Explanation

(a) Reasons for sparse population of the West Coast of South Africa

(i) Harsh climate. Temperatures of desert type ranging from 21°C to 40°C.

(ii) The influence of cold Benguela Current leading to low precipitation.

(iii) Few rivers to encourage irrigation.

(iv) Poor development of mineral resources.

(v)The presence of Kalahari and Namib deserts discourage farming and settlement.

(vi) Poor soil discourages farming to support population growth.

(vii) The area falls in the rain shadow areas of Drakensberg Mountains.

(viii) Political instability

(ix) Poor infrastructural development.

(x) Pests and diseases

(xi) Low commercial and industrial activities.

(b) Disadvantages of the low population density of the West Coast of South Africa

(i) Low population density discourages development

(ii) Inadequate labour to run the few industries

(iii) Shortage of food

(iv) Under-utilization of natural resources

(v) Low income/low standard of living

(vi) Low level of investments / industrialisation

(vii) Low revenue to government

(viii) Low production output

(ix) Low population hampers developmental planning.

(x) Deterioration of unused facilities.

(c) Ways of increasing the population of the West Coast of South Africa

(a) The desert areas should be irrigated for food production.

(b) Encourage the establishment of industries

(c) Provision of infrastructure.

(d) Government should revolutionize agriculture.

(e) Provision of social amenities.

(f) Creation of job opportunities

(g) Development of new towns

(h) Improved transportation.

(i) Use of fertilizers to enrich poor soil.

Correct Answer: B. supremacy of the law

Correct Answer: B. Decrease in productivity

Correct Answer: A. sucrose content is reduced

Correct Answer: D. Farming

Correct Answer: B)

CaCO\(_3\)(s) \(\iff\) CaO(s) + CO\(_2\)(g)

Explanation

When the pressure on a system is increased, the equilibrium shifts in the direction that has fewer moles of gas. This is because the side with fewer moles of gas will occupy a smaller volume, thereby reducing the pressure.

In this case, we need to identify the reaction that will have the equilibrium shift to the left when the pressure is increased:

Option A: C(s) + O\(_2\)(g) \(\iff\) CO\(_2\)(g)

In this reaction, there is 1 mole of gas on both sides, so increasing the pressure will not cause a shift in equilibrium.

Option B: CaCO\(_3\)(s) \(\iff\) CaO(s) + CO\(_2\)(g)(Correct)

In this reaction, there are no moles of gas on the left side and 1 mole of gas on the right side. Increasing the pressure will cause the equilibrium to shift to the left, which has fewer moles of gas.

Option C: 2Mg(s) + O\(_2\)(g) \(\iff\) 2MgO(s)

In this reaction, there is 1 mole of gas on the left side and no moles of gas on the right side. Increasing the pressure will cause the equilibrium to shift to the right, which has fewer moles of gas.

Option D: 2H\(_2\)(g) + O\(_2\)(g) \(\iff\) 2H\(_2\)O(g)

In this reaction, there are 3 moles of gas on the left side and 2 moles of gas on the right side. Increasing the pressure will cause the equilibrium to shift to the right, which has fewer moles of gas.

Therefore, the correct answer is Option B: CaCO\(_3\)(s) \(\iff\) CaO(s) + CO\(_2\)(g), as the equilibrium will shift to the left when the pressure is increased.

Correct Answer: D. 1960 and 1966

Correct Answer: B. 3.5cm

Explanation

Using V = \(\frac{3}{1} \pi r^2h\),

so, 38\(\frac{1}{2} = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 3\)

\(\frac{77}{2} = \frac{22}{7} \times r^2\)

r² = \(\frac{77 \times 7}{2 \times 22}\)

r² = \(\frac{49}{4}\)

Hence, r = \(\sqrt{\frac{49}{4}}\)

= 3\(\frac{1}{2}\)

Correct Answer: D. receptiveness

Correct Answer: A. an input device