Hem V

1. Differentiate between resultant and equilibrant forces.

2. Two forces of magnitude [tex]8 N[/tex] and [tex]6 N[/tex] act perpendicularly to each other. Calculate the magnitude of their resultant.

3. Distinguish between the electromotive force (e.m.f.) of a cell and the potential difference across the same cell.

4. Two identical cells, each with an e.m.f. of [tex]1.5 V[/tex], are connected in parallel with an internal resistance of [tex]r[/tex]. If a current of [tex]0.6 A[/tex] passes through a [tex]2.0 \Omega[/tex] resistor connected across the cells, calculate the value of [tex]r[/tex].

5. Distinguish between reflection and refraction of light.

6. A ray of light incident at an angle of [tex]30^{\circ}[/tex] to the normal is deviated through an angle of [tex]10.35^{\circ}[/tex] in a rectangular glass prism. Calculate the refractive index of the glass.



Answer :

Sure! Let's break down the problems provided step by step:

### Resultant Force vs. Equilibrant Force

Resultant Force:
- The resultant force is the single force which represents the combined effect of two or more forces acting on a body.
- If multiple forces act on a body, the body behaves as if a single force, called the resultant force, is acting on it. This force can be calculated using vector addition.

Equilibrant Force:
- The equilibrant force is the force which, if applied to a system of forces, brings the system into equilibrium.
- The equilibrant force is equal in magnitude and opposite in direction to the resultant force. Essentially, it negates the effect of the resultant force and stops any acceleration.

Example Problem:
Given two forces of magnitude 8 N and 6 N acting perpendicularly to each other:
- The magnitude of the resultant force \( R \) can be calculated using the Pythagorean theorem:
[tex]\[ R = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ N} \][/tex]

### Electromotive Force vs. Potential Difference

Electromotive Force (e.m.f.):
- The electromotive force (e.m.f) of a cell is the energy provided by the cell per coulomb of charge passing through it.
- It is the total energy provided by the cell per charge that is not dependent on the resistance in the circuit.
- e.m.f. is the open-circuit voltage when no current is flowing.

Potential Difference:
- The potential difference across a cell is the energy given to the charges (or the work done by the charges) as they move through a circuit.
- The potential difference accounts for the drop in voltage due to the resistance within the circuit.
- It is measured when the cell is connected to a circuit and current flows through the internal resistance.

### Calculating Internal Resistance in Parallel Connection

Given:
- Each cell has an e.m.f of 15 V.
- Internal resistance of each cell is \( r \).
- A total current of 0.6 A passes through a 2.0 Ω resistor.

Cells are connected in parallel, so:
- The total effective e.m.f remains 15 V because it is common for both cells.
- The internal resistance of two identical resistors in parallel is given by \( \frac{r}{2} \).

Using Ohm’s Law:
- The total voltage \( V \) across the external resistor and the effective internal resistance is given by:
[tex]\[ V = I ( R + \frac{r}{2} ) \][/tex]
where \( V = 15 \) V, \( I = 0.6 \) A, and \( R = 2.0 \) Ω.

Now, solving for the internal resistance \( r \):
- Set up the equation:
[tex]\[ 15 = 0.6 \times (2.0 + \frac{r}{2}) \][/tex]
- Simplify:
[tex]\[ 25 = 2.0 + \frac{r}{2} \][/tex]
- Isolate \( r \):
[tex]\[ 23 = \frac{r}{2} \][/tex]
[tex]\[ r = 46 \text{ Ω} \][/tex]

So the internal resistance of each cell is 46 Ω, and the effective internal resistance in parallel configuration would be:
[tex]\[ \frac{46}{2} = 23 \text{ Ω} \][/tex]

### Reflection vs. Refraction of Light

Reflection:
- Reflection occurs when light bounces off a surface. The angle of incidence is equal to the angle of reflection.
- There is no change in the medium during reflection.

Refraction:
- Refraction occurs when light passes from one medium to another, changing speed and direction.
- The amount of bending depends on the refractive indices of the two media.

Example Problem:
Given:
- A ray of light is incident at an angle of \( 30^\circ \) to the normal.
- It is deviated through an angle of \( 10.35^\circ \).
- The refractive index \( n \) of the glass is to be found.

Using Snell's Law:
[tex]\[ n = \frac{\sin(i)}{\sin(r)} \][/tex]
where \( i = 30^\circ \) and \( r = 30^\circ - 10.35^\circ = 19.65^\circ \).

[tex]\[ n = \frac{\sin(30^\circ)}{\sin(19.65^\circ)} = \frac{0.5}{0.3368} \approx 1.485 \][/tex]

Therefore, the refractive index of the glass is approximately 1.485.

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