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Charges & Magnetism Test - 16

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Charges & Magnetism Test - 16
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  • Question 1
    1 / -0

    In a moving coil galvanometer the deflection (ϕ) on the scale by a pointer attached to the spring is

    Solution

    Since magnetic torque on the coil,
    τ = NIAB
    This torque is balanced by counter torque
    ∴ kϕ = 
    where k is torsional constant. It is a scalar quantity having dimension of torque or energy i.e. [ML2T−2]

  • Question 2
    1 / -0

    Two moving coil meters, M1 and M2 ave thhe following particulars:
    R1 = 10 Ω, N1 = 30, A1 = 3.6 × 10–3 m2, B1 = 0.25 T
    R2 = 14 Ω, N2 = 42, A2 = 1.8 × 10–3 m2, B2 = 0.50 T
    The spring constants are identical for the two meters. What is the ratio of  current sensitivity and voltage sensitivity of M2 and M1.

    Solution

    For meter M1, R1 = 10 Ω, N1 = 30, A1 = 3.6 × 10–3 m2, B1 = 0.25 T, k= k 
    For meter M2, R2 = 14 Ω, N2 = 42, A2 = 1.8 × 10–3 m2, B2 = 0.50 T, k2 = k 
    As, current sensitivity Is = NBA/k.

    Voltage sensitivity Vs
    So, 




    = 1

  • Question 3
    1 / -0

    If the current sensitivity of a galvanometer is doubled, then its voltage sensitivity will be

    Solution

    Current sensitivity of galvanometer is deflection per unit current i.e

    Similarly voltage sensitivity is deflection per unit voltage
    i.e. 

    From (i) and (ii)
    Voltage sensitivity = current sensitivity × 1/resistance
    Now if current sensitivity is doubled, then the resistance in the circuit will also be doubled since it is proportional to the length of the wire, then voltage sensitivity

    = (current sensitivity) × 1 / (resistance)
    Hence, voltage sensitivity will remain unchanged

  • Question 4
    1 / -0

    A moving coil galvanometer can be converted into an ammeter by:

    Solution

    To utilise a galvanometer (G) as an ammeter, there is the difficulty in measurement of current due to the sensitivity of galvanometer and also the connection of galvanometer with a very large resistance in series that may change the value of current in the circuit.
    To overcome these difficulties one attaches a small resistance rs called shunt resistance in parallel with the galvanometer coil as shown in the figure.
    solution

  • Question 5
    1 / -0

    A galvanometer of resistance 70Ω, is converted to an ammeter by a shunt resistance rs = 0.03Ω. The value of its resistance will become

    Solution


    Here, RG = 70Ω, rs = 0.03Ω
    ∴ R = 
    = 0.02998 
    = 0.03Ω

  • Question 6
    1 / -0

    If the galvanometer current is 10mA, resistance of the galvanometer is 40Ω and shunt of 2Ω is connected to the galvanometer, the maximum current which can be measured by this ammeter is

    Solution



    = 0.21A

  • Question 7
    1 / -0

    A  galvanometer of resistance 40Ω gives a deflection of 5 divisions per mA. There are 50 divisions on the scale. The maximum current that can pass through it when a shunt resistance of 2Ω is connected is

    Solution

    IG = 50/5 = 10mA; R= 40Ω, Rs = 2Ω
    Maximum current,
    I = 

    = 210 mA

  • Question 8
    1 / -0

    In the given circuit, a galvanometer with a resistance of 70Ω is converted to an ammeter by a shunt resistance of 0.05Ω total current measured by this device is
    image

    Solution


    The equivalent circuit of given circuit is given by
    Here,

    ∴ RS = 0.0499Ω = 0.05Ω
    The total resistance in the circuit
    R = RS + 5Ω = 0.05 + 5 = 5.05Ω
    The current measured by the device

    = 0.99 A

  • Question 9
    1 / -0

    A galvanometer having a resistance of 50Ω, gives a full scale deflection for a current of 0.05A. The length (in metres) of a resistance wire of area of cross section 3 × 10−2cm2 that can be used to convert the galvanometer into an ammeter which can read a maximum of 5A current is
    (Specific resistance of the wire ρ = 5 × 10−7Ωm)

    Solution






    ∵ S = 
    ∴ l  = 
    = 3.0m

  • Question 10
    1 / -0

    The conversion of a moving coil galvanometer into a voltmeter is done by

    Solution

    Galvanometer is a very sensitive instrument, therefore it can not measure high potential difference.
    In order to convert a Galvanometer into voltmeter, a very high resistance known as "series resistance" is connected in series with the galvanometer.

  • Question 11
    1 / -0

    A galvanometer of resistance 10Ω gives full-scale deflection when 1mA current passes through it. The resistance required to convert it into a voltmeter reading upto 2.5V is

    Solution

  • Question 12
    1 / -0

    A  voltmeter which can measure 2V is constructed by using a galvanometer of resistance 12Ω and that produces maximum deflection for the current of 2mA, then the resistance R is
    image

    Solution


    Putting V = 2V,
    Ig = 2mA
    = 2 × 10−3A,
    G = 12 Ω,
    ∴ R= 2 / 2 × 10−3 −12 = 1000 − 12

    = 988Ω

  • Question 13
    1 / -0

    The value of current in the given circuit if the ammeter is a galvanometer with a resistance R= 50Ω is
    949865

    Solution


    Req = Rg +3 = 50 + 3 = 53Ω

    solution

  • Question 14
    1 / -0

    A galvanometer coil has a resistance of 15Ω and the metre shows full scale deflection for a current of 4mA. To convert the meter into a voltmeter of range 0 to 18V, the required resistance is

    Solution

    The galvanometer can be transformed into a voltmeter by applying a high resistance in series. If the series resistance is R, then total resistance in circuit  = (15 + R)Ω
    Now 

    = 4.5 × 10
    ∴ R  = 4500−15
    = 4485Ω

  • Question 15
    1 / -0

    A galvanometer of resistance 50Ω is connected to a battery of 3V along with a resistance of 2950Ω in series. A full-scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be:

    Solution

    Total initial resistance
    = G + R = 50Ω + 2950Ω = 3000Ω
    Current,I = 

    = 1 × 10−3 A = 1mA
    If the deflection has to be reduced to 20 divisions, then current

    Let x be the effective resistance of the circuit,
    3V = 3000Ω × 1mA =  
    or x = 
    ∴  Resistance to be added =(4500Ω − 50Ω) = 4450Ω 

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