Lakshya Education MCQs

Question: A 1Ω resistance is in series with an ammeter which is balanced by 75 cm of potentiometer wire. A standard cell of 1.02 V is balanced by 50 cm. The ammeter shows  a reading of 1.5A. The error in the ammeter reading
Options:
A.0.002 A
B.0.03 A
C.1.01 A
D.no error
Answer: Option B
: B

ip=Vl=1.0250
Voltage across ammeter = Vll1
=1.0250×75=1.022×3=0.51×3
The true current I1A=VAR=1.531I1A=1.53AError=I1AIA=1.531.5=0.03A

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Question 1. Two resistances X and Y are in the left and right gaps of a meter bridge. The balance point is 40 cm from left.Two resistances of 10Ω each are connected in series with X and Y separately. The balance point is 45 cm. The value of X and Y are
  1.    12Ω,8Ω
  2.    4Ω,6Ω
  3.    8Ω,12Ω
  4.    12Ω,16Ω
Answer: Option C
: C

xy=4010040=4060=23(1)x+10y+10=4555=91123y+10y+10=91123×11y+110=9y+9020=9y223y20×3=5yy=12Ωx=23×12=8Ω
Question 2. The potential difference across the terminals of a battery is 50V when 6A of current is drawn and 60V when 1A is drawn. The emf and internal resistance of the battery are
  1.    62V, 2Ω
  2.    63V, 1Ω
  3.    61V, 1Ω
  4.    64V, 2Ω
Answer: Option A
: A

V= E - Ir
50 = E - 6r (1)
60 = E - 1r (2)
(2) - (1) 10 = 5r; r = 2Ω
E = 62V
Question 3. The scale of a galvanometer of resistance 100Ω  contains 25 divisions. It gives a deflection of one division on passing a current of 4×104 A. The resistance in ohms to be added to it, so that it may become a voltmeter of range 2.5 volt is
  1.    100
  2.    150
  3.    250
  4.    300
Answer: Option B
: B

Current sensitivity of galvanometer =4×104 Amp/div.
So full scale deflection current (ig)=Currentsensitivity×Totalnumberofdivision=4×104×25=102A
To convert galvanometer in to voltmeter, resistance to be put in series is
R=VigG=2.5102100=150Ω

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