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Given: - Original resistance \( R_1 = 5 \, \Omega \) - The new length \( L_2 = 2L_1 \) (where \( L_1 \) is the original length) - The new cross-sectional area \( A_2 = \frac{1}{2} A_1 \) (where \( A_1 \) is the original area) Step 1: Recall the formula for resistance The resistance \( R \) of a wire is given by: \(R = \rho \frac{L}{A}\) where: - \( \rho \) is the resistivity, - \( L \) is the length, - \( A \) is the cross-sectional area. Step 2: Calculate the new resistance The new resistance \( R_2 \) can be calculated using the new length and new area: \(R_2 = \rho \frac{L_2}{A_2}\) Substituting for \( L_2 \) and \( A_2 \): \(R_2 = \rho \frac{2L_1}{\frac{1}{2}A_1} = \rho \frac{2L_1 \cdot 2}{A_1} = \rho \frac{4L_1}{A_1}\) Step 3: Relate \( R_1 \) and \( R_2 \) From the original resistance \( R_1 \): \(R_1 = \rho \frac{L_1}{A_1}\) Now, we can express \( R_2 \) in terms of \( R_1 \): \(R_2 = 4 \left(\rho \frac{L_1}{A_1}\right) = 4R_1\) Step 4: Substitute the value of \( R_1 \) Now substitute \( R_1 = 5 \, \Omega \): \(R_2 = 4 \times 5 \, \Omega = 20 \, \Omega\) Therefore, the new resistance is: \(\boxed{20 \, \Omega}\)

In general, if a charged particle is accelerated between two metal plates/a region by a potential difference V volts, the work done on the charge, given by the product of the charge and the potential difference, is equal to the kinetic energy of the charge Thus eV = K.E 1.6 x 10\(^{-19}\) x V = 4.8 x 10\(^{-17}\) ∴ V = \(\frac{4.8 \times 10^{-17}}{1.6 \times 10^{-19}}\) = 300V

A charge in motion always experienced a force on entering a magnetic field

From the above diagram the widest separation is given by Blue and red

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

At minimum deviation, it is observed that light passes symmetrically through, the prism such that the angle of incidence (io) is angle to the angle of emergency (eo).(i/e Io = eo)

116 days = 4 half - lives Thus, after 4 half lives the fraction left of any radioactive materials is given by (1/2)4 = 1/16 ∴ Count rate / amount left = 1/16 x 800/1 = 50 count / min

The induced voltage \(\epsilon\) in a coil = N* B* A * \(\omega\) where: N = number of loops (50 loops),B = magnetic field strength (0.4 T),A = area of the loop (0.4 m\(^2\)),\(\omega\) = angular velocity (10 rad/s) \(\epsilon\) in a coil = N* B* A * \(\omega\) = 50 x 0.4 x 0.4 x 10 = 80V Therefore, the amplitude of the induced voltage is 80V

the force on a moving charge in a magnetic field is given by: F = q V β sin θ and this force is maximum when θ = 90o, during which sin 90o = 1: and for sinθ = 0.5, θ = 30o

The stopping potential is the potential required to stop the most energetic photo-emitted electron. From the relation e Vs = K.E,(max) where e is the electron charge, and Vs is the stopping potential => e x Vs = 0.34eV ∴ Vs = (0.34eV) / e = 0.34V ∴ stopping potential = 0.34V

To find the power dissipated in a resistor due to an alternating current (AC), we use the formula for average power: \(P = I_{\text{rms}}^2 R\) where: - \( P \) is the average power, - \( I_{\text{rms}} \) is the root mean square (RMS) value of the current, - \( R \) is the resistance. The current is given by: \(I(t) = 10 \sin(120\pi t)\) The peak current is: \(I_0 = 10 \, \text{A}\) The RMS value of a sinusoidal current is given by: \(I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, \text{A} \approx 7.07 \, \text{A}\) Using the resistance \( R = 12 \, \Omega \): \(P = I_{\text{rms}}^2 R\) Substituting the values: \(P = (5\sqrt{2})^2 \cdot 12\) Calculating \( (5\sqrt{2})^2 \): \((5\sqrt{2})^2 = 25 \cdot 2 = 50\) Now substituting back into the power formula: \(P = 50 \cdot 12 = 600 \, \text{W}\) Thus, the power dissipated in the resistor is: \(600 \, \text{W}\)

In a semiconductor junction diode, applying forward bias to the depletion or barrier layer causes the layer to B. narrow. This occurs because the applied voltage lowers the potential barrier, facilitating the recombination of charge carriers and reducing the width of the depletion region.

Experiment shows that increase in temperature increases the resistance of a pure metal but decreases the resistance of pure semiconductor

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

For a concave mirror to produce an image at infinity, the image has to be place at the principal focus of the mirror