33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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Telegram @unacademyplusdiscounts50 NEET-AIPMT Chapterwise Topicwise Solutions Physics73. (c) : Here, resistance of the galvanometer = GCurrent through the galvanometer,IG = 02 I = 02 .I = 1. % ofI100 500\ Current through the shunt,1 499IS= I− IG= I− I = I500 500As shunt and galvanometer are in parallel\ I G G = I S S⎛ 1 ⎞ 499GI G S S⎝⎜500 ⎠⎟ = ⎛ ⎝ ⎜ ⎞500 ⎠⎟ or =499Resistance of the ammeter R A is1 1 1 1 1 500= + = + =RAG S G G G499RA = 1 G500Vg74. (a) : S =( I − Ig)Neglecting I gV−3 g 25 × 10 V∴ S = == 0. 001ΩI 25 A75. (d) : Let resistance R is to be put in series withgalvanometer G to keep the main current in the circuitunchanged.GS∴ + R=GG + S22GS G + GS−GSGR= G− ⇒ R ==G + SG+S G+S76. (a) : Here, Resistance of galvanometer,G = 100 WCurrent for full scale deflection, I g = 30 mA= 30 × 10 –3 ARange of voltmeter, V = 30 VTo convert the galvanometer into an voltmeter of a givenrange, a resistance R is connected in series with it as shownin the figure.From figure, V = I g (G + R)or R = V −GIg30= − 100 Ω= 1000 − 100 = 900 Ω−330 × 1077. (c) : iG = (I – i)S where G is the galvanometerresistance and S is the shunt used with the ammeter.1.0 × 60 = (5 – 1)S where S is the shunt used to read a 5 Acurrent when the galvanometer can stand by 1 A.10 . × 60S = = 15 Ω in parallel.478. (b) : Total initial resistance= R G + R 1 = (50 + 2950) W = 3000 We = 3 V3V∴ Current = = 1× 10 −3A=1mA3000 ΩvvvIf the deflection has to be reduced to 20 divisions, current2i = 1 mA × as the full deflection scale for 1 mA = 303divisions.23 V= 3000 Ω× 1 mA =xΩ×mA3⇒ x = 3000 × 1 × 3 = 4500 Ω2But the galvanometer resistance = 50 WTherefore the resistance to be added= (4500 – 50)W = 4450 W.79. (a) : Let the shunt resistance be S.Given: I = 750 A,I g = 100 A, R G = 13 WFrom the figure,I g R G = (I – I g )Sor 100 × 13 = [750 – 100]Sor 1300 = 650 S\ S = 1300/650 = 2 W80. (d) : The total current shown by the galvanometer is25 × 4 × 10 –4 A.\ I g = 10 –2 AThe value of resistance connected in series to convertgalvanometer into voltmeter of 25 V isRV 25= − G = − 50 = 2450 Ω.I−2g 1081. (a) : Voltmeter is used to measure the potentialdifference across a resistance and it is connected inparallel with the circuit. A high resistance is connectedto the galvanometer in series so that only a small fraction(I g ) of the main circuit current (I) passes through it.If a considerable amount of current is allowed to passthrough the voltmeter, then the reading obtained bythis voltmeter will not be close to the actual potentialdifference between the same two points.82. (b) : The shunt and galvanometer are in parallel.1 1 1 18Therefore, = + or R q =R eq 9 2 e Ω11Using Ohm’s law, V = IR eq = 1 × 18 1811= 11V.\ Current through shunt = V R s18 / 11 9= = 2 11 08 . amp83. (a) : To convert a galvanometer into ammeter, oneneeds to connect a low resistance in parallel so thatmaximum current passes through the shunt wire andammeter remains protected.
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33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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