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ScienceClass 10CBSE

1. Which of the following phenomena of light are involved in the formation of a rainbow? (a) Reflection, refraction and dispersion (b) Refraction, dispersion and total internal reflection (c) Refraction, dispersion and internal reflection Dispersion, scattering and total internal reflection

Answer is (c) Refraction, dispersion and internal reflection Explanation: Dispersion of light leads to scattering of white light into different color to an angle to cause internal reflection.rs. Refraction bends incident light leading to the formation of rainbow.

ScienceClass 10CBSE

1. At noon the sun appears white as (a) light is least scattered (b) all the colours of the white light are scattered away (c) blue colour is scattered the most red colour is scattered the most

Soln: Answer is (b) all the colours of the white light are scattered away Explanation: This is due to dispersion of light by the atmosphere.

ScienceClass 10CBSE

1. A student sitting on the last bench can read the letters written on the blackboard but is not able to read the letters written in his text book. Which of the following statements is correct? (a) The near point of his eyes has receded away (b) The near point of his eyes has come closer to him (c) The far point of his eyes has come closer to him The far point of his eyes has receded away

Soln: Answer is (a) The near point of his eyes has receded away Explanation: Near point of eye move away for 25 cm in hypermetropia. Hence person should keep the book 25 cm apart to read properly.

ScienceClass 10CBSE

1. A person cannot see distinctly objects kept beyond 2 m. This defect can be corrected by using a lens of power (a) + 0.5 D (b) – 0.5 D (c) + 0.2 D (d) – 0.2 D

Soln: Answer is (b) – 0.5 D Explanation: The person is Myopic and he need a concave mirror hence the power would be in negative. P= 1= 1 = 0.5 D

ScienceClass 10CBSE

1.      What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life.

Soln: According to Joules heating effect heat produces in a resistor is Directly proportional to square of current for the given Directly proportional to resistance for a given current, Directly proportional to the time of current flowing through the This can be expressed as H = I2Rt H is heating effect, I is electric current, R is resistance and t is time. Experiment to demonstrate Joules law of heating Take a water heating immersion rod and connect to a socket which is connected to It Is important to recall that a regulator controls the amount of current flowing through a device. Keep the pointer of regulator on minimum and count the time taken by immersion rod to heat a certain amount of water. Increase the pointer of regulator to next Count the time taken by immersion rod to heat the same amount of water. Repeat above step for higher levels on regulator to count the time. Observation: It is seen that with increased amount of electric current, less time is required o heat the same amount of water. This shows Joule's Law of Heating. Application: Electric toaster, oven, electric kettle and electric heater etc. work on the basis of leafing effect of current.

ScienceClass 10CBSE

1. Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit 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.

Resistance of the bulbs in series will be three times the resistance of single Hence, the current in the series combination will be one-third compared to current in each bulb in parallel combination. The parallel combination bulbs will glow more brightly. The bulbs in series combination will stop glowing as the circuit is broken and current is However the bulbs in parallel combination shall continue to glow with the same brightness.

ScienceClass 10CBSE

1. B1 , B2 and B3 are three identical bulbs connected as shown in Figure 12.8. When all the three bulbs glow, a current of 3A is recorded by the ammeter A. (i) What happens to the glow of the other two bulbs when the bulb B1 gets fused? (ii) What happens to the reading of A1 , A2 , A3 and A when the bulb B2 gets fused? (iii) How much power is dissipated in the circuit when all the three bulbs glow together?

Potential difference does not get divided in parallel Hence glowing of other bulbs will not get affected when bulb one is fused. Ammeter A shows a reading of This means each of the Al. A2, and A3 show IA reading. R= V= 5 V = 1.5Ω I 3A Now P= I2R = (3A)2 x 1.5 Ω = 13.5 W

ScienceClass 10CBSE

1.      Why is parallel arrangement used in domestic wiring?

Parallel arrangement used in domestic wiring because it provides the same potential difference across each electrical appliance.

ScienceClass 10CBSE

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

Soln: Let R be the resistance of the electric In series total resistance = 5 + R I = v/r 1 = 10/5+R R = 5 ohm V across Lamp + conductor = 10 V V acoess Lamp = I × R = 1 * 5 = 5 Volt

ScienceClass 10CBSE

1.      What is the commercial unit of electrical energy? Represent it in terms of joules.

Commercial unit of electrical energy is kilowatt/hr 1 kw/hr = 1 kW h = 1000 W × 60 × 60s = 3.6 × 106 J

ScienceClass 10CBSE

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

Property of the conductor which resists the flow of electric current is called resistivity. Resistance for a particular material is unique. Resistance is directly proportional to length of conductor and inversely proportional to current flow. When length is doubled resistance becomes double and current flow reduces to half. This is the reason for the decrease in ammeter reading.

ScienceClass 10CBSE

1.      How does use of a fuse wire protect electrical appliances?

Fuse wire has great resistance than the main wiring. When there is significant increase in the electric current. Fuse wire melts to break the circuit. This prevents damage of electrical appliance.

ScienceClass 10CBSE

1. Unit of electric power may also be expressed as (a) volt ampere (b) kilowatt hour (c) watt second joule second

Soln: Answer is (b) kilowatt hour Explanation: Volt-ampere (VA) is the unit used for the apparent power in an electrical circuit. A watt second (also watt-second, symbol W s or W. s) is a derived unit of energy equivalent to the joule. The joule-second is the unit used for Planck's constant.

ScienceClass 10CBSE

1. Two resistors of resistance 2 Ω and 4 Ω when connected to a battery will have (a) same current flowing through them when connected in parallel (b) same current flowing through them when connected in series (c) same potential difference across them when connected in series different p

Soln: Answer is (b) same current flowing through them when connected in series Explanation: In series combination current does not get divided into branches because resistor receives a common current.

ScienceClass 10CBSE

1. An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it? (a) 1 A (b) 2 A (c) 4 A (d) 5 A

Soln: Answer is (d) 5 A Explanation: P=V x I Or 1000 w = 220v x I I = 1000𝑤 220𝑣 = 5 A = 4.54 A

ScienceClass 10CBSE

1. In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be (a) 5 J (b) 10 J (c) 20 J (d) 30 J

Answer is (c) 20 J Explanation: Equivalent resistance of the circuit is R = 4+2 = 6Ω current, I= V/R) 6/6= 1A the heat dissipated by 4 ohm resistor is, H = IRt = 20J

ScienceClass 10CBSE

1. In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness? (a) Brightness of all the bulbs will be the same (b) Brightness of bulb A will be the maximum (c) Brightness of bulb B will be more than that of A Brightness of bulb C will be less than that of B

Soln: Answer is (c) Brightness of bulb B will be more than that of A Explanation: Bulbs are connected in parallel so resistance of combination would be less than arithmetic sum of resistance of all the bulbs. So. there will be no negative effect on flow of current. As a result, bulbs would glow according to their wattage.

ScienceClass 10CBSE

1. The resistivity does not change if (a) the material is changed (b) the temperature is changed (c) the shape of the resistor is changed both material and temperature are changed

Soln: Answer is (c) the shape of the resistor is changed

ScienceClass 10CBSE

1. If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be (a) 100 % (b) 200 % (c) 300 % (d) 400 %

Soln: Answer is (c) 300 % Explanation: Heat generated by a resistor is directly proportional to square of current. Hence, when current becomes double, dissipation of heat will multiply by 2 =4. This means there will be an increase of 300%.

ScienceClass 10CBSE

1. A cylindrical conductor of length l and uniform area of crosssection A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross section (a) A/2 (b) 3A/2 (c) 2A (d) 3A

Soln: Answer is (c) 2A Explanation: P=𝑅𝐴 𝑙 When Length doubles P=𝑅𝐴 2𝑙 𝑅𝐴=𝑅𝐴 𝑙 2𝑙 A=2A

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