Here you get the CBSE Class 10 Science chapter 12, Electricity: NCERT Exemplar Problems and Solutions (Part-III). This part of the chapter includes solutions for Question No. 29 to 35 from the NCERT Exemplar Problems for Class 10 Science Chapter: Electricity. These questions include only the Long Answer Type Questions framed from various important topics in the chapter. Each question is provided with a detailed explanation.
NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Science Board Examination 2017-2018 as well as other competitive exams.
Find below the NCERT Exemplar problems and their solutions for Class 10 Science Chapter, Electricity:
Long Answer Type Questions
Question 29. Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another set of three bulbs of the same wattage are connected in parallel to the same source.
(a) Will the bulb in the two circuits glow with the same brightness? Justify your answer.
(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.
(a) The two situations given in questions is shown in the figure given below:
Let us assume that the resistance of each bulb is R and potential difference is V
Equivalent resistance in series combination = Req = R + R + R = 3R.
Let current through each bulb in series combination be I1.
So, brightness of each bulb in parallel combination will increase. Each bulb will glow 3 times brighter to that of each bulb in series combination.
(b) If one bulb gets fused in series combination then, circuit gets broken and current stops flowing and remaining bulb don't glow.
If one bulb gets fused in series combination then, same voltage continue to act on the remaining voltage and hence, other bulbs continue to glow with same brightness.
Question 30. State Ohm's law? How can it be verified experimentally? Does it hold good under all conditions? Comment.
According to Ohm's law the electric current flowing through a conductor is directly proportional to the potential difference applied across its ends provided the physical conditions such as temperature etc remains unchanged.
Let V be the potential difference applied across the ends of conductor through which current I flows, then according to Ohm's law.
Ohm’s law can be verified experimentally by the activity given below:
(i) Firstly, Set up a circuit as shown in figure given below, consisting of a nichrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each. (Nichrome is an alloy of nickel,chromium, manganese, and iron metals.)
(ii) First use only one cell as the source in the circuit. Note the reading in the ammeter I, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in the Table given.
(iii) Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.
(iv) Repeat the above steps using three cells and then four cells in the circuit separately.
(v) Calculate the ratio of V to I for each pair of potential difference V and current I.
(vi) Plot a graph between V and I. The graph will be a straight line as shown below.
(vii) This verifies the Ohm’s law.
Ohm's does not hold under all conditions as it is basically not a fundamental law which means it has exceptions. Following are the conditions when Ohm’s law does not hold:
(i) It is not obeyed when physical conditions of conductors like temperature keep on changing.
(ii) It is not obeyed by a lamp filament, junction diode, thermistor etc.
(iii) It is not obeyed in case of superconductors whose resistance is equal to zero.
Question 31.What is electrical resistivity of a material? What is its unit? Describe an experiment to study the factors on which the resistance of conducting wire depends.
If l is the length of the conductor, A its area of cross section and R its total resistance then,
Where, ρ is a constant of proportionality and is called the electrical resistivity of the material of the conductor. The SI unit of resistivity is ohm meter.
Experiment to study the factors on which the resistance of conducting wires depend
- Complete an electric circuit consisting of a cell, an ammeter, a nichrome wire of length l [say, marked (1)] and a plug key, as shown in figure given below.
- Now, plug the key. Note the current in the ammeter
- Replace Replace the nichrome wire by another nichrome wire of same thickness but twice the length, that is 2l [marked (2) in the Fig. above]
- Note the ammeter reading
- Now replace the wire by a thicker nichrome wire, of the same length l [marked (3)]. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit.
- Instead of taking a nichrome wire, connect a copper wire [marked (4) in Fig. above] in the circuit. Let the wire be of the same length and same area of cross-section as that of the first nichrome wire [marked (1)]. Note the value of the current.
- Note the difference in the current in all cases
- We will find that the current depend on the length of the conductor
- We will also find that the current depend on the area of cross-section of the wire used
Question 32. How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery?
Suppose 3 resistances are R1, R2 and R3. We will connect these resistances in series with an ammeter, key and a battery of known voltage as shown in the figure given below.
Now, we will switch the position of ammeter i.e., between R1 and R2, between R2 and R3, after R3 and note the reading in each case by closing and opening the key. We will find that the reading in each case will be same in all cases.
Question 33. How will you conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery?
Ans. Suppose 3 resistances are R1, R2 and R3. We will connect these resistances in parallel with
an ammeter (A), a voltmeter (V), a plug key (K)and a battery of known voltage as shown in figure given below.
We will close the key K and record the ammeter and voltmeter readings.
Now we will open the key and switch the position of voltmeter and ammeter as show in the figure given below.
We will close the switch and note the readings. Similarly, we will switch the position of voltmeter to resistances R2 and R3 and observe the readings.
We will find that readings of ammeter keeps on changing but readings of voltmeter in all cases almost remain same in all cases.
Question 34. What is Joule's heating effect? How can it be demonstrated experimentally? List its four applications in daily life.
When an electric current is passed through a high resistance wire, like nichrome wire, the resistance wire becomes very hot and produces heat. This effect is known as heating effect of current or Joule’s law of heating.
It states that the heat H produced by a resistor of resistance R due to current flowing through it for time t is H = I2Rt
A simple experiment to demonstrate heating effect of current is that if we switch on the bulb for a long period of time then it will become hot.
Four applications of Joule's heating effect in daily life are:
(i) Electric fuse is a safety circuit devices work on this principle.
(ii) Electric iron we use to iron our clothes works on this principle.
(iii) Electric kettle used to boil water also works on this principle.
(iv) Electric toaster to make bake breads works on the same principle.
Question 35. Find out the following in the electric circuit given in figure:
(a) Effective resistance of two 8Ω resistors in the combination
(b) Current flowing through 4Ω resistors
(c) Potential difference across 4Ω resistances
(d) Power dissipated in 4 Ω resistors
(e) Difference in ammeter readings, if any
(a) Here, two 8 Ω resistances are connected in parallel so their effective resistances are
Here, current through resistor will be same as current from battery = 1 Ampere
(c) Potential difference (V’) across 4resistors (R’) is given by V’ = IR’ = 1 × 4 = 4 V
(d) Power dissipated across 4Ω resistor, P = I2R =12 × 4 = 4 W
(e) There is no difference in the readings of ammeters A1 and A2 as same amount of current flows through the ammeter.