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To change a milliammeter to an ammeter, a low resistance known as shunt must be connected to it in parallel.

The depolarizing agent in a Leclanché cell is manganese dioxide (MnO\(_2\)). This compound helps to reduce the polarization effect and improves the efficiency of the cell during operation.

(vii) E(V) V(V) I(A) V\(^{-1}\)(V\(^{-1}\)) I\(^{-1}\)(I\(^{-1}\)) 3.00 2.86 1.43 0.350 0.699 4.50 4.28 2.14 0.234 0.467 6.00 5.70 2.86 0.175 0.350 7.50 7.14 3.57 0.140 0.280 9.00 8.57 4.28 0.117 0.234 (viii) SEE THE GRAPH ABOVE (ix) Slope = \(\frac{\Delta I^{-1}}{\Delta V^{-1}}\) = \(\frac{0.6 - 0.3}{0.3 - 0.15}\) = \(\frac{0.3}{0.15}\) = 2 (x) Intercept, c = 0.01 (xi) Slope, s represents external resistance of the circuit (xii) I. I ensured tight connections II. I avoided error due to parallax on the ammeter/voltmeter b(i) I. Use of cell II. hydropower (ii) Ohm's law states that the current passing through a metallic conductor is directly proportional to the potential difference applied at both ends provided temperature remains constant.

Effective voltage in an RLC circuit is given by V = \(\sqrt{V_R^2 + (V_L - V_C)^2}\) V = \(\sqrt{3^2 + (6 - 2)^2}\) = \(\sqrt{3^2 + 4^2}\) = \(\sqrt{9 + 16}\) = 5V

X\(_c\) = \(\frac{1}{2\pi FC}\) X\(_c\) = \(\frac{1}{2 \times 3.142 \times 60 \times 2 \times 10^{-6}}\) X\(_c\) = 1326.1192\(\Omega\) I = \(\frac{\text{V}}{X_c}\) I = \(\frac{150}{1326.1192}\) = 0.113A

Cathode rays are streams of electrons emitted from the cathode (negative electrode) in a vacuum tube. Cathode rays can be deflected by magnetic and electric fields, demonstrating their charge and nature as electrons.

E = hf + W\(_o\) E = 1.12 x 10\(^{-18}\)J Max. Energy = 8.0 x 10\(^{-19}\)J. W\(_o\) = E - Max. Energy = 1.12 x 10\(^{-18}\)J - 8.0 x 10\(^{-19}\)J. = 3.2 x 10\(^{-19}\)J. But W\(_o\) = hf\(_o\) f\(_o\) = \(\frac{W_o}{h}\) = \(\frac{3.2 \times 10^{-19}}{6.626 \times 10^{-34}}\) = 4.83 x 10\(^{14}\)Hz

(a) electrons are only emitted when the photon frequency exceeds the threshold frequency, - the energy of each photon is proportional to its frequency, - the emission of photoelectrons depends on the frequency of incident light on the metal, - the intensity of light does not affect the energy of the electrons emitted/photoelectrons, - the energy of the photoelectron depends on the frequency of incident radiation. (b) Given: P = 6 x 10\(^{-3}\) W, no of photons per seconds = 1 x 10\(^{16}\), m\(_e\) = 9.1 x 10\(^{-31}\) kg, e = 1.6 x 10\(^{-19}\) C, W\(_o\) = ? Photon Energy E = \(\frac{\text{power}}{\text{no of photons per second}}\) = \(\frac{\text{P}}{\text{n}}\) = \(\frac{6.0 \times 10^{-3}}{1 \times 10^{16}}\) = 6 x 10\(^{-19}\)J But K.E = E + W\(_o\) W\(_o\) = E - K.E = E - eV W\(_o\) = 6 x 10\(^{-19}\) - 2.2 x (1.6 x 10\(^{-19}\)) = 2.48 x 10\(^{-9}\)J (c) Factors on which radioactive activity depends are: (i) decay constant, (ii) number of unstable nuclei present at a given time (iii) time of the decay process (iv) environmental conditions, (v) type of radioisotope, etc (d) Given: A = 2.0 x 10\(^6\)Bq, t\(_{\frac{1}{2}}\) = 1.8 x 10\(^8\)s A = \(\lambda\)N, where \(\lambda\) = decay constant, N = number of radioactive nuclei, A = activity. But \(\lambda\) = \(\frac{In(2)}{t_{\frac{1}{2}}}\) \(\lambda\) = \(\frac{In(2)}{1.8 \times 10^8}\) ≈ \(\frac{0.693}{1.8 \times 10^8}\) ≈ 3.85 x 10\(^{-9}\)s\(^{-1}\) Recall, A = \(\lambda\)N N = \(\frac{A}{\lambda}\) = \(\frac{2.0 \times 10^6}{3.85 \times 10^{-9}}\) = 5.2 x 10\(^{14}\) nuclei

X-rays are typically used for medical imaging and other applications related to examining structures, but they are not used to determine the depth of a mine

X-rays are usually produced by electronic transitions within atoms, whereas gamma rays are released from an atom's nucleus during radioactive decay.

The correct sequence of ionization power from highest to lowest is: Alpha particle > Beta particle > Gamma ray.

To show that \(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\) The de broglie wavelength \(\lambda\), is given by \(\lambda\) = \(\frac{\text{h}}{\text{p}}\) where p = mv \(\lambda\) = \(\frac{\text{h}}{\text{mv}}\) When an electron is accelerated through a potential difference V\(_0\), it gains kinetic energy given K.E.= eV\(_0\) The kinetic energy can also be expressed in terms of the mass m of the electron and its speed v: K.E = \(\frac{1}{2}\)mv\(^2\) K.E = eV\(_0\) = \(\frac{1}{2}\)mv\(^2\) v = \(\sqrt{\frac{2eV_o}{m}}\) Substituting v = \(\sqrt{\frac{2eV_o}{m}}\) into the de Broglie equation \(\lambda\) = \(\frac{\text{h}}{m \times \sqrt{\frac{2eV_o}{m}}}\) = \(\frac{\text{h}}{\sqrt{\frac{2eV_o m^2}{m}}}\) \(\lambda\) = \(\frac{\text{h}}{\sqrt{2meV_o}}\) Note: use the fact that \(\sqrt{m}\) x \(\sqrt{m}\) = m

When an object is placed in front of a plane mirror, the image formed is at the same distance from the mirror as the object but on the opposite side. Thus, the total distance between the object and the image is 1.5 m + 1.5 m = 3.0m

When an object is placed at the centre of curvature of a concave mirror, an image that is the same as the object will be formed. Therefore, the height of the image is 2.0 cm.

The main disadvantage of a convex mirror as a driving mirror is that it distorts the perception of distance, making them appear farther away than they are.