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

     

(3)   

         Given, R0=8Ω and R100=10Ω

          ∴     R100=R0(1+α∆T)

          ∴      10=8(1+α×100)⇒100α=14

           Also, R400=R0(1+α∆T)

           ∴    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

     

(3)     

           Rnew=V/I=80Ω

            Rnew=Rold[1+α∆T]

             80=60[1+2×10-4∆T]

             ∆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

     

(2)   

         Series: Req=R1+R2

         2R(1+αeq∆θ)=R(1+α1∆θ)+R(1+α2∆θ)

         2R(1+αeq∆θ)=2R+(α1+α2)R∆θ

         ⇒αeq=α1+α22

        Parallel: 1Req=1R1+1R2

        1R2(1+αeq∆θ)=1R(1+α1∆θ)+1R(1+α2∆θ)

         21+αeq∆θ=11+α1∆θ+11+α2∆θ

           21+αeq∆θ=1+α2∆θ+1+α1∆θ(1+α1∆θ)(1+α2∆θ)

            2[(1+α1∆θ)(1+α2∆θ)]

            =[2+(α1+α2)∆θ][1+αeq∆θ]

             2[1+α1∆θ+α2∆θ+α1α2∆θ]

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

             Neglecting small terms

            2+2(α1+α2)∆θ=2+2αeq∆θ+(α1+α2)∆θ

            (α1+α2)∆θ=2αeq∆θ⇒αeq=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=lBAB⇒lAlB=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=ρlA⇒R∝lr2

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

              Vi=Vf⇒πr2l=πr24lf⇒lf=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+α∆T)⇒∆RR0=α∆T

              Case-I:

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

               Case-II:

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

                 ⇒t100=0.950.2=475oC⇒t=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+α∆T)

                  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

 



Q 8 :

In the given circuit, the current in resistance R3 is             [2024]

  • 2.5 A

     

  • 1 A

     

  • 1.5 A

     

  • 2 A

     

(2)

Req=2Ω+2Ω+1Ω=5Ω

i=VReq=105=2A

Current in Resistance R3=2×(44+4)=2×48=1A

 



Q 9 :

A wire of resistance 160 Ω is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be      [2023]

  • 10 Ω

     

  • 640 Ω

     

  • 40 Ω

     

  • 16 Ω

     

(1)

Volume = Constant

A1L1=A2L2

A1L=A2L4

4A1=A2

R1=ρL1A1,   R2=ρL2A2

R2R1=L2A1A2L1=14A14A1L

R2=116R1=10 Ω



Q 10 :

The length of a metallic wire is increased by 20% and its area of cross section is reduced by 4%. The percentage change in resistance of the metallic wire is ______. [2023]



(25)

R=ρℓA be the initial resistance new resistance

R'=ρ1.2ℓ0.96A=1.25ρℓA=1.25R

Percentage change=1.25R-RR×100=25%



Q 11 :

A rectangular parallelopiped is measured as 1 cm×1 cm×100 cm. If its specific resistance is 3×10-7 Ωm, then the resistance between its two opposite rectangular faces will be _______×10-7 Ω.              [2023]



(3)

R=ρlA=3×10-7×(1×10-2)100×1×10-4=3×10-7Ω



Q 12 :

The current flowing through a conductor connected across a source is 2 A and 1.2 A at 0°C and 100°C respectively. The current flowing through the conductor at 50°C will be _______ ×102 mA.            [2023]



(15)

i0R0=i100R100  [For same source]

⇒2R0=1.2R0[1+100α]                            ...(i)

⇒1+100α=53

⇒50α=13

∴  i50R50=i0R0

⇒i50=i0R0R50=2×R0R0(1+50α)=21+13=1.5 A

⇒i50=15×102 mA



Q 13 :

A 6 volt battery is connected to the terminals of a three metre long wire of uniform thickness and resistance of 100 ohm. The difference of potential between two points on the wire separated by a distance of 50 cm will be

  • 2 volt

     

  • 3 volt

     

  • 1 volt

     

  • 1.5 volt

     

(3)

VAB=E

∴    Potential gradient K=VABL=EL

∴     Potential difference across length ℓ is V=Kℓ