Question Library

Capillarity is the tendency of a liquid to rise or fall in a narrow tube and it explains why a blotting paper absorbs ink

When two mirror are inclined at an angle \theta, the number of images formed, n = \(\frac{360}{\theta}\) − 1 ∴ since \(\theta\) = 90o ∴ \(\frac{360}{\theta}\) − 1 = 4 − 1 = 3

When silicon is doped with a trivalent atom. A p-type semi-conductor will be formed when silicon is doped with a pentavalent atom such as arsenic, a n-type semi-conductor will be formed Phosphorous is a pentavalent element, so it will produce a o -type semi-conductor

For a refractive index () = \(\frac{\sin\frac{1}{2} (A + D)}{\sin\frac{1}{2}A}\) D = angle of minimum deviation A = refractive angle of the prism 1.5 = \(\frac{\sin\frac{1}{2} (60 + D)}{\sin\frac{1}{2} \times 60}\) 1.5 = \(\frac{\sin\frac{1}{2} (60 + D)}{\sin 30}\) Sin \(\frac{1}{2}\) (60 + D) = 1.5 * Sin 30 Sin \(\frac{1}{2}\) (60 + D) = 0.75 \(\frac{1}{2}\) (60 + D) = Sin-1 (0.75) but Sin-1 (0.75) = 49o \(\frac{1}{2}\) (60 + D) = 49 60 + D = 2 * 49 = 98o D = 98o - 60o D = 38o

Seven 40W lamps = 7 \times 40 = 280W Five 80W lamps = 5 \times 80 = 400W Total power = 280W + 400W = 680W = \(\frac{680KW}{1000}\) = 0.68KW The kilowatt hour = 0.68 KW \times 12 hrs = 8.16 KWh Cost is N 7 per KWh So the cost of running them = 8.16KWh \times N 7/KWh = 57.12

V = IR from ohms law I = \(\frac{V}{R}\) Since the resistances are in parallel \(\frac{1}{Reff}\) = \(\frac{1}{18}\) + \(\frac{1}{12}\) + \(\frac{1}{6}\) = \(\frac{2 + 3 + 6}{36}\) = \(\frac{11}{36}\) R\(_{eff}\) = \(\frac{36}{11}\) = 3.27 I = I\(_1\) + I\(_2\) + I\(_3\) = \(\frac{V}{Reff}\)= \(\frac{12}{3.27}\) = 3.669A I = 3.7A

u = , H = 20, g = 10 From equation of motion under gravity H = ut ± \(\frac{1}{2}\)gt2, It is falling so g is +ve H = 0 + \(\frac{1}{2}\)gt2 H = \(\frac{1}{2}\)gt2 2H = gt2 \(\frac{2H}{g}\) = t2 t = \(\frac{\sqrt{2H}}{g}\) = \(\frac{\sqrt {2 \times 20}}{10}\) = √4 = 2s

Mechanical waves require material medium for their propagation only radio waves among the options docs require a material medium

Consider an inclined plane shown below, as the effort moves along OB, the load moves is lifted up through a vertical height AB V.R = \(\frac{\text{distance moved by effort}}{\text{distance moved by load}}\) = \(\frac{OB}{AB}\) But Sinθ = \(\frac{opp}{hyp}\) = \(\frac{AB}{OB}\) therefore; = \(\frac{OB}{AB}\) = \(\frac{1}{\sin \theta}\)

Using boyle's law P1V1 = P2V2 V2 = \(\frac{P_1 V_1}{200}\) = \(\frac{400 \times 10}{200}\) = 20cm3

An atom can be represented thus, \(^{Z}_{A}X\) where Z = mass number A = atomic number A = number of proton = number of electrons in a free state = 28 Z = number of protons + number of neutrons = 28 + 30 = 58 \(^{58}_{28}X\)is the representation

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

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

Impulse = Force \(\times\) time Force has the dimension \(\frac{ML}{T^{2}}\), hence Impulse= \(\frac{ML}{T^{2}} \times T = \frac{ML}{T}\)

Given that \(u = 0ms^{-1}, a = 2ms^{-2}, s = 9m\) We can use the equation, \(v^{2}=u^{2} +2as\) so that we have, \(v^{2} = 0^{2} + 2\times 2\times 9\) \(v^{2} = 36; v= 6.0ms^{-1}\) Note; u = Initial velocity v = final velocity a = acceleration s = distance covered