Physics 2018 Jamb Examination

Since the image is erect and magnified, the lens must be a converging (convex) lens, and the image must be virtual. For a virtual image in a convex lens, the image distance \( v \) is negative.
Given magnification \( M = 3 \) and \( f = 15 \; \text{cm} \):\[ M = \left| \frac{v}{u} \right| \Rightarrow 3 = \frac{-v}{u} \Rightarrow v = -3u \]Using the lens formula:\[ \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \]\[ \frac{1}{15} = \frac{1}{u} + \frac{1}{-3u} \]\[ \frac{1}{15} = \frac{3 - 1}{3u} = \frac{2}{3u} \]\[ 3u = 30 \Rightarrow u = 10 \; \text{cm} \]Then, \( v = -3(10) = -30 \; \text{cm} \).
The distance between the object and the image is the absolute difference between their positions from the lens:\[ \text{Distance} = |v| - u = 30 - 10 = 20 \; \text{cm} \]

A photo cell (or photoelectric cell) operates based on the photoelectric effect. This phenomenon occurs when electromagnetic radiation, such as light, strikes the surface of a material (usually a metal), causing the emission of electrons from that surface. These emitted electrons are known as photoelectrons. The cell converts light energy directly into electrical energy by creating a current of these electrons.

A metre bridge works on the principle of the Wheatstone bridge. The ratio of the resistances is equal to the ratio of the lengths of the wire segments at the balance point. Let \( R = 2.0 \; \Omega \) at the 0.0cm end, and \( X \) be the unknown resistor at the other end. The balance length \( l_1 = 55.0 \; \text{cm} \), so the remaining length \( l_2 = 100 - 55 = 45.0 \; \text{cm} \).\[ \frac{R}{X} = \frac{l_1}{l_2} \]\[ \frac{2.0}{X} = \frac{55}{45} \]\[ 55X = 2.0 \times 45 \]\[ 55X = 90 \]\[ X = \frac{90}{55} \approx 1.636 \; \Omega \]

Electrical energy (W) dissipated in a resistor can be calculated using the formula derived from power and time:\[ W = VIt \]Given:
Voltage \( V = 12 \; \text{V} \)
Current \( I = 2 \; \text{A} \)
Time \( t = 5 \; \text{minutes} = 5 \times 60 = 300 \; \text{seconds} \)

Substituting the values:\[ W = 12 \times 2 \times 300 \]\[ W = 24 \times 300 \]\[ W = 7,200 \; \text{J} \]

Relative density (R.D.) of a solid is the ratio of its weight in air to the weight of an equal volume of water. According to Archimedes' principle, the weight of an equal volume of water is equal to the upthrust experienced by the solid when fully immersed in water.\[ \text{Upthrust} = \text{Weight in air} - \text{Weight in water} \]\[ \text{Upthrust} = 45 \text{N} - 15 \text{N} = 30 \text{N} \]Now, calculate Relative Density:\[ R.D. = \frac{\text{Weight in air}}{\text{Upthrust}} \]\[ R.D. = \frac{45}{30} = 1.50 \]

When an echo is heard, the sound has travelled from the source to the obstacle and back to the source. Therefore, the total distance travelled by the sound is twice the distance (\( d \)) between the boy and the hill.
The formula is:\[ 2d = v \times t \]Where:
\( v = 340 \; \text{ms}^{-1} \)
\( t = 0.8 \; \text{s} \)
\[ 2d = 340 \times 0.8 \]\[ 2d = 272 \]\[ d = \frac{272}{2} = 136 \; \text{m} \]

The critical angle (\( c \)) is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium (air) is 90°. The relationship between the refractive index (\( n \)) and the critical angle is given by:\[ n = \frac{1}{\sin c} \]Given \( c = 34^\circ \):\[ n = \frac{1}{\sin 34^\circ} \]Using the value \( \sin 34^\circ \approx 0.5592 \):\[ n = \frac{1}{0.5592} \approx 1.788 \]Rounding to two decimal places, the refractive index is 1.79.