{"id":521,"date":"2021-07-02T15:30:02","date_gmt":"2021-07-02T10:00:02","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=521"},"modified":"2021-08-02T15:47:19","modified_gmt":"2021-08-02T10:17:19","slug":"kseeb-sslc-class-10-science-solutions-chapter-12","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-science-solutions-chapter-12\/","title":{"rendered":"KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity"},"content":{"rendered":"

KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity<\/strong>\u00a0are part of KSEEB SSLC Class 10 Science Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 12 Electricity.<\/p>\n

Karnataka SSLC Class 10 Science Solutions Chapter 12 Electricity<\/h2>\n

KSEEB SSLC Class 10 Science Chapter 12 Intext Questions<\/h3>\n

Text Book Part I Page No. 94<\/p>\n

KSEEB Solutions For Class 10 Science Question 1.<\/strong>
\nWhat does an electric circuit mean?
\nAnswer:
\nA continuous and closed path of an electric current is called an electric circuit.<\/p>\n

KSEEB Solutions For Class 10 Science Electricity Question 2.<\/strong>
\nDefine the unit of current.
\nAnswer:
\nThe SI unit of electric current is ampere (A).
\nWhen I coulomb of electric charge flows through any cross. Section of a conductor in I second, the electric current flowing through it is said to be 1 ampere.
\n\u2235 1 ampere = 1C\/1s<\/p>\n

Electricity Class 10 KSEEB Solutions Question 3.<\/strong>
\nCalculate the number of electrons constituting one coulomb of charge.
\nAnswer:
\nThe SI unit of electric charge is column, which is equivalent to the charge contained in nearly 6 \u00d7 1018<\/sup> electrons.<\/p>\n

Text Book Part I Page No. 96<\/p>\n

Electricity Class 10 Notes KSEEB Question 1.<\/strong>
\nName a device that helps to maintain a potential difference across a conductor.
\nAnswer:
\nVoltmeter.<\/p>\n

Electricity KSEEB Solutions Question 2.<\/strong>
\nWhat is meant by saying that the potential difference between two points is 1 V?
\nAnswer:
\nOne Volt is the potential difference between two points in a current-carrying conductor when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.
\n\"KSEEB
\n1 V = 1 Jc-1<\/sup><\/p>\n

KSEEB Solutions For Class 10 Electricity Question 3.<\/strong>
\nHow much energy is given to each coulomb of charge passing through a 6 V battery?
\nAnswer:
\nW = VQ
\n= 6 \u00d7 1 = 6 joules
\nHence 6 joules of energy is given to each coulomb of charge passing through a 6 V battery.<\/p>\n

Text Book Part I Page No. 103<\/p>\n

KSEEB Science Solutions Question 1.<\/strong>
\nOn what factors does the resistance of a conductor depend?
\nAnswer:
\nThe resistance of the conductor depends<\/p>\n

    \n
  1. on its length<\/li>\n
  2. on its area of cross section and<\/li>\n
  3. on the nature of its material.<\/li>\n<\/ol>\n

    KSEEB Solutions Class 10 Science Question 2.<\/strong>
    \nWill current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
    \nAnswer:
    \nResistance, \\(\\mathrm{R} \\alpha \\frac{1}{\\mathrm{A}}\\). The resistance of a conductor is inversely proportional to its area of cross-section. A thick wire has a greater area of cross-section whereas a thin wire has a smaller area of cross-section. Thus, thick wire has less resistance and a thin wire has more resistance therefore current will flow more easily through a thick wire.<\/p>\n

    KSEEB Solutions Science Class 10 Question 3.<\/strong>
    \nLet the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
    \nAnswer:
    \nAs per ohm\u2019s law, V = IR
    \n\\(\\mathrm{I}=\\frac{\\mathrm{V}}{\\mathrm{R}}\\) —— (i)
    \nPotential difference is half
    \n\"KSEEB
    \n\u2234 Current flowing is also half of ts former value.<\/p>\n

    KSEEB Solutions For Class 10th Science Question 4.<\/strong>
    \nWhy are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
    \nAnswer:
    \na) Resistivity of iron = 10.0 \u00d7 10-8<\/sup>
    \nResistivity of Mercury = 94.0 \u00d7 10-8<\/sup>
    \nResistivity of an alloy is greater than iron. By this we conclude that, Iron is good conductor of heat comparing to Mercury.
    \nb) Resistivity of silver is less, hence it is a good conductor of heat.<\/p>\n

    Text Book Part I Page No. 107<\/p>\n

    KSEEB Solutions For Class 10 Science Pdf Download Question 1.<\/strong>
    \nDraw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 \u2126 resistor, an 8\u2126 resistor, and a 12\u2126 resistor, and a plug key, all connected in series.
    \nAnswer:
    \nRs<\/sub> = R1<\/sub> + R2<\/sub> + R3<\/sub>
    \n= 2V + 2V + 2V
    \n= 6W
    \n\"KSEEB<\/p>\n

    KSEEB Class 10 Science Solutions Question 2.<\/strong>
    \nRedraw the circuit of Question 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12\u2126 resistor. What would be the readings in the ammeter and the voltmeter?
    \nAnswer:
    \n\"KSEEB
    \nResistance, R = 5 + 8 + 12 = 25 \u2126
    \n\\(\\mathrm{I}=\\frac{\\mathrm{V}}{\\mathrm{R}}=\\frac{6}{25}=0.24 \\mathrm{A}\\)
    \nAs per ohm’s law =0.24A
    \nV1<\/sub> = IR
    \nPotential difference
    \nV1<\/sub> = IR = 0.24 \u00d7 12 = 2.88 V.
    \nreading in ammeter = 0.24A
    \nreading in voltmeter = 2.88 v.<\/p>\n

    Text Book Part I Page No. 110<\/p>\n

    KSEEB 10th Science Solutions Question 1.<\/strong>
    \nJudge the equivalent resistance when the following are connected in parallel –
    \n(a) 1 \u2126 and 106<\/sup> \u2126,
    \n(b) 1 \u2126 and 103<\/sup> \u2126, and 106<\/sup> \u2126.
    \nAnswer:
    \n(a) If two resistances are connected in parallel
    \n\"KSEEB
    \nequivalent resistance = 1\u2126.
    \n(b) If three resistances 1\u2126, 103<\/sup>\u2126 and 106<\/sup>\u2126 are connected in parallel,
    \n\"KSEEB
    \nequivalent resistance = 0.999\u2126<\/p>\n

    Question 2.
    \nAn electric lamp of 100\u2126, a toaster of resistance 50\u2126 and a water filter of resistance 500\u2126 are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
    \nAnswer:
    \nResistance of Electric 1 amp R1<\/sub> = 100\u2126
    \nResistance of Tosser, R2<\/sub> = 50\u2126
    \nResistance of water filter, R3<\/sub> = 50\u2126
    \nPotential difference, V = 220V
    \nWhen these are connected in parallel,
    \n\"10th
    \n\"Science
    \n7.04A of electricity is obtained by three appliances
    \nResistance of an electric iron connected to the same source that takes as much current as all three appliances
    \n\"KSEEB
    \n\u2234 Resistance of iron box = 31.25\u2126
    \nElectricity flowing through this = 7.04A.<\/p>\n

    Question 3.
    \nWhat are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
    \nAnswer:
    \n(i) In parallel circuit, if one electrical appliance stops working due to some defect, then all other appliances keep working normally. In series circuit, if one electrical appliance stops working due to some defect, then all other appliances also stop working.<\/p>\n

    (ii) In parallel circuits, each electrical appliance gets the same voltage as that of the power supply line. In series circuit, appliances do not get the same voltage, as that of the power supply line.<\/p>\n

    (iii) In the parallel connection of electrical appliances, the overall resistance of the household circuit is reduced due to which the current from the power supply is high. In the series connection, the overall resistance of the circuit increases too much due to which the current from the power supply is low.<\/p>\n

    Question 4.
    \nHow can three resistors of resistances 2\u2126, 3\u2126, and 6\u2126 be connected to give a total resistance of (a) 4\u2126, (b) 1\u2126?
    \nAnswer:
    \nIn this diagram 2 resistors of resistance 3W and 3 W are connected in parallel.
    \n\"KSEEB
    \nIf 2\u2126 and 2\u2126 are connected in series 2\u2126 + 2\u2126 = 4\u2126
    \n\u2234 Total resistance = 4\u2126
    \nIf resistors are connected in series
    \n\"KSEEB
    \n\u2234 Total resistance = 1\u2126<\/p>\n

    Question 5.
    \nWhat is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4\u2126, 8\u2126, 12\u2126, 24\u2126?
    \nAnswer:
    \n(a) If four resistors are connected in series then resistance
    \n= 4 + 8 + 12 + 24
    \n= 48\u2126
    \n(b) If these are connected in parallel
    \n\"KSEEB
    \n\u2234 We can get lowest resistance.<\/p>\n

    Text Book Part I Page No. 112<\/p>\n

    Question 1.
    \nWhy does the cord of an electric heater not glow while the heating element does?
    \nAnswer:
    \nIn cord of an electric heater, as current flows these become hot arid glows but in case of electric heater this will not happen.<\/p>\n

    Question 2.
    \nCompute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
    \nAnswer:
    \nAs per Joule’s law H = VIt.
    \nHere V = 50 V.
    \n\"KSEEB<\/p>\n

    Question 3.
    \nAn electric iron of resistance 20\u2126 takes a current of 5 A. Calculate the heat developed in 30 s.
    \nAnswer:
    \nAs per Joule’s law
    \nH = VIt
    \nH = IR = 5A \u00d7 20\u2126 = 100V
    \nI = 5A, t = 30 sec.
    \n\u2234 H = 100 \u00d7 5 \u00d7 30 J
    \n= 15000 J = 1.5 \u00d7 104<\/sup> J.<\/p>\n

    Text Book Part I Page No. 114<\/p>\n

    Question 1.
    \nWhat determines the rate at which energy is delivered by a current?
    \nAnswer:
    \nThe rate at which electric energy is dissipated or consumed in an electric circuit is termed as electric power.
    \nP = VI.<\/p>\n

    Question 2.
    \nAn electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.
    \nAnswer:
    \nP = VI
    \nV = 220V, and I = 5A.
    \nPower P = 220 \u00d7 5 = 1100 W
    \nPower of the motor = P \u00d7 t
    \nP = 1100 W.
    \nt = 2 Hr 2 \u00d7 60 \u00d7 60 W
    \n= 7200 S
    \n\u2234 Energy consumed, E = 1100 \u00d7 7200 J
    \n= 7920000
    \n= 7.92 \u00d7 106<\/sup>J.<\/p>\n

    KSEEB SSLC Class 10 Science Chapter 12 Textbook Exercises<\/h3>\n

    Question 1.
    \nA piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R’, then the ratio \\(\\frac { R }{ R` } \\) is –
    \n(a) \\(\\frac { 1 }{ 25 }\\)
    \n(b) \\(\\frac { 1 }{ 5 }\\)
    \n(c) 5
    \n(d) 25
    \nAnswer:
    \n(d) 25<\/p>\n

    Question 2.
    \nWhich of the following terms does not represent electrical power in a circuit?
    \n(a) I2<\/sup>R
    \n(b) IR2<\/sup>
    \n(c) VI
    \n(d) \\(\\frac { { V }^{ 2 } }{ R } \\)
    \nAnswer:
    \n(b) IR2<\/sup><\/p>\n

    Question 3.
    \nAn electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be –
    \n(a) 100 W
    \n(b) 75 W
    \n(c) 50 W
    \n(d) 25 W
    \nAnswer:
    \n(d) 25 W<\/p>\n

    Question 4.
    \nTwo conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be-
    \n(a) 1 : 2
    \n(b) 2 : 1
    \n(c) 1 : 4
    \n(d) 4 : 1
    \nAnswer:
    \n(c) 1 : 4<\/p>\n

    Question 5.
    \nHow is a voltmeter connected in the circuit to measure the potential difference between two points?
    \nAnswer:
    \nVoltmeter is connected in parallel in the circuit to measure the potential difference between two points.<\/p>\n

    Question 6.
    \nA copper wire has diameter 0.5 mm and resistivity 1.6 10-8<\/sup>\u2126 m. What will be the length of this wire to make its resistance 10\u2126? How much does the resistance change if the diameter is doubled?
    \nAnswer:
    \nArea of the copper wire A \\(=\\pi\\left(\\frac{\\mathrm{D}}{2}\\right)^{2}\\)
    \ndiameter = 0.5 mm = 0.0005 m
    \nResistance, R = 10W.
    \n\"KSEEB
    \nDiameter of copper wire is doubled, then diameter = 2 \u00d7 0.5 = 1 mm = 0.001 m
    \n\"KSEEB
    \n\u2234 Length of wire = 122.7 m
    \nResistance of wire = 2.5 W<\/p>\n

    Question 7.
    \nThe values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below –<\/p>\n\n\n\n\n
    I (amperes)<\/td>\n0.5<\/td>\n1.0<\/td>\n2.0<\/td>\n3.0<\/td>\n4.0<\/td>\n<\/tr>\n
    V (volts)<\/td>\n1.6<\/td>\n3.4<\/td>\n6.7<\/td>\n10.2<\/td>\n13.2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    Plot a graph between V and I and calculate the resistance of that resistor.
    \nAnswer:
    \nIn the following graph, voltage is taken along the x-axis and the current is taken along the y-axis. Different values are as follows:<\/p>\n\n\n\n\n
    V (volts)<\/td>\n1.6<\/td>\n3.4<\/td>\n6.7<\/td>\n10.2<\/td>\n13.2<\/td>\n<\/tr>\n
    I (amperes)<\/td>\n0.5<\/td>\n1.0<\/td>\n2.0<\/td>\n3.0<\/td>\n4.0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

    \"KSEEB
    \nSlope indicates Resistance
    \n\"KSEEB
    \n\u2234 Resistance of the resistor = 3.4\u2126.<\/p>\n

    Question 8.
    \nWhen a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
    \nAnswer:
    \nHere I = 2.5 mA = 2.5 \u00d7 10-3<\/sup> A.
    \n\u2234 \\(R=\\frac{12}{2.5 \\times 10^{-3}}\\)
    \n= 4.8 \u00d7 103<\/sup>\u2126
    \n= 4.8 K cal.<\/p>\n

    Question 9.
    \nA battery of 9 V is connected in series with resistors of 0.2\u2126, 0.3\u2126, 0. 4\u2126, 0.5\u2126 and 12\u2126, respectively. How much current would flow through the 12\u2126 resistor?
    \nAnswer:
    \nA battery of 9V is connected in series with resistors 0.2\u2126, 0.3\u2126, 0.4\u2126,
    \nR = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4\u2126.
    \nV = 9V
    \n\\(I=\\frac{9}{13.4}=0.671 \\mathrm{A}\\)
    \n\u2234 12 V battery, 0.671 A current flows.<\/p>\n

    Question 10.
    \nHow many 176\u2126 resistors (in parallel) are required to carry 5 A on a 220 V line?
    \nAnswer:
    \nLet the resistors be x’
    \nResistance = 176\u2126
    \nAs per ohm’s law,
    \nV = IR
    \n\"KSEEB
    \n\"KSEEB
    \n\u2234 Four resistors (in parallel) are required to carry 5A on a 220 V line.<\/p>\n

    Question 11.
    \nShow how you would connect three resistors, each of resistance 6\u2126, so that the combination has a resistance of (i) 9\u2126, (ii) 4\u2126.
    \nAnswer:
    \nIf resistors are connected in series 6\u2126 + 6\u2126 + 6\u2126 =18\u2126 This ia not correct
    \nWhen they are connected in parallel
    \n\\(\\frac{1}{6}+\\frac{1}{6}+\\frac{1}{6}=3\\) This is also wrong,
    \ni) When they are connected in parallel
    \n\"KSEEB
    \nTwo 6\u2126 resistors are connected in parallel
    \n\"KSEEB
    \nIf 3rd resistor of 6\u2126 and 3\u2126 are connected in series, it becomes 6\u2126 + 3\u2126 = 9\u2126.
    \nii) When they are connected in series
    \n\"KSEEB
    \nResistance = 6\u2126 + 6\u2126
    \n= 12\u2126
    \nIf 3rd resistor 6\u2126 is connected to 12\u2126 in parallel
    \n\"KSEEB
    \nTotal resistance = 4\u2126.<\/p>\n

    Question 12.
    \nSeveral electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?
    \nAnswer:
    \nHere V = 220V
    \nI – 5A
    \n\"KSEEB
    \n\u2234 110 lamps can be connected in parallel with each other across the two wires.<\/p>\n

    Question 13.
    \nA hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24\u2126 resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?
    \nAnswer:
    \n(i) If coils are connected separately V = 220 V
    \nResistance R1<\/sub> = 240
    \nAs per ohm’s law, V = IR
    \n\"KSEEB
    \n\u2234 If coils are connected separately 9.16A electricity flows in the coil.
    \n(ii) If coils are connected in series
    \nResistance R2<\/sub> = 24\u2126 + 24\u2126 = 48\u2126
    \nAs per ohm’s law V = IR
    \n\"KSEEB
    \n\u2234 If coils are connected in series 4.58A electricity flows.<\/p>\n

    Question 14.
    \nCompare the power used in the 20 resistors in each of the following circuits:
    \n(i) a 6 V battery in series with 1\u2126 and 2\u2126 resistors, and
    \n(ii) a 4 V battery in parallel with 12\u2126 and 2\u2126 resistors.
    \nAnswer:
    \n(i) Potential Difference V = 6V
    \nIf 1\u2126 and 2\u2126 resistors are connected in series, then Resistance
    \nR = 1 + 2 = 3W.
    \nAs per ohm’s law
    \n\\(\\mathrm{I}=\\frac{6}{3}=2 \\mathrm{A}\\)
    \nP(I2<\/sup>)R = (2)2<\/sup> \u00d7 2 = 8W.
    \n(ii) Potential difference V = 4V
    \nIf 12\u2126 and 2\u2126 resistors are connected in parallel, voltage is equal
    \nVoltage of resistance 2 W is 4 volts
    \n\"KSEEB
    \n\u2234 Power of 2\u2126 is 8 W.<\/p>\n

    Question 15.
    \nTwo lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
    \nAnswer:
    \nBoth lamps are connected in parallel potential difference = 220 V
    \nPower = V \u00d7 I.
    \n\"KSEEB<\/p>\n

    Question 16.
    \nWhich uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
    \nAnswer:
    \n250W TV set in used in 1 Hr, then its energy
    \n= 250 \u00d7 3600 = 9 \u00d7 105<\/sup>
    \nenergy of Toaster = 1200 \u00d7 600
    \nIf it is used in 10 minutes, then its power = 1200 \u00d7 600 \u00d7 7.2 \u00d7 105<\/sup> J
    \n\u2234 Energy of 250 W TV set is used in 1 Hr is greater than 1200 W toaster used in 10 minutes.<\/p>\n

    Question 17.
    \nAn electric heater of resistance 8\u2126 draws 15 A from the service mains 2 hours.
    \nCalculate the rate at which heat is developed in the heater.
    \nAnswer:
    \nP = I2<\/sup>R
    \nR = 8\u2126, I = 15A
    \nP= (15)2<\/sup> \u00d7 8 = 1800 J\/s.
    \n\u2234 Rate at which heat is developed in the heater
    \n= 1800 J\/s.<\/p>\n

    Question 18.
    \nExplain the following.
    \n(a) Why is the tungsten used almost exclusively for filament of electric lamps?
    \nAnswer:
    \nTungstan is a strong and metal having high melting point. This will not melt in high temperature. Because of this tungsten is used almost exclusively for filament of electric lamps.<\/p>\n

    (b) Why are the conductors of electric heating devices, such as bread- toasters and electric irons, made of an alloy rather than a pure metal?
    \nAnswer:
    \nThe resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (burn) readily at high temperatures. Because of this reason the conductors of electric heating devices, such as bread-toasters and electric irons made of an alloy rather than a pure metal.<\/p>\n

    (c) Why is the series arrangement not used-for domestic circuits?
    \nAnswer:
    \nIn case of series connection, when one component fails, the circuit it broken and none of the components work. But in case of parallel connection, circuit divides the current throughout the electrical gadgets. Because of this reason series arrangement is not used for domestic purposes.<\/p>\n

    (d) How does the resistance of a wire vary with its area of cross-section?
    \nAnswer:
    \nResistance is inversely proportional to its cross section. As area is increasing resistance is less. Thus resistance of wire is changing with area of cross section.<\/p>\n

    (e) Why are copper and aluminium wires usually employed for electricity transmission?
    \nAnswer:
    \nCopper and Aluminium are good conductors of electricity and they have less resistance. Hence they are used in electricity transmission.<\/p>\n

    KSEEB SSLC Class 10 Science Chapter 12 Additional Questions and Answers<\/h3>\n

    Question 1.
    \nDraw a Neat diagram of Electric circuit for studying ohm’s law.
    \nAnswer:
    \n\"KSEEB<\/p>\n

    Question 2.
    \nWhat is the S.I. Unit of resistivity?
    \nAnswer:
    \n\u2126M<\/p>\n

    Question 3.
    \nHow is fuse wire made of?
    \nAnswer:
    \nIt consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron, lead etc.<\/p>\n

    Question 4.
    \nWrite the formula of Electric power.
    \nAnswer:
    \n\\(P\\quad =\\quad \\frac { { V }^{ 2 } }{ R } \\)<\/p>\n

    Question 5.
    \nGive examples where we find heating of electric current.
    \nAnswer:
    \nElectric heater, electric iron etc.<\/p>\n

    We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 12 Electricity, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

    KSEEB SSLC Class 10 Science Solutions Chapter 12 Electricity\u00a0are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 12 Electricity. Karnataka SSLC Class 10 Science Solutions Chapter 12 Electricity KSEEB SSLC Class 10 Science Chapter 12 Intext Questions Text Book Part I Page No. 94 …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/521"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=521"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/521\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=521"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=521"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=521"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}