=\frac{2}{2}=1$
$\text{R}^1=1\text{ohm.}$
$R^{\prime \prime}\left(R_3\right.$ and $R_4$ are also in parallel $)=10 h m$. Now, $R^1$ and $R^{\prime \prime}$ are in series so, $R^{\text {III }}=R^I+R^{\| l}=1+1=2$ ohm Now, $R^{\text {III }}$ $\left(R_1, R_2, R_3, R_4\right)$ and $R^{\text {III }}\left(R_5, R_6, R_7, R_8\right)$ Are in the parallel, (Total resistance) $\frac{1}{R}=\frac{1}{R^{|l|}}+\frac{1}{R^{\text {III }}}$
$=\frac{1}{2}+\frac{1}{2}$
$=\frac{1}{\text{R}}+\frac{2}{2}=1$
$\text{R}=1\text{ohm}$

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Question 183 Marks

Consider the circuit given below where A, B and C are three identical light bulbs of constant resistance.

List the bulbs in order of increasing brightness.
If C burns out, what will be the brightness of A now compared with before?
If B burns out instead, what will be the brightness of A and C compared with before?

Answer

C will be the brightest. Voltage will be distributed equally between A and B, so they will have equal brightness but lesser than that of C.
A gets the same voltage as before, so its brightness remains the same.
If B burns put, A will also stop glowing because it is connected in series with B. However, brightness of C remains the same.

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Question 193 Marks

Five resistors are connected in a circuit as shown. Find the ammeter reading when circuit is closed.

Answer

$R_1$ and $R_2$ are in serie $s_{,} R_5 1=R_1+R_2=3+3=6 \Omega R_5 1$ and $R_3$ are in parallel. $\therefore \frac{1}{R_p}=\frac{1}{R_5 1}+\frac{1}{R_3}$
$=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}$
$\Rightarrow\text{R}_\text{p}=2\Omega$
$R _4, R _{ p }$ and $R _5$ are in series, $\therefore R _5= R _4+ R _{ P }+ R _5$
$=0.5+2+0.5$
$=3\Omega$
Then, current,$\text{I}=\frac{\text{V}}{\text{R}_5}=\frac{3}{3}=1\text{A}$

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Question 203 Marks

The electrical resistivities of four materials P, Q, R and S are given below:$\begin{matrix}\text{P}&6.84\times10^{-8}\Omega\text{m}\\\text{Q}&1.70\times10^{-8}\Omega\text{m}\\\text{R}&1.0\times10^{15}\Omega\text{m}\\\text{S}&11.0\times10^{-7}\Omega\text{m}\end{matrix}$
Which material will you use for making:

Heating element of electric iron.
Connecting wires of electric iron.
Covering of connecting wires?

Give reason for your choice in each case.

Answer

S; because it has high resistivity of $\frac{11}{10000000}\text{ohm\ m}$ (it is actually nichrome).
Q; because it has very low resistivity of $\frac{1.7}{100000000}\text{ohm\ m}$ (it is actually copper).
R; because it has very very high resistivity of$1.0\times100000000000000\ \text{ohm\ m}$ (it is actually rubber).

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Question 213 Marks

Calculate the resistance of a copper wire 1.0km long and 0.50mm diameter if the resistivity of copper is $1.7\times10\Omega\ \text{m}.$

Answer

I = 1km = 1000m$\text{r}=\frac{\text{d}}{2}=\frac{0.5}{2}\text{mm}=0.25\text{mm}=0.25\times10^{-3}\text{m}$
$\rho=1.7\times10^{-8}\Omega\text{m}$
$\text{R}=\rho\frac{\text{I}}{\text{A}}=\rho\frac{\text{l}}{\pi\text{r}^2}$
$\text{R}=1.7\times10^{-8}\times\frac{1000}{3.14\times(0.25\times10^{-3})^2}=86.6\Omega$

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Question 223 Marks

A $4 Ω$ coil and a $2 Ω$ coil are connected in parallel. What is their combined resistance? A total current of 3A passes through the coils. What current passes through the $2 Ω$ coil?

Answer

$4\Omega$ and $2\Omega$ coil are connected in parallel.Combined resistance is R
$\frac{1}{\text{R}}=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$
$\text{R}=\frac{4}{3}\Omega$
Total current $\text{I}=\frac{\text{V}}{\text{R}}=3\Omega$
$\frac{\text{V}}{\frac{4}{3}}=3$
$\text{V}=3\times\frac{4}{3}=4\text{V}$
Current through $2\Omega$ coil $=\frac{\text{V}}{2}=\frac{4}{2}=2\text{A}$

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Question 233 Marks

When a current of 4.0A passes through a certain resistor for 10 minutes, 2.88 × 10J of heat are produced. Calculate:

The power of the resistor.
The voltage across the resistor.

Answer

Given: $I=4 amp , t =10 min=10 \times 60=600 sec , H =2.88 \times 10^4 J$
a. We have
$K=I^2 R T$
$28800=4^2 \times R \times 600$
$R=3 \text { ohm }$
We know that
$P=1^2 \times R$
$=4^2 \times 3$
$P=48 W$
b. $V=$ ?
We know that
$V=I R$
$V=4 \times 3$
$V=12 V$

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Question 243 Marks

A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5Ω when connected to a 10V battery. Calculate the resistance of the electric lamp.
Now if a resistance of 10Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5Ω conductor and potential difference across the lamp will take place? Give reason.

Answer

Total resistance of circuit can be calculated as follows:$\text{R}=\frac{\text{V}}{\text{I}}=\frac{10\text{V}}{1\text{A}}=10\Omega$
Since lamp and conductor are in series so resistance of lamp,
$=10\Omega-5\Omega=5\Omega$
The new resistance in parallel to earlier combination has same value, i.e. 10Ω as the resistance of series combination. This means that the amount of current would be equally divided into two branches. Hence, 0.5A current will flow through 5Ω conductor.
Now, resistance remains the same but current has become half. Using Ohm formula, potential difference across the lamp can be calculated as follows:$\text{V}=\text{IR}=0.5\text{A}\times5\Omega=2.5\text{V}$

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Question 253 Marks

Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W . How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A ?

Answer

Resistance $R _1$ of the bulb is given by the expression, Supply voltage, $V =220 V$ Maximum allowable current, $I =5 A$ Rating of an electric bulb $P =10$ watts Because $R= v ^2 / P R_1=\frac{(220)^2}{10}=4840 \Omega$
According to Ohm's law, $V= R$ Let $R$ is the total resistance of the circuit for x number of electric bulbs $R=V / I=\frac{220}{5}=44 \Omega$ resistance of each electric bulb, $R _1=4840 \Omega \frac{1}{ R }=\frac{1}{ R _1}+\frac{1}{ R _2}+\ldots$ upto $\times$ times
$\frac{1}{R}=\frac{1}{R_1} \times x$
$x=\frac{R_1}{R}=\frac{4840}{44}=110$

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Question 263 Marks

A wire is 1.0m long, 0.2mm in diameter and has a resistance of $10Ω.$ Calculate the resistivity of its material?

Answer

l = 1m$\text{r}=\frac{\text{d}}{2}=\frac{0.2}{2}\text{mm}=0.1\text{mm}=0.0001\text{m}$
R = 10 ohm We know that,$\text{R}=\text{P}\frac{\text{I}}{\text{A}}$
$\text{P}=\frac{\text{RA}}{\text{I}}$
$=\frac{10\times\pi\times(0.0001)^2}{1}$
$=31.4\times10^{-8}\Omega\text{m}$

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Question 273 Marks

Two resistances when connected in parallel give resultant value of 2 ohm; when connected in series the value becomes 9 ohm. Calculate the value of each resistance.

Answer

Two resistance when connected in series, resultant value is 9 ohms. Two resistance when connected in parallel, resultant values is 2 ohms. Let the two resistance be $R_1$ and $R_2$ If connected in series, then $9=R_1+R_2 R_1=9-R_2$ If connected in parallel, then $\frac{1}{2}=\frac{1}{R_1}+\frac{1}{R_2}$
From aboves equations we get that $\frac{1}{2}=\frac{\left( R _1+ R _2\right)}{ R _1 R _2}$
$\frac{1}{2}=\frac{9}{\left(9-R_2\right) R_2}$
$9 R_2-R_2^2=18$
$R_2^2-9 R_2+18=0$
$\left(R_2-6\right)\left(R_2-3\right)=0$
$R_2=6,3$
So if $R_2 6$ ohms, then $R_1=9-6=3$ ohms. If $R_2=3$ ohms, then $R_1=9-3=6$ ohms.

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Question 283 Marks

A resistance of 40 ohms and one of 60 ohms are arranged in series across 220 volt supply. Find the heat in joules produced by this combination of resistances in half a minute.

Answer

Given: $R 1=40$ ohms, $R 2=60$ ohms (in series), $V=220 V, t =30 sec$ we know that Total resistance, $R=40+60=100$ ohms By ohm's law $V=I R$
$I=\frac{V}{R}$
$I=\frac{220}{100}=2.2 amp$
Putting the value of $I , R$ and t in eq. $H = I ^2 RT H =2.2^2 \times 100 \times 30 H =14520 J$.

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Question 293 Marks

How much energy is consumed when a current of 5 amperes flows through the filament (or element) of a heater having resistance of 100 ohms for two hours? Express it in joules.

Answer

I = 5 amp, R = 100 ohms, t = 2h
We know that
Electric energy consumed = P x t = I * I * Rt
= 25 × 100 × 2
= 5000Wh
= 5kwh
We know that 1kwh = 3.6 × 106J
Therefore, 5kwh = 5 × 3.6 × 106J = 18 × 106J.

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Question 303 Marks

A p.d. of 6V is applied to two resistors of $3 Ω$ and $6 Ω$ connected in parallel. Calculate:
The combined resistance.
The current flowing in the main circuit.
The current flowing in the $3 Ω$ resistor.

Answer

$V=6 V, R_1=3$ ohm, $R_2=6$ ohm (in parallel)
Combined resistance, $\frac{1}{\text{R}}=\frac{1}{\text{R}_1}+\frac{1}{\text{R}_2}$$\frac{1}{\text{R}}=\frac{1}{3}+\frac{1}{6}=\frac{3}{6}=\frac{1}{2}$
R = 2 ohm Current flowing in the main circuit, $\text{I}=\frac{\text{V}}{\text{R}}=\frac{6}{2}=3\text{A}$ Current flowing in 3 ohm resistor $=\frac{\text{V}}{\text{R}_1}=\frac{6}{3}=2\text{A}$

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Question 313 Marks

Calculate the work done in moving a charge of 4 coulombs from a point at 220 volts to another point at 230 volts.

Answer

$V_1=220 V, V _2=230 V$, Charge moved $=4 CTh$ us, the potential difference $= V _2- V _1=230-220=10$.
We know that,
Work done = Potential difference × Charge moved
= 10 × 4
Work done = 40 joules.

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Question 323 Marks

Why, we do not use series combination of connecting electric appliances in household circuit?

Answer

In a house, there are many electrical appliances that have to run independent of each other. if the appliances were wired in series, the potential difference across each appliance would vary depending on the resistance of the appliance. This would make it very difficult to provide the right power to flow through the appliances. When house wiring is done in parallel this problem does not arise as the potential difference across each appliance is the same and equal to the potential difference being provided by the power company.

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Question 333 Marks

Write the advantages of connecting electrical appliances in parallel and disadvantages of connecting them in series in a household circuit.

Answer

parallel connection:
Advandages:

Every unit that is connected in a parallel circuit gets equal amount of voltage.
It becomes easy to connect or disconnect a new element without affecting the working of other elements.
If any fault happened to the circuit, then also the current is able to pass through the circuit through different paths.

Series Connection:

We do not use series combination for connecting electrical appliances in household circuit as whenever there will be a damage/breakage in the circuit of any one appliance of the household then, due to the series connection all other connections.

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Question 343 Marks

Distinguish between resistance and resistivity.

Answer

a. Resistance is the property of the conductor, while resistivity is the property of the material of the conductor.
b. Resistance of a conductor is the opposition to the flow of electric current through it. resistivity of a substance is the opposition to the flow of electric current by a rod of that substance which is 1 m long and $1 m^2$ in cross section.
c. Resistivity of a substances depends on the nature of the substance and temperature.
d. Resistivity of a substance on the nature of the substance and its temper. it does not depend the length or thickness of the conductor.

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Question 353 Marks

Show how you would connect two 4 ohm resistors to produce a combined resistance of:

2 ohms.
8 ohms.

Answer

By connecting in parallel: Since equivalent resistance will be$\frac{1}{\text{R}}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$
Therefore. R = 2 ohm By connecting in series: since equilaent resistance will be R = 4 ohm + 4 ohm = 8 ohm.

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Question 363 Marks

Calculate the resistance of an aluminium cable of length 10km and diameter 2.0mm if the resistivity of aluminium is $2.7\times10 Ω\ \text{m}.$

Answer

I = 10km = 10000m d = 2mm
$r =1 mm=10^{-3} m$
$\rho=2.7\times10^{-8}\Omega\ \text{m}$
$\text{R}=\rho\frac{\text{I}}{\text{A}}$
$=2.7\times10^{-8}\times\frac{10000}{3.14\times(10^{-3})^2}$
$=0.859\times10^2\Omega$
$=86\Omega$

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Question 373 Marks

You are given one hundred $1 Ω $ resister. What is the smallest and largest resistance you can make in a circuit using these?

Answer

Given: n = 100, R = 1 ohm
For obtaining the smallest resistance, these resistance are connected in parallel:
Equivalent resistance $=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}....100\ \text{times}=\frac{100}{1}$
$R_eq$ $=\frac{1}{100}=0.01\ \text{ohm}$
For obtaining the largest resistance, these are connected in series:
Equivalent reisistance = 1 + 1 + 1 .........100 Times = 100
$R_eq$ = 100 ohm

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Question 383 Marks

Calculate the cost of operating a heater of 500W for 20hours at the rate of? ₹3.90 per unit.

Answer

Given P = 500W = 0.5KW, t = 20hr
We know that
Energy consumed = P × t = 0.5 × 20
= 10KWh
Total cost = 10 × cost per unit
cost per unit = Rs. 3.9 per unit
Therefore, total cost = 10 × 3.9 = Rs. 39

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Question 393 Marks

Give three reasons why different electrical appliances in a domestic circuit are connected in parallel.

Answer

Different electrical appliances in a domestic circuit are connected in parallel because of the following advantages:

If one electrical appliance stops working due to some defect, then all other appliances keep working properly.
Each electrical appliance has its own switch due to which it can be turned on or turned off independently, without affecting other appliances.
Each electrical appliance gets the same voltage as that of the power supply line.

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Question 403 Marks

Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2Ω in series with a combination of two resistors (4Ω each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2Ω resistor be the same as that across the parallel combination of 4Ω resistors? Give reason.

Answer

Equivalent resistance of two resistors (4Ω each) connected in parallel is given by
$\frac{1}{\text{R}}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\text{ or R}=2\Omega$
Therefore, the above circuit can be redrawn as given below:

Therefore, potential difference across 2Ω resistor will be same; as that of across the parallel combination of 4Ω resistors. V = IR.
As R and I in both the cases is same so V = same.

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Question 413 Marks

What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?

Answer

The inherent property of a conductor because of which it resists the flow of electric current is called resistivity. Resistivity for a particular material is unique.
Resistance varies directly as length of the conductor.
Current varies inversely as resistance.
So, when length of the wire is doubled, its resistance becomes double. When resistance becomes double, current becomes half.
This explains why the reading of ammeter decreases to half when the length of the wire is doubled.

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Question 423 Marks

What will be the length of a nichrome wire resistance $5.0\Omega$ if the length of similar wire of 120cm has resistance of $2.5\Omega$? Why?

Answer

Here,$\text{I}_{2}=120\text{cm}=1.2\text{cm}$
$\text{R}_{1}5.0\Omega$
$\text{R}_{2}2.5\Omega$
$\therefore\frac{\text{R}_{1}}{\text{R}_{2}}=\frac{\rho\times\frac{\text{I}_{1}}{\text{A}}}{\rho\times\frac{\text{I}_{2}}{\text{A}}}$
$\frac{\text{R}_{1}}{\text{R}_{2}}=\frac{\text{I}_{1}}{\text{I}_{2}}=\frac{\text{I}_{1}}{120\times10^{-2}}$
$\frac{5.0}{2.5}=\frac{\text{I}_{1}}{120\times10^{-2}}$
$\text{I}_{1}2\times120\times10^{-2}\text{m}$
$\text{I}_{1}=24\times10^{-1}\text{m}$
$\therefore\text{I}_{1}=240\text{cm}$

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Question 433 Marks

Two exactly similar electric lamps are arranged (i) in parallel, and (ii) in series. If the parallel and series combination of lamps are connected to 220V supply line one by one, what will be the ratio of electric power consumed by them?

Answer

Let resistance of each lamp = R ohms. Case1: Parellel connection Resultant resistance $=\frac{1}{\frac{1}{\text{R}}+\frac{1}{\text{R}}}=\frac{\text{R}}{2}$ Electric power consumed $\text{P}_1=\frac{\text{V}_2}{\text{R}}=\frac{220^2}{\frac{\text{R}}{2}}=\frac{96800}{\text{R}}$ Case2: Series connection Resultant resistance = R + R = 2R Electric Power consumed $\text{P}_2=\frac{\text{V}^2}{2\text{R}}=\frac{24200}{\text{R}}$$\therefore\frac{\text{P}_1}{\text{P}_2}=\frac{\frac{96800}{\text{R}}}{\frac{24200}{\text{R}}}=\frac{4}{1}$

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Question 443 Marks

Find the value of current I in the circuit given below:

Answer

$R_{A C}$ abd $R_{E D}$ are in parellel, so $\frac{1}{ R _{ p }^{\prime}}=\frac{1}{ R _{ AC }}+\frac{1}{ R _{ ED }}=\frac{1}{30}+\frac{1}{30}=\frac{1}{15}$ $\Rightarrow R _{ p }^{\prime}=15 \Omega$
Now $R _{ p }^{\prime}$ and $R _{ BC }$ are in series, so $R _{ s }= R _{ p }+ R _{ BC }=15+15=30 \Omega$
Again, $R _{ AB }$ and $R _{ p }^{\prime}$ are in parallel, so $\frac{1}{ R ^{\prime \prime}}=\frac{1}{ R _{ AB }}+\frac{1}{ R _{ s }^{\prime}}$
$=\frac{1}{15}+\frac{1}{30}=\frac{1}{10}$
$\therefore R^{\prime \prime}{ }_{p}=10 \Omega$
So, current flowing through the circuit is $I =\frac{ V }{ R " p }=\frac{3}{10}=0.3 A$

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Question 453 Marks

A piece of wire of resistance $20 Ω$ is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.

Answer

We know that the resistance of a conductor is given by:$\text{R}=\rho\frac{1}{\text{A}}=20\Omega$
Where $\rho=\text{resistivity}$ l = length of the conductor A = area of cross-section of the conductor Now the length is increased to twice the original length. Then let the new resistance be denoted by R'.$\text{R}'=\rho\frac{2\text{I}}{\frac{\text{A}}{2}}=4\rho\frac{1}{\text{A}}$
$\text{R}'=4\text{R}=4\times20=80\Omega$
Thus, the new resistance will become four times.

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Question 463 Marks

Bulb is rated at 200V, 100W. Calculate its resistance. Five such bulbs are lighted for 4 hours daily. Calculate the units of electrical energy consumed per day. What would be the cost of
using these bulbs per day at the rate of Rs. 4.00 per unit?

Answer

Using,$\text{P} = \frac{\text{V} 2 }{\text{R}}$ we get $\text{R} = \frac{\text{V} 2}{ \text{P}}$$= 200\frac{2}{100}$
$= 400 $
Energy consumed = P × t = 100 × 4 = 400 Wh = 0.4kWh Cost at the rate of 50 paise per unit = Rs (4 × 0.4) = Rs 1.6

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Question 473 Marks

The electrical resistivities of five substances A, B, C, D and E are given below:$\begin{matrix}\text{B}&110\times10^{-8}\Omega\text{ m}\\\text{C}&2.60\times10^{-8}\Omega\text{ m}\\\text{D}&10.0\times10^{-8}\Omega\text{ m}\\\text{E}&1.70\times10^{-8}\Omega\text{ m}\end{matrix}$

Answer

E is best conductor of electricity due to its least electrical resistivity.
C, because its resistivity is lesser than that of A.
B, because it has the highest electrical resistivity.
C and E, because of their low electrical resistivities.

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Question 483 Marks

A wire of given material having length l and area of cross-section A, has a resistance of $2\Omega$. Find the resistance of another wire of same material having length 2l and are of cross-section
$\frac{\text{A}}{2}$.

Answer

Let the resistivity of the material is s. now Its length is given = l Area of cross section = A Resistan a $=4\Omega$ We know $\text{R}=\text{s}\frac{1}{\text{a}}$
Now in case = ii Resistance = s Lenght = 21 Area of section$ = \frac{\text{A}}{2}$ So, Resistance$={\text{p}}\frac {1}{\text{a}}$[p is constant] For firest wire length = l and area of cross section = a Resistance $= \frac{p}{a} = 4\text{ohm}$ [given] For second wire lenght = 2L and area of cross section $= \frac{\text{A}}{2}$ Resistance $= \text{p } 2\text{L }\frac{\text{A}}{2}$$= \text{P} \frac{\text{L}}{\text{A}}$
$= \text{P} \frac{\text{L}}{\text{A }} [\text{p }\frac{\text{L}}{\text{a}} = 4 \text{ohm}]$
$= 4 \text{ohm}$
The resistance of second wire = 4 ohm.

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Question 493 Marks

What is the resistance between A and B in the figure given below?

Answer

The three resistance of 20 ohm, 10 ohm and 20 ohm on the extreme right side are in series. So, the resultant of these three resistance = 20 + 20 + 10 = 50 ohms. This 50 ohms is in parallel with 30 ohms. so resultant of these two will be$\frac{1}{\text{R}}=\frac{1}{30}+\frac{1}{50}$
$\frac{1}{\text{R}}=\frac{80}{1500}$
R = 18.75 ohms Now, the resistance 10 ohms, 18.75 ohms and 10 ohms are in series. Therefore, resultant resistance = 18.75 + 10 + 10 = 38.75 ohms.

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Question 503 Marks

Define resistance. Write the SI unit of resistance and define it. Match the correct range of resistivity with the materials given.
a. Conductors (i) $10^{-6} Nm$
b. Alloys (ii) $10^{12}$ to $10^{17} \Omega$
c. Insulators (iii) $10^{-6}$ to $10^{-8} m$

Answer

Electrical resistance of a conductor may be considered as a measure of the opposition offered by it for the flow of electric charge through it. SI unit of resistance is ohm.

$\text{Conductors } — 10^{-6} \text{ to } 10^{-8}\Omega$
$\text{Alloys } — 10^{-6} \Omega$
$\text{Insulators} — 10 ^{12} \text{ to } 10^{17}Ωm. $

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[3 Mark Questions] - Science STD 10 Questions - Vidyadip