Solution of questions of class test held in suryoday
Group A (1 mark questions)
1 On which factors does the heat content of a body depend? [1]
Answer: The heat content of a body depends on the number of molecules contained in the body and the kinetic energy carried by the molecules.
2 What is specific heat capacity? [1]
Answer: Specific heat capacity of a substance is defined as the amount of heat required to raise the temperature of 1 kilogram of the substance through 1°C.
3 Why concave lens is called a diverging lens? [1]
Answer: Concave lens is called a diverging lens because it diverges a parallel beam of light after refraction through it.
4 What are electrodes? [1]
Answer: The metal plates or wires that are dipped into the electrolytic solution to make an electric connection are known as electrodes.
5 What is temperature? [1]
Answer: Temperature is defined as the degree of hotness or coldness of a body.
6 What is long-sightedness? [1]
Answer: The defect of vision in which a person can see the distant object clearly but cannot see distinctly the near objects is called long sightedness.
7 What is electrolysis? [1]
Answer: The conduction of an electric current through a substance by its chemical decomposition is called electrolysis.
8 What is principal axis? [1]
Answer: The straight line passing through the optical centre and perpendicular to the optical plane is known as principal axis.
9 What is the principle of caloriemetry? [1]
Answer: The amount of heat gained is equal to the amount of heat lost, this relation is called principle of caloriemetry.
10 What is alternating current? [1]
Answer: The current which changes its direction a number of times in one second is called alternating current.
Group B (2 marks questions)
11 Water is not used as a thermometric liquid. Why? [2]
Answer: Water is not used as a thermometric liquid because water has not uniform expansion as it contracts on heating from 0oC to 4oC and then expands on further heating from 4oC.
12 Write down the two causes of energy losses in a transformer. [2]
Answer: The causes of energy losses in a transformer are as followings:
I. Heat loss
Heat loss is the most common form of energy loss in a transformer. Since there is an electrical current in both the primary and secondary coils of the transformer, energy will be used to overcome resistance in the wires. This energy is transformed into heat which escapes from the coils resulting in the energy loss.
II. Eddy currents
The changing magnetic field not only induces currents in the secondary coil but also currents in the iron core itself. These currents flow in little circles in the iron core and are called eddy currents. The eddy currents cause heat loss.
13 Why does water overflow from the beaker when the beaker full of water at 4oC is heated? [2]
Answer: This is because water at 4oC has minimum volume and when the beaker full of water at 4oC is heated, water expands for which it overflows from the beaker.
14 Write any two differences between heating effect and lighting effect of electricity. [2]
Answer; Differences between heating effect and lighting effect of electricity are as followings.
Heating effect
1. When electric current passes through some electric appliances, they change electrical energy into heat energy. This is called heating effect of electricity.
2. Electrical appliances having heating effect have a coil of wire that converts electrical energy into heat energy. Such coil is termed as heating element.
Lighting effect
1. When electric current is passed through some electrical appliances they change electrical energy into light energy. This is called lighting effect of electricity.
2. In filament lamp tungsten coil is used, which is termed as filament.
15 In the given diagram, if 1000 V line is connected to the transformer, what is its output voltage?
Answer: Here, we have to find the output voltage in the given transformer. For that, let us first collect all the given information. The information given are; Total number of turns in primary coil, n1 = 400 Input voltage (V1) = 1000 V Total number of turns in secondary coil, n2 = 100 Output voltage (V2) =? According to the law of transformer, we have, |
V2/V1=n2/n1
The value of V2 is unknown.
So, let us arrange the equation with respect to V2.
V2=n2/n1×V1
Let us now substitute the respective values.
V2=100400×1000
V2= 250
This is our final answer.
So, let us write the unit of voltage, Volt.
V2 = 250 V
16 "The use of alternating current would be limited, if transformer was not invented." Prove the statement with two clues. [2]
Answer: The use of alternating current would be limited, if transformer was not invented due to the following reasons:
I. If transformer were not invented, then sending current over long distance while keeping constant voltage would cause loss of power in form of heat.
II. To use alternating current, there would be need of generation/power plant everywhere. But due to the invention of transformer, there is no need of establishing power plant everywhere.
17 What is heating device? Give two examples of this. [2]
Answer: When electric current passes through some electric appliances, they change electrical energy into heat energy. Those electrical appliances are termed as heating device. Heating devices have a coil of wire that converts electrical energy into heat energy. Such coil is called heating element, heating element is usually made of nichrome.
Examples of heating device-
I. Electric iron
II. Toaster
Group C (3 marks questions)
18 10 electric bulbs of 100 watt each are used for 6 hrs and 2 heaters of 2 kW are used for 4 hrs daily, calculate the consumptions of electricity in one day. [3]
Solution
Here,
Given,
For bulb,
Power (P1) = 100 watt = (100 / 1000) kW = 0.1 kW
Number (N1) = 10
Time (t1) = 6 hrs.
Energy consumptions in 1 day = P1 х N1 х t1= 0.1 х 10 х 6 = 6 kWh
Again,
For Iron,
Power (P1) = 2 kW
Number (N2) = 2
Time (t2) = 4 hrs.
Energy consumptions of 30 days = P2 х N2 х t2 = 2 х 2 х 4 = 16 kWh
Again,
Total energy consumption in 1 day = (16 + 6) kWh = 22 unit
Hence, units of electricity consumed in 1 day 22 units.
19 10 bulbs of 60 watts each two heaters of 1500 watt are used two hours daily. Find the units of electricity consumed in 30 days. [3]
Answer
Here,
Given,
For bulb,
Power (P1) = 60 watt = (60 / 1000) kW = 0.06 kW
Number (N1) = 10
Time (t1) = 2 х 30 hrs. = 60 hrs.
Energy consumptions of 30 days = P1 х N1 х t1= 0.06 х 10 х 60 = 36 kWh
For Iron,
Power (P1) = 1500 watt = (1500 / 1000) kW = 1.5 kW
Number (N2) = 2
Time (t2) = 2 х 30 hrs. = 60 hrs.
Energy consumptions of 30 days = P2 х N2 х t2 = 1.5 х 2 х 60 = 180 kWh
Again,
Total energy consumption for 1 month = (180 + 36) kWh = 216 unit
Hence, units of electricity consumed in 30 days is 216 units.
20 On what factors does the quantity of heat of an object depend? How much heat is required to raise the temperature of 500 gm. of water from 5°C to 30°C? [3]
Do yourself
21 If 50 kJ of heat is transferred to 10 kg of water, what is the rise in its temperature? (Specific heat capacity of water is 4200 Jkg-1 °C-1) [3]
Do yourself
Group D (4 marks questions)
22 A convex lens has a focal length of 3 cm. If a candle is placed at twice of its focal length then:
I. Draw a clear ray diagram to show formation of image.
II. Write nature of image.
III. Calculate the magnification of image.
IV. What is the power of lens?
[4]
Answer
I. In the figure below, F is the principal focus of convex lens, O is the optical centre. Thus, the line OF represents focal length (f). The point 2F on the principal axis is at a distance of 2f from the optical centre O. If f = 3 cm, the point 2F is at a distance of 6 cm from O.
II. When an object is placed at twice of its focal length (2F), its image is formed at 2F on the other side of the object in the convex lens. The image shows following characteristics:
a) Image is real.
b) Image is inverted.
c) Image is same size as that of object.
III. In this case, the size of image is same as that of object. So, the magnification of the image is 1.
IV. Here,
Given,
Focal length (f) = 3 cm
Power (p) =?
We have,
p = 1focal length of lens in metre
= 13100cm
= 33.33dioptre (D)
Hence, the power of lens is 33.33 D.
23 A student sitting on the last bench of the class can see things written on the board but wears spectacles while reading a book. On the basis of this, answer the following questions:
I. What is the type of his defect?
II. What are the causes of a defect of vision?
III. What type of spectacles should be used to remove his defect of vision?
IV. Draw a diagram to show the remedy of this defect? [4]
Answer
I. The type of defect he has is hypermetropia or long sightedness. It is defined as the defect of vision in which a person cannot see nearby objects clearly but can see the distant object distinctly.
II. The causes of this type of defect of vision are as following:
a. As the eye lens becomes thin and the ciliary muscles fail to press the lens for the required thickness, it results in an increase in the focal length of the eye lens.
b. If the eyeball gets shortened, the distance between the eyes lens and retina becomes less. This also results in an increase in the focal length of the eye lens.
III. For the correction of this defect, a convex lens is suitable focal length is used in spectacles.
IV.
Fig: Use of the convex lens for the cure of hypermetropia.
24 Draw ray for the formation of the image if an object is placed before a concave lens at the following positions and also write the nature of the image.
I. The object is placed at infinity.
II. The object is placed between F and 2F.
[4]
Answer
I. In the figure below, F is the principal focus of the concave lens, O is the optical centre. Thus, the line OF represents focal length (f). The point 2F on the principal axis is at a distance of 2f from the optical centre O.
When the object is at infinity, the image is formed at the focus (F). The characteristics of the image are as followings:
a. The image is virtual.
b. The image is erect.
c. The image is highly diminished.
Fig: The ray diagram for the concave lens when an object is placed at infinity.
Where,
O = Optical centre
F = Focus
II. In the figure below, F1 is the principal focus of the concave lens, O is the optical centre. Thus, the line OF represents focal length (f). The point 2F1 on the principal axis is at a distance of 2f from the optical centre.
When the object is placed between F1and 2F1 of a concave lens the image is formed between focus (F1) and the optical centre. The image is: The characteristics of the image are as followings:
a. The image is virtual.
b. The image is erect.
c. The image is diminished.
Fig: The ray diagram for a concave lens when an object is placed between F and 2F.
