SUBJECT - PHYSICS
01 Calculate the potential difference between two terminals of a battery if 100 joules of work is done to
transfer 20 coulomb from one terminal to another.
Q2 Calculate the current in a circuit if 500 C of charge pass on through it in 10 minutes.
Q3 Calculate the amount of charge that would flow in 2 hours through an element of an electric bulb
drawing a current of 0.25 A
Q4 Define electric circuit . Distinguish between open and closed electric circuits.
Q5 A piece of wire of resistance 20 ohm is drawn out so that its length is increased to twice its original
length. Calculate the resistance of the wire in the new situation.
Q6 Resistance of a metal wire of length 1m is 26 ohm at 20 degree Celsius. If the diameter of the wire is O
mm, what will be the resistivity of the metal at that temperature ?
Q7 A toaster of resistance 100 ohm is connected to 220 V line. Calculate the current drawn by the toaster
Q8 State the factors on which resistance of a conductor depends.
Q9 Define resistance of a material. Write its Sl unit.
Q10. State ohms law. write mathematical form of ohm's law.
can somebody answer my questions at once please it will help me alot.
thx.
Answers
Answer:
question no.2answer is 50j current is passed in a circuit
Answer:
Answered all questions below however Q6 is wrong
Explanation:
1. Potential difference = Work done ÷ Total Charge
∴Potential Difference = 100 J ÷ 20 C = 5 volts or 5 V
2. Current = Charge ÷ Time (in seconds)
∴Current = 500 ÷ (10×60) = 500 C ÷ 600 s ≈ 0.83 Amperes or 0.83 A
3. Charge = Current × Time (in seconds) = 0.25 * (2×60×60) = 0.25 × 7200 s
∴ Charge = 1800 Coulombs or 1800 C
4. A closed and continuous path for the flow of electric current is called an electric current. A closed circuit has a complete path for current to flow, whereas an open circuit doesn't have an complete path, which means that it's not functional and is considered as open.
5. R = ρL/A
When the length of the wire is doubled its area of cross-section should change in such a way that the volume of the wire remains constant.
Let old resistance = R₁ = 20Ω, old length = L and old area = A
∴New lengh = 2×L and new area = A/2 So, the new resistance is,
R₂ = ρ×(2L)(A/2) = 4 ρL/A
=> R₁ = 4R₂
Since, R₁ = 20 Ω
R₂ = 80 Ω
6. Insufficient Information in the question (Diameter is given as "O")
7. Voltage = Resistance × Current
220 V = 100Ω × Current
∵ Current = 220/100 = 2.2 Amperes or 2.2 A
8. The factors affecting resistance of a conductor are :
1. Length of the Conductor (Directly proportional)
2. Area of cross section (Inversely proportional)
3. Material of the conductor / resistivity (higher resistivity means higher resistance)
4. Temperature (resistance increases with increase in temperature)
9. Resistance of a material is known as its resistivity. It is the characteristic property of a material to oppose the flow of electric current. Its SI unit is Ohm meter denoted by Ωm
10. According to Ohms Law, the current flowing through a conductor is directly proportional to the potential difference across its ends provided temperature remains constant.
V ∝ I , V = IR where R is a constant and is called the Resistance of the conductor.
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