Correct Answer: A. drawee

Correct Answer: A. a rod of almond

Correct Answer: D. tenure of government is always not stable

Explanation

In a cabinet system of government, one of the low points is that the tenure of government is not always stable. This is because the government is dependent on the confidence of the parliament, and a vote of no confidence can lead to the dissolution of the government.

Correct Answer: B. best

Explanation

(a) Range = \(\frac{U^2 sin 2\theta}{g}\)

for first attempy, Range = \(\frac{8^2 \sin 60}{10} = \frac{64 x\times 0.8660}{10}\)

= 5.5424m

for second attemt; Range = \(\frac{8^2 \sin 90}{10} = \frac{64 \times 1}{10}\)

= 6.4m

gain 6.40 - 5.54 = 0.8576m

(b)(i) Hooke's law states that provided the limit of proportionality is not exceeded, the force is proportional to the extension in an elastic material.

(ii) Description: Suspend the spiral spring from a rigid support with a pointer attached at the free end. Suspend a stretching weight from the free end, and note the position of the pointer. Suspend a known mass, note the new position and determine the extension. Repeat the experiment using various known masses. Repeat the above procedure while unloading and determine mean extension. Plot a graph of load against mean extension. A straight line graph through the origin verifies Hooke's law.

Precautions: (i) Loading beyond elastic limit should be avoided

(ii) Pointer to be rigidly attached to the free end and made perpendicular to the spring.

(iii) Avoid parallax error while taking readings.

(c) mg = Ke, m = total mass If m = mass of scale pan then;

(m + 40)g = 1.8k-----------(i)

(m + 60) = 2.05k --------- (ii)

Dividing (i) by (ii) \(\frac{m + 40}{m + 60} = \frac{1.8}{2.05}\)

2.05(m + 40) = 1.8 (m + 60)

M = \(\frac{26}{0.25}\) = 104gm

Correct Answer: A. N1.50

Correct Answer: B. segmentation

Correct Answer: D. payments to foreign shipping companies

Correct Answer: C. outrageous

Correct Answer: D) 150 cm³

Explanation

\(\begin{array}{llllcr}

& \mathrm{2NO} & + & \mathrm{O_2} & \longrightarrow & \mathrm{2NO_2} \\

\text{Initially} & \mathrm{150cm^3} & & \mathrm{100cm^3} & & \\

\text{Gay Lussac's Law} & \text{2 mols} & & \text{1 mol} & & \text{2 mols} \\

\text{Vol. used up for reac.} & \mathrm{150cm^3} & & \mathrm{75cm^3} & & \mathrm{150cm^3} \\

\end{array}\)

Vol. of oxygen left after reac. \(= \mathrm{100cm^3} - \mathrm{75cm^3} = \mathrm{25cm^3}\)

Vol. of nitrogen (IV) oxide prod. = \(\mathrm{150cm^3}\)

Note: According to Gay Lussac's Law, 2 volumes of nitrogen II oxide combine with 1 volume of oxygen to form 2 volumes of nitrogen IV oxide.