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+3t⇒dq=(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×(22–12)+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 1.7×10-8 Ω-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 5 Ω. 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)=0⇒t=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×10-3 Ω. If the resistivity of the material is 2.4×10-8 Ω 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 2 V 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=μVℓ

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



Q 16 :

A potential difference is applied across the ends of a metallic wire. If the potential difference is doubled, then the drift velocity:

  • will be doubled

     

  • will be halved

     

  • will be quadrupled

     

  • will remain unchanged

     

(1)

Vd=μE=μVd⇒Vd∝V    If V (potential difference)



Q 17 :

The current i in the network is _______ ×10-1A



(3)

Both the diodes are reverse biased, so, there is no flow of current through 5Ω and 20Ω resistances. Now, two resistors of 10Ω and two resistors of 5Ω are in series.

Hence, current I through the network = 0.3A.