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Important-Questions-For-Class-10-Science-Chapter-6-Life-Processes.pdf

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Solutiona. Since Bulb are identical and connected in parallel with Voltage. Same current will flow through each bulb. Since the total current is 6 A.Individual current will be 2 Ab. If the Bulb B3 is blown away, The potential difference across other bulb still remains same, So same current will flow and they will glow as it is. No changec. when Bulb B1 goes down, the current in that part become zero.So reading of Ammeter A1 becomes zeroReading of Ammeter A2 will remain same i.e. 2 AReading of Ammeter A3 will remain same i.e. 2 AReading of Ammeter A will be = 2+2=4 Ad. As shown above, the reading of Ammeter A will changeQuestion 8Give the formula for each1Ohm’s Law2Resistance in terms of Length,Area,resistivity3Current in terms of Resistance and Voltage4Equivalent Resistance for Resistors in Series5Equivalent Resistance for Resistors in Parallel6Power produced in the resistanceSolution1Ohm’s Law[Math Processing Error]V=IR2Resistance in terms of Length,Area,resistivity[Math Processing Error]R=ρLA3Current in terms of Resistance and Voltage[Math Processing Error]I=VR4Equivalent Resistance for Resistors in SeriesR=R1+R2+R3…..5Equivalent Resistance for Resistors in Parallel[Math Processing Error]1R=1R1+1R26Power produced in the resistance[Math Processing Error]P=I2RQuestion 9Two students (P) and (Q) connect their two given resistors [Math Processing Error]R1 and [Math Processing Error]R2 in the manners shown below :Student (P) connects the terminals marked ([Math Processing Error]b1) and ([Math Processing Error]c1) while student (Q) connects the terminals marked ([Math Processing Error]d2) and ([Math Processing Error]c2) in their respective circuits at the points marked X and Y.Which one of the following is correct in relation to the above arrangements?a. both the students will determine the equivalent resistance of the series combination of the two resistors.b. both the students will determine the equivalent resistance of the parallel combination of the two resistorsc. student (P) will determine the equivalent resistance of the series combination while student (Q) will determine the equivalent resistance of the parallel combination of the two resistors.d. student (P) will determine the equivalent resistance of the parallel combination while student (Q) will determine the equivalent resistance of the series combination of the two resistors.SolutionIn combination made by student A, only one end of the resistors are in contact so they are connected in series. But, in the combination made by student B, both the ends of both the resistances are connected together, so they are connected in parallel.So, Student P will measure the equivalent resistance for series combination and Student Q will measure for parallel combinationHence, the correct answer is option c.Question 10which one of the below circuit is properly connected with the electrical componentsa. Pb. Qc. Rd. SSolutionAmmeter should be in series and Voltmeter in parallel. Also current flows from + to -. So (b) is the correct optionQuestion 11Judge the equivalent resistance when the following are connected in parallel(a) 1 Ω and [Math Processing Error]106 Ω(b) 1 Ω, [Math Processing Error]103 Ω and [Math Processing Error]106 Ω.Solution(a) [Math Processing Error]1R= 11+ 1106 = 106+1106[Math Processing Error]=1000000+11000000= 10000011000000[Math Processing Error]=10000001000001=0.9Ω(b) [Math Processing Error]1R= 11+ 1103+ 1106= 106+103+1106[Math Processing Error]=10010011000000[Math Processing Error]R=10000001001001= 0.9ΩQuestion 12How can three resistors of resistances 2 Ω, 3 Ω and 6Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?Solution(a) for 4 Ω reistance:3Ω and 6 Ω resistances are connected in parallel and this combination is connected in series to 2 Ω resistor.(b) For 1 Ω resistance:All these three resistances are connected in parallel to get equivalent resistance of 1 ΩQuestion 13What is (a) the highest, (b) the lowest total resistance that can be secured by co

mbinations of four coils of resistances 4 Ω, 8 Ω, 12 Ω, 24 Ω?Solution(a) For the highest resistance, resistors are connected in series:[Math Processing Error]R=R1+R2+R3+R4= 4 + 8 + 12 + 24R = 48 Ω(b) The lowest resistance will be obtained when these are connected in parallel:[Math Processing Error]1R= 1R1+ 1R2+ 1R3+ 1R4[Math Processing Error]1R= 14+ 18+ 112+ 124

Question 4We have four resistors A ,B ,C and D of resistance 4 ohm,8 ohm ,12 ohm and 24 ohm respectively?1Lowest resistance which can be obtained by combining these four resistors2highest resistance which can be obtained by combining these four resistorsSolutionLowest resistance is obtained in parallel combination1R=14+18+112+1241R=14+18+112+124Or R=2ΩHighest resistance is obtained in series combinationR=4+8+12+24=48ΩR=4+8+12+24=48ΩQuestion 5Three resistors 5Ω, 10Ω and 30Ω are connected in parallel with the battery of Voltage 6V?S.noQuestions1The value of current across each resistor2The value of Potential difference across each resistor3Total current in the circuit4Effective resistance of the circuitSolutionPotential difference remains same across parallel combinationSo current in each resistor is calculated asI1=V/R1=6/5=1.2 AI2=V/R2=6/10=.6 AI3=V/R3=6/30=.2 ATotal current in the circuitI=I1+I2+I3=1.2+.6+.2=2 AEffective resistance1R=15+110+1301R=15+110+130R= 3ΩQuestion 6An electric bulb draws a current of .8 A and works on 250 V on the average 8 hours a day.a. Find the power consumed by the bulbb. If the electric distribution company changes Rs 5 for 6 KWH, what is the monthly bill for 60 daysSolution Power of the electrical bulb is given byP=V×I=.8×250=200W=.2KWP=V×I=.8×250=200W=.2KWTotal energy consumption by the bulb in 60 daysE=P×t=.2×8×60=96KWHE=P×t=.2×8×60=96KWHSo, cost will be = 5×966=Rs805×966=Rs80Question 7A1 ,A2 ,A3 and A are ammeters connected in the circuitB1 ,B2 and B3 are three identical bulbsThey all are connected to Voltage source as shown in FigureWhen the three bulb are working good and glowing ,the current recorded in Ammeter A is 6 AAnswer Following questionsa. Same amount of current will go through each Bulb. And the value is 2 A .True or Falseb. If the Bulb B3is blown away, the bulb B1 and B2 will start glowing more. True or Falsec. What will happen to all the ammeter reading if Bulb B1 is blown away

Question 1A wire of length 3 m and area of cross-section 1.7 × 10-6 m2 has a resistance 3 × 10-2 ohm.a. What is the formula for resistivity of the wire and what is the unit of itb. Calculate the resistivity of the wireSolutiona. Resistivity of the wire is given byρ=RALρ=RALAnd It unit is Ohm-mb. In this caseL=3 mA=1.7×10−6A=1.7×10−6 m2R=3×10−2R=3×10−2 ohmSoρ=1.7×10−8ρ=1.7×10−8 Ohm-mQuestion 2The table given below shows the resistivity of three Material X, Y and Z?SamplesXYZResistivity3 X 10-911.1 X 0-618X10-17a. Arrange the samples in increasing order of conductivityb. Which of these is best conductor?c. Which are these is best insulator?Solutiona.Conductivity is inversely proportional to resistivitySoY< X< Zb. Z is the best conductor as it has least resistivityc. Y is the best insulator as it has highest resistivityQuestion 3.There are m resistor each of resistance R. First they all are connected in series and equivalent resistance is X. Now they are connected in parallel and equivalent resistance is Y. What is the ratio of X and Y?SolutionSeries combinationX=R+R+R+......mtimes=mRX=R+R+R+......mtimes=mRParallel combination1Y=1R+1R+1R+...=mR1Y=1R+1R+1R+...=mROr Y=RmY=RmSo X : Y=m2 : 1Question 4We have four resistors A ,B ,C and D of resistance 4 ohm,8 ohm ,12 ohm and 24 ohm respectively?1Lowest resistance which can be obtained by combining these four resistors2highest resistance which can be obtained by combining these four resistorsSolutionLowest resistance is obtained in parallel combination1R=14+18+112+1241R=14+18+112+124Or R=2ΩHighest resistance is obtained in series combinationR=4+8+

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