Science Chapter 12 Electricity to solve different types of questions in the exam.
Question 1.
A current of 10 A flows through a conductor for two minutes.
(i) Calculate the amount of charge passed through any area of cross section of the conductor.
(ii) If the charge of an electron is 1.6 × 10-19 C, then calculate the total number of electrons flowing. (Board Term I, 2013)
Answer:
Given that: I = 10 A, t = 2 min = 2 × 60 s = 120 s
(i) Amount of charge Q passed through any area of cross-section is given by I = Qt
or Q = I × t ∴ Q = (10 × 120) A s = 1200 C
(ii) Since, Q = ne
where n is the total number of electrons flowing and e is the charge on one electron
∴ 1200 = n × 1.6 × 10-19
or n = 12001.6×10−19 = 7.5 × 1021
Question 2.
Define electric current. (1/5, Board Term 1,2017)
Answer:
Electric current is the amount of charge flowing through a particular area in unit time.
Question 3.
Define one ampere. (1/5, Board Term 1,2015)
Answer:
One ampere is constituted by the flow of one coulomb of charge per second.
1 A = 1 C s-1
Question 4.
Name a device that you can use to maintain a potential difference between the ends of a conductor. Explain the process by which this device does so. (Board Term I, 2013)
Answer:
A cell or a battery can be used to maintain a potential difference between the ends of a conductor. The chemical reaction within a cell generates the potential difference across the terminals of the cell, even when no current is drawn from it. When it is connected to a conductor, it produces electric current and, maintain the potential difference across the ends of the conductor. Question 5.
Draw the symbols of commonly used components in electric circuit diagrams for
(i) An electric cell
(ii) Open plug key
(iii) Wires crossing without connection
(iv) Variable resistor
(v) Battery
(vi) Electric bulb
(vii) Resistance (4/5, Board Term 1,2017)
Question 6.
A student plots V-I graphs for three samples of nichrome wire with resistances R1, R2 and R3. Choose from the following the statements that holds true for this graph. (2020)
(a) R1 = R2 = R3
(b) R1 > R2 > R3
(c) R3 > R2 > R1
(d) R2 > R1 > R3
Answer:
(d) : The inverse of the slope of I-V graph gives the resistance of the material. Here the slope of -Rj is highest. Thus, R2 > R1 > R3
Question 7.
State Ohms law. (AI 2019)
Answer:
It states that the potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. Mathematically,
V ∝ I
V = RI
where R is resistance of the conductor.
Question 8.
A V-I 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. (2020)
Answer:
As graph is a straight line, so it is clear from the graph that V ∝ I.
The shape of the graph obtained by plotting potential difference applied across conductor against the current flowing v. llmuigh il will be a straight line.
According to ohms law,
V = IR or R = VI
So, the slope of V’-/ graph at any point represents the resistance of the given conductor.
Question 9.
Study the V-I graph for a resistor as shown in the figure and prepare a table showing the values of I (in amperes) corresponding to four different values V (in volts). Find the value of current for V = 10 volts. How can we determine the resistance of the resistor from this graph? (Board Term I, 2016)
Answer:
Since, the graph is straight line so we can either extrapolate the data or simply mark the value from graph as shown in figure.
| Current, I(A) | Voltage, V(V) |
| 0 | 0 |
| 1 | 2 |
| 2 | 4 |
| 3 | 6 |
| 4 | 8 |
Hence, the value of current for V = 10 volts is 5 amperes (or 5 A).
From Ohm’s law, V = IR,
We can write, R = VI
At any point on the graph, resistance is the ratio of values of V and I. Since, the given graph is straight line (ohmic conductor) so, the slope of graph will also give the resistance of the resistor
R = 10V5A = 2Ω
Alternately, R = (8−2)V(4−1)A = 6V3A = 2 Ω
Question 10.
V-I graph for a conductor is as shown in the figure
(i) What do you infer from this graph?
(ii) State the law expressed here. (Board Term I, 2014)
Answer:
(i) Refer to answer 8.
(ii) Refer to answer 7.
Question 11.
State Ohm’s law. Draw a labelled circuit diagram to verify this law in the laboratory. If you draw a graph between the potential difference and current flowing through a metallic conductor, what kind of curve will you get? Explain how would you use this graph to determine the resistance of the conductor. (Board Term I, 2016)
Answer:
Refer to answer 7 and 8.
Question 12.
State and explain Ohm’s law. Define resistance and give its SI unit. What is meant by 1 ohm resistance? Draw V-I graph for an ohmic conductor and list its two important features. (Board Term I, 2014)
Answer:
Ohm’s law: Refer to answer 7.
Resistance : It is ihe properly of a conductor lo resist the How of charges through it.
Its SI unit is ohm (Ω). If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is 1 ohm (1 Ω). lvolt
1 ohm = 1volt1ampere
V-I graph for an ohmic conductor can be drawn as given in figure.
Important feature of V-I graph are:
(i) It is a straight line passing through origin.
(ii) Slope of V-I graph gives the value of resistance of conductor slope = R = VI
Question 13.
Assertion (A) : The metals and alloys are good conductors of electricity.
Reason (R) : Bronze is an alloy of copper and tin and it is not a good conductor of electricity.
(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true. (2020)
Answer:
(c) : Metals and alloys are good conductors of electricity. Bronze is an alloy of copper and tin which are metals and thus is a good conductor of electricity.
Question 14.
A cylindrical conductor of length ‘l’ and uniform area of cross section ‘A’ has resistance ‘R’. The area of cross section of another conductor of same material and same resistance but of length ‘2l’ is (2020)
(a) A2
(b) 3A2
(c) 2A
(d) 3A
Answer:
(c) : The resistance of a conductor of length!, and area of cross section, A is
R = ρlA
where ρ is the resistivity of the material.
Now for the conductor of length 21, area of cross-section A’ and resistivity ρ.
R’ = ρl′A′ = ρ2lA′
But given, R = R’ ⇒ ρlA = ρ2lA or A’ = 2A
Question 15.
Assertion (A) : Alloys are commonly used in electrical heating devices like electric iron and heater.
Reason (R): Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points then their constituent metals.
(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true. (2020)
Answer:
(a)
Question 16.
How is the resistivity of alloys compared with those of pure metals from which they may have been formed? (Board Term I, 2017)
Answer:
The resistivity of an alloy is generally higher than that of its constituent metals.
Question 17.
(i) List three factors on which the resistance of a conductor depends.
(ii) Write the SI unit of resistivity. (Board Term 1, 2015)
Answer:
(i) Resistance of a conductor depends upon the following factors:
(1) Length of the conductor : (Treater the length (I) of the conductor more will be the resistance (R).
R ∝ I
(2) Area ol cross section of the conductor: (Ireater the cross-sectional area of the conductor, less will be the resistance.
R ∝ 1A
(3) Nature of conductor.
(ii) SI unit of resistivity is Ω m.
Question 18.
Calculate the resistance of a metal wire of length 2m and area of cross section 1.55 × 106 m², if the resistivity of the metal be 2.8 × 10-8 Ωm. (Board Term I, 2013)
Answer:
For the given metal wire,
length, l = 2 m
area of cross-section, A = 1.55 × 10-6 m²
resistivity of the metal, p = 2.8 × 10-8 Ω m
Since, resistance, R = ρlA
So R = (2.8×10−8×21.55×10−6)Ω
= 5.61.55 × 10-2 Ω = 3.6 × 10-2Ω or R = 0.036Ω
Question 19.
(a) List the factors on which the resistance of a conductor in the shape of a wire depends.
(b) Why are metals good conductors of electricity whereas glass is a bad conductor of electricity ? Give reason.
(c) Why are alloys commonly used in electrical heating devices ? Give reason. (2018)
Answer:
(a) Refer to answer 17 (i).
(b) Metal have very low resistivity and hence they are good conductors of electricity.
Whereas glass has very high resistivity so glass is a bad conductor of electricity.
(c) Alloys are commonly used in electrical heating devices due to the following reasons
(i) Alloys have higher resistivity than metals
(ii) Alloys do not get oxidised or burn readily.
Question 20.
Calculate the resistivity of the material of a wire of length 1 m, radius 0.01 cm and resistance 20 ohms. (Board Term I, 2017)
Answer:
We are given, the length of wire, l = 1 m, radius of wire, r = 0.01 cm = 1 × 10-4 m and resistance, R = 20Ω As we know,
R = ρlA, where ρ is resistivity of the material of the wire.
∴ 20Ω.= ρlπr2 = ρ1 m3.14×(10−4)2 m2
∴ ρ = 6.28 × 10-7 Ω m
Question 21.
A copper wire has diameter 0.5 mm and resistivity 1.6 × 10-8 Ω m. Calculate the length of this wire to make it resistance 100 Ω. How much does the resistance change if the diameter is doubled without changing its length? (Board Term I, 2015)
Answer:
Given; resistivity of copper = 1.6 × 10-8 Ω m, diameter of wire, d = 0.5 mm and resistance of wire, R = 100 Ω
Radius of wire, r = d2 = 0.52 mm
= 0.25 mm = 2.5 × 10-4 m
Area of cross-section of wire, A = nr²
∴ A = 3.14 × (2.5 × 10-4)²
= 1.9625 × 10-7 m²
= 1.9 × 10-7 m²
As, R = ρlA
∴ 100 Ω = 1.6×10−8Ωm×l1.9×10−7 m2
l = 1200 m
If diameter is doubled (d’ = 2d), then the area of cross-section of wire will become
A’ = πr² = π(d′2)² = π(2d2)² = 4A
Now R ∝ 1A, so the resistance will decrease by four times or new resistance will be
R’ = R4 = 1004 = 25Ω
Question 22.
The resistance of a wire of 0.01 cm radius is 10 Ω. If the resistivity of the material of the wire is 50 × 10-8 ohm meter, find the length of the wire. (Board Term I, 2014)
Answer:
Here, r = 0.01 cm = 10-4 m, ρ = 50 × 10-8 Ω m and R = 10 Ω
As, R = ρlA
or l = RAρ=Rρ(πr2)
so l = 1050×10−83.14×(10−4)2
= 0.628 m = 62.8 cm
Question 23.
A wire has a resistance of 16 Ω. It is melted and drawn into a wire of half its original length. Calculate the resistance of the new wire. What is the percentage change in its resistance? (Board Term I, 2013)
Answer:
When wire is melted, its volume remains same, so,
V’ = V or A’l’ = Al
Here, l’ = l2
Therefore, A’ = 2 A
Resistance, R = ρlA = 16 Ω
Now, R’ = ρl′A′=ρ(l/2)2A=14ρlA
So, R’ = R4 = 164 = 4 Ω
Percentage change in resistance,
= (R−R′R)×100=(16−416) × 100 = 75%
Question 24.
If the radius of a current carrying conductor is halved, how does current through it change? (2/5 Board Term I, 2014)
Answer:
If the radius of conductor is halved, the area of cross-section reduced to (14) of its previous value.
Since, R ∝ 1A, resistance will become four times
From Ohm’s law, V = IR
For given V, I ∝ 1R
So, current will reduce to one-fourth of its previous value.
Question 25.
Define resistance of a conductor. State the factors on which resistance of a conductor depends. Name the device which is often used to change the resistance without changing the voltage source in an electric circuit. Calculate the resistance of 50 cm length of wire of cross sectional area 0.01 square mm and of resistivity 5 × 10-8 Ω m. (Board Term I, 2014)
Answer:
Resistance is the property of a conductor to resist the flow of charges through it.
Factors affecting resistance of a conductor:
Refer to answer 17(i)
Rheostat is the device which is often used to change the resistance without changing the voltage source in an electric circuit.
We are given, length of wire, l = 50 cm = 50 × 10-2 m cross-sectional area, A = 0.01 mm²
= 0.01 × 10-6 m²
and resistivity, ρ = 5 x 10-8 Ω m.
As, resistance, R = ρlA
∴ R = (5×10−8×50×10−20.01×10−6) Ω
= 2.5 Ω
Question 26.
If a person has five resistors each of value 15 Ω, then the maximum resistance he can obtain by connecting them is
(a) 1 Ω
(b) 5 Ω
(c) 10 Ω
(d) 25 Ω (2020)
Answer:
(a) The maximum resistance can be obtained from a group of resistors by connecting them in series. Thus,
Rs = 15+15+15+15+15 1 Ω
Question 27.
The maximum resistance which can be made using four resistors each of 2 Ω is
(a) 2 Ω
(b) 4 Ω
(c) 8 Ω
(d) 16 Ω (2020)
Answer:
(c) : A group of resistors can produce maximum resistance when they all are connected in series.
∴ Rs = 2 Ω + 2 Ω + 2 Ω + 2 Ω = 8 Ω
Question 28.
The maximum resistance which can be made using four resistors each of resistance 12 Ω is
(a) 2 Ω
(b) 1 Ω
(c) 2.5 Ω
(d) 8 Ω (2020)
Answer:
(a) The maximum resistance can be produced from a group of resistors by connecting them in series.
Thus, Rs = 12 Ω + H 12 Ω + 12 Ω + 12 Ω = 2 Ω Question 29.
Three resistors of 10 Ω, 15 Ω and 5 Ω are connected in parallel. Find their equivalent resistance. (Board Term I, 2014)
Answer:
Here, R1 = 10 Ω, R2 =15 Ω, R3 = 5 Ω.
In parallel combination, equivalent resistance, (Req) is given by
Question 30.
List the advantages of connecting electrical devices in parallel with an electrical source instead of connecting them is series. (Board Term I, 2013)
Answer:
(a) When a number of electrical devices are connected in parallel, each device gets the same potential difference as provided by the battery and it keeps on working even if other devices fail. This is not so in case the devices are connected in series because when one device fails, the circuit is broken and all devices stop working.
(b) Parallel circuit is helpful when each device has different resistance and requires different current for its operation as in this case the current divides itself through different devices. This is not so in series circuit where same current flows through all the devices, irrespective of their resistances.
Question 31.
Show how would you join three resistors, each of resistance 9 Ω so that the equivalent resistance of the combination is (i) 13.5 Ω, (ii) 6 Ω (2018)
Answer:
(i) The resistance of the series combination is higher than each of the resistances. A parallel combination of two 9 Ω resistors is equivalent to 4.5 Ω. We can obtain 13.5 Ω by coupling 4.5 Ω and 9 Ω in series. So, to obtain 13.5 Ω, the combination is as shown in figure (a).
(ii) To obtain a equivalent resistance of 6 Ω, we have to connect two 9 Ω resistors in series and then connect the third 9 Ω resistor in parallel to the series combination as shown in the figure (b).
Question 32.
Three resistors of 3 Ω each are connected to a battery of 3 V as shown. Calculate the current drawn from the battery. (Board Term I, 2017)
Answer:
As given in circuit diagram, two 3 Ω resistors are connected in series to form R1; so R1 = 3 Ω + 3 Ω = 6 Ω
And, R1 and R2 are in parallel combination, Hence, equivalent resistance of circuit (Req) given by
Req = 2 Ω
Using Ohm’s law, V = IR
We get,
3 V = I × 2 Ω
or I = 32 A = 1.5 A
Current drawn from the battery is 1.5 A.
Question 33.
Two identical resistors are first connected in series and then in parallel. Find the ratio of equivalent resistance in two cases. (Board Term I, 2013)
Answer:
Let resistance of each resistor be R.
For series combination,
Rs = R1 + R2
So, Rs = R + R = 2R
For parallel combination,
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