Q 1 :

The electric current through a wire varies with time as I=I0+βt where I0=20A and β=3A/s. The amount of electric charge crossed through a section of the wire in 20 sec is                           [2024]

  • 80 C

     

  • 1000 C

     

  • 800 C

     

  • 1600 C

     

(2)       

           Given that, current I=I0+βt

            Also, I0=20A and β=3A/s

            Hence, dq/dt=20+3tdq=(20+3t)dt

            0qdq=020(20+3t)dt

            q=[20t+3t22]020=1000C

 



Q 2 :

The current in a conductor is expressed as I=3t2+4t3, where I is in Ampere and t is in second. The amount of electric charge that flows through a section of the conductor during t=1s to t=2s is _________ C.               [2024]



(22)       I=3t2+4t3

              dQ=Idt

              Q=12(3t2+4t3)dt

                   =[t3+t4]12

                    =(8+16)-(2)=22C

 



Q 3 :

Which of the following resistivity (ρ) v/s temperature (T) curve is most suitable to be used in wire bound standard resistors?          [2025]

  •  

  •  

  •  

  •  

(1)

For bound standard resistors the resistivity is independent of temperature and remain nearly constant.



Q 4 :

The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have          [2025]

  • low thermal conductivity and low electrical conductivity

     

  • high thermal conductivity and high electrical conductivity

     

  • low thermal conductivity and high electrical conductivity

     

  • high thermal conductivity and low electrical conductivity

     

(3)

Material should have low thermal conductivity and high electrical conductivity.



Q 5 :

Current passing through a wire as function of time is given as I(t) = 0.02 t + 0.01 A. The charge that will flow through the wire from t = 1s to t = 2s is:          [2025]

  • 0.06 C

     

  • 0.02 C

     

  • 0.07 C

     

  • 0.04 C

     

(4)

q=idt=12(0.02t+0.01)dt

q=[0.02t22+0.01t]12

     =0.022×(2212)+0.01×1=0.04 C



Q 6 :

A uniform metallic wire carries a current of 2 A, when a 3.4 V battery is connected across it. The mass of the uniform metallic wire is 8.92×10-3 kg, density is 8.92×103 kg/m3 and resistivity is 1.7×10-8Ω-m. The length of the wire is              [2023]

  • l=10 m

     

  • l=100 m

     

  • l=5 m

     

  • l=6.8 m

     

(1)

mass=8.92×10-3 kg

Density, d=8.92×103 kg/m3

ρ=1.7×10-8 Ω-m

R=Vi=3.42=1.7 Ω

R=ρlA=ρl2vol=ρl2dmass (d=massvol)

l=Rmρd=1.7×8.92×10-31.7×10-8×8.92×103=10 m



Q 7 :

The resistance of a wire is 5Ω. Its new resistance in ohm, if stretched to 5 times of its original length, will be              [2023]

  • 125

     

  • 5

     

  • 25

     

  • 625

     

(1)

 Volume of wire is constant in stretching

      Vi=Vf

Aili=Aflf

Al=A'(5l)

A'=A5

Rf=ρlfAf=ρ(5l)(A5)=25(ρlA)=25×5=125 Ω



Q 8 :

The charge flowing in a conductor changes with time as Q(t)=αt-βt2+γt3, where α, β and γ are constants. Minimum value of current is            [2023]

  • α-3β2γ

     

  • α-γ23β

     

  • β-α23γ

     

  • α-β23γ

     

(4)

Q=(αt-βt2+γt3)

i=dQdt=(α-2βt+3γt2)

didt=(3γt-2β)=0t=β3γ

i=(α-2βt+3γt2)=(α-β23γ)



Q 9 :

The drift velocity of electrons for a conductor connected in an electrical circuit is Vd. The conductor is now replaced by another conductor with the same material and the same length but double the area of cross-section. The applied voltage remains the same. The new drift velocity of electrons will be            [2023]

  • Vd

     

  • Vd2

     

  • Vd4

     

  • 2Vd

     

(1)

i=nAVde

i1=(VR) and i2=(2VR)

So,  i1i2=12=(AVd)1(AVd)2=Vd(Vd)2×(12)

12×Vd(Vd)2=12(Vd)2=Vd



Q 10 :

In a metallic conductor, under the effect of applied electric field, the free electrons of the conductor           [2023]

  • drift from higher potential to lower potential.

     

  • move in the curved paths from lower potential to higher potential.

     

  • move with the uniform velocity throughout from lower potential to higher potential.

     

  • move in the straight line paths in the same direction.

     

(2)

Move in curved path

I=neAVd



Q 11 :

A hollow cylindrical conductor has length of 3.14 m, while its inner and outer diameters are 4 mm and 8 mm respectively. The resistance of the conductor is n×10-3Ω. If the resistivity of the material is 2.4×10-8Ω m, the value of n is ________.                 [2023]



(2)

R=ρlA, the cross-sectional area is π(b2-a2)

R=ρlπ(b2-a2)=2.4×10-8×3.143.14×(42-22)×10-6=2×10-3 Ω

n=2



Q 12 :

If a copper wire is stretched to increase its length by 20%, the percentage increase in resistance of the wire is ______ %.           [2023]



(44)

As volume is constant.

So resistance(length)2

 % change in resistance=20+20+400100=44%



Q 13 :

A current of 2 A flows through a wire of cross-sectional area 25.0 mm2. The number of free electrons in a cubic meter are 2.0×1028. The drift velocity of the electrons is ____ ×10-6ms-1 (given, charge on electron =1.6×10-19 C).             [2023]



(25)

Drift velocity

vd=IneA=22×1028×1.6×10-19×25×10-6

                    =25×10-6 ms-1



Q 14 :

The number density of free electrons in copper is nearly 8×1028 m-3. A copper wire has its area of cross section =2×10-6 m2 and is carrying a current of 3.2 A. The drift speed of the electrons is _______ ×10-6 ms-1.              [2023]



(125)

n=8×1028 m-3,  Area=2×10-6 m2,  I=3.2 A

I=neAvd

vd=IneA=125×10-6 m/s



Q 15 :

A cylindrical conductor of length 2 m and area of cross-section 0.2 mm2 carries an electric current of 1.6 A when its ends are connected to a 2V battery. Mobility of electrons in the conductor is α×10-3 m2/V.s. The value of α is :       

(electron concentration=5×1028/ m3 and electron charge=1.6×10-19 C)        [2026]



(1)

Vd=μE=μV

I=neAVd

Vd=IneA

μ=IlNneA

μ=1.6×32×5×1026×1.6×10-19×2×10-7

μ=1×10-3 m2/V s

α=1