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T = \(\frac{0.692}{\lambda} = \frac{0.692}{0.077}\) = 9s

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

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

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 C.

Table of values/observation \(l_o\) = 0.8A S/N L(cm) l(A) (L\(^{-1}\)cm\(^{-1}\)) 1 25.0 1.20 0.040 2 40.0 1.00 0.025 3 55.0 0.95 0.018 4 70.0 0.85 0.014 5 85.0 0.8 0.012 When 1\(^{-1}\) =0, l = 0.66A \(\frac{|_{o}}{|} = \frac{0.8}{0.66}\) = 1.212 Precautions: The Key was open when readings were not being taken. Tight connections were ensured. Parallax errors were avoided in reading the meter-rule Zero error was noted in reading metre-rule The jockey was not rolled on or allowed to scratch the wire (b) The p.d between two points is the work done (in joules) in moving an electric charge of one coulomb across the two points. ii. R = \(\frac{PL}{A} = \frac{4pL}{\pi d^{2}}\) d = 0.2m, \(\pi\) = \(\frac{22}{7}\), R = 7\(\Omega\), p = 8.8 x 10\(^{-7} \pi \), L = ? \(\therefore\) 7 = \(\frac{4 \times 8.8 \times 10^{-7} \times L}{\frac{22}{7} \times (0.2)^{2}}\) 7 = \(\frac{4 \times 8.8 \times 10^{-7} \times L \times 7}{22 \times 0.04} = \frac{28 \times 8.8 \times 10^{-7} \times L}{0.88}\) L = \(\frac{7 \times 0.88}{28 \times 8.8 \times 10^{-7}}\) L = \(\frac{6.16}{246.4}\) x 10\(^{-7}\) L = 0.025 x 10\(^{7}\) L = 2.5 x 10\(^{-2}\) x 10\(^{7}\) L = 2.5 x 10\(^{5}\)

Work done by friction = Friction force x Displacement from the relation, f = uR where u = coefficient of friction; R= Normal reaction F = u mg (R =mg) = 0.1 x 0.01 x 10 work done = F x Displacement = 0.1 x 0.01 x 10 x 0.2 = 0.002 or \( 2 \times 10^{-3}J \)

For the equilateral glass prism, A = 60º n = \(\frac{sin \frac{Dm + A}{2}}{sin \frac{A}{2}}\) \(\frac{3}{2}\) = \(\frac{sin \frac{Dm + 60}{2}}{sin \frac{60}{2}}\) = \(\frac{sin \frac{Dm + 60}{2}}{sin 30}\) \(\frac{3}{2}\) x \(\frac{1}{2}\) = sin\(\frac{Dm + 60}{2}\) \(\frac{3}{4}\) = 0.75 = sin\(\frac{Dm + 60}{2}\) sin\(^{-1}\)(0.75) = \(\frac{Dm + 60}{2}\) 48.590 = \(\frac{Dm + 60}{2}\) 48.590 x 2 = Dm + 60 97.180 = Dm + 60 Dm = 97.180 - 60 = 37.18 ≈ 37 .2º

From faraday's law of electrolysis M = I t Z 0.990 = I x (40 x 60) x 3.3 x 10-4 \( \implies \frac{0.990}{2400 \times 3.3 \times 10^{-4}} = 1.25A \\ \text{Thus the correction to be made } = 1.25 - 1.20 = 0.05A \)

Atomic emission and absorption of energy is usually in packets referred to as photon or quantum of energy of magnitude E =hf, where f is the frequency and h is the plancks constant Energy emitted = hf = 6.6 x 10-34 x 8.0 x 10-14 = 5.28 x 10-19J

From Newton's second law: \( F = \frac{mV - mµ}{t} \) Let the final velocity = V; and since the football is initially at rest, its initial vel µ = 0, M = 0.8kg t = 0.8s \(\implies 100 = \frac{0.8V - 0.8 \times 0}{0.8}\) \(\text{Therefore } V = \frac{100 \times 0.8}{0.8} = 100ms^{-1}\)

The speed of sound in a material is given as: V = \(\sqrt{\frac{E}{P}}\), Where E = Young's modulus P = Density Therefore, V = \(\sqrt{\frac{7.0 \times 10^{10}}{2.7 \times 10^3}}\) = 5091.75 = 5.1 \(\times 10^3ms^{-1} \)

From the uncertainty principle the product of the position and momentum of a sub-atomic particle is equal or greater than planck's constant. i.e \( \Delta x. \Delta p \geq h.\) \(\text{Thus} 5 \times 10^{-10} \times \Delta p = 6.6 \times 10^{-34}\) \(\Delta p = \frac{6.6 \times 10^{-34}}{5 \times 10^{-10}} = 1.32 \times 10^{-24}Ns\) \(\text{Therefore Uncertainty in the momentum } = 1.32 \times 10^{-24}Ns \)

In general, the energy released during nuclear fission is given by Einstein's mass-energy equation, given by \( E = \Delta MC^2 \) Where \( \Delta m\) is the mass defect, and C is the speed of light. Therefore \( \Delta m = 0.01% \text{of} 1.0g = \frac{0.01}{100}\times 1.0g \) = 1.0 \(\times 10^{-4} \) = 1.0 \(\times 10^{-7}kg\) \(\text{Energy Released } = \Delta MC^2 \) = 1.0 \(\times 10^{-7} \times (3.0 \times 10^8)^2\) = 1.0 \(\times 10^{-7} \times 9.0 \times 10^{16} \) = 9.0 \(\times 10^9 J \)

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