answer eight questions in all, five questions from part i and three questions from part ii
part i
15 marks
answer any five questions from this part
all questions carry equal marks.
1. the youngs modulus of an elastic material is represented by the equation, $\frac{kl}{a}=m^{x}l^{y}t^{z}$ where k, l and a represent the force constant, length and area respectively. determine the values of x, y and z.
3 marks
2. fig. 1 illustrates a body projected at point, o, with an initial velocity, u, to the horizontal.
(a) at what point is the velocity of the projectile momentarily at rest?
(b) state the value of $\theta$ when the distance ob is maximum.
(c) write an equation for the time, t, taken by the projectile to reach maximum height.
3 marks
3. state three conditions under which a laser will produce laser beam.
3 marks
4. a spherical ball is dropped in a cylinder that contains viscous fluid. list three forces that act on the ball as it floats through the fluid.
3 marks
5. a simple pendulum bob is displaced at an angle of 10° and then released after which it undergoes oscillation on the vertical plane. the pendulum is 100 cm long and has a mass of 60 g. calculate the maximum potential energy attained by the bob in oscillation. g = 10 m s⁻²
3 marks
6. state three uses of a diode.
3 marks
7. an elastic material of force constant 750 nm⁻¹ has a length of 200 cm. if the cross - sectional area of the material is 5 mm², calculate the youngs modulus of the material.
3 marks

Question

Accepted Answer

### Question 1
# Explanation:
## Step1: Write dimensions of each quantity
Dimensions of $K$ (force constant) is $[MLT^{- 2}]$, $L$ is $[L]$, $A$ is $[L^{2}]$, and $M^{x}L^{y}T^{z}$ represents the right - hand side. The left - hand side $\frac{KL}{A}$ has dimensions $\frac{[MLT^{-2}][L]}{[L^{2}]}=[MT^{-2}]$.
## Step2: Equate dimensions
Equating $[MT^{-2}]$ with $[M^{x}L^{y}T^{z}]$, we get $x = 1,y=0,z = - 2$.
# Answer:
$x = 1,y = 0,z=-2$

### Question 2(a)
# Explanation:
The vertical component of velocity $v_y=u_y - gt$. At the maximum height of the projectile's motion, the vertical component of velocity is zero. Since the horizontal component of velocity $v_x$ remains constant throughout the motion, the velocity of the projectile is momentarily at rest at the maximum height.
# Answer:
At the maximum height of the projectile's motion.

### Question 2(b)
# Explanation:
The horizontal range $R=\frac{u^{2}\sin2\theta}{g}$. The range is maximum when $\sin2\theta = 1$, i.e., $2\theta=90^{\circ}$, so $\theta = 45^{\circ}$.
# Answer:
$\theta = 45^{\circ}$

### Question 2(c)
# Explanation:
At the maximum height, the vertical component of velocity $v_y = 0$. Using the equation $v_y=u_y - gt$, where $u_y = u\sin\theta$. Solving for $t$ gives $t=\frac{u\sin\theta}{g}$.
# Answer:
$t=\frac{u\sin\theta}{g}$

### Question 3
# Explanation:
1. Population inversion: There must be more atoms in the excited state than in the ground state.
2. Resonant cavity: A structure (like two parallel mirrors) to provide feedback and enhance stimulated emission.
3. Suitable gain medium: A material that can amplify light through stimulated emission.
# Answer:
1. Population inversion.
2. Resonant cavity.
3. Suitable gain medium.

### Question 4
# Explanation:
1. Gravitational force $F_g=mg$ acting downwards.
2. Buoyant force $F_b=
ho_{fluid}gV$ acting upwards.
3. Viscous drag force $F_d$ acting in the opposite direction of motion of the ball through the fluid.
# Answer:
1. Gravitational force.
2. Buoyant force.
3. Viscous drag force.

### Question 5
# Explanation:
## Step1: Calculate the height $h$
The length of the pendulum $L = 100\ cm=1\ m$, and the angle $\theta = 10^{\circ}$. The height $h = L(1 - \cos\theta)$. $\cos10^{\circ}\approx0.985$, so $h = 1\times(1 - 0.985)=0.015\ m$.
## Step2: Calculate potential energy
The mass $m = 60\ g = 0.06\ kg$, and using $U = mgh$ with $g = 10\ m/s^{2}$, we have $U=0.06\times10\times0.015 = 9\times10^{-3}\ J$.
# Answer:
$9\times 10^{-3}\ J$

### Question 6
# Explanation:
1. Rectification: Converting alternating current to direct current.
2. Signal detection: Detecting radio - frequency signals.
3. Light - emitting diodes (LEDs) are used for producing light.
# Answer:
1. Rectification.
2. Signal detection.
3. Light - emitting (in LEDs).

### Question 7
# Explanation:
## Step1: Identify the values
$K = 750\ N/m$, $L = 200\ cm = 2\ m$, $A=5\ mm^{2}=5\times10^{-6}\ m^{2}$.
## Step2: Calculate Young's modulus $Y$
Using $Y=\frac{KL}{A}$, we substitute the values: $Y=\frac{750\times2}{5\times10^{-6}}=3\times10^{8}\ Pa$.
# Answer:
$3\times 10^{8}\ Pa$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Write dimensions of each quantity

Step2: Equate dimensions

Answer:

Question 2(a)