Physics 2006 Jamb Examination

Nuclear fission produces more neutrons than the initiating neutron (I is true), there is a mass defect (II is true), and energy is released (III is true). Statement IV describes fusion, not fission (false). Therefore I, II and III are correct.

An n-type semiconductor is produced by doping with pentavalent (group V) elements which donate extra electrons. Arsenic and antimony are group V elements, so they produce n-type material.

A p–n junction diode allows current to pass primarily in one direction and is commonly used to convert alternating current (AC) to direct current (DC). Thus it is used as a rectifier in an a.c. circuit.

For a concave mirror, the radius of curvature $R=2f=2\times5\;\text{cm}=10\;\text{cm}$. An object placed at $2f$ produces an image at $2f$ (real, inverted, same size). Therefore the image is at the radius of curvature (10 cm).

Apparent depth $=\dfrac{\text{real depth}}{\mu}$.
Initial apparent depth = $\dfrac{80}{\dfrac{4}{3}}=80\times\dfrac{3}{4}=60\;\text{m}$.
Final apparent depth = $\dfrac{60}{\dfrac{4}{3}}=60\times\dfrac{3}{4}=45\;\text{m}$.
Change in apparent depth = $60-45=15\;\text{m}$.

Non-luminous objects do not produce their own light. They are visible only because they reflect light from luminous sources into our eyes.

In a closed pipe, the fundamental wavelength $\lambda_1 = 4L = 4\times 33 = 132\;\text{cm}$. The first overtone corresponds to the 3rd harmonic: $\lambda_3 = \dfrac{4L}{3} = \dfrac{132}{3} = 44\;\text{cm}$.

The least distance of distinct vision required = 25 cm. \(\therefore\) For the person to see an object clearly at \(25 \, \text{cm}\) from the eye, the image must be formed at \(75 \, \text{cm}\) on the same side of the lens, at his near point. \\ \(\therefore U = 25 \, \text{cm}, \quad V = -75 \, \text{cm} \; (\text{image should be virtual})\) \[ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} \] \[ = \dfrac{1}{-75} + \dfrac{1}{25} \] \[ = -\dfrac{1}{75} + \dfrac{3}{75} = \dfrac{2}{75} \] \[ \therefore f = \dfrac{75}{2} = 37.5 \, \text{cm} \]

Resistivity $\rho = R\dfrac{A}{L}$. Diameter $=1.00\;\text{mm} = 1.0\times 10^{-3}\;\text{m}$, radius $r = 0.5\times 10^{-3}\;\text{m}$.
Area $A = \pi r^2 = \pi(0.5\times 10^{-3})^2 = 7.85\times 10^{-7}\;\text{m}^2$.
$\rho = 0.1\times\dfrac{7.85\times 10^{-7}}{2.0} = 3.93\times 10^{-8}\;\Omega\,\text{m}$.

In the stress - strain graph shown above, X represents the breaking point of the wire