NEET Relation between Current and Drift Velocity MCQs & PYQs

To find the drift velocity

$v_d$

of the electrons, we use the formula for current in terms of drift velocity:

$$I = n A e v_d$$

Where:

• 
  $I$ 

  is the current (1.2 A),

• 
  $n$ 

  is the number of free electrons per unit volume ( $5 \times 10^{28} \, \text{m}^{-3}$ 

  ),

• 
  $A$ 

  is the cross-sectional area of the conductor ( $10^{-4} \, \text{m}^2$ 

  ),

• 
  $e$ 

  is the charge of an electron ( $1.6 \times 10^{-19} \, \text{C}$ 

  ),

• 
  $v_d$ 

  is the drift velocity of the electrons (which we need to calculate).

Step 1: Rearranging the formula to solve for $v_d$

$$v_d = \frac{I}{n A e}$$

Step 2: Substituting the given values

$$v_d = \frac{1.2}{(5 \times 10^{28}) \times (10^{-4}) \times (1.6 \times 10^{-19})}$$

Step 3: Performing the calculation

$$v_d = \frac{1.2}{(5 \times 10^{28}) \times (10^{-4}) \times (1.6 \times 10^{-19})}$$

$$v_d = \frac{1.2}{8 \times 10^{6}}$$

$$v_d = 1.5 \times 10^{-7} \, \text{m/s}$$

Final Answer:

The drift velocity of the electrons is

$\boxed{1.5 \times 10^{-7} \, \text{m/s}}$

.

When a wire is stretched, its length increases, and its cross-sectional area decreases. The resistance of a wire is given by the formula:

$$R = \rho \frac{L}{A}$$

Where:

• 
  $R$ 

  is the resistance,

• 
  $\rho$ 

  is the resistivity of the material (constant),

• 
  $L$ 

  is the length of the wire,

• 
  $A$ 

  is the cross-sectional area of the wire.

When the wire is stretched:

1. Length increases by 10%: The new length
   $L'$ 

   is given by: 

   $$L' = L + 0.1L = 1.1L$$ 

2. Volume remains constant: The volume of the wire before and after stretching remains the same. Volume is the product of length and area: 

   $$\text{Volume before} = L \times A$$ 

   $$\text{Volume after} = L' \times A' = 1.1L \times A'$$Since the volume remains constant:

    

   $$L \times A = 1.1L \times A'$$Solving for

   $A'$, the new cross-sectional area:

    

   $$A' = \frac{A}{1.1}$$ 

Step 2: New Resistance

The new resistance

$R'$

of the stretched wire is given by:

$$R' = \rho \frac{L'}{A'} = \rho \frac{1.1L}{\frac{A}{1.1}} = \rho \frac{1.1L \times 1.1}{A} = \rho \frac{1.21L}{A}$$

So, the new resistance

$R'$

is 1.21 times the original resistance

$R$

.

Step 3: Conclusion

The resistance increases by 21% when the wire is stretched by 10%.

Specific resistance is property of substance it doesn't depend on any other physical factor

Let's break this down step by step.

1. Resistances connected in parallel

When

$n$

resistors, each of resistance

$r$

, are connected in parallel, the total or equivalent resistance

$R_{\text{parallel}}$

is given by the formula:

$$\frac{1}{R_{\text{parallel}}} = \frac{1}{r} + \frac{1}{r} + \cdots + \frac{1}{r} \quad \text{(n terms)}$$

This simplifies to:

$$\frac{1}{R_{\text{parallel}}} = \frac{n}{r}$$

So, the equivalent resistance is:

$$R_{\text{parallel}} = \frac{r}{n}$$

We're told that the resultant resistance when connected in parallel is

$x$

, so:

$$x = \frac{r}{n}$$

2. Resistances connected in series

When the same

$n$

resistors, each of resistance

$r$

, are connected in series, the total or equivalent resistance

$R_{\text{series}}$

is simply the sum of the individual resistances:

$$R_{\text{series}} = r + r + \cdots + r \quad \text{(n terms)}$$

So:

$$R_{\text{series}} = n \times r$$

3. Relating the two scenarios

From the parallel connection, we know that:

$$x = \frac{r}{n}$$

Thus, the resistance when the resistors are connected in series is:

$$R_{\text{series}} = n \times r = n \times \left( x \times n \right) = r \times n \times x$$

Therefore, the resistance when these

$n$

resistors are connected in series is:

$$\boxed{r \times n \times x}$$

The resistance

$R$

of a wire is given by the formula:

$$R = \rho \frac{l}{A}$$

Where:

• 
  $\rho$ 

  is the resistivity of the material (which is constant for copper),

• 
  $l$ 

  is the length of the wire,

• 
  $A$ 

  is the cross-sectional area of the wire.

Given:

• Wire 1: Length =
  $l$ 

  , Area = $A$ 

• Wire 2: Length =
  $2l$ 

  , Area = $\frac{A}{2}$ 

• Wire 3: Length =
  $\frac{l}{2}$ 

  , Area = $2A$ 

We will calculate the resistance for each wire.

Step 1: Calculate the resistance for each wire

Wire 1: Length = $l$

, Area = $A$

$$R_1 = \rho \frac{l}{A}$$

Wire 2: Length = $2l$

, Area = $\frac{A}{2}$

$$R_2 = \rho \frac{2l}{\frac{A}{2}} = \rho \frac{2l \times 2}{A} = \rho \frac{4l}{A}$$

Wire 3: Length = $\frac{l}{2}$

, Area = $2A$

$$R_3 = \rho \frac{\frac{l}{2}}{2A} = \rho \frac{l}{4A}$$

Step 2: Compare the resistances

• 
  $R_1 = \rho \frac{l}{A}$ 
• 
  $R_2 = \rho \frac{4l}{A}$ 
• 
  $R_3 = \rho \frac{l}{4A}$ 

Clearly,

$R_3$

is the smallest resistance because

$\frac{l}{4A}$

is the smallest fraction compared to

$\frac{l}{A}$

and

$\frac{4l}{A}$

.

Step 3: Conclusion

The wire with the minimum resistance is Wire 3, which has a cross-sectional area of

$2A$

.

Thus, the answer is Wire 3.

The current I is calculated by:

Thus, the correct answer is 0.8 A (Option 4)

Since \(I = n e A v_d = n e \pi r^2 v\), the new current is \(I' = n e \pi \left(\frac{r}{2}\right)^2 (2v) = \frac{1}{2} n e \pi r^2 v = I/2\).

Assertion (A) is true because \( I = JA \) and \( E = \rho J \) imply \( E = \frac{\rho I}{A} \). Reason (R) is true as it's the microscopic form of Ohm's law. Reason correctly explains assertion.