12th Class Physics Sample Paper Physics Sample Paper-6

  • question_answer
    (i) The following graph shows the variation of terminal potential difference V, across a combination of three cells in series to a resistor versus current I.
      Graph of terminal potential difference
    (a) Calculate the emf of each cell.
    (b) For what current I, will the power dissipation of the circuit be maximum?
    (ii) Determine the current drawn from a 12 V supply with internal resistance \[0.5\,\Omega \] by the infinite network shown in given figure. Each resistor has \[1\,\Omega \] resistance.
    (i) Derive an expression for drift velocity of free electrons in a conductor in terms of relaxation time.
    (ii) At room temperature \[(27{}^\circ C),\] the resistance of a heating element in \[100\,\Omega .\] What is the temperature of the element if the resistance is found to be \[117\,\Omega ,\] given that the temperature coefficient of resistance of the material of the resistor is \[1.70\times {{10}^{-4}}/{}^{o}C.\]


    (i) (a) Let E be emf and r the internal resistance of each call V          \[={{E}_{eq}}-Ir\Rightarrow V=3E=Ir\] When,               I = 0,    V = 6 V \[\Rightarrow \]               \[6=3E-0\Rightarrow E=\frac{6}{3}=2\,V\] Emf of each call is 2 V. (b) For maximum power dissipation, the effective internal resistance (r) of cells must be equal to external resistance (R). When,   V = 0,  I = 2 A \[\because \]       \[V={{E}_{eq}}-Ir\] \[\therefore \]      \[0=3E-2\times r\] or,        \[0=3\times 2-2r\] or, \[2r=6\Rightarrow r=3\,\Omega \] For maximum power dissipation current in circuit,             \[r=3\,\Omega =R\] i.e. internal resistance = external resistance.             \[I=\frac{{{E}_{eq}}}{R+r}\therefore I=\frac{3\times 2}{3+3}=1\,A\] (ii) Let the effective resistance of the network be x. If one part of the network has resistance \[(1\,\Omega ,\,\,1\,\Omega ,\,\,1\,\Omega )\] is separated as shown, the effective resistance remains \[\times \] (as it is infinite network). Here, x and \[1\,\Omega \] are in parallel. \[\frac{1}{{{R}_{P}}}=\frac{1}{x}+\frac{1}{1}=\frac{1+x}{x}\Rightarrow {{R}_{P}}=\frac{x}{1+x}\] Now, resistances \[{{R}_{p}},\] \[1\,\Omega \] and \[1\,\Omega \] are in series. So, the resultant resistance, \[R={{R}_{P}}+1+1=\frac{x}{1+x}+1+1=\frac{x}{1+x}+2\] ?(i) In case of infinite resistances, the value of R remains x. \[\therefore \]      \[x=\frac{x}{1+x}+2\] \[x(x+1)=x+2+2x\Rightarrow {{x}^{2}}-2x-2=0\]             \[x=\frac{-(-2)\pm \sqrt{4+8}}{2}=\frac{2\pm \sqrt{2}}{2}=1\pm \sqrt{3}\] The value of resistance cannot be negative. So, the resistance of network             \[x=1+\sqrt{3}=1+1.732=2.732\,\Omega \] Total resistance of the circuit  \[=2.732+0.5=3.232\,\Omega \] Current drawn from the supply,             \[I=\frac{V}{3.232}=\frac{12}{3.232}=3.72\,A\] Or (i) When a conductor is subjected to an electric field E, each electron experiences a force \[F=-eE\]and free electron acquires an acceleration \[a=\frac{F}{m}=-\frac{eE}{m}\]                 ?(i) where, m = mass of electron, e = electronic charge and E = electric field. Free electron starts accelerating and gains velocity and collide with atoms and molecules of the conductor. The average time difference between two consecutive collisions is known as relaxation time of electron and \[\overline{\tau }=\frac{{{\tau }_{1}}+{{\tau }_{2}}+...{{\tau }_{n}}}{n}\]                                   ?(ii) where, \[{{\tau }_{1}},{{\tau }_{2}},....,{{\tau }_{n}}\] are the average time difference between 1st, 2nd, ..,, nth collisions. \[\therefore \,\,{{v}_{1}},{{v}_{2}},...,{{v}_{n}},\] are velocities gained by electron in 1st, 2nd ?, nth collisions with initial thermal velocities \[{{u}_{1}},{{u}_{2}},....,{{u}_{n}},\] respectively. \[\therefore {{v}_{1}}={{u}_{1}}+a{{\tau }_{1}},\] \[{{v}_{2}}={{u}_{2}}+a{{\tau }_{2}},\] \[{{v}_{n}}={{u}_{n}}+a{{\tau }_{2}}\] The drift speed \[{{v}_{d}}\]may be defined as             \[{{v}_{d}}=\frac{{{v}_{1}}+{{v}_{2}}+...+{{v}_{n}}}{n}\] or \[{{v}_{d}}=\frac{({{u}_{1}}+{{u}_{2}}+...+{{u}_{n}})+a({{\tau }_{1}}+{{\tau }_{2}}+...+{{\tau }_{n}})}{n}\] or \[{{v}_{d}}=\frac{({{u}_{1}}+{{u}_{2}}+...+{{u}_{n}})}{n}+\frac{a({{\tau }_{1}}+{{\tau }_{2}}+...+{{\tau }_{n}})}{n}\] or         \[{{v}_{d}}=0+a\tau \] [\[\because \]Average thermal velocity in n collisions = 0] \[{{v}_{d}}=-\left( \frac{eE}{m} \right)\tau \]                              [from Eq. (i)] This is the required expression or drift speed of free electrons. (ii) Given, \[{{R}_{27}}=100\,\Omega ,\] \[{{R}_{1}}=117\,\Omega ,\] \[t=?,\] \[\alpha =1.70\times {{10}^{-4}}/{}^{o}C\] Temperature coefficient, \[\alpha =\frac{{{R}_{t}}-{{R}_{27}}}{{{R}_{27}}(t-27)},\] temperature, t is unknown. \[\Rightarrow t-27=\frac{{{R}_{t}}-{{R}_{27}}}{{{R}_{27}}.\alpha }=\frac{117-100}{100\times 1.70\times {{10}^{-4}}}=1000\] \[\Rightarrow \] \[t=1000+27=1027{}^{o}C\]

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