Question 1: A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams in following figure. The current recorded in the ammeter will be
Answer: (d) the same in all the cases
Question 2. In the following circuits, heat produced in the resistor or combination of resistors connected to a 12 V battery will be
Answer: (c) Maximum in case (ii)
Explanation: Heat produced is directly proportional to the resistance. In case two the total effective resistance is equal to 4Ω which is maximum in all the given three conditions. Thus, in this case heat produced would be maximum.
Question 3. Electrical resistivity of a given metallic wire depends upon
Answer: (d) nature of the material.
Question 4. A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly
Answer: (a) 1020
Explanation: Electric current, I = 1A,
Time, t = 16 second
Charge, Q =?
⇒Therefore, `Q=16` coulomb
Since, `1.6xx10^(-19)` coulomb of charge contains 1 electron
Thus,1 C of charge have number of electrons equal `= 1/(1.6xx10^(-19))`
Therefore,16 coulomb of charge contains `1/(1.6xx10^(-19))xx16` electrons
`=(1xx16)/(16xx10^(-20))` electrons = 1020 electrons
Question 5. Identify the circuit in which the electrical components have been properly connected.
Answer: (d) (ii)
Question 6. What is the maximum resistance which can be made using five resistors each of 1/5 Ω?
Answer: (d) 1Ω
Explanation: When resistors are connected in series, the effective resistance becomes the sum resistance of all resistors which is maximum.
The total effective resistance when connected in series `= 1/5 Ω+1/5 Ω+1/5 Ω+1/5 Ω+1/5 Ω`
`=(1+1+1+1+1)/5 Ω= 5/5=1Ω`
Question 7. What is the minimum resistance which can be made using five resistors each of `1/5` Ω?
Answer: (b) `1/(25)`Ω
Explanation: When resistors are connected in parallel, the total effective resistance is reduced and that comes minimum.
Let total effective resistance = R
Therefore,`1/R=5/1 Ω+5/1 Ω+5/1 Ω+5/1 Ω+5/1 Ω``= (5+5+5+5+5)/1 Ω=25/1 Ω=25Ω`
Thus,`R= 1/(25)` Ω
Question 8. The proper representation of series combination of cells (given figure) obtaining maximum potential is
Answer: (a) (i)
Question 9. Which of the following represents voltage?
Answer: (a) (Work done) ÷ (Current × Time)
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