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Physics Test - 35

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Physics Test - 35
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  • Question 1
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    In the absence of intermolecular forces of attraction, the observed pressure P will be

    Solution
    In the absence of intermolecular forces, there will be no stickiness of molecules. Hence,pressure will increase.
    KEY CONCEPTS
    Surroundings thermodynamics
    Thermodynamic system may be of three types
    (a) Open system: It exchange both energy and matter with the surrounding.
    (b) Closed system: It exchange only energy (not matter) with the surroundings.
    (c) Isolated system : It exchange neither energy nor matter with the surrounding.
    Surroundings:
    In thermodynamics, the system is the part of the universe being studied, while the surroundings are the rest of the universe that interacts with the system.
    Potential Energy Due to molecular forces Solids and liquids exist due to the inter molecular potential energy (molecular "stickiness') of atoms. If molecules were not "sticky" all matter would be gases or solids. Thus, the principles of molecular potential energy are important for developing a molecular perspective on the nature of liquids, solids, and non-ideal gases. Potential energy is associated with the "Work" of moving a system some distance through a force field. On the macroscopic scale, we are well aware of the effect of gravity. As an example, the Earth and the moon are two spherical bodies which are attracted by a force which varies as \(r^{-2}\). The potential energy represents the work of moving the two bodies closer together or farther apart, which is simply the integral of the force over distance. (The force is the neqative derivative of potential with respect to distance.) Thus, the potential function varies as \(\mathrm{r}-1 .\) Potential energies are similar at the microscopic level except that the forces vary with position according to different laws. The gravitational attraction between two individual atoms is insignificant because the masses are so small. Rather, the important forces are due to the nature of the atomic orbitals. For a rigorous description, the origin of the inter molecular potential is traced back to the solution of Schrodinger's quantum mechanics for the motions of electrons around nuclei. However, we do not need to perform quantum mechanics to understand the principles.
    Potential Energy Due to inter atomic forces When two wet glass plates are pressed together, they cannot be separated easily. This is because of the force of attraction existing between the atoms and molecules, and are electrical in origin.
    All mechanical systems have to acquire a state of minimum potential energy, to attain stability. This is achieved by the force of attraction and repulsion acting between the atoms. This can be understood by the following graph showing the variation of potential energy with interatomic distances.
    Interatomic potential energy \(\mathrm{V}(\mathrm{R})\) between two identical atoms as a function of separation \(\mathrm{R}\) between their nuclet
    As the two isolated Hydrogen atoms approach each other, the nucleus of one atom attracts the valence electrons of the other and vice versa. As the hydrogen atoms come closer, the attractive force tends to decrease the potential energy of the system. When the potential energy reaches a minimum value, sharing of electrons take place between the two, resulting in a covalent bond. This stage is represented by the distance 'ro' in the graph. It is interesting to note that when the distance between the two atoms becomes even lesser, their nuclei repel each other, thus preventing them from collapsing. In other words, the force of repulsion increases the potential energy of the system.
    At distances equal to \(r_{0}\), the two forces balance each other, resulting in a molecule. Similarly, forces of attraction exist between molecules, binding them together. These are inter molecular forces. The inter molecular forces are weaker than interatomic forces, as the former forces are the Van der waals forces.
  • Question 2
    1 / -0

    The variation of photoelectric current given by the photocell with the intensity of light, is given by a graph, which is a straight line with

    Solution

    Photoelectric current ∝ intensity of incident light. Therefore, the graph is a straight line having positive slope passing through origin

    KEY CONCEPTS

    Einstein's Theory Of Photoelectric effect

    According to classical electromagnetic theory, this effect can be attributed to the transfer of energy from the light to an electron in the metal. From this perspective, an alteration in either the amplitude or wavelength of light would induce changes in the rate of emission of electrons from the metal. Furthermore, according to this theory, a sufficiently dim light would be expected to show a lag time between the initial shining of its light and the subsequent emission of an electron. However, the experimental results did not correlate with either of the two predictions made by this theory.

    Instead, as it turns out, electrons are only dislodged by the photoelectric effect if light reaches or exceeds a threshold frequency, below which no electrons can be emitted from the metal regardless of the amplitude and temporal length of exposure of light. To make sense of the fact that light can eject electrons even if its intensity is low, Albert Einstein proposed that a beam of light is not a wave propagating through space, but rather a collection of discrete wave packets (photons), each with energy hf. This shed light onMax Planck's previous discovery of the Planck relation (E = hf) linking energy (E) and frequency (f) as arising from quantization of energy. The factor h is known as the Planck constant.

    Photoelectric Effect ​& Electromagnetic Waves

    The photo-electric effect is the emission of electrons called photo-electrons when light strikes a surface. To escape from the surface, the electron must absorb enough energy from the incident radiation to overcome the attraction of positive ions in the material of the surface. The photoelectric effect is based on the principle of conservation of energy.

    The photoelectric effect is the observation that many metals emit electrons when Electromagnetic Waves incident on them. Electrons emitted in this manner can be called photoelectrons. The phenomenon is commonly studied in electronic physics, as well as in fields of chemistry, such as quantum chemistry or electrochemistry.

  • Question 3
    1 / -0

    The Kα X-ray emission line of tungsten occurs at λ= 0.021 nm. The energy difference between K and L levels in this atoms is about

    Solution
    \(\lambda_{k \alpha}=0.021 \mathrm{nm}=0.21 \mathrm{A}\)
    since, \(\lambda_{k \alpha}\) corresponds to the transition of an electron from \(L\) -shell to \(K\) -shell, therefore
    \(E_{L}-E_{K}=(\text { in } \mathrm{eV})=\frac{12375}{\lambda(\mathrm{inA})}=\frac{12375}{0.21} \approx 58928 \mathrm{eV}\)
    or \(\Delta E=59 \mathrm{keV}\)
    KEY CONCEPTS
    Production of Characteristic X rays
    Learn this Concept
    Characteristic X-rays are produced when an element is bombarded with high-energy electrons. When a highenergy electron (the incident electron) strikes a bound electron (the target electron) in an atom, the target electron is ejected from the inner shell of the atom.
    In some collisions an electron coming with high velocity penetrate through the outer shells of the electrons and strike against an electron in the inner shell, and knockout an electron from the inner shell of the atom. The incident and the dislodged electrons both escape from the atom causing a vacancy in the innermost K shell. An electron from the outer shell (say L shell) in higher energy state jump to the K shell to fill the vacancy. This gives rise to an \(\mathrm{X}\) -ray photon of energy hi. This causes a vacancy in the \(\mathrm{L}\) shell, which is again filled up by another electron jumping from one of the outer shells along with the emission of another X-ray photon of different energy.. This process continues till the outer-most shell is reached, thus resulting in the emission of a series of characteristic \(X\) - ray photons. This gives rise to sharp lines or peaks superimposed on the continuous line
  • Question 4
    1 / -0

    An infinitely long conductor PQR is bent to form a right angle as shown in figure. A current I flows through PQR . The magnetic field due to this current at the point M is H1 Now, another infinitely long straight conductor QS is connected at Q , so that current is I/2 in QR as well as in QS. The current in PQ remaining unchanged. The magnetic field at M is now H2 . The ratio H1 is given by

    Solution
    \(\mathrm{H}_{1}=\) Magnetic field at \(\mathrm{M}\) due to \(\mathrm{PQ}+\) Magnetic field at \(\mathrm{M}\) due to QR But magnetic field at \(\mathrm{M}\) due to \(\mathrm{QR}=\mathrm{O}\)
    \(\therefore\) Magnetic field at \(\mathrm{M}\) due to \(\mathrm{PQ}\) (or due to current \(I\) in \(\mathrm{PQ}\) ) \(=\mathrm{H}_{1}\)
    Now \(\mathrm{H}_{2}=\) Magnetic field at \(\mathrm{M}\) due to \(\mathrm{PQ}\) (current \(I\) ) \(+\) magnetic field at M due to QS (current \(\frac{I}{2}\) ) \(+\) magnetic field at M due to QR
    \(=\mathrm{H}_{1}+\frac{\mathrm{H}_{1}}{2}+0=\frac{3}{2} \mathrm{H}_{1}\)
    \(\frac{\mathrm{H}_{1}}{\mathrm{H}_{2}}=\frac{2}{3}\)
    Note Magnetic field at any point lying on the current carrying straight conductor is zero.
  • Question 5
    1 / -0

    Which one among the following shows particle nature of light ?

    Solution

    Particle nature of light was established by photoelectric effect.

    KEY CONCEPTS

    Einstein's Theory Of Photoelectric effect

    According to classical electromagnetic theory, this effect can be attributed to the transfer of energy from the light to an electron in the metal. From this perspective, an alteration in either the amplitude or wavelength of light would induce changes in the rate of emission of electrons from the metal. Furthermore, according to this theory, a sufficiently dim light would be expected to show a lag time between the initial shining of its light and the subsequent emission of an electron. However, the experimental results did not correlate with either of the two predictions made by this theory.

    Instead, as it turns out, electrons are only dislodged by the photoelectric effect if light reaches or exceeds a threshold frequency, below which no electrons can be emitted from the metal regardless of the amplitude and temporal length of exposure of light. To make sense of the fact that light can eject electrons even if its intensity is low, Albert Einstein proposed that a beam of light is not a wave propagating through space, but rather a collection of discrete wave packets (photons), each with energy hf. This shed light onMax Planck's previous discovery of the Planck relation (E = hf) linking energy (E) and frequency (f) as arising from quantization of energy. The factor h is known as the Planck constant.

  • Question 6
    1 / -0

    The wavelength of maximum energy, released during an atomic explosion was 2.93 × 10−10 m. Given that the Wien's constant is 2.93 × 10−3 m-K, the maximum temperature attained must be of the order of

    Solution
    Wien's displacement law is given by
    \[
    \begin{array}{l}
    \left.\lambda_{m} T=\text { constant (say } b\right) \\
    \text { Given, b=Wien's constant=2.93 } \times 10^{-3} \mathrm{m}-\mathrm{K} \\
    \lambda_{m}=2.93 \times 10^{-10} \mathrm{m}
    \end{array}
    \]
    Substituting the values, we obtain
    \[
    T=\frac{b}{\lambda_{m}}
    \]
    \[
    =\frac{2.93 \times 10^{-3}}{2.93 \times 10^{-10}}=10^{7} \mathrm{K}
    \]
  • Question 7
    1 / -0

    In the phenomenon of diffraction of light, when blue light is used in the experiment instead of red light, then

    Solution
    Width of central maximum is given by
    \[
    w=\frac{2 f \lambda}{a} \quad \ldots(i)
    \]
    Where \(f\) is focal length of lens, a is width of slit and \(\lambda\) is wavelength of light used.
    From Eq. (i), it is clear that fringe width
    \[
    w \propto \lambda
    \]
    So, when blue light is used in the experiment instead of red light, the fringes will become narrower.
  • Question 8
    1 / -0

    Eight point charges (can be assumed as small spheres uniformly charged and their centres at the corner of the cube) having values q each are fixed at vertices of a cube. The electric flux through square surface ABCD of the cube is

    Solution
    Net charge enclosed inside the cube \(=8\left(\frac{q}{8}\right)=q\)
    net electric flux through cube \(=\frac{q}{\varepsilon_{0}}\)
    \(\therefore\) charges are symmetrically placed so the flux through each face \(=\frac{q}{6 \varepsilon_{0}}\)
    KEY CONCEPTS
    Electric flux
    (a) \(\mathrm{d} \phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dS}}\)
    (b) \(\phi \int \quad \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dS}}=E A \cos \theta\) (If electric field is constant over the whole surface)
    (c) Unit of \(\phi=\left(\mathrm{Nm}^{2} / \text { Coulomb }\right)=\mathrm{J} . \mathrm{m} /\) Coulomb
    (d) \([\phi]=\left[\mathrm{M}^{1} \mathrm{L}^{3} \mathrm{T}^{-3} \mathrm{A}^{-1}\right]\)
    Electric Flux In electromagnetism, electric flux is the measure of flow of electric field through a given area. Electric flux is proportional to the number of electric field lines going through a normally perpendicular surface. If the electric field is uniform, the electric flux passing through a surface of vector area is
    \(\Phi_{E}=E . S=E S \cos \theta\)
    Where \(\mathrm{E}\) is the electric field (having units of \(\frac{V}{m}\) ), \(E\) is its magnitude, \(S\) is the area of the surface, and \(\theta\) is the angle between the electric field lines and the normal (perpendicular) to
    For a non-uniform electric field, the electric flux \(d \Phi \boldsymbol{E}\) through a small surface area \(d \boldsymbol{S}\) is given by
    \(d \Phi_{E}=E . d S\)
    (the electric field, \(E\), multiplied by the component of area perpendicular to the field). The electric flux over a surface \(S\) is therefore given by the surface integral:
    \(\Phi_{E}=\iint_{S} E \cdot d S\)
    Where \(E\) is the electric field and \(d S\) is a differential area on the closed surface \(S\) with an outward facing surface normal defining its direction.
    For a closed Gaussian surface, electric flux is given by:
    \(\Phi_{E}=\iint_{S} E . d S=\frac{Q}{\epsilon_{0}}\)
    where
    \(\mathrm{E}\) is the electric field
    S is any closed surface,
    \(\mathrm{Q}\) is the total electric charge inside the surface \(\mathrm{S}\)
    \(\varepsilon_{0}\) is the electric constant (a universal constant, also called the "permitivity of free space")
    \(\left(\varepsilon_{0} \approx 8.854187817 \ldots x 10-12 \text { faradays per meter }\left(F \cdot m^{-1}\right)\right)\)
    This relation is known as Gauss law for electric field in its integral form and it is one of the four Maxwell's equations. While the electric flux is not affected by charges that are not within the closed surface, the net electric field, \(E\), in the Gauss' Law equation, can be affected by charges that lie outside the closed surface. While Gauss' Law holds for all situations, it is only useful for "by hand" calculations when high degrees of symmetry exist in the electric field. Examples include spherical and cylindrical symmetry.Electrical flux has SI units of volt meters (V \(\mathrm{m}\) ), or, equivalently, newton metres squared percoulomb \(\left(\mathrm{N} \mathrm{m}^{2} \mathrm{C}^{-1}\right)\) Thus, the SI base units of electric flux are \(k g \cdot m^{3} \cdot s^{-3} \cdot A^{-1}\). Its dimensional formula is \(\left[L^{3} M T^{-3}\right]\)
    Gauss Law Gauss's law can be stated using either the electric field E or the electric displacement field D. This section shows some of the forms with \(\mathrm{E}\); the form with \(\mathrm{D}\) is below, as are other forms with E.
    Integral form
    Gauss's law may be expressed as:
    \(\Phi_{E}=\frac{Q}{\epsilon_{0}}\)
    where \(\Phi_{E}\) is the electric flux through a closed surface \(\$\) enclosing any volume \(V\), \(Q\) is the total charge enclosed within \(S\), and \(\varepsilon_{0}\) is the electric constant. The electric flux \(\Phi E\) is defined as a surface integral of the electric field:
    \(\Phi_{E}=\iint_{S} E . d A\)
    where \(\mathrm{E} \text { is the electric field, } \mathrm{d} \mathrm{A} \text { is a vector representing an infinitesimal element of area, [note } 1]\) and \(\cdot\) represents the dot product of two vectors.
    since the flux is defined as an integral of the electric field, this expression of Gauss's law is called the integral form.
    Applying the integral form
    If the electric field is known everywhere, Gauss's law makes it quite easy, in principle, to find the distribution of electric charge:
    The charge in any given region can be deduced by integrating the electric field to find the flux.
    However, much more often, it is the reverse problem that needs to be solved: The electric charge distribution is known, and the electric field needs to be computed. This is much more difficult, since if you know the total flux through a given surface, that gives almost no information about the electric field, which (for all you know) could go in and out of the surface in arbitrarily complicated patterns.
    An exception is if there is some symmetry in the situation, which mandates that the electric field passes through the surface in a uniform way. Then, if the total flux is known, the field itself can be deduced at every point. Common examples of symmetries which lend themselves to Gauss's law include cylindrical symmetry, planar symmetry, and spherical symmetry. See the article Gaussian surface for examples where these symmetries are exploited to compute electric fields.
    Differential form
    By the divergence theorem, Gauss's law can alternatively be written in the differential form:
    \(\nabla \cdot E=\frac{\rho}{\varepsilon_{0}}\)
    where \(\nabla \cdot E\) is the divergence of the electric field, \(\varepsilon 0\) is the electric constant, and \(\rho\) is the total electric charge density (charge per unit volume).
    Equivalence of integral and differential forms
    The integral and differential forms are mathematically equivalent, by the divergence theorem. Here is the argument more specifically.
  • Question 9
    1 / -0

    If two mirrors are kept at 60o to each other, then the number of images formed by them is

    Solution

  • Question 10
    1 / -0

    Two identical wires have the same fundamental frequency of 400 Hz when kept under the same tension. If the tension in one wire is increased by 2%, the number of beats produced will be

    Solution

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