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v = f\(\lambda\) = 825 x 0.4 = 330.0ms-1

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 D.

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

L = \(\frac{5\lambda}{4}\) \(\lambda = \frac{4L}{5}\) But \(\lambda = f\lambda\) f = \(\frac{v}{\lambda} = \frac{v}{\frac{4L}{5}}\) = \(\frac{5v}{4L}\)

(a) Critical angle is defined as the angle of incidence in the denser medium for which the angle of refraction in the less dense medium is 90^\circ. (b) Antinodes are created in a stationary wave when an incident and its reflected waves are superposed and 180^\circ out of phase. (c) n = \(\frac{Sin\frac{(A + Dm)}{2}}{\sin (\frac{A}{2})}\) = \(\frac{Sin\frac{(60 + 46.2)}{2}}{\sin (\frac{60}{2})}\) = \(\frac{sin 53.1}{sin 30}\) = 1.6 (d) No image is obtained on the screen because - Object, screen and mirror are not coaxial; - Object is placed at focal point of mirror; - Object is placed between the focus and the pole; - Screen is located at a distance less than or greater than actual image distance. (e) (i) \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\) \(\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{30} - \frac{1}{7}\) = \(\frac{5 - 2}{150} = \frac{3}{150}\) U = 75 + 25 = 100cm \(\frac{1}{u} + \frac{1}{v} = \frac{1}{f}\) \(\frac{1}{100} + \frac{1}{50} = \frac{1}{50} = \frac{1}{50}\) f = 50cm

a) Dielectric strength is the maximum potential gradient or electric field strength an insulator can withstand without breaking down. It is measured in Vmm\(^{-1}\) or Vm\(^{-1}\). (b)(i) The wave is infra-red wave. (ii) Suitable detector can be -Infra-red camera; -Thermopile; -Bolometer or -Photographic film. (iii) Source of the wave can be - Hot bodies or sun (c) E = \(\frac{F}{q} = \frac{W}{q}\) W =mF = Eq = 1200 x 1.0 x 10\(^{-19}\) = 1200 x 1.0 x 10\(^{-19}\)N (d) X\(_L\) = 1.2\(\pi fL\) = 2 x 3.142 x 50 x 0.50 = 15.7\(\Omega\) X\(_c\) \(\frac{1}{2 \pi fL}\) = \(\frac{1}{2 \times 3.14 \times 50 \times 25 \times 10^{-16}}\) = 127.4 \(\Omega\) (i) Z = \(\sqrt{R^2 + (X_1 - X_c)^2}\) = \(\sqrt{100^2 + (127.4 - 15.7)^2}\) = 149.9\(\Omega\) (ii) I \(\frac{V}{Z} = \frac{220}{149.9}\) = 1.47A

Stroboscope is a scientific instrument used to make a wave appears stationary when used

The appearance of pool of water on road surfaces is called MIRAGE His observation show that light from the sun ray refracted through a layer of hot air(layer) which is closed to the ground, passes from a hot air layer to the cold layer

As object and image are usually at equal distance apart from the mirror /p> Distance between Abioye (a) and hip image (I) = AI i.e. AI = 4.7 + 47 = 9.4m

Primary colours of light are red, blue and green Secondary colours of light are cyan, magneta and yellow

The vibrating particles behave like a spiral spring with a series of compression region and space–out region.

(Neglect of Air resistance) Note: Both objects reach the ground at the same time

\(Wavelength (\lambda) = 10^{-13}m\) Speed of X-ray = speed of light \(C = 3\times10^{8}m/s\) Frequently (f) = ? \(Using V = f\lambda (or (– f \pi for electromagnetic wave)\) \(f = \frac{c}{\lambda} = 3 \times 10^{8} \div 10^{-13} = 3 \times 10^{8} \times 10^{13}\) = \(3 \times 10^{8} \times 10^{13}\) = \(3 \times 10^{21}Hz\)

Convection take place commonly in liquids (water) and slightly in gasses