Question 31
The terminal velocity of a ball-bearing falling through a viscous fluid is reached when the
As a ball-bearing falls through a viscous fluid, it initially accelerates due to gravity. As its velocity increases, viscous drag and upthrust also increase. Terminal velocity is achieved when the downward force of gravity equals the sum of the upward viscous and buoyant forces. At this point, the net force becomes zero, causing the ball to move with a constant or uniform velocity. Therefore, terminal velocity corresponds to the stage when velocity becomes uniform.
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Question 32
The pressure of two moles of an ideal gas at a temperature of 27°C and volume 10-2 m3 is
[R = 8.313 J mol-1K-1]
[R = 8.313 J mol-1K-1]
From the ideal gas equation:\[PV = nRT\]\[P = \dfrac{nRT}{V}\]\(n = 2\), \(R = 8.313\), \(T = 27 + 273 = 300\, K\), and \(V = 10^{-2}\).\[P = \dfrac{2 \times 8.313 \times 300}{10^{-2}} = \dfrac{4987.8}{10^{-2}}\]\[P = 4.99 \times 10^{5}\, \text{Nm}^{-2}\]Therefore, the pressure of the gas is 4.99 × 105 Nm-2.
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Question 33
The value of x, y and z respectively in the expression MxLyTz for the universal gravitational constant G are
From Newton’s law of gravitation:\[F = \dfrac{Gm_1m_2}{r^2}\]\[G = \dfrac{Fr^2}{m_1m_2}\]\(F\) has the dimension \(MLT^{-2}\).
Substituting gives:\[G = (MLT^{-2}) \times L^2 \times M^{-2}\]\[G = M^{-1}L^{3}T^{-2}\]\[\Rightarrow x = -1, \; y = 3, \; z = -2\]Thus, the dimensional expression is \(M^{-1}L^{3}T^{-2}\).
Substituting gives:\[G = (MLT^{-2}) \times L^2 \times M^{-2}\]\[G = M^{-1}L^{3}T^{-2}\]\[\Rightarrow x = -1, \; y = 3, \; z = -2\]Thus, the dimensional expression is \(M^{-1}L^{3}T^{-2}\).
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Question 34
When a pentavalent impurity is added to a pure semiconductor, it becomes
Adding a pentavalent impurity (such as phosphorus or arsenic) introduces extra electrons into a semiconductor lattice. These additional electrons act as majority charge carriers, enhancing conductivity. Such doping produces an n-type semiconductor because the current is mainly carried by negatively charged electrons rather than holes.
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Question 35
Cancerous cells can be destroyed by
X-rays are a form of high-energy electromagnetic radiation that can penetrate deep into tissues. They are used in radiotherapy to destroy cancerous cells by damaging their DNA and disrupting their ability to reproduce. Healthy cells can repair themselves more effectively, which makes targeted X-ray treatment effective in cancer management.
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Question 36
If the decay constant of a radioactive substance is 0.231 s-1, the half-life is?
The relationship between half-life \(t_{1/2}\) and decay constant \(\lambda\) is:\[t_{1/2} = \dfrac{0.693}{\lambda}\]\[t_{1/2} = \dfrac{0.693}{0.231} = 3.00\,\text{s}\]Hence, the half-life of the substance is 3.00 seconds.
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Question 37
A piece of radioactive material contains 1020 atoms.
If the half-life of the material is 20 seconds, the number of disintegrations in the first second is
If the half-life of the material is 20 seconds, the number of disintegrations in the first second is
The activity \(A\) of a radioactive sample is given by \(A = \lambda N\), where \(\lambda\) is the decay constant.
Half-life \(t_{1/2} = 20\,s\), thus:\[\lambda = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{20} = 0.03465\,\text{s}^{-1}\]\(N = 10^{20}\). Therefore:\[A = 0.03465 \times 10^{20} = 3.47 \times 10^{18}\, \text{disintegrations per second}\]Hence, there are approximately 3.47 × 1018 disintegrations in the first second.
Half-life \(t_{1/2} = 20\,s\), thus:\[\lambda = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{20} = 0.03465\,\text{s}^{-1}\]\(N = 10^{20}\). Therefore:\[A = 0.03465 \times 10^{20} = 3.47 \times 10^{18}\, \text{disintegrations per second}\]Hence, there are approximately 3.47 × 1018 disintegrations in the first second.
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Question 38
Thermionic emission is the process whereby
Thermionic emission occurs when a metal is heated to a high temperature, providing sufficient energy for electrons to overcome the work function and escape from the surface. This phenomenon is the basis of vacuum tubes and cathode-ray tubes, where electrons emitted from a hot cathode are accelerated toward an anode. It is purely thermal and does not involve nuclear or chemical reactions.
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Question 39
Which of the following can conduct electricity owing to the presence of free mobile electrons?
Copper is a metal with a large number of free mobile electrons in its conduction band. These electrons move freely when a potential difference is applied, allowing copper to conduct electricity efficiently. In contrast, germanium and silicon are semiconductors with limited conduction at room temperature, and grapefruit is an insulator.
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Question 40
A progressive wave is represented by \( y = 10 \sin (1000\pi t - \pi x / 34) \). Two layers of the wave separated by 153 cm have a phase difference of:
The given wave equation is \( y = 10 \sin (1000\pi t - \pi x / 34) \).
Comparing with the general form \( y = A \sin (\omega t - kx) \):
\( k = \pi / 34 \).
The phase difference \( \Delta \phi \) between two points separated by a distance \( \Delta x \) is given by:\[ \Delta \phi = k \Delta x \] Substituting the values:\[ \Delta \phi = \dfrac{\pi}{34} \times 153 = 4.5\pi \; \text{radians} \] Converting radians to degrees:\[ 4.5\pi \times \dfrac{180}{\pi} = 810^\circ \] Since phase is periodic, \( 810^\circ \) is equivalent to \( 810 - 720 = 90^\circ \).
Therefore, the phase difference is 90°.
Comparing with the general form \( y = A \sin (\omega t - kx) \):
\( k = \pi / 34 \).
The phase difference \( \Delta \phi \) between two points separated by a distance \( \Delta x \) is given by:\[ \Delta \phi = k \Delta x \] Substituting the values:\[ \Delta \phi = \dfrac{\pi}{34} \times 153 = 4.5\pi \; \text{radians} \] Converting radians to degrees:\[ 4.5\pi \times \dfrac{180}{\pi} = 810^\circ \] Since phase is periodic, \( 810^\circ \) is equivalent to \( 810 - 720 = 90^\circ \).
Therefore, the phase difference is 90°.
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