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Physics Test - 11

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Physics Test - 11
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  • Question 1
    3 / -1
    Two masses \(m_{1}\) and \(m_{2}\) are suspended together by a mass-less spring of spring constant \(K\). When the masses are in equilibrium, \(m_{1}\) is removed without disturbing the system. The angular frequency and amplitude of oscillation of \(m_{2}\) are
    Solution
    Angular Frequency: \(\omega=\sqrt{\frac{k}{m_{2}}}\)
    Amplitude \(: y=A \sin \sqrt{\frac{k}{m_{2}}}\)
    Step-by-step explanation:
    Mass \(1=m_{1}\)
    Mass \(2=m_{2}\)
    Spring constant = \(k\)
    Angular frequency \(=?\)
    Amplitude \(=?\)
    Solution:
    As it is given that masses are in equilibrium so initially there is no motionwhich means that there will be no angular velocity in the start
    when \(m_{1}\) is removed equilibrium will be disturbed and an angular velocity is produced and it will revolve in a circle
    We know the formula for angular frequency of an object
    \(\omega=\sqrt{\frac{k}{m}}\)
    Where \(k\) is spring constant and \(m\) is the mass of object so for the given
    Angular frequency will be
    \(\omega=\sqrt{\frac{k}{m_{2}}}\)
    Now for amplitude we know the formula
    \(y=A \sin \omega t\)
    Here \(t\) is the time period, \(A\) is amplitude and \(\omega\) is angular frequency
    Putting the value of \(\omega\) here,
    \(y=A \sin \left(\sqrt{\frac{k}{m_{2}}}\right) t\)
  • Question 2
    3 / -1

    A thief stole a box with valuable article of weight W and jumped down a wall of height h. Before he reach the ground he experienced a load of-

    Solution

    While the thief along with the box was in air, the box and the thief experienced same amount of acceleration due to gravity g in downward direction. Thus the box did not exert any force on the thief due to which he did not experience any load due to the box.

  • Question 3
    3 / -1

    Find the amount of work done in rotating a dipole of dipole moment \(3 \times 10^{-3} {~cm}\) from its position of stable equilibrium to the position of unstable equilibrium, in a uniform electric field of intensity \(10^{4} {NC}^{-1}\).

    Solution

    Given that:

    Dipole Moment, \(P=3 \times 10^{-3} {~cm}\)

    Electric Field Intensity, \(E=10^{4} {NC}^{-1}\)

    We know that,

    In rotating the dipole from the position of stable equilibrium by an angle \(\theta\), the amount of work done is given by,

    \(W=P E(1-\cos \theta)\)

    For unstable equilibrium, \(\theta=180^{\circ}\)

    \(\therefore W=P E\left(1-\cos 180^{\circ}\right) \quad\left[\because \cos 180^{\circ}=-1\right]\)

    \(=2 P E\)

    \(=2 \times 3 \times 10^{-3} \times 10^{4} J\)

    \(=60 ~J\)

  • Question 4
    3 / -1

    Why can't we explain the photoelectric effect by the wave theory of light?

    Solution

    The phenomena of interference, diffraction, and polarisation were explained in a natural and satisfactory way by the wave picture of light.

    According to this picture, light is an electromagnetic wave consisting of electric and magnetic fields with a continuous distribution of energy over the region of space over which the wave is extended.

    According to the wave picture of light, the maximum kinetic energy of the photoelectrons on the surface of the metal is expected to increase with the increase in intensity.Also, no matter what the frequency of radiation is, a sufficiently intense beam of radiation (over sufficient time) should be able to impart enough energy to the electrons so that they exceed the minimum energy needed to escape from the metal surface.

    A threshold frequency, therefore, should not exist.But the experimental results show that the maximum kinetic energy of the photoelectrons does not depend on the intensity of radiation and the minimum frequency is required for the photoelectric effect.The minimum frequency that is required for the photoelectric effect is called the threshold frequency.

    The experimental results show that the maximum kinetic energy of the photoelectrons does not depend on the intensity of radiation. Further, we should note that in the wave picture, the absorption of energy by electrons takes place continuously over the entire wavefront of the radiation. Since a large number of electrons absorb energy, the energy absorbed per electron per unit time turns out to be small.

    Explicit calculations estimate that it can take hours or more for a single electron to pick up sufficient energy to overcome the work function and come out of the metal. But the experimental results show that the photoelectric emission is instantaneous. From the above explanation, we can say that the wave picture is unable to explain the most basic features of photoelectric emission.

  • Question 5
    3 / -1

    A paramagnetic sample shows a net magnetisation of \(6 {A} / {m}\) when it is placed in an external magnetic field of \(0.4 {T}\) at a temperature of \(4 {K}\). When the sample is placed in an external magnetic field of \(0.3 {T}\) at a temperature of \(24 {K}\), then the magnetisation will be:

    Solution

    Given,

    \(T_{1}=4K,~T_{2}=24K\)

    For paramagnetic material,

    According to curies law,

    \(M=\chi H, \quad \chi=\frac{C}{T}\)

    Where

    \(\chi>0\) is the (volume) magnetic susceptibility,

    \(M\) is the magnitude of the resulting magnetization (A/m),

    \(H\) is the magnitude of the applied magnetic field \((\mathrm{A} / \mathrm{m})\)

    \(T\) is absolute temperature \((\mathrm{K})\),

    \(C\) is a material-specific Curie constant \((\mathrm{K})\).

    \(\chi \propto \frac{1}{{T}}\)

    \( \Rightarrow \chi_{1} {T}_{1}=\chi_{2} {T}_{2}\)

    \(\Rightarrow \frac{6}{0.4} \times 4=\frac{{I}}{0.3} \times 24\)

    \({I}=\frac{0.3}{0.4}=0.75 {A} / {m}\)

  • Question 6
    3 / -1

    The stopping potential for material \(A\) is more than material \(B\) when a light of frequency \(\mathrm{V}\) is incident on both the materials. If the work function of material \(A\) and material \(B\) is \(\phi_{A}\) and \(\phi_{B}\) respectively, then:

    Solution

    When the stopping potential is \(V_{0}\), the maximum kinetic energy of an electron is given as,

    \( KE_{\max }=\mathrm{eV}_{0}\).....(1)

    So Einstein's photoelectric equation can be written as:

    \(e V_{0}=h v-\phi_{0}\)

    \(\Rightarrow \mathrm{eV}_{0}+\phi_{0}=\mathrm{hv}\).....(2)

    For material A,

    \(\mathrm{eV}_{\mathrm{OA}}+\phi_{\mathrm{A}}=\mathrm{hv} \)......(3)

    For material B,

    \(\mathrm{eV}_{\mathrm{OB}}+\phi_{\mathrm{B}}=\mathrm{hv}\).....(4)

    \(\because V_{O A}>V_{O B}\)

    So by equation 3 and equation 4 ,

    \(\phi_{A}<\phi_{B}\)

  • Question 7
    3 / -1

    If F1 is the force exerted by the earth on the moon and F2 is the force exerted by the moon on earth. Which force is greater according to newton’s law of gravitation?

    Solution

    Given,

    \(F_{1}=\) Force exerted by the earth on the moon

    \(F_{2}=\) Force exerted by the moon on earth

    The magnitude of the gravitational force acting on the moon by the earth is

    \(F_{1}=G \times \frac{M_{\text{moon}} M_{\text {earth }}}{r^{2}}\)...(1)

    The magnitude of the gravitational force acting on earth by the moon is

    \(F_{2}=G \times \frac{M_{\text {earth }} M_{\text {moon }}}{r^{2}}\)...(2)

    Dividing equation (1) by (2), we get

    \(\frac{F_{1}}{F_{2}}=\frac{G \times \frac{M_{\text {moon}} M_{\text {earth }}}{r^{2}}}{G \times \frac{M_{\text {earth } M_{\text {moon }}}}{r^{2}}}\)

    \(\Rightarrow \frac{F_{1}}{F_{2}}=1\)

    \(\therefore F_{1}=F_{2}\)

    So, according to Newton's law of gravitation, the force of attraction between two bodies on each other is the same in magnitude (irrespective of shape and size of the body).

  • Question 8
    3 / -1

    The reflecting surface is given by \(\mathrm{y}=\frac{10 \mathrm{~L}}{\pi} \sin \frac{\pi \mathrm{x}}{5 \mathrm{~L}}\). The co-ordinates of the point where a horizontal ray becomes vertical after reflection is:

    Solution

    Given:

    \(\mathrm{y}=\frac{10 \mathrm{~L}}{\pi} \sin \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right) \).....(i)

    According to the question co-ordinates of the point where a horizontal ray becomes vertical after reflection.

    Differentiating equation (i) w.r.t. to \(x\). we get:

    \(\frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}=\frac{\ 10 \mathrm{~L}}{\pi} \cos \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right) \times \frac{ \pi}{ \mathrm{5L}} \times 2\)

    \(=2 \cos \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right) \)

    Since the horizontal ray becomes vertical \(\frac{\mathrm{dy}}{\mathrm{dx}}=1\)

    So,

    \(2 \cos \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right)=1 \)

    \(\Rightarrow \cos \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right)=\frac{1}{2} \)

    \(\Rightarrow \cos\left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right)=\cos \frac{\pi}{3} \)

    \(\Rightarrow \left(\frac{\pi \mathrm{x}}{5 \mathrm{~L}}\right)=\frac{\pi}{3} \)

    \(\mathrm{x}=\frac{5 \mathrm{~L}}{3} \)

    Put the value of \(x\) in equation (i),

    \(\mathrm{y}=\frac{10 \mathrm{~L}}{\pi} \operatorname{sin}\left(\frac{\pi \times 5 \mathrm{~L}}{3 \times 5 \mathrm{~L}}\right) \)

    \(y=\frac{5 \sqrt{3}L}{\pi}\)

  • Question 9
    3 / -1

    Three thermodynamic systems are at a temperature of 60C. What can be concluded about them?

    Solution

    The three bodies are at the same temperature and hence, there will be no heat flow between them.Therefore, three bodies are in thermal equilibrium with each other.

    The zeroth law of thermodynamics explains the conditions for systems to be in thermal equilibrium with each other. So, the given systems obey zeroth law as well.

    Therefore, the correct answer is All of the above.

  • Question 10
    3 / -1

    The ratio of magnetic field and magnetic moment at the centre of a current carrying circular loop is X. When both the current and radius is doubled the ratio will be

    Solution

    Given,

    The ratio of the magnetic field and magnetic moment at the center of a current-carrying circular loop = X

    We have to find the ratio when both the current and radius is doubled.

    Magnetic field at the centre of current carrying loop is given by

    \(B=\frac{\mu_{0} I}{2 r}\)....(i)

    where: \(I\)=current

    r=radius

    Let M be the magnetic moment of the current carrying loop, so

    \(M=I A=I\left(\pi {r}^{2}\right)\).....(ii) \(\quad\left[\text { as } {A}=\pi {r}^{2}\right]\)

    Dividing eq. (i) by eq. (ii), we get \(\frac{B}{M}=\frac{\mu_{0} I}{2 r}\left[\frac{1}{I\left(\pi r^{2}\right)}\right]\)

    \(\Rightarrow X=\frac{B}{M}=\frac{\mu_{0}}{2 \pi r^{3}}\)

    \(X \alpha \frac{1}{r^{3}}\)

    Thus, the ratio is independent of current and inversely proportional to the cube of radius. When the radius is doubled i.e. \(r^{\prime}=2 r,\) then the new ratio is

    \(\Rightarrow X^{\prime} \alpha \frac{1}{8 r^{3}}\)

    \(\Rightarrow X^{\prime}=\frac{X}{8}\)

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