Unit Test (Solutions): Electricity | Science Class 10 PDF Download

Time: 1 hour

M.M. 30

Attempt all questions.

• Question numbers 1 to 5 carry 1 mark each.
• Question numbers 6 to 8 carry 2 marks each.
• Question numbers 9 to 11 carry 3 marks each.
• Question number 12 & 13 carry 5 marks each.

Q1: When electric current is passed, electrons move from: (1 Mark)
(a) high potential to low potential
(b) low potential to high potential
(c) in the direction of the current
(d) against the direction of the current

Ans:  (d)

Electrons, which are negatively charged, move from the negative terminal to the positive terminal of the power source, opposite to the direction of conventional current flow. Hence, they move against the direction of the current.

Q2: The instrument used for measuring electric current is: (1 Mark)
(a) Ammeter
(b) Galvanometer
(c) Voltmeter
(d) Potentiometer

Ans: (a)

An ammeter is the instrument specifically designed to measure electric current in a circuit. It is connected in series with the circuit so that the entire current flows through it, allowing the ammeter to measure the current accurately.

Q3: The unit of potential difference is __________________. (1 Mark)

Ans: Volt

Q4: What is the maximum resistance which can be made using five resistors each of 1/5 Ω? (1 mark)
(a) 5 Ω
(b) 10 Ω
(c) 1/5 Ω
(d) 1 Ω

Ans: (d)

Resistance is maximum when resistors are connected in series.
R= 1/5+1/5+1/5+1/5+1/5
R = 5/5
R = 1Ω

Q5: A piece of wire of resistance R is cut into five equal parts. These parts are then arranged in parallel. If the equivalent resistance of this combination is R′, then the ratio R/R′ is _____. (1 Mark)
(a)  5
(b) 1/5
(c)1/25
(d) 25

Ans: (d)

The resistance is divided into five halves, each of which has a resistance of R/5.
Since we are aware that each component is linked to the others in parallel, we can compute the equivalent resistance as follows:
R’ = 5/R + 5/R + 5/R + 5/R +5/R
R’ = (5 + 5+ 5+ 5+ 5)/R = 25/R
RR’ = 25
The ratio of R/R′ is 25.

Q6: What does an electric circuit mean? (2 Marks)

Ans: An electric circuit is a continuous, closed path or loop composed of electronic components through which an electric current flows. Conductors, cells, Switch, and Load are the components of a simple circuit.

Q7: Write the relation between the resistance (R) of the filament of a bulb, its power (P) and a constant voltage V applied across it. (2 Marks)

Ans: The relation between resistance (R) of the filament of a bulb, its power (P) and a constant voltage V applied across it can be represented as follows:

P = V²/R

Q8: How does the use of a fuse wire protect electrical appliances? (2 Marks)

Ans: Compared to the main wiring, the fuse wire has a high resistance. Whenever there is an abrupt surge in electric current, the circuit is broken by melting fuse wire. This keeps electrical equipment from being damaged.

Q9:  List the three factors on which the resistance of a conductor depends. (3 Marks)

Ans: A conductor’s resistance is influenced by the following factors:

(1) Length of the conductor: The resistance (R) will increase as the conductor’s length (I)  increases.
R ∝ I
(2) Area of the cross-section of the conductor: (as the cross-sectional area of the conductor increases, the resistance decreases.
R ∝ 1A
(3) Nature of conductor

Q10: What is the commercial unit of electrical energy? Represent it in terms of joules. (3 marks)

Ans: The commercial unit of electrical energy is kilowatt/hr.

1 kW/hr = 1 kW h

= 1000 W × 60 × 60s

= 3.6 × 106 J 

Q11: If an electric heater rated 800 W operates 6h/day. Find the Cost of energy to operate it for 30 days at ₹3.00 per unit of consumption. (3 Marks)

Ans: 

Here, the Power of the heater, P = 800 W
Time, t = 6 hour/day
No. of days, n = 30
Cost per unit = ₹3.00

Thus, Energy Consumed Per Day (in kWh)

And, Energy consumed in 30 days
Total energy consumed (in kWh) = Energy consumed per day × Number of days
$4.8kWh/day\times 30days=144kWh=\; 4.8\; \backslash text\{\; kWh/day\}\; \backslash times\; 30\; \backslash text\{\; days\}\; =\; 144\; \backslash text\{\; kWh\}$=4.8 kWh/day×30 days=144 kWh 

Now, the Cost of 1 unit = ₹3

Therefore, Cost of 144 units = 3 × 144 = ₹432

Q12: What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life. (5 Marks)

Ans: According to the Joule’s heating effect, the heat produced in a resistor is known to be
(i) Directly proportional to the square of current for the given resistor.

(ii) Directly proportional to the resistance for a given current,

(iii) Directly proportional to the time of current flowing through the resistor.

It can be expressed as H = I²Rt
‘H’ is the heating effect, ‘I’ is the electric current, ‘R’ is resistance, and ‘t’ is time.

Experiment to demonstrate Joule’s law of heating

1. Take an immersion rod for water heating and attach it to a regulator-connected socket. It’s crucial to keep in mind that a regulator regulates how much current flows through a gadget.
2. Keep the regulator’s pointer at the lowest setting and time how long it takes the immersion rod to heat a specific volume of water.
3. Increase the regulator’s pointer to the following level. Time the same quantity of water heating with an immersion rod.
4. To measure higher amounts of the regulator, repeat the previous step.

Observation: It has been observed that it takes less time to heat the same amount of water with an increasing electric current. This illustrates the Joule’s Law of Heating.

Application: Electric appliances like toasters, ovens, kettles, and heaters operate using the leafing effect of current.

Q13: A hot plate of an electric oven connected to a 220 V supply line has two resistance coils such as A and B, each of 24 Ω resistance, which may be used separately, in series, or in parallel. What will be the currents in the three cases? (5 Marks)

Ans:

Case (i) When coils are used separately
By using Ohm’s law, we will be able to find the current flowing through each coil as follows:
I = VR
Substituting the values, we get
I = 220 V24 = 9.166 A
When used individually, each resistor allows 9.166 A of current to pass through it.

Case (ii) When the coils are connected in series
The total resistance is 24 Ω + 24 Ω = 48 Ω in the series circuit
The current flowing through this series circuit is calculated as follows:
I = VR = 220 V48 = 4.58 A
Therefore, a current of 4.58 A will flow through the circuit in series.

Case (iii) When the coils are in parallel connection, the equivalent resistance is calculated as follows:
R = 24 X 2424 + 24 = 57648 = 12
By using Ohm’s law, the current flowing through the parallel circuit is given by
I = VR = 22012 = 18.33 A
The current is 18.33 A in the parallel circuit.