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Current Electricity Test - 21

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Current Electricity Test - 21
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  • Question 1
    1 / -0

    Consider a current carrying wire (current I) in the shape of a circle.

    Solution

    As we know, electric current per unit area

    I/A, is called current density j i.e., j=IA

    The SI units of the current density are A/m2. The current density is also directed along E and is also a vector and the relationship is

    j=σE

    Current density changes due to electric field produced by charges accumulated on the surface of wire.

  • Question 2
    1 / -0

    A resistance R is to be measured using a meter bridge, student chooses the standard resistance S to be 100Ω. He finds the null point at l1 = 2.9 cm. He is told to attempt to improve the accuracy. Which of the following is a useful way?

    Solution

    Adjusting the blance point near the middle of the bridge, i.e., when l1 is close to 50 cm. requires a suitable choice of S, R is unknown resistance:

    Since, RS=Rl1R(100l1)

    RS=l1(100l1) or R = S[l1(100l1)]

    R = S[2.997.1]

    So, here, R : S = 2.9 : 97.1 implies that the S is nearly 33 times to that of R. In order to make this ratio 1 : 1 it is necessary to reduce the value of S nearly 133 times i.e., nearly 3Ω

  • Question 3
    1 / -0

    Two cells of emfs approximately 5 V and 10 V are to be accurately compared using a potentiometer of length 400 cm

    Solution

    The potential drop across wires of potentiometer should be more than emfs of primary cells. Here, values of emfs of two cells are given as 5V and 10V, so the potential drop along the potentiometer wire must be more than 10V. So battery should be of 15V and about 4V potential is dropped by using variable resistance.

  • Question 4
    1 / -0

    A metal rod of length 10 cm and a rectangular cross-section of 1cm x 12cm is connected to a battery across opposite faces. The resistance will be

    Solution

    As we known that the resistance of wire is

    R = ρlA

    For maximum value of R, l must be higher and A should be lower and it is possible only when the battery is connected across area of cross section

    = 1cm x (12)cm.

  • Question 5
    1 / -0

    Which of the following characteristics of electrons determines the current in a conductor?

    Solution

    We know that the relationship between current and drift speed is

    I = ne Avd

    Where, I is the current and Vd is the drift velocity.

    So IVd

    Hence, only drift velocity determines the current in a conductor.

  • Question 6
    1 / -0

    Kirchhoff’s junction rule is a reflection of

    (a) conservation of current density vector

    (b) conservation of charge

    (c) the fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction

    (d) the fact that there is no accumulation of charges at a junction

    Solution

    According to Kirchhoff’s junction rule or current law the algebraic sum of the currents flowing towards any point in an electric network is zero i.e., charges are conserved in an electric network, and no any charges accumulate at junction as the sum of entering and outgoing charge are equal at any time interval.

    So, Kirchhoff’s junction rule is the reflection of conservation of charge.

  • Question 7
    1 / -0

    Temperature dependence of resistivity ρ(T) of semiconductors, insulators and metals is significantly based on the following factors

    (a) number of charge carriers can change with temperature T

    (b) time interval between two successive collisions can depend on T

    (c) length of material can be a function of T

    (d) mass of carriers is a function of T

    Solution

    As we know that, the resistivity of a metallic conductor is

    (ρ)=mne2τ

    where n is number of charge carriers per unit volume which can change with temperature T and τ is time interval between two successive collisions which decreases with the increase of temperature, so it will affect (ρ) resistivity.

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