{"id":36499,"date":"2022-01-10T18:10:43","date_gmt":"2022-01-10T12:40:43","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=36499"},"modified":"2022-01-11T10:44:01","modified_gmt":"2022-01-11T05:14:01","slug":"mcq-questions-for-class-10-science-chapter-12","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-10-science-chapter-12\/","title":{"rendered":"MCQ Questions for Class 10 Science Chapter 12 Electricity"},"content":{"rendered":"
Question 1.<\/p>\n
(A) \\(\\frac{A}{2}\\)
\n(B) \\(\\frac{3 A}{2}\\)
\n(C) 2A
\n(D) 3A
\nAnswer:
\n(C) 2A<\/p>\n
Explanation:
\nResistivity of the conductor in the first case, p = \\(\\frac{\\mathrm{RA}}{l}\\)
\nResistivity of the conductor in second case, p = \\(\\rho=\\frac{\\mathrm{RA}^{\\prime}}{2 l}\\)
\nSince, both conductors are of same material and are at same temperature, so the resistivity of both the conductors will be same.
\nTherefore, from equations, (i) and (ii), we have :
\n\\(\\Rightarrow \\frac{\\mathrm{RA}}{l}=\\frac{\\mathrm{RA}^{\\prime}}{2 l}\\)
\n=>A’ = 2A<\/p>\n
<\/p>\n
Question 2.<\/p>\n
(A) 2 \u2126
\n(B) I\u2126
\n(C) 2.5 \u2126
\n(D) 8 \u2126
\nAnswer:
\n(A) 2 \u2126<\/p>\n
Explanation:
\nMaximum resistance in series = 41 x l\/2 = 2 Ohm.<\/p>\n
Question 3.<\/p>\n
(A) 4 \u2126
\n(B) 40 \u2126
\n(C) 400 \u2126
\n(D) 0.4 \u2126
\nAnswer:
\n(B) 40 \u2126<\/p>\n
Explanation:
\nV=IR, V = 4 V, I = 100 mA = 0.1 A
\nHence, R=\\(=\\frac{V}{l}\\)=\\(\\frac{4}{0.1}\\) = 40\u2126<\/p>\n
Question 4.<\/p>\n
(A) Volt-ampere
\n(B) Kilowatt-hour
\n(C) Watt-second
\n(D) Joule-second JR[
\nAnswer:
\n(A) Volt-ampere<\/p>\n
Explanation:
\nUnit of electric power is volt- ampere.<\/p>\n
Question 5.<\/p>\n
(A) maximum in (i).
\n(B) maximum in (ii).
\n(C) maximum in (iii).
\n(D) the same in all the cases.
\n
\nAnswer:
\n(D) the same in all the cases.<\/p>\n
Explanation:
\nIn series connections, the order of elements in the circuit will not affect the amount of current flowing in the circuit.<\/p>\n
<\/p>\n
Question 6.<\/p>\n
(A) its length.
\n(B) its thickness.
\n(C) its shape.
\n(D) nature of the material.
\nAnswer:
\n(D) nature of the material.<\/p>\n
Explanation:
\nThe resistivity of a material is constant for a particular material at a constant temperature. It only depends on the temperature. Resistivity of material does not depend on length, thickness, and shape of the A material.<\/p>\n
Question 7.<\/p>\n
(A) 1020<\/sup> Explanation: n = 1020<\/sup><\/p>\n Question 8.<\/p>\n (A) the material is changed. Explanation: Question 9.<\/p>\n (A) 0.4 A Explanation: Question 10.<\/p>\n (A) \\(\\frac{1}{5}\\)\u03a9 Explanation: <\/p>\n Question 11.<\/p>\n (A) \\(\\frac{1}{5}\\)\u2126 Explanation: => \\(\\frac{1}{R}\\) = \\(\\frac{25}{1}\\) Question 12.<\/p>\n (A) \\(\\frac{\\text { Work done }}{\\text { Current } \\times \\text { Time }}\\) Explanation: Question 13.<\/p>\n (A) 100% Explanation: Question 14.<\/p>\n (A) Brightness of all the bulbs will be the same. Explanation: <\/p>\n Question 15.<\/p>\n (A) 1 A Explanation: Assertion and Reason Based MCQs<\/span><\/p>\n Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. Question 1.<\/p>\n Answer: Explanation: Question 2.<\/p>\n Answer: Explanation: Question 3.<\/p>\n Answer: Explanation: <\/p>\n Question 4.<\/p>\n Answer: Explanation: Question 5.<\/p>\n Answer: Explanation: Question 6.<\/p>\n Answer: Explanation: Question 7.<\/p>\n Answer: Explanation: <\/p>\n Question 8.<\/p>\n Answer: Explanation: Question 9.<\/p>\n Answer: Explanation: Question 10.<\/p>\n Answer: Explanation: Case- Based MCQs<\/span><\/p>\n Attempt any 4 sub-parts from each question. Each sub-part carries 1 mark. Question 1.<\/p>\n Expianation: (A) 30\u2126 Explanation: Question 3.<\/p>\n Explanation: Question 4.<\/p>\n (A) Ammeter Explanation: Question 5.<\/p>\n (A) Ammeter Explanation: <\/p>\n III. Study the given table and answer any four questions from Question 1. to Question 5.<\/p>\n <\/p>\n [alloy of Cu, Mn and Ni]<\/td>\n [alloy of Ni, Cr, Mn and Fe]<\/td>\n Question 1.<\/p>\n (A) Chromium Explanation: Question 2.<\/p>\n (A) Iron Explanation: Question 3.<\/p>\n (A) Copper Explanation: <\/p>\n Question 4.<\/p>\n (A) Resistance increases Explanation: Question 5.<\/p>\n (A) 1\/4th of original resistance Explanation: So, resistance = \\(\\frac{\\mathrm{R}}{4}\\) IV. Observe the following table and answer any four questions from Question 1. to Question 5. Electrical resistivities of some substances, at 20\u00b0C are given as follows:<\/p>\n Question 1.<\/p>\n (A) Silver Explanation: Question 2.<\/p>\n (A) Iron Explanation: <\/p>\n Question 3.<\/p>\n (A) It has high resistivity Explanation: Question 4.<\/p>\n (A) Current drawn is less Explanation: Question 5.<\/p>\n (A) Series Explanation: V. Based on the given diagram, answer any of the four questions form Question 1. to Question 5. Question 1.<\/p>\n Explanation: Question 2.<\/p>\n (A) Ammeter reading will increase Explanation: <\/p>\n Question 3.<\/p>\n (A) 1 : 2 Explanation: Question 4.<\/p>\n (A) Ammeter Explanation: Question 5.<\/p>\n (A) Nichrome wire Explanation: VI. Study the given circuit diagram and answer any of the four questions form Question 1. to Question 5. In this circuit, three identical bulbs B1<\/sub>, B2<\/sub> and B3 <\/sub>are connected in parallel with a battery of 4.5 V. Question 1.<\/p>\n (A) They will also stop glowing Explanation: Question 2.<\/p>\n (A) 1.0A Explanation: Question 3.<\/p>\n (A) 1.00 Explanation: Question 4.<\/p>\n (A) 2 resistors Explanation: <\/p>\n Question 5.<\/p>\n (A) The voltage of each component is same Explanation: VII. Study the given passage and answer any of the four questions from Question 1. to Question 5. Question 1.<\/p>\n <\/p>\n (D) none of these Explanation: Question 2.<\/p>\n (A) Series arrangement Explanation: Question 3.<\/p>\n (A) Series arrangement Explanation: Question 4.<\/p>\n (A) 15 \u2126 Explanation: Question 5.<\/p>\n (A) 15 \u2126 Explanation: VIII. Study the given circuit diagram and answer any of the four questions given below from Question 1. to Question 5. Question 1.<\/p>\n (A) 40 Explanation: <\/p>\n Question 2.<\/p>\n (A) 1A Explanation: Question 3.<\/p>\n (A) 1V Question 4.<\/p>\n (A) 4W Explanation: <\/p>\n Question 5.<\/p>\n (A) will become double Explanation: Electricity Class 10 MCQ Questions With Answers Question 1. A cylindrical conductor of length T and uniform area of cross section A’ has resistance ‘R’. The area of cross section of another conductor of same material and same resistance but of length ’21’ is (A) (B) (C) 2A (D) 3A Answer: (C) 2A Explanation: Resistivity …<\/p>\n
\n(B) 1020<\/sup>
\n(C) 1018<\/sup>
\n(D) 1019<\/sup>
\nAnswer:
\n(A) 1020<\/sup><\/p>\n
\nGiven, 1=1 Amp.
\nt = 16 s
\nAsl = \\(\\frac{Q}{t}\\) = \\(\\frac{n e}{t}\\)
\nn = \\(\\frac{\\left(1^{t} \\times T\\right)^{t}}{e}\\) = \\(\\frac{n e}{t}\\) \\(\\left(1^{t} \\times \\mathrm{T}\\right)^{t}\\) = \\(\\frac{1 \\times 16}{1.6 \\times 10^{-19}}\\)<\/span><\/p>\nThe resistivity does not change if :<\/h2>\n
\n(B) the temperature is changed.
\n(C) the shape of the resistor is changed.
\n(D) both material and temperature are changed.
\nAnswer:
\n(C) the shape of the resistor is changed.<\/p>\n
\nResistivity always varies with change in temperature, nature of material. But resistivity cannot be change in any shape of conductor.<\/p>\nTwo bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1 A. The current through the 40 W bulb will be:<\/h2>\n
\n(B) 0.6 A
\n(C) 0.8 A
\n(D) 1 A
\nAnswer:
\n(D) 1 A<\/p>\n
\nIn a series connection, the current through each device remains the same. Therefore, the current through the 40 W bulb will also be 1 A.<\/p>\nWhat is the maximum resistance which can be made using five resistors each of \u00a0\\(\\frac{1}{5}\\)\u03a9 ?<\/h2>\n
\n(B) 10\u2126
\n(C) 5\u2126
\n(D) 1 \u2126
\nAnswer:
\n(D) 1 \u2126<\/p>\n
\nThe highest resistance is always given by connecting the resistors in series. Here, the highest resistance would be 5 x – =1 \u2126. 5 Therefore, the maximum resistance is 1 ohm.<\/p>\nWhat is the minimum resistance which can be made using five resistors each of \u00a0\\(\\frac{1}{5}\\)\u03a9 ?<\/h2>\n
\n(B) \\(\\frac{1}{25}\\)\u2126
\n(C) \\(\\frac{1}{10}\\)\u2126
\n(D) 25 \u2126
\nAnswer:
\n(B) \\(\\frac{1}{25}\\)\u2126<\/p>\n
\nMinimum resistance is obtained when resistors are connected in parallel combination. Thus, equivalent resistance obtained by connecting five resistors of resistance \\(\\frac{1}{5} \\Omega\\) each, parallel to each other :
\n\\(\\frac{1}{R}=\\frac{1}{1_{5}}+\\frac{1}{1_{5}}+\\frac{1}{1_{5}}+\\frac{1}{1_{5}}+\\frac{1}{1 \/ 5} \\Rightarrow \\frac{1}{R}=\\frac{5}{1 \/ 5}\\)<\/p>\n
\n=> R =(B) \\(\\frac{1}{25}\\)\u2126<\/p>\nWhich of the following represents voltage?<\/h2>\n
\n(B) Work done \u00d7 Charge
\n(C) \\(\\frac{\\text { Work done } \\times \\text { Time }}{\\text { Current }}\\)
\n(D) \\(\\frac{\\text { Work done } \\times \\text { Charge }}{\\text { Time }}\\)<\/p>\n
\nAs we know that,
\nWork done = Charge x Potential difference =>Work done = (Current x Time) x Potential Difference [\u2234 Charge = Current x time]
\n=> Potential difference \\(=\\frac{\\text { Work done }}{\\text { Current } \\times \\text { Time }}\\)<\/p>\nIf the current through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be<\/h2>\n
\n(B) 200%
\n(C) 300%
\n(D) 400%
\nAnswer:
\n(C) 300%<\/p>\n
\nIf is current and R is resistance then,
\nPower, P =I2<\/sub>R
\nPower in first case, P1<\/sub> = I2<\/sub>R 100% increase in current means that current becomes 21
\nPower in second case, P2 = (2I)2<\/sub>R = 4I2<\/sub>R
\nNow, increase in dissipated power = P2 – P1
\n= 4I2<\/sub>R – I2<\/sub>R = 3I2R
\nPercentage increase in dissipated power = \\(\\frac{3 P_{1}}{P_{1}} \\times 100\\) = 300%<\/p>\nIn an electrical circuit three incandescent bulbs A, B, and C of rating 40 W, 60 W, and 100 W, respec\u00actively are connected in parallel to an electric source. Which of the following is likely to happen regard\u00acing their brightness?<\/h2>\n
\n(B) Brightness of bulb A will be the maximum.
\n(C) Brightness of bulb B will be more than that of A.
\n(D) Brightness of bulb C will be less than that of B.
\nAnswer:
\n(C) Brightness of bulb B will be more than that of A.<\/p>\n
\nWe know that power is defined as rate of doing work. A bulb consumes electric energy and produces heat and light. Now, bulb with more power rating will produce more heat and light or we can say that power rating of bulb is directly proportional to the brightness produced by bulb. Therefore, brightness of bulb B with power rating 60 W will be more than the brightness of bulb A having power rating as 40 W.<\/p>\nAn electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?<\/h2>\n
\n(B) 2 A
\n(C) 4 A
\n(D) 5 A
\nAnswer:
\n(D) 5 A<\/p>\n
\nGiven that,
\npower = P = 1 kW = 1000 W
\nVoltage = V = 220
\nN0w,I = \\(\\frac{P}{V}\\) = \\(\\frac{1000}{220}\\) = 4.5A Now rating of fuse wire must be slightly greater than 4.5 A, that is, 5 A.<\/p>\n
\n(A) Both A and R are true and R is the correct explanation of A.
\n(B) Both A and R are true but R is NOT the correct explanation of A.
\n(C) A is true but R is false.
\n(D) A is false and R is true.<\/p>\nAssertion (A): A conductor has + 3.2 x 1019<\/sub> C charge.
\nReason (R): Conductor has gained 2 electrons.<\/h2>\n
\n(C) A is true but R is false.<\/p>\n
\nConductor has positive charge, so it has lost two electrons.<\/p>\nAssertion (A): The resistivity of conductor increases with the increasing of temperature.
\nReason (R): The resistivity is the reciprocal of the conductivity.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A.<\/p>\n
\nThe resistivity of the conductors is directly proportional to temperature.<\/p>\nAssertion (A): Bending a wire does not affect electrical resistance.
\nReason (R): Resistance of wire is proportional to resistivity of material.<\/h2>\n
\n(A) Both A and R are true and R is the correct explanation of A.<\/p>\n
\nResistance of wire R = p \\(\\left(\\frac{l}{A}\\right)\\) Where p is resistivity of material which does not depend on the geometry of wire. Since when wire is bent its, resistivity, length and area of cross-section do not change, therefore resistance of wire also remains same.<\/p>\nAssertion (A): Two resistance having value R each. Their equivalent resistance is\\(\\frac{R}{2}\\).
\nReason (R): Given Resistances are connected in parallel.<\/h2>\n
\n(A) Both A and R are true and R is the correct explanation of A.<\/p>\n
\nWhen two resistances R1 and R2 connected in parallel than their equivalent resistance will be R = \\(\\frac{R_{1} R_{2}}{R_{1}+R_{2}}\\)<\/p>\nAssertion (A): Alloys are commonly used in electrical heating devices like electric iron and heater.
\nReason (R): Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points then their constituent metals.<\/h2>\n
\n(C) A is true but R is false.<\/p>\n
\nAlloys have high resistivity and high melting point as compared to pure metals. So alloys cannot easily bum or oxidize at higher temperature. Now as we want higher temperature in heating devices so we use alloys in heating devices.<\/p>\nAssertion (A): In a simple battery circuit the point of lowest potential is positive terminal of the battery.
\nReason (R): The current flows towards the point of the lower potential as it flows in such a circuit from the positive to the negative terminal.<\/h2>\n
\n(D) A is false and R is true.<\/p>\n
\nIn a simple battery circuit, the point of lowest potential is the negative terminal of the battery and the current flows from higher potential to lower potential.<\/p>\nAssertion (A): Electric appliances with metallic body have three connections, whereas an electric bulb has a two pin connection.
\nReason (R): Three pin connections reduce heating of connecting wires.<\/h2>\n
\n(C) A is true but R is false.<\/p>\n
\nThe metallic body of an electrical appliances is connected to the third pin which is connected to the earth. This is a safety precaution and avoids eventual electric shock. By doing this the extra charge flowing through the metallic body is passed to earth and avoid shocks. There is nothing such as reducing of the heating of connecting wires by three pin connections.<\/p>\nAssertion (A): The electric bulbs glow immediately when switch is ON.
\nReason (R): The drift velocity of electrons in a metallic wire is very high.<\/h2>\n
\n(A) Both A and R are true and R is the correct explanation of A.<\/p>\n
\nIn a conductor there are large numbers of free electrons. When we close the circuit, the electric field is established instantly with the speed of electromagnetic wave which causes electron drift at every portion of the circuit. Due to which the current is set up in the entire circuit instantly. The current which is set up does not wait for the electrons flow from one end of the conductor to another end. It is due to this, the bulb glows immediately when switch is ON.<\/p>\nAssertion (A): Copper is used to make electric wires.
\nReason (R): Copper has very low electrical resistance.<\/h2>\n
\n(A) Both A and R are true and R is the correct explanation of A.<\/p>\n
\nA low electrical resistance of copper makes it a good electric conductor. So, it is used to make electric wires.<\/p>\nAssertion (A): Silver is not used to make electric wires.
\nReason (R): Silver is a bad conductor.<\/h2>\n
\n(C) A is true but R is false.<\/p>\n
\nSilver is a good conductor of electricity but it is not used to make electric wires because it is expensive.<\/p>\n
\nI. Read the passage and answer any four questions from Question 1. to Question 5.
\nThree resistors of 5 Si, 10 Q and 15 Q are connected in series and the combination is connected to the battery of 30 V. Ammeter and voltmeter are connected in the circuit.<\/p>\nWhich of the following is the correct circuit diagram to connect all the devices in proper correct order.<\/h2>\n
\n
\nAnswer:
\n<\/p>\n
\nThe correct circuit diagram is:
\n
\nQuestion 2.<\/p>\nHow much is the total resistance?<\/h2>\n
\n(B) 20\u2126
\n(C) \\(\\frac{11}{30} \\Omega\\)
\n(D) \\(\\frac{30}{11} \\Omega\\)
\nAnswer:
\n(A) 30\u2126<\/p>\n
\nR = R1 + R2 + R3 = 30\u2126<\/p>\nTwo students perform experiments on two given resistors R1<\/sub> and R2<\/sub> and plot the following V-I graphs. If R1<\/sub> > R2<\/sub> which of the diagrams correctly<\/h2>\n
\n
\nAnswer:
\n<\/p>\n
\nDiagram 1 is correct as R, is large so the slope of V-I graph (V\/I) is greater in diagram and is correctly represented as R1<\/sub>.
\n<\/p>\nThe device used to measure the current:<\/h2>\n
\n(B) Galvanometer
\n(C) Voltmeter
\n(D) None of these
\nAnswer:
\n(A) Ammeter<\/p>\n
\nAmmeter is used to measure the current.<\/p>\nWhich of the following is connected in series in circuit:<\/h2>\n
\n(B) Voltmeter
\n(C) Both of these
\n(D) None of these
\nAnswer:
\n(A) Ammeter<\/p>\n
\nAmmeter is used to measure the current. It is connected in series in the circuit.<\/p>\n\n\n
\n <\/td>\n Material<\/td>\n Resistivity (\u2126 m)<\/td>\n<\/tr>\n \n Conductors<\/td>\n Silver<\/td>\n 1.60 x 10-8<\/sup><\/td>\n<\/tr>\n \n <\/td>\n Copper<\/td>\n 1.62 x 10-8<\/sup><\/td>\n<\/tr>\n \n <\/td>\n Aluminium<\/td>\n 2.63 x 10-8<\/sup><\/td>\n<\/tr>\n \n <\/td>\n Tungsten<\/td>\n 5.20 x 10-8<\/sup><\/td>\n<\/tr>\n \n <\/td>\n Nickel<\/td>\n 6.84 x 10-8<\/sup><\/td>\n<\/tr>\n \n <\/td>\n Iron<\/td>\n 10.0 x l0-8<\/sup><\/td>\n<\/tr>\n \n <\/td>\n Chromium<\/td>\n 12.9 x 10-8<\/sup><\/td>\n<\/tr>\n \n <\/td>\n Mercury<\/td>\n 94.0 x 10-8<\/sup><\/td>\n<\/tr>\n \n <\/td>\n Manganese<\/td>\n 1.84 x 10-6<\/sup><\/td>\n<\/tr>\n \n Alloys<\/td>\n Constantan<\/td>\n 49 x l0-6<\/sup><\/td>\n<\/tr>\n \n <\/td>\n [alloy of Cu and Ni]<\/td>\n <\/td>\n<\/tr>\n \n <\/td>\n Manganin<\/p>\n 44 x 10-6<\/sup><\/td>\n<\/tr>\n \n <\/td>\n Nichrome<\/p>\n 100 x 10-6<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Which is a better conductor ?<\/h2>\n
\n(B) Nickel
\n(C) Mercury
\n(D) Iron
\nAnswer:
\n(B) Nickel<\/p>\n
\nNickel is better conductor because its resistivity value is lower than others.<\/p>\nElement used to make heating element of electric geyser:<\/h2>\n
\n(B) Silver
\n(C) Nichrome
\n(D) Tungsten
\nAnswer:
\n(C) Nichrome<\/p>\n
\nNichrome is used to make the heating element of an electric geyser.<\/p>\nElement used to make filament of incandescent bulb:<\/h2>\n
\n(B) Silver
\n(C) Nichrome
\n(D) Tungsten
\nAnswer:
\n(D) Tungsten<\/p>\n
\nTungsten is used to make filament of incandescent bulb.<\/p>\nWhat happens to resistance of a conductor when its area of cross section is increased?<\/h2>\n
\n(B) Resistance decreases
\n(C) No change
\n(C) Resistance doubles
\nAnswer:
\n(B) Resistance decreases<\/p>\n
\nElectrical resistance is directly proportional to the length (L) of the conductor and inversely proportional to the cross sectional area (A).<\/p>\nA given length of a wire is doubled and this process is repeated once again. The resistance of wire becomes:<\/h2>\n
\n(B) 16 times of original resistance
\n(C) Double the original resistance
\n(D) Half of original resistance
\nAnswer:
\n(B) 16 times of original resistance<\/p>\n
\nLet resistance of the wire be R. When the wire is doubled and the process is repeated then the length of the wire reduced by 1\/4 times.<\/p>\n
\nThe total resistance of the wire when the resistance are arranged in parallel,
\n\\(\\frac{1}{R_{p}}=\\frac{1}{\\frac{R}{4}}+\\frac{1}{\\frac{R}{4}}+\\frac{1}{\\frac{R}{4}}+\\frac{1}{\\frac{R}{4}}+\\frac{1}{\\frac{R}{4}} \\Rightarrow \\frac{16}{R}\\)
\n\\(\\frac{1}{R_{p}}=\\frac{16}{R}\\)
\nThe resistance of the wire reduced by 16 times<\/p>\n\n\n
\n Silver<\/td>\n 1.60 x 10-8<\/sup>\u2126.m<\/td>\n<\/tr>\n \n Copper<\/td>\n 1.62 x 10-8 <\/sup>\u2126.m<\/td>\n<\/tr>\n \n Tungsten<\/td>\n 5.2 x 10-8 <\/sup>\u2126.m<\/td>\n<\/tr>\n \n Mercury<\/td>\n 94 x 10-8 <\/sup>\u2126.m<\/td>\n<\/tr>\n \n Iron<\/td>\n 10 x 10-8 <\/sup>\u2126.m<\/td>\n<\/tr>\n \n Nichrome<\/td>\n 10 x 10-6 <\/sup>\u2126.m<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Which is a better conductor of electric current ?<\/h2>\n
\n(B) Copper
\n(C) Tungsten
\n(D) Mercury
\nAnswer:
\n(A) Silver<\/p>\n
\nSilver is a better conductor because it has lower resistivity.<\/p>\nWhich element will be used for electrical transmission lines ?<\/h2>\n
\n(B) Copper
\n(C) Tungsten
\n(D) mercury
\nAnswer:
\n(B) Copper<\/p>\n
\nCopper, because it is economical and has low resistivity.<\/p>\nNichrome is used in the heating elements of electric heating device because:<\/h2>\n
\n(B) It does not oxidise readily at high temperature
\n(C) Both of the above
\n(D) None of the above
\nAnswer:
\n(C) Both of the above<\/p>\n
\nNichrome as it has very high resistivity as it is an alloy, it does not oxidize readily at high temperature.<\/p>\nSeries arrangement is not used for domestic circuits because:<\/h2>\n
\n(B) Current drawn is more
\n(C) Neither of the above
\n(D) Both of the above
\nAnswer:
\n(A) Current drawn is less<\/p>\n
\nIn series arrangement, same current will flow through all the appliances which is not required and the equivalent resistance becomes higher, hence the current drawn becomes less.<\/p>\nIf the resistance is to be increased, then the resistors are to be increased in:<\/h2>\n
\n(B) Parallel
\n(C) Mixed arrangement
\n(D) None of the above
\nAnswer:
\n(A) Series<\/p>\n
\nThe effective resistance in the series is the sum of individual resistances. So, the resistances should be connected in series in order to increase the effective resistance.<\/p>\n
\nIn the given circuit, connect a nichrome wire of length ‘L’ between points X and Y and note the ammeter reading.<\/p>\nWhen this experiment is repeated by inserting another nichrome wire of the same thickness but twice the length (2L), what changes are observed in the ammeter reading ?<\/h2>\n
\n(A) Ammeter reading will increase
\n(B) Ammeter reading will decrease
\n(C) Will show double the increase
\n(D) No change in ammeter reading
\nAnswer:
\n(A) Ammeter reading will increase<\/p>\n
\nThe ammeter reading will decrease (becomes half). This is because with the increase in length, resistance of the circuit increases, hence current decreases.<\/p>\nState the changes that are observed in the ammeter reading if we double the area of cross section without changing the length in the above experiment.<\/h2>\n
\n(B) Ammeter reading will decrease
\n(C) Will decrease to half
\n(D) No change in ammeter reading
\nAnswer:
\n(B) Ammeter reading will decrease<\/p>\n
\nThe ammeter reading will increase (becomes two times). This is because as area increases, resistance decreases and hence current increases.<\/p>\nIn a circuit two resistors of 5\u2126 and 10 \u2126 are connected in series. Compare the current passing through the two resistors.<\/h2>\n
\n(B) 1 : 3
\n(C) 2 : 1
\n(D) 1 : 1
\nAnswer:
\n(D) 1 : 1<\/p>\n
\nIn a series connection of resistors, same current passes through all the resistors. Hence, current will be same. Ratio of the currents will be 1:1.<\/p>\nThe instrument used to measure current is ……..<\/h2>\n
\n(B) Voltmeter
\n(C) Galvanometer
\n(D) manometer
\nAnswer:
\n(A) Ammeter<\/p>\n
\nAmmeter is used to measure current.<\/p>\nWhen nichrome and copper wire of same length and same radius are connected in series and current is passed through them. Which wire gets heated up more?<\/h2>\n
\n(B) Copper wire
\n(C) Both will heat up at the same temperature
\n(D) None of the wire will get heated up.
\nAnswer:
\n(A) Nichrome wire<\/p>\n
\nIn series combination current is same in nichrome and copper wires. Thus nichrome wire will produce more heat.<\/p>\n
\n<\/p>\nWhat will happen to the other two bulbs if the bulb B3<\/sub> gets fused ?<\/h2>\n
\n(B) Other bulbs will glow with same brightness
\n(C) They will glow with low brightness
\n(D) They glow with more brightness
\nAnswer:
\n(B) Other bulbs will glow with same brightness<\/p>\n
\nOther bulbs will glow with same brightness<\/p>\nIf the wattage of each bulb is 1.5 W, how much readings will the ammeter A show when all the three bulbs glow simultaneously.<\/h2>\n
\n(B) 2A
\n(C) 1.5W
\n(D) None of the above
\nAnswer:
\n(A) 1.0A<\/p>\n
\nWhen the bulbs are in parallel, wattage will be added (4,5 W) and the ammeter reading would be,
\nI = p\/v = \\(\\frac{4.5}{4.5}\\) = 1A<\/p>\nFind the total resistance of the circuit:<\/h2>\n
\n(B) 4.50
\n(C) 1.50
\n(D) 2.00
\nAnswer:
\n(B) 4.50<\/p>\n
\nAmmeter reading =l.OA
\nV = 4.5V
\nR = V\/I
\n= 4.5 = 4.5 \u2126<\/p>\nHow many resistors of 88 W are connected in parallel to carry 10 A current on a 220 V line ?<\/h2>\n
\n(B) 1 resistors
\n(C) 3 resistors
\n(D) 4 resistors
\nAnswer:
\n(D) 4 resistors<\/p>\n
\nEquivalent resistance,
\n= \\(\\frac{1}{R_{p}}\\) = \\(\\frac{n}{88}\\)
\nRp = \\(\\frac{88}{n}\\)
\nV = IR
\nR = \\(\\frac{88}{n}\\)
\n\\(\\frac{88}{n}\\) = \\(\\frac{220}{10}\\)
\nn = 4 resistors<\/p>\nFind the statement which does not justify parallel connection:<\/h2>\n
\n(B) Overall resistance is lower than the resistance of single component
\n(C) Automobile headlight is connected in parallel
\n(D) There is a single path for the flow of electrons charge
\nAnswer:
\n(D) There is a single path for the flow of electrons charge<\/p>\n
\nThere is path of flow of electron change through each resistor. So there are several path.<\/p>\n
\nTwo conductors A and B of resistances 5 O and 10 C2 respectively are first joined in parallel and then in series. In each case the voltage applied is 20 V.<\/p>\nWhich of these circuit diagram shows the correct connection when A and B are joined in parallel?<\/h2>\n
\nAnswer:
\n<\/p>\n
\nThe diagram (A) correctly shows the combination of these conductors in each case.<\/p>\nIn which combination will the voltage across the conductors A and B be the same ?<\/h2>\n
\n(B) Parallel arrangement
\n(C) Both of the above
\n(D) None of the above
\nAnswer:
\n(B) Parallel arrangement<\/p>\n
\nVoltage across A and B will be same in parallel arrangement.<\/p>\nIn which arrangement will the current through A and B be the same ?<\/h2>\n
\n(B) Parallel arrangement
\n(C) Both of the above
\n(D) None of the above
\nAnswer:
\n(A) Series arrangement<\/p>\n
\nCurrent in A and B will be same in series arrangement.<\/p>\nEquivalent resistance in parallel combination is:<\/h2>\n
\n(B) 3.33 \u2126
\n(C) 0.3 \u2126
\n(D) None of the above
\nAnswer:
\n(B) 3.33 \u2126<\/p>\n
\n\\(\\frac{R_{1} R_{2}}{R_{1}+R_{2}}\\) = \\(\\frac{5 \\times 10}{5+10}\\)
\n= 3.33 \u2126<\/p>\nEquivalent resistance in series combination is:<\/h2>\n
\n(B) 3.33 \u2126
\n(C) 0.3 \u2126
\n(D) None of the above
\nAnswer:
\n(A) 15 \u2126<\/p>\n
\nSeries combination Rs = R1<\/sub> + R2<\/sub> = 5 + 10 = 15 \u2126<\/p>\n
\n<\/p>\nEffective resistance of two 8 \u2126 resistors in the combination.<\/h2>\n
\n(B) 160
\n(C) 80
\n(D) lO
\nAnswer:
\n(A) 40<\/p>\n
\n\\(\\vec{a}\\) = \\(\\vec{a}\\) + \\(\\vec{a}\\) = \\(\\vec{a}\\)
\nR = 4\u2126<\/p>\nCurrent flowing through 4\u2126 resistor is :<\/h2>\n
\n(B) 2A
\n(C) 8A
\n(D) 4A
\nAnswer:
\n(A) 1A<\/p>\n
\nCurrent flowing through the circuit = Current flowing through 4\u2126 Equivalent resistance of the circuit = 4\u2126 + 4\u2126 = 8\u2126
\nCurrent flowing in the circuit = \\(\\frac{V}{R}\\) = \\(\\frac{8 \\mathrm{~V}}{8 \\Omega}\\) = 1A<\/p>\nPotential difference across 4\u2126 resistor is :<\/h2>\n
\n(B) 2V
\n(C) 4V
\n(D) 8V
\nAnswer:
\n(C) 4V
\nExplanation:
\nPotential difference across 4\u2126 = V1<\/sub>
\nV1<\/sub> = IR1<\/sub>
\nV1<\/sub> = 1A x 4\u2126 = 4V<\/p>\nPower dissipated in 4\u2126 resistor is : A<\/h2>\n
\n(B) 2W
\n(C) 1W
\n(D) 8W
\nAnswer:
\n(A) 4W<\/p>\n
\nPower dissipated in 4\u2126=p
\np = VI = 4V x 1A = 4 watt<\/p>\nDifference in reading of ammeter A1<\/sub> and A2<\/sub><\/h2>\n
\n(B) will become half
\n(C) remains the same
\n(D) 8A
\nAnswer:
\n(C) remains the same<\/p>\n
\nSince the resistance are in series combination, hence same current will flow through each resistor. Hence ammeter reading will be same.<\/p>\nMCQ Questions for Class 10 Science with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"