DC Pandey Test: Current Electricity - JEE MCQ

J= {J0[(x/R)-1] 0 ≤ x ≤ R/2,     {J0(x/R) R/2 ≤ x ≤ R
 We know that,
J=current/area
⇒ current=J Area
Current= J1A1 +J2A2
=J0{(x/R)−1}x(2πxdx)+J0 (x/R)(2πxdx)
=∫0R/2J0{(x/R)-1} x (2πxdx) + =∫0R/2J0 x 2π (x2/R)dx
= J0×2π∫0R/2 {(x2/R)−x}dx + (J0×2π/R) ∫RR/2 x2dx
⇒ i=J0×2π [(x3/3R)-(x2/2)] R/2 + (J0×2π/R) [x3/3]R2/R
⇒i=J0×2π [(R2/24)−(R2/8)]+ (J0×2π/R)[(R3/3)-(R3/24)]
=J0×2π(−2R2/24)(J0×2π/R)[7R3/24]
i =(5/12)πJ0R2
 

The current through the circuit before the battery of emf E2​ is short circuited, is
i_1​=(E₁​+E₂)/r₁​+r₂​+R​​             ...(i)
After short circuiting the battery of emf E_2​, current through the resistance R would be
i2​= E₁/R+r₁​ ​​            ...(ii)
Since, from equation (i) and (ii), we get
i2​>i1​
∴ E₁/R+r₁​ ​​> E₁​+E₂​​/R+r_1​+r₂​
This gives E₁​r₂​>E₂​(R+r₁​)

Wire of resistance =1Ω
Keeping in view that, ABCD is a square where each side is uniform wire of resistance 1Ω. If a uniform wire of resistance 1Ω is connected across AE and a potential difference is applied across A and C, the points B and E will be equipotential; a point E on CD is CE​/ ED =√2/1
∴    CE:ED= √2​:1

A battery is of emf E is being charged from a charger and hence current inside the cell is from anode to cathode 
I=V−E/r​
V=E+Ir
Therefore, when a cell is being charged the potential difference across its terminals is greater than emf of a cell. Also in charging the positive terminal is connected to anode of the cell and negative terminal to cathode.
Here, positive terminal of the battery is connected to terminal A of charger and negative terminal of the battery is connected to terminal B of charger.
Hence, Potential difference across points A and B must be more than E, A must be at higher potential than B and In battery, current flows from positive terminal to the negative terminal
 

Resistance, resistivity and drift velocity varies with relaxation time which is dependent on temperature.
Number of free electrons in a conductor remains invariant even if its temperature changes.
Hence correct option is option D.

The answer provided in the forum is wrong,the correct answer would be option A,B,C,D.
Solution:
Current is constant across the cross-section.
E ∝1/A ;     r∝1/A
=>i2r∝1/A
∴heat at Q>heat a P.
∴A,B,C,D are correct.

Drift speed and current density are inversely proportional to the area of cross-section of a wire. Thus, they are dependent on the cross-section. The charge crossing in a given time interval is independent of the area of cross-section of the wire. Free electron density is the total number of free electrons per unit volume of the wire. The density of free electrons depends on the distribution of the free electrons throughout the volume of the wire. It does not depend on the cross-section of the wire.

When a resistor is added parallel to other resistor in circuit it results in the reduction of overall resistance, since there are multiple pathways by which charge can flow. adding another resistor in a separate branch provides another pathway by which to direct charge through the main area of resistance within the circuit. This decreased resistance resulting from increasing the number of branches will have the effect of increasing the rate at which charge flows i.e. the current.

R_AD=1/(1/10)+(1/3+4+3)=5Ω
R_CB=1/(1/10)+(1/3+5+2)= 5Ω
i=28/5+5+4=2A
current through 5Ω is 2A
R_CADB=R_CB
⇒i_CA=i_CB=i/2=1A
V_CA=1x3=3V
V_CB=1x10=10V
V_AB=VCB-VCA=10-3
V_A-V_B=7V

The equivalent resistance of the group of resistances is R.
case 1:
a resistance r(say) is connected in parallel to the group, its new equivalent resistance R₁ is
R₁​=Rr/R+r​
hence, R₁​<R.
case 2:
a resistance r is connected in series to the group, its new equivalent resistance R₂ is
R₂​=R+r
hence, R₂​>R.

The power dissipated in circuit is given by
P= V²/R​
For the resistances in parallel the equivalent resistance is less than smallest resistance in combination of resistors hence power dissipation is maximum in this case. 
For the resistances in series the equivalent resistance is addition of resistors hence power dissipation is minimum in this case.
Hence, to avoid maximum power dissipation in parallel circuit R must be equal to zero so the equivalent of parallel resistors will be
 
R_p​=(0)(8)/8+0​=0Ω 
Since, R_p​=0Ω voltage across it is zero.
Therefore, power dissipated in 2Ω resistor is
P=V²​/R
P= (12)²​/2 
P=72W
 

The electric field does not exist inside the current carrying conductor. Hence, statement 1 is false.
When the wires are in series their cross sectional area is same hence current density will be equal i.e.current through the wires is equal. Thus, statement 2 is correct

Statement-1: Potential difference across the terminals of a battery is always less than its emf.

• This statement is true. The potential difference across the terminals of a battery (V) is given by V=emf−Ir,  where III is the current and r is the internal resistance of the battery. Because Ir is always positive when there is a current, the terminal potential difference is always less than the emf.

Statement-2: A battery always has some internal resistance.

• This statement is true. All real batteries have some internal resistance, which is inherent in the materials and construction of the battery.

Since the internal resistance of the battery causes a voltage drop (Ir) when current flows, this is why the potential difference across the terminals is less than the emf, making Statement-2 a correct explanation for Statement-1. Thus, the correct answer is: Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

When the battery is supplying power, inside the battery positive charge moves opposite to the electric field. So work done by electrostatic forces is negative.

Hence, option D is the correct answer.
 

On increase in temperature, the number of free electrons increases. But also the collisions will increase. Hence conductivity decreases.

Given that the amount of current passing is 0.5 A. In 10 s time the amount of charge is,
Q=I×t=0.5×10=5 C
Therefore, the number of Cu²⁺ ions are,
According to quantization,
Q=nq; Where q is the charge of each Cu²⁺
Therefore,
n=Q/q=5/2e=5/(2×1.6×10⁻¹⁹)=5×10¹⁹/3.2
⇒n=1.56×10¹⁹

A wire has a length =2.0m
resistance =5Ω
Electric field existing inside the wire I=10A
V=I×R
    =10×5
    =50V
∴   E=50V/2​=25V/m

Applying kvl 
first in loop ABCDE
−3i−6(i−i_1​) + 4.5=0.                    -------(1)
then in loop ABCMNE
−3i−10i₁​−3+4.5=0.                     --------(2)
From equation (1) and (2) WE GET,
i₁​=0
hence current passing through 10-ohm resistor is ZERO