2003 - JAMB Physics Past Questions and Answers - page 5

Work done by friction = Friction force x Displacement from the relation, f = uR

where u = coefficient of friction; R= Normal reaction

F = u mg (R =mg)

= 0.1 x 0.01 x 10

work done = F x Displacement

= 0.1 x 0.01 x 10 x 0.2

= 0.002 or ( 2 \times 10^{-3}J )

For the equilateral glass prism A = 60⁰

( \frac{3}{2} \text{Sin} \frac{\left( \frac{Dm + 60}{2}\right)}{\left(\frac{60}{2}\right)}= \frac{\left( \text{ Sin }\frac{Dm + 60}{2} \right)}{0.5} \

\implies Dm = 37.2^{\circ} )

From faraday's law of electrolysis

M = I t Z

0.990 = I x (40 x 60) x 3.3 x 10^-4

( \implies \frac{0.990}{2400 \times 3.3 \times 10^{-4}} = 1.25A \

\text{Thus the correction to be made } = 1.25 - 1.20 = 0.05A )

Atomic emission and absorption of energy is usually in packets referred to as photon or quantum of energy of magnitude E =hf, where f is the frequency and h is the plancks constant

Energy emitted = hf

= 6.6 x 10^-34 x 8.0 x 10^-14

= 5.28 x 10^-19J

From Newton's second law:

( F = \frac{mV - mµ}{t} )

Let the final velocity = V; and since the football is initially at rest, its initial vel µ = 0, M = 0.8kg

( t = 0.8s \

\implies 100 = \frac{0.8V - 0.8 \times 0}{0.8} \

\text{Therefore } V = \frac{100 \times 0.8}{0.8} = 100ms^{-1} )

The speed of sound in a material is given as:

( V = \sqrt{\frac{E}{P}}, Where\

E = \text{Young's modulus} \

P = \text{Density }\

\text{Therefore } V = \sqrt{\frac{7.0 \times 10^{10}}{2.7 \times 10^3}} = 5091.75\

= 5.1 \times 10^3MS^{-1} )

From the uncertainty principle the product of the position and momentum of a sub-atomic particle is equal or greater than planck's constant.

i.e ( \Delta x. \Delta p \geq h. \

\text{Thus} 5 \times 10^{-10} \times \Delta p = 6.6 \times 10^{-34} \

\Delta p = \frac{6.6 \times 10^{-34}}{5 \times 10^{-10}} = 1.32 \times 10^{-24}Ns\

\text{Therefore Uncertainty in the momentum } = 1.32 \times 10^{-24}Ns )

In general, the energy released during nuclear fission is given by the Einstein's mass-energy equation, given by

( E = \Delta MC^2 )

Where ( \Delta m) is the mass defect, and C is the speed of light.

Therefore

( \Delta m = 0.01% \text{of} 1.0g = \frac{0.01}{100}

\times 1.0g \

= 1.0 \times 10^{-4} \

= 1.0 \times 10^{-7}kg\

\text{Energy Released } = \Delta MC^2 \

= 1.0 \times 10^{-7} \times (3.0 \times 10^8)^2\

= 1.0 \times 10^{-2} \times 9.0 \times 10^{16} \

= 9.0 \times 10^9 J )