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seer; /ˈsiə,/ spear; /spɪə/

For words ending with ''ical'' the primary stress is placed on the syllable immediately before “-ic-. po-LIT-i-cal

In pronouncing the word satisfactory, the third syllable is more pronounced than the rest. Hence we have sa-tis-FAC-to-ry

This is stressed as cap-TI-vi-ty

In responding to questions on emphatic stress, Interrogative sentence that contradicts the emphasized word is the correct option. This means that, the statement that does not have the emphasized word in it is the correct answer.

In responding to questions on emphatic stress in examinations, interrogative sentences will be provided as options. The interrogative sentence that contradicts the emphasized word is the correct option. ''sailed'' is the emphasized word here. ''fly'' in option A contradicts the emphasized word

In responding to questions on emphatic stress in examinations, interrogative sentences will be provided as options. The interrogative sentence that contradicts the emphasized word is the correct option. LEATHER is the emphasized word here. ''polythene'' in option B contradicts the emphasized word

- Alpha radiation consists of helium- He\(^{4}_{2}\) 2 Alpha decay --> 2 X He\(^{4}_{2}\) = Element\(^{8}_{4}\) - Beta particles are high energy electrons- e\(^{0}_{-1}\) 3 Beta particles --> 3 X e\(^{0}_{-1}\) = Element\(^{0}_{-3}\) - Gamma rays are waves of electromagnetic energy, or photons. the mass and atomic number of the resulting element = U\(^{235}_{92}\) - Elements( \(^{8}_{4}\) + \(^{0}_{-3}\) ) --> Element\(^{227}_{91}\)

The container is filled 'only with ice' to the brim (no liquid water mentioned). Ice is less dense than water (it expands when freezing). When the ice melts, the same mass turns into water that occupies less volume. Result: The water level drops below the brim.

Dew point is the temperature at which the water vapour present in air is just sufficient to saturate it. Dew is formed when air is cooled below its dew point

The velocity of sound in air, V∝\(\sqrt{T}\), where T is the absolute temperature Thus V = K\(\sqrt{T}\) or K = \(\frac{\text{V}}{\sqrt{T}}\) V is independent of pressure and density at a fixed temperature. ∴ \(\frac{V_1}{\sqrt{T_1}}\) = \(\frac{V_2}{\sqrt{T_2}}\) T\(_1\) = 16 + 273 = 289K, V\(_1\) = 340ms\(^{-1}\) T\(_2\) = 127 + 273 = 400K, V\(_2\) = ? \(\frac{340}{\sqrt{289}}\) = \(\frac{V_2}{\sqrt{400}}\) ∴ V\(_2\) = \(\frac{340 \sqrt{400}}{\sqrt{289}}\) = 400m/s

1/R = 1/8 + 1/8 = 2/8 ∴ R = 4Ω l/l1 = R/R1 (x)/(100-x) = 4/8 4(100-x) = 8x 400-4x = 8x 400 = 8x + 4x = 12x x = 400/12 = 33.3cm

1/R = 1/4 + 1/4 + 1/4 = 3/4 R = 4/3 I = V/R = 164/3 = 16 x 3_4 = 12 A

No expanded working has been attached for this question yet. The verified answer is Option A.

No expanded working has been attached for this question yet. The verified answer is Option C.