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Physics Test - 13

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Physics Test - 13
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  • Question 1
    3 / -1

    A ball is rolling on a table without slipping. Its part of energy associated with rotational motion will be:

    Solution

    Kinetic Energy: The energy associated with the body in motion is called kinetic energy.

    Linear Kinetic Energy: The kinetic energy associated with the linear speed of an object is called linear kinetic energy. It is given by

    \(K_{l}=\frac{1}{2} m v^{2}\)

    \(\mathrm{m}\) is mass and \(\mathrm{v}\) is the speed of the body

    Rotational Kinetic Energy: The kinetic energy associated with the rotational motion of the body is called rotational kinetic energy.

    \(K_{r}=\frac{1}{2} I \omega^{2}\)

    \(I\) is the moment of inertia, and \(\omega\) is the angular velocity.

    Moment of Inertia: The inertia associated with the rotational motion of an object is called the moment of inertia. It is measured along a particular axis.

    Moment of Inertia of a sphere with radius \(r\) is given as

    \(I=\frac{2}{5} m r^{2}-(1)\)

    Angular velocity: The rate of angular displacement with respect to time is called angular velocity. For a body in rotational motion with radius \({r}\), angular velocity is given as

    \(w=\frac{v}{r} \cdots(2)\)

    Rotational Kinetic Energy of ball

    \(K_{R}=\frac{1}{2} I_{\omega}^{2}\)

    \(K_{R}=\frac{1}{2} \frac{2}{5} m r^{2} \times\left(\frac{v}{r}\right)^{2} \text { using Eq (1) and (2) }\)

    \(K_{R}=\frac{1}{5} m v^{2}-(3)\)

    The total energy of the body in rotational motion is equal to the sum of liner kinetic energy and rotational kinetic energy.

    \(E=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}-(4)\)

    Putting Equation (3) in (4)

    \(E=\frac{1}{2} m v^{2}+\frac{1}{5} m v^{2}\)

    \(E=\frac{5+2}{10} m v^{2}=\frac{7}{10} m v^{2}\)

    So, the total energy associated with the ball is

    \(E=\frac{7}{10} m v^{2}-(5)\)

    Now, to get the energy ratio \(\mathrm{R}\) of rotational kinetic energy to the total energy we divide (3) by (5)

    \(R=\frac{\frac{1}{5} m v^{2}}{\frac{7}{10} m v^{2}}\)

    \(R=\frac{1}{5} \times \frac{10}{7}\)

    \(R=\frac{2}{7}\)

    So, the required fraction is \(\frac{2}{7}\).

  • Question 2
    3 / -1

    In a series resonant circuit, the AC voltage across resistance \(\mathrm{R}\), inductor \(\mathrm{L}\) and capacitor \(\mathrm{C}\), are \(5 \mathrm{~V}, 10 \mathrm{~V}\) and \(10 \mathrm{~V}\) respectively. The AC voltage applied to the circuit will be:

    Solution
    Given, Voltage across resistor, \(V_{R}=5 V\)
    Voltage across inductor, \(V_{L}=10 V\)
    Voltage across capacitor, \(V_{C}=10 {V}\)
    the \(\mathrm{AC}\) voltage applied to the circuit is given as
    \(V=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}\)
    Substituting the given values, we get,
    \(=\sqrt{(5)^{2}+(10-10)^{2}}\) \(=5 V\)
  • Question 3
    3 / -1

    Gauss's law for magnetism states that, the net _______ through any closed surface is zero.

    Solution

    Gauss's law for magnetism: The net magnetic flux of the magnetic field must always be zero over any closed surface.

    This means that as many field lines should seem to enter the surface as the number of lines leaving it.

    Net magnetic flux \(\left(\phi_{{B}}\right)=\sum {B} \cdot \Delta {S}=0\)

    Integral form: \(\int B . n d s=0\)

    Where \(B\) is the magnetic field and \(S\) is surface area.

  • Question 4
    3 / -1

    Work energy theorem is the “Work done by ______ force acting on the particle is equal to change in kinetic energy”.

    Solution

    Work energy theorem: Work done by net force acting on the particle is equal to change in kinetic energy So,

    Work done \(=(\mathrm{KE})_{\text {final }}-(\mathrm{KE})_{\text {initial }}=\Delta \mathrm{KE}\)

    From the above definition it is clear that work done by net force is the reason to change the kinetic energy of particle.

  • Question 5
    3 / -1
    A compound lens is formed by combining one convex and one concave lens of the same focal length \(20 \mathrm{~cm}\). This compound lens is used to see a \(5 \mathrm{~cm}\) high object situated at a distance of \(20 \mathrm{~cm}\). The image will be:
    Solution

    Given:

    Focal length of the concave lens, \(f_{1}=-20 \mathrm{~cm}\)

    Focal length of the convex lens, \(f_{2}=20 \mathrm{~cm}\)

    The effective focal length \({f}\) of a combination of two lenses having focal length \(f_{1}\) and \(f_{2}\)

    \( \frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

    The lens maker's formula for thin lens is

    \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

    Where \(f\) is the focal length of the lens, \(u\) is the object distance from the lens and \(v\) is the image distance from the lens.

    The magnification produced by a lens or combination of the lens is given by \(m=\frac{v}{u}\)

    Where \(u\) is the object distance from the lens and \(v\) is the image distance from the lens.

    Therefore, the effective focal length of the concavo-convex lens combination is:

    \( \frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

    \(\Rightarrow \frac{1}{f}=\frac{1}{-20}+\frac{1}{20}\)

    \(\Rightarrow \frac{1}{f}=0\)

    \(\Rightarrow f=\infty\)

    So, the effective focal length is infinite.

    The object is kept at a distance of \(20 \mathrm{~cm}\) from the lens combination.

    So, \(u=20 \mathrm{~cm}\)

    So, the image distance is equal to the object distance, that is, \(v=20 \mathrm{~cm}\).

    Therefore, magnification produced by the lens is given by,

    \( m=\frac{v}{u}\)

    \(\Rightarrow m=\frac{20}{20}\)

    \(\Rightarrow m=1\)

    Therefore, the image is of the same size as the object and is erect.

  • Question 6
    3 / -1

    Value of forbidden energy gap for semi conductor is:

    Solution

    Germanium and Silicon are the most preferable material whose electrical properties lie in between semiconductors and insulators. 

    The energy band diagram of the semiconductor is shown where the conduction band is empty and the valence band is completely filled but the forbidden gap between the two bands is very small which is about 1eV.

  • Question 7
    3 / -1

    ________ is defined as the rate of doing work.

    Solution

    Power is defined as the rate of doing work or the rate of transfer of energy. The standard unit is the watt, symbolized by W.

    Energy is the capacity for doing work. All forms of energy are either kinetic or potential.Force is any interaction that, when unopposed, will change the motion of an object.Momentum refers to the quantity of motion that an object has.

  • Question 8
    3 / -1
    The relationship between relative magnetic permeability '\(\mu_{r}\) of a substance and its magnetic susceptibility \(\chi_{m}\)' is given by:
    Solution

    Magnetic susceptibility \(\chi_{m}\) indicates the intensity or the degree of magnetization (I) of a material in response to the intensity of the magnetizing force \((\mathrm{H})\) of the applied magnetic field (B).

    It is given as: \(\chi_{m}=\frac{I}{H}\)

    Magnetic permeability is a property of the material that characterizes the amount of magnetizing force (H) that a material experiences under the influence of an applied magnetic field (B).

    It is given as: \(\mu=\frac{B}{H}\)

    Relative permeability is the ratio of the permeability of the material \((\mu)\) to that of free \(\operatorname{space}\left(\mu_{0}\right)\).

    It is given as: \(\mu_{r}=\frac{\mu}{\mu_{0}}\)

    The net field will be a resultant of magnetization due to applied field \(\left(B_{0}\right)\) and magnetization due to induced magnetism \(\left(B_{i}\right)\)

    \(B=B_{0}+B_{i}\)

    Where, \(B_{0}=\mu_{0} H\) and \(B_{i}=\mu_{0} I\)

    \( B=\mu_{0}(H+I) \)

    \(\Rightarrow \frac{B}{H}=\mu_{0}\left(1+\frac{I}{H}\right)\)

    Substituting \(\mu=\frac{B}{H}\) and \(\chi_{m}=\frac{I}{H}\) in the above equation, we get:

    \(\mu=\mu_{0}\left(1+\chi_{m}\right) \)

    \(\Rightarrow \frac{\mu}{\mu_{0}}=\left(1+\chi_{m}\right) \)

    \(\Rightarrow \mu_{r}=\left(1+\chi_{m}\right)\)

    The relationship between relative magnetic permeability '\(\mu_{r}\) of a substance and its magnetic susceptibility \(\chi_{m}\)' is given by\(\mu_{r}=\left(1+\chi_{m}\right)\).

  • Question 9
    3 / -1

    A force applied on a body is represented as\(\vec{F}=6 \hat{i}-8 \hat{j}+10 \hat{k}\)and accelerates it at \(1 {~m} / {s}^{2}\). The mass of the body is:

    Solution

    Given,

    A force vector applied on a mass is represented as:

    \(\vec{F}=6 \hat{i}-8 \hat{j}+10 \hat{k}\)

    Here,

    \({a}=1 {~m} / {s}^{2}\)

    \(\vec{F}\) is the force vector.

    The acceleration of the body, \(a=1 {~ms}^{-2}\)

    Newton’s second law of motion,

    \(F=m \times a\)...(1)

    Here we need to find the mass of the body, so keep mass \(m\) in one side and other terms in other side,

    \(m=\frac{F}{a}\)......(2)

    Substitute the force vector value and the acceleration value in the equation (2),

    Then,

    \(m=\frac{6 \hat{i}-8 \hat{j}+10 \hat{k}}{1}\)

    Now the above equation is written as,

    \( m=6 \hat{i}-8 \hat{j}+10 \hat{k}\)

    Now the equation is in vector form, to remove the vector form, take modulus on both sides,

    \(|m|=|6 \hat{i}-8 \hat{j}+10 \hat{k}|\)

    On further,

    \(|m|=\sqrt{(6)^{2}+(-8)^{2}+(10)^{2}}\)

    By using the square inside the square root,

    \(|m|=\sqrt{36+64+100}\)

    By adding the terms inside the square root,

    \(|m|=\sqrt{200}\)

    For easy simplification of square root, the above equation can be written as,

    \(|m|=\sqrt{100 \times 2}\)

    On further simplification,

    \(|m|=10 \sqrt{2}\)

    Then, the mass of the body, \(m=10 \sqrt{2} \mathrm{~kg}\).

    Hence, the correct option (B).

  • Question 10
    3 / -1

    Two capillary tubes are dipped in a liquid. If water level in capillary \(A\) is \(\frac{2}{3}\) rd of capillary B. Find ratio of their radius \(r_{A}\) : \(r_{B}\)

    Solution

    Given,

    \(h_{A}=\frac{2}{3} h_{B}\) i.e. \(\frac{h_{A}}{h_{B}}=\frac{2}{3}\)

    In a capillary tube, the rise in water is:

    \(h=\frac{2 T \cos \theta}{r d g}\)

    \(h \propto \frac{1}{r}\) (all other parameters are the same for both the capillaries)

    \(\frac{h_{A}}{h_{B}}=\frac{r_{B}}{r_{A}} \)

    \(\frac{2}{3}=\frac{r_{B}}{r_{A}} \)

    \(\frac{r_{A}}{r_{B}}=\frac{3}{2}\)

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