Question Library

Given Data: u = 150 ms\(^{-1.}\), Ø = 30^\circ Time of Flight = \(\frac{2usinØ}{g}\) → \(\frac{2 * 150 * sin30^\circ}{10}\) = 15s t = time taken to reach maximum height = ? t = usin\theta \divg t = [150 \times sin(30)] \div10 t = 7.5secs.

A photoresistor or light dependent resistor is an electronic component that is sensitive to light. When light falls upon it, then the resistance changes. Used where there is a need to sense the presence and absence of light is necessary. These resistors are used as light sensors and the applications of LDR mainly include alarm clocks, street lights, light intensity meters, burglar alarm circuits.

(a) The circuit is half-wave rectifier. (b)

In a gas, the distance between molecules, whether monatomic or polyatomic, is very large compared with the size of the molecules; thus gases have a low density and are highly compressible. Cohesive Force is negligible in case of gases.

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

The dissolution of sugar in tea is accompanied by positive change in entropy. When there is a change in the state of a substance from solid to liquid, there is an increase in entropy. In general, if a substance changes its structure from a condensed state to a less condensed state, then we say there is an increase in entropy.

Metal oxides which react with both acids as well as bases to produce salts and water are known as amphoteric oxides. Many metals (such as zinc, tin, lead, aluminium, and beryllium) form amphoteric oxides or hydroxides.

Percentage birthrate = \(\frac{62}{6250} \times 100%\) = \(\frac{6200}{6250}\) = 0.992% \(\approx\) 1.0%.

1¼ ÷ [ 2 ÷ ¼] of 28 Apply BODMAS rules \( \frac{5}{4}\) ÷ [ 2 ÷ ¼ × 28 ] \( \frac{5}{4} ÷ \frac{2}{7} \) \( \frac{5}{4} × \frac{7}{2} \) \( \frac{35}{8}\) = \(4 \frac{3}{8}\)

\( x^2 + 9x + 20 \) Find the two numbers whose product is 20 and its sum is 9} \( 5x \times 4x = 20x^2\) \(5x + 4x = 9x\) \((x^2 + 5x) + (4x + 20)\) \( x(x + 5)+ 4(x + 5) = (x + 5)(x + 4)\)

Total of their ages: 18 + 20 + 22 = 60 The second staff will get \(\frac{20}{60} \times 120,000\) = N40,000.00K

n(X ∪ Y) = n(X) + n(Y) − n(X ∩ Y) = 15 + 12 − 7 ∴ n(X ∪ Y) = 20

Starting from 3x − z \(\leq\) 9 x = 0, substitute the value of x (i.e. x = 0) into the equation 3x − z \(\leq\) 9 3(0) − z \(\leq\) 9 z \(\leq\) − 9 If z = 0 Then, 3x − 0 \(\leq\) 9 3x \(\leq\) 9 x \(\leq\) 3 Intercept of x and z i.e (x,z) = (3, -9)

log101.5 + 3 log102 − log100.3 log101.5 + log1023 − log100.3 [log10(1.5 × 23) ÷ log100.3] [log10(15/10 × 8 ) ÷ log10( 3/10)] log10(15/10 × 8 × 10/3 ) log1040

The total surface area of a cylinder = 2\pirl + 2\pir2 = 2\pir(l + r) = 2 \times 3.14 x 5(7+5) 2 \times 3.14 \times 12 x 5 = 377.1cm (1DP)