Question

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

1. 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

1. state three conditions under which a laser will produce laser beam.

3 marks

1. 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

1. 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

1. state three uses of a diode.

3 marks

1. 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

Explanation:

1. Question 1 Solution:

• Given \(\frac{KL}{A}=M^{x}L^{y}T^{z}\), where \(K\) has dimensions of force per unit length (\(\frac{F}{L}\)), \(L\) is length, and \(A\) is area.
• The dimension of force \(F = MLT^{- 2}\), so the dimension of \(K=\frac{MLT^{-2}}{L}=MT^{-2}\).
• Substitute the dimensions into the left - hand side of the equation: \(\frac{KL}{A}=\frac{MT^{-2}\times L}{L^{2}}=ML^{-1}T^{-2}\).
• Comparing with \(M^{x}L^{y}T^{z}\), we have \(x = 1\), \(y=-1\), \(z = - 2\).

1. Question 2 Solution:

• (a): The velocity of a projectile is momentarily at rest at the maximum - height point.
• (b): The range \(R=\frac{u^{2}\sin2\theta}{g}\), and the range is maximum when \(\sin2\theta = 1\), so \(2\theta=90^{\circ}\), then \(\theta = 45^{\circ}\).
• (c): The vertical component of the initial velocity is \(u_y=u\sin\theta\). At the maximum height, the vertical velocity \(v_y = 0\). Using the equation \(v = u+at\) (where \(v = 0\), \(u = u_y\), \(a=-g\)), we get \(0 = u\sin\theta−gt\), so \(t=\frac{u\sin\theta}{g}\).

1. Question 3 Solution:

• Three conditions for a laser to produce a laser beam are:
• Population inversion: There must be more atoms in the excited state than in the ground state.
• Resonant cavity: A pair of mirrors (one partially - transmitting) to provide feedback and enhance the stimulated emission.
• Stimulated emission: The process by which an excited atom emits a photon that has the same frequency, phase, and direction as the incident photon.

1. Question 4 Solution:

• Three forces acting on a spherical ball floating through a viscous fluid are:
• Gravitational force \(F_g=mg\) (downward), where \(m\) is the mass of the ball and \(g\) is the acceleration due to gravity.
• Buoyant force \(F_b=

ho_{fluid}gV\) (upward), where \(
ho_{fluid}\) is the density of the fluid and \(V\) is the volume of the ball.

• Viscous drag force \(F_d = 6\pi\eta rv\) (opposite to the direction of motion), where \(\eta\) is the dynamic viscosity of the fluid, \(r\) is the radius of the ball, and \(v\) is the velocity of the ball.

1. Question 5 Solution:

• The maximum potential energy of a simple pendulum is given by \(U = mgh\).
• The height \(h = L(1 - \cos\theta)\), where \(L = 100\ cm=1\ m\), \(\theta = 10^{\circ}\), and \(m = 60\ g = 0.06\ kg\), \(g = 10\ m/s^{2}\).
• \(\cos10^{\circ}\approx0.985\), so \(h = 1\times(1 - 0.985)=0.015\ m\).
• \(U=mgh=0.06\times10\times0.015 = 9\times10^{-3}\ J\).

1. Question 6 Solution:

• Three uses of a diode are:
• Rectification: Converting alternating current (AC) to direct current (DC).
• Signal modulation and demodulation: In communication systems for processing signals.
• Voltage - regulation: In voltage - regulator circuits to maintain a constant output voltage.

1. Question 7 Solution:

• Given \(K = 750\ N/m\), \(L = 200\ cm = 2\ m\), \(A=5\ mm^{2}=5\times10^{-6}\ m^{2}\).
• The Young's modulus \(Y=\frac{KL}{A}\).
• Substitute the values: \(Y=\frac{750\times2}{5\times10^{-6}}=3\times10^{8}\ Pa\).

Answer:

1. \(x = 1\), \(y=-1\), \(z = - 2\)
2. (a): Maximum - height point; (b): \(45^{\circ}\); (c): \(t=\frac{u\sin\theta}{g}\)
3. Population inversion, resonant cavity, stimulated emission
4. Gravitational force, buoyant force, viscous drag force
5. \(9\times10^{-3}\ J\)
6. Rectification, signal modulation and demodulation, voltage - regulation
7. \(3\times10^{8}\ Pa\)