12th Class Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Question Bank MCQ - Current Electricity

  • question_answer
        Four wires of the same diameter are connected, in turn, between two points maintained at a constant potential difference. Their resistivities and lengths are; \[\rho \] and L (wire 1), \[1.2\text{ }\rho \] and 1.2 L (wire 2), \[0.9\text{ }\rho \] and 0.9 L (wire 3) and \[\rho \] and  1.5 L (wire 4). Rank the wires according to the rates at which energy is dissipated as heat, greatest first:

    A) 4>3>1>2        

    B) 4>2>1>3             

    C) 1>2>3>4       

    D) 3>1>2>4             

    Correct Answer: D

    Solution :

    (d) 3>1>2>4                                  Resistance of a wire, \[R=\frac{\rho l}{A}\] Rate of energy dissipated as heat is \[H=\frac{{{V}^{2}}}{R}=\frac{{{V}^{2}}A}{\rho l}\] For wire 1, \[{{H}_{1}}=\frac{{{V}^{2}}A}{\rho L}\] For wire \[2,\,{{H}_{2}}=\frac{{{V}^{2}}A}{(1.2\rho )(1.2L)}=\frac{0.694{{V}^{2}}A}{\rho L}=0.694{{H}_{1}}\] For wire 3, \[{{H}_{3}}=\frac{{{V}^{2}}A}{(0.9\rho )(0.9L)}=\frac{1.23{{V}^{2}}A}{\rho L}=1.23{{H}_{1}}\] For wire 4, \[{{H}_{4}}=\frac{{{V}^{2}}A}{(\rho )(1.5L)}=\frac{0.666{{V}^{2}}A}{\rho L}=0.666{{H}_{1}}\] \[\therefore {{H}_{3}}>{{H}_{1}}>{{H}_{2}}>{{H}_{4}}\]  


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