Q 1 :    

The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are  8Ω and 10Ω respectively. After inserting in a hot bath of temperature 400°C, the resistance of platinum wire is :                     [2024]

  • 8Ω

     

  • 10Ω

     

  • 16Ω

     

  • 2Ω

     

(C)    Given, R0=8Ω and R100=10Ω

               R100=R0(1+αT)

                10=8(1+α×100)100α=14

           Also, R400=R0(1+αT)

               R400=8(1+400α)=8(1+1)=16Ω

 



Q 2 :    

An electric toaster has resistance of 60Ω at room temperature (27°C). The toaster is connected to a 220 V supply. If the current flowing through it reaches 2.75 A, the temperature attained by toaster is around   

(if α=2×10-4/Co)                   [2024]

  • 694°C

     

  • 1235°C

     

  • 1694°C

     

  • 1667°C

     

(C)      Rnew=V/I=80Ω

            Rnew=Rold[1+αT]

             80=60[1+2×10-4T]

             T=1666.67

              T-27=1666.67

              T=1693.66=1694oC

 



Q 3 :    

Two conductors have the same resistances at 0oC but their temperature coefficients of resistance are α1 and α2. The respective temperature coefficients for their series and parallel combinations are                                          [2024]

  • α1+α2,α1+α22

     

  • α1+α22,α1+α22

     

  • α1+α22,α1+α2

     

  • α1+α2,α1α2α1+α2

     

(B)    Series: Req=R1+R2

         2R(1+αeqθ)=R(1+α1θ)+R(1+α2θ)

         2R(1+αeqθ)=2R+(α1+α2)Rθ

         αeq=α1+α22

        Parallel: 1Req=1R1+1R2

        1R2(1+αeqθ)=1R(1+α1θ)+1R(1+α2θ)

         21+αeqθ=11+α1θ+11+α2θ

           21+αeqθ=1+α2θ+1+α1θ(1+α1θ)(1+α2θ)

            2[(1+α1θ)(1+α2θ)]

            =[2+(α1+α2)θ][1+αeqθ]

             2[1+α1θ+α2θ+α1α2θ]

             =2+2αeqθ+(α1+α2)θ+αeq(α1+α2)θ2

             Neglecting small terms

            2+2(α1+α2)θ=2+2αeqθ+(α1+α2)θ

            (α1+α2)θ=2αeqθαeq=α1+α22

 



Q 4 :    

Two wires A and B are made up of the same material and have the same mass. Wire A has radius of 2.0 mm and wire B has radius of 4.0 mm. The resistance of wire B is 2 Ω. The resistance of wire A is _______ Ω.        [2024]



(32)      Given: rA=2×10-3m, rB=4×10-3m, RB=2Ω

             We know R=ρlA

              RARB=ρAlAAA×ABρBlB=lAlBABAA                  ...(i)

              The volume of wire remains constant, VA=VB

               lAAA=lBABlAlB=ABAA                             ...(ii)

              from equation (i) and equation (ii)

             RA=RB(ABAA)2=2(π×(4×10-3)2π×(2×10-3)2)2=2×16=32Ω

                 RA=32Ω

 



Q 5 :    

A wire of resistance R and radius r is stretched till its radius became r/2. If new resistance of the stretched wire is x R. then value of x is _______ .           [2024]



(16)      We know R=ρlARlr2

              As we stretch the wire, its length will increase but its radius will decrease keeping the volume constant.

              Vi=Vfπr2l=πr24lflf=4l

              RnewRold=(4lr24)r2l=16

               Rnew=16R

              x=16

 



Q 6 :    

Resistance of a wire at 0°C, 100°C and t °C is found to be 10 Ω, 10.2 Ω and 10.95 Ω respectively. The temperature t in Kelvin scale is _______ .     [2024]



(748)    R=R0(1+αT)RR0=αT

              Case-I:

              10.2-1010=α(100-0)                  ...(i)

               Case-II:

                10.95-1010=α(t-0)                     ...(ii)

                 t100=0.950.2=475oCt=475+273=748K

 



Q 7 :    

At room temperature (27°C) the resistance of a heating element is 50 Ω. The temperature coefficient of the material is 2.4×10-4o C-1. The temperature of the element, when its resistance is 62 Ω, is ______ Co.     [2024]



(1027)       R0=50Ω; α=2.4×10-4o C-1; R=62Ω

                   R=R0(1+αT)

                  62=50(1+2.4×10-4×T)

                   1.24=1+2.4×10-4×T

                   0.24=2.4×10-4×T

                   T=0.242.4×10-4=1000

                    Final temperature, Tf=1000+27=1027oC