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Current Electricity Test - 77

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Current Electricity Test - 77
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  • Question 1
    1 / -0
    Two cells of emf E1E_1 and E2E_2 (E1,>E2)(E_1, > E_2) are connected as shown in figure. When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300cm. When the same potentiometer is connected between A and C, the balancing length is 100cm. The ratio of E1E_1 and E2E_2 is

    Solution
    For balance potentiometer : EabEac=lablac\dfrac{E_{ab}}{E_{ac}}=\dfrac{l_{ab}}{l_{ac}}
    or E1E1E2=300100\dfrac{E_1}{E_1-E_2}=\dfrac{300}{100}
    or 3E13E2=E13E_1-3E_2=E_1
    or 2E1=3E22E_1=3E_2 or E1/E2=3/2E_1/E_2=3/2
    or E1:E2=3:2E_1:E_2=3:2
  • Question 2
    1 / -0
    Read the following statements carefully :
    Y : The resistivity of a semiconductor decreases with increases of temperature.
    Z : In a conducting solid, the rate of collision between free electrons and ions increases with
    increase of temperature. Select the correct statement from the following :

    Solution

  • Question 3
    1 / -0
    In the figure, the potentiometer wire AB of length L and resistance 9r9r is joined to the cell D of emf ε \varepsilon  and internal resistance r. The cell Cs emf is ϵ /2\epsilon  /2 and its internal resistance is 2r2r. The galvanometer G will show no deflection when the length AJ is

    Solution

  • Question 4
    1 / -0
    In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell, the balance point shifts to 63 cm, then emf of the second cell is
    Solution
    Given : E1=1.25V,l1=35cm,l2=63cm\textbf{Given : } E_1=1.25V, l_1=35 cm, l_2= 63 cm 

    Step 1: \textbf{Step 1: }
    The potential gradient of wire, K=ElK=\dfrac{E}{l} will remain constant, Therefore EMF of external cell is proportional to the balance length
     
     E1E2=l1l2\therefore \dfrac{E_1}{E_2}=\dfrac{l_1}{l_2}                  ....(1)....(1)

    Step 2: Substituting the values in equation (1)\textbf{Step 2: Substituting the values in equation (1)}

    So, 1.25E2=3563\dfrac{1.25}{E_2}=\dfrac{35}{63} or E2=2.25 VE_2=2.25  V
  • Question 5
    1 / -0
    In a Wheatstone bridge, three resistances P, Q and R are connected in the three arms and the fourth arm is formed by two resistances S1S_1 and S2S_2 connected in parallel. The condition for the bridge to be balanced will be
    Solution
    The balanced condition for Wheatstone bridge is  PQ=RS\dfrac{P}{Q}=\dfrac{R}{S}
    Here S=S1S2S1+S2S=\frac{S_1S_2}{S_1+S_2}  (as S1S_1 and S2S_2 are connected in parallel in S arm)
    so, PQ=R(S1+S2)S1S2\dfrac{P}{Q}=\dfrac{R(S_1+S_2)}{S_1S_2}
  • Question 6
    1 / -0
    Figure shows a potentiometer with a cell of 2.0 V and internal resistance 0.40Ω\Omega maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V gives a balance point at 67.3 cm length of the wire. The standard cell is then replaced by a cell of unknown emf E and the balance point, similarly turns out to be at 82.3 cm, length of the wire. Then the E is

    Solution
    The balance condition for a potentiometer : E1E2=l1l2\dfrac{E_1}{E_2}=\dfrac{l_1}{l_2}
    Here E1=1.02V, l1=67.3cm, l2=82.3 cmE_1=1.02 V,  l_1=67.3 cm,   l_2=82.3  cm and E2=?E_2=?
    so, 1.02E2=67.382.3\dfrac{1.02}{E_2}=\dfrac{67.3}{82.3} or E2=1.25VE_2=1.25 V
  • Question 7
    1 / -0

    Directions For Questions

    The length of the potentiometer wire AB is 600 cm, and it carries a constant current of 40 mA from A to B. For a cell of emf 2 V and internal resistance 10ω\omega, the null point is found at 500 cm from A. When a voltmeter is connected across the cell, the balancing length of the wire is decreased by 10 cm.

    ...view full instructions

    Potential gradient along AB is
    Solution
    Since 2 V is balanced across a length of 500 cm == 5m, so the potential gradient k=2/5Vm1k = 2/5 Vm^{-1}

    Reading of voltmeter == potential difference across voltmeter

    =VAVD=k(4.9)= V_A - V_D = k(4.9)

    Reading of the voltmeter =25×4.9=1.96V= \displaystyle \frac{2}{5} \times 4.9 = 1.96 V

    Terminal potential difference of the battery will also be 1.96 V. So

    1.96=2i×101.96 = 2 - i \times 10 or i=4×103Ai = 4 \times 10^{-3} A

    So, resistance of the voltmeter isRV=1.96/(4×103)=490ωR_V = 1.96/(4 \times 10^{-3}) = 490 \omega

    RV=1.96/(4×103)=490ωR_V = 1.96/(4 \times 10^{-3}) = 490 \omega
  • Question 8
    1 / -0
    In the potentiometer arrangement shown in figure, the driving cell D has emf E and internal resistance r. The cell C, whose emf is to measured, has emf E2\displaystyle \dfrac{E}{2} and internal resistance 2r. The potentiometer wire is 100cm long. If balance is obtained, the length AJ is

    Solution
    In the measurement of external emf by potentiometer, the internal resistance of external battery does not matter since no current flows through it at the null point.
    Due to internal resistance of cell D, the potential drop across AB is less than EE.
    Hence, to achieve the potential difference of E/2E/2 across AJ, the length AJ must be more than half the total length.
  • Question 9
    1 / -0
    How many positions are there on wire ABAB such that on touching KK to those positions current from AA to BB remain same throughout if V>EV>E and internal resistance of primary circuit battery is zero?

  • Question 10
    1 / -0
    If the filament of bulb D breaks then does bulb C give more light, less light, the same light or no light?

    Solution
    It is assumed that every bulb is identical and has same resistance(RR).
    So C and D are in parallel.And A , B and combination of (C and D) all these three are in series.
    Initially resistance here was R/2R/2 (C and D both have RR in parallel).So resistances in series are RR ,RR and R/2R/2. Since resistance across third element is less than first two elements so less portion of cell voltage will be shared here.

    Now bulb D breaks.So combination of C and D will reduced to only C.Now only C is there so net resistance here is RR .So now A (RR resistance) , B (RR resistance) and C (RR resistance) are in series and have equal resistances so voltage across C will be 1/31/3 of cell voltage.

    In other words after breaking of D voltage across C has increased from before hence it will glow more brightly than before.
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