Physics 2010 Jamb Examination

Each alpha particle emission decreases the mass number by 4 and the atomic number by 2.
Therefore, emission of two alpha particles results in a decrease of 8 in mass number and 4 in atomic number.
Each beta particle emission increases the atomic number by 1, without affecting the mass number. Hence, emission of three beta particles increases the atomic number by 3.
Gamma radiation has no effect on either the mass or atomic number.
Thus:
New mass number = 235 − (2 × 4) = 227
New atomic number = 92 − 4 + 3 = 91
The resulting element has a mass number of 227 and an atomic number of 91, which corresponds to protactinium (Pa).

Mass is an intrinsic property of matter and remains constant irrespective of location. However, weight depends on the acceleration due to gravity, which varies slightly across the Earth's surface due to its shape and rotation. Therefore, while the object's mass remains constant, its weight changes with the local gravitational field strength. Hence, mass is constant while weight varies.

To rationalize, multiply numerator and denominator by the conjugate of the denominator:\[\dfrac{(2\sqrt{3} + 5)(\sqrt{5} + \sqrt{3})}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})}\]The denominator becomes \(5 - 3 = 2\).
Expanding the numerator:\[2\sqrt{3}\sqrt{5} + 2\sqrt{3}\sqrt{3} + 5\sqrt{5} + 5\sqrt{3}\]\[= 2\sqrt{15} + 6 + 5\sqrt{5} + 5\sqrt{3}\]Collect like terms to simplify:\[= (2\sqrt{15} + 5\sqrt{5} + 5\sqrt{3} + 6)\]Hence, divide by 2:\[= \dfrac{3\sqrt{15} + 11}{2}\]\[\Rightarrow \text{Final answer: } \dfrac{3\sqrt{15} + 11}{2}\]

The increase in length \(\Delta L\) is proportional to the product of the original length \(L\), the linear expansivity \(\alpha\), and the temperature change \(\Delta T\):\[\Delta L \propto \alpha L\]Therefore, the ratio of the increases in lengths of P to Q is:\[\dfrac{\Delta L_P}{\Delta L_Q} = \dfrac{\alpha_P L_P}{\alpha_Q L_Q}\]\[= \dfrac{2 \times 3}{3 \times 4} = \dfrac{6}{12} = 1:2\]Hence, the ratio of increase in lengths is 1:2.

When the south poles of two magnets stroke a steel bar from both ends towards the centre, the ends being stroked acquire polarities opposite to the stroking poles due to magnetic induction. Since both ends are stroked with south poles, both T and V will become north poles, leaving the centre as south. Therefore, both T and V are north.

According to Faraday’s first law of electrolysis, the mass of a substance deposited is directly proportional to the quantity of electricity passed through the electrolyte:\[m = ZQ\]where \(Z\) is the electrochemical equivalent. Hence, when a graph of mass \(m\) is plotted against charge \(Q\), the slope represents the electrochemical equivalent \(Z\).

In parallel connection, the total capacitance is the sum of individual capacitances:\[C_T = 2 + 4 + 8 = 14 \; μF = 14 × 10^{-6} F\]The energy stored in a capacitor is given by:\[E = \dfrac{1}{2}CV^2\]\[= \dfrac{1}{2} (14 × 10^{-6})(6)^2\]\[= 0.5 × 14 × 10^{-6} × 36 = 252 × 10^{-6} = 2.52 × 10^{-4} J\]Therefore, the total energy stored is \(2.52 × 10^{-4}\; J\).

The current in a circuit is given by Ohm’s law:\[I = \dfrac{E}{R + r}\]\[= \dfrac{12}{2 + 4} = \dfrac{12}{6} = 2 A\]Therefore, the current flowing through the circuit is 2 A.

For a balanced metre bridge, the ratio of resistances equals the ratio of the bridge wire segments:\[\dfrac{R_1}{R_2} = \dfrac{L_1}{L_2}\]If \(L_1 = x\) and \(L_2 = 100 - x\):\[\dfrac{R_1}{R_2} = \dfrac{x}{100 - x}\] Where $R_1$ is the effective resistance of the parallel combinaton 8//8. $$ R_1 = \dfrac{8\times8}{8+8} = \dfrac{64}{16} = 4 ohms.$$ And $R_2$ is the resistance after the galvanometer.\[\dfrac{4}{8} = \dfrac{x}{100 - x}\] \[\dfrac{1}{2} = \dfrac{x}{100 - x}\] 
100−x=2x
Add x to both sides:
100=3x
\(x = 33.3\; cm\)

For resistors in parallel, the reciprocal of the total resistance is the sum of the reciprocals of the individual resistances:\[\dfrac{1}{R_T} = \dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{3}{4}\]\[\Rightarrow R_T = \dfrac{4}{3} \; \Omega\]Using Ohm’s law:\[I = \dfrac{V}{R_T} = \dfrac{16}{4/3} = 16 \times \dfrac{3}{4} = 12\; A\]Hence, the total current is 12 A.