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a)(i) Nuclear fission is a nuclear reaction in which a heavy nucleus is split into different lighter nuclei emitting neutrons with the release of a large amount of energy. (ii) Nuclear fusion is a nuclear reaction between two light nuclei at high temperatures formic a heavier nucleus with the release of a large amount of energy. (b)(i) Applications of atomic energy are: - for the production of nuclear weapons for warfare - in the generation of electric power - for health diagnosis and radiotherapy - to detect leakages. In underground pipes carrying oils or gases. (ii) Half life is the time required for radioactive substance to decay to half of its original quantity. (iii) T = \(\frac{0.693}{\lambda}\) (c)(i) \(^{226}_{88} X \) ---> \(^{222}_{88} Y + ^4_2He\) By the emission of \(\alpha\) the mass number of X decreases by 4 the atomic number by 2 to give daughter nucleus Y (ii) \(^{222}_{86} Y \) ---> \(^{222}_{87} Z + ^0_{-1} e\) By the emission of \(\beta\) particle the mass number of Y is unchanged while the atomic number of Y is unchanged while the atomic number is increased by 1. (iii) \(^{226}_{87} Z \) ---> \(^{222}_{87} Z + ^0_0\gamma\) By emission of \(\gamma\) radiation, the mass and atomic number of Z remain unchanged.

(p - 760)/(0 - 760) = (h - 0)/(80 - 0) ∴ (p - 760)/-760 = (20 - 0)/80 80(p-760) = -760(20-0) ∴80p - 60800 = -15200 80p = 60800 - 15200 80p = 45600 ∴ p = 45600/80 p = 570mmHg

At the height of 10m, its total energy is potential, given by P.E. = mgh = 1 x 10 x 10 = 100J. But as the stone descends, its potential energy decreases, while its kinetic energy increases. Half way during the fall, (h = 5m), its P.E. = 50J, and consequently, its K.E. should be 50J, such that from the principle of conservation of energy, the total energy of the stone at any instant should be conserved.

To find the ratio of the final volume to the initial volume of a gas when both pressure and temperature change, we can use the Ideal Gas Law, which states that \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) P\(_1\) = 300N/m\(^2\) V\(_1\) = ?, V\(_2\) = ?, T\(_1\) = 127^\circC =127+273 = 400K, T\(_2\) = −73^\circC = −73 + 273 = 200K P\(_2\) = 120N/m\(^2\) Using the Ideal Gas Law, we can rearrange it to find the ratio of volumes: = \(\frac{V_2}{V_1} = \frac{P_1T_2}{P_2T_1}\) = \(\frac{V_2}{V_1} = \frac{300 \times 200}{120 \times 400}\) = \(\frac{5}{4}\) Therefore, V\(_2\) : V\(_1\) = 5 : 4.

The resultant of two forces acting on an object is maximum when the angle between them is zero, during which the two forces are parallel and pointing in the same direction

In real-world machines, due to factors such as friction, deformation, and other losses, the mechanical advantage is always less than the velocity ratio: MA < VR This means that the work output (useful work done) is always less than the work input (energy supplied), leading to efficiency being less than 100%.

Velocity: Vector quantity (has both magnitude and direction). Displacement: Vector quantity (change in position). Reaction: Typically refers to a reaction force, which is a vector quantity.(OPTION A)

An absolute thermometer, such as a gas thermometer, often uses helium as its thermometric substance because it has favorable properties for measuring temperature accurately over a wide range. Helium behaves nearly ideally at low temperatures and pressures, making it suitable for measuring temperatures close to absolute zero. While mercury is useful for many traditional thermometers due to its liquid state and thermal properties, it is not suitable for absolute thermometers, which require a substance that can accurately measure extremely low temperatures and behave ideally across a wide range of conditions. Thus, helium is the correct choice for an absolute thermometer.

If F = 85.5Hz, V = 342ms-1, Then wavelength = V/F = 342/85.5 = 4m Then if wavelength = 4m => wavelength/2 = 4/2 = 2m = NN, Thus NA = wavelength/4 = 4/4 = 1m.

For a string fixed at both ends and plucked at the middle, the fundamental frequency is given by f0 = v/λ, where v = velocity of sound, λ = wave length. At its fundamental, note, the the length of the string is given l = λ/2, => λ0 = 2l, where λ0 = wave length fundamental note. for the first over tone, λ1 = l. For the second over tone, λ2 = 21/3 = (2 x 24)/3 = 16cm. i.e. λ2 = 16cm.

Efficiency = \(\frac{power output}{power input}\) = \(\frac{R}{R + r}\) Given efficiency = 75% = 0.75, R = 6.0\Omega, r = ? Efficiency = \(\frac{R}{R + r}\) 0.75 = \(\frac{6.0}{6.0 + r}\) 0.75(6.0 + r) = 6.0 4.5 + 0.75r = 6.0 0.75r = 6.0 - 4.5 = 1.5 r = \(\frac{1.5}{0.7}\) = 2\Omega The internal resistance r of the cell is 2.0 ohms.

Electrostatic force = F\(_E\) = \(\frac{e^2}{4\piε_od^2}\) Gravitational force = F\(_G\) = \(\frac{Gm^2}{d^2}\) \(\frac{F_E}{F_G}\) = \(\frac{\frac{e^2}{4\piε_od^2}}{\frac{Gm^2}{d^2}}\) = \(\frac{e^2}{4\piε_oGm^2}\) NOTE: (d\(^2\)s canceled out)

Electricity conduction through gases is usually done under low pressure (I) and high voltage (III). Thus OPTION B is correct.

Efficiency = \(\frac{\text{power output}}{\text{power Input}}\) => \(\frac{80}{100} = \frac{\text{power output}}{1.5\times 250}\) i.e power output = \(\frac{1.5\times250\times80}{100}\) = 300W

Total power in watt = \(60\times5 + 4\times100\) = 300 + 400 = 700W Total power in kilo watt = 700/1000 = 0.7kW Unit of electricity in kilowatt-hour = 0.7 * 20hrs = 14 okwh i.e If I kwh costs # 10.00 14kwh = \(10.00 \times14\) = #140