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Physics Test-27

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Physics Test-27
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  • Question 1
    4 / -1

    Bernoulli’s equation is applied to ____________.

    Solution

    Bernoulli’s equation refers to the conservation of energy.

    All of the above are the measuring devices like Venturimeter, Orifice meter, and Pitot tube meter works on the Bernoulli’s theorem.

    Orifice Meter:

    A pipe-based orifice meter measures the flow rate of moving fluids (liquids or gases) in the pipeline. It is a topic of fluid machinery.

    Venturi meter:

    Venturi meters measure the rate of liquid flow by converting pressure energy into kinetic energy. Clemans Herchel, an American engineer, created it, and Giovanni Venturi, an Italian physicist, named it. It builds upon Bernoulli's Equation as its foundation.

  • Question 2
    4 / -1

    Which of the following statement regarding electric field lines is correct?

    Solution

    If two lines cross each other and we take tangents line on each curve at that point then we will get two directions of the electric field at that same point which is not possible. Since an electric field can have only one direction at a single point so the two electric field lines never cross each other.

  • Question 3
    4 / -1

    In an unbiased \(n-p\) junction electrons diffuse from \(n\)-region to \(p\) region because :

    Solution

    In p-type materials holes are majority carriers and electrons are majority carriers in n-type materials. When the two types of semiconductor materials are joined together, the electrons from the \(\mathrm{n}\)-type material diffuse into p-type material and combines with holes as their concentration is higher in \(n\)-type layer. This creates a layer of negative ions near the junction in p-type material. Negative ions are formed because the trivalent impurities (e.g. Aluminum) now has an extra electron from the ntype material. Similarly, the holes from the p-type material diffuse into \(n\)-type material resulting in a layer of positive ions in the \(n-\) type material.

    These negative ions creates an electric field in the direction from \(\mathbf{n}\)-type to p-type. As more electrons diffuse into p-type material, the electric field strength goes on increasing. The electrons from n-type material now diffusing into p-type material will have to overcome the electric field due to negative ions. At one point, the electric field becomes sufficiently strong to stop further diffusion of electrons.

    Hence, the correct option is (B)

  • Question 4
    4 / -1

    A permanent retarding force of \(50 N\) is applied on a body of mass \(20 kg\) moving with an initial speed of \(15 ms^{-1}\). How long will it take for the body to stop?

    Solution

    Given,

    Retarding force, \(F=-50 N\)

    Mass of the body, \(m=20 kg\)

    Initial velocity, \(u=15 ms ^{-1}\)

    Final velocity, \(v=0\)

    Acceleration in the body, \(a=\frac{F}{m}\)

    \(=\frac{-50}{20}=-2.5 ms ^{-2}\)

    From the first equation of motion

    \(v=u+a t\)

    \(0=15+(-2.5) \times t\)

    \(t=\left(\frac{15}{2.5}\right)\)

    \(t=6 s\)

  • Question 5
    4 / -1

    Total heat is the heat required to _________.

    Solution

    Total heat is the heat required to Convert water into steam and superheat it.

    Total Heat is the thermal equivalent of the energy required to convert unit mass of a liquid at one temperature (as the melting point of the substance) into saturated vapor at any other given temperature.

  • Question 6
    4 / -1

    In Young's double slit experiment, the two equally bright slits are coherent, but of phase difference \(\frac{\pi}{3}\). If maximum intensity on the screen is \(I_0\), the intensity at the point on the screen equidistant from the slits is:

    Solution

    Given: \(\phi=\frac{\pi }{ 3}\)

    \(I=I_{\max } \cos ^2\left(\frac{\phi}{2}\right)\)

    \(I=I_0 \cos ^2\left(\frac{\frac{\pi }{ 3}}{2}\right)\)

    \(= I_0 \cos ^2(\frac{\pi }{ 6})\)

    \(= I_0\left(\frac{\sqrt{3}}{2}\right)^2\left(\because \cos \frac{\pi }{ 6}=\frac{\sqrt{3}}{2}\right)\)

    \(= I _0 \frac{3}{4}\)

    So, intensity on the screen equidistant from the slats is \(\frac{3}{4} I_0\).

    Hence, the correct option is (C).

  • Question 7
    4 / -1

    Which of the following pairs of physical quantities does not have same dimensional formula?

    Solution

    Option (C):

    Tension (force) \(=\left[M L T^{-2}\right]\)

    Surface tension \(=\frac{\text { force }}{\text { length }}\)

    \(=\frac{\left[M L T^{-2}\right]}{[L]}=\left[M L^{0} T^{-2}\right]\)

    Option (A):

    Work \(=F \times \Delta x\)\(=\left[\mathrm{MLT}^{-2}[\mathrm{~L}]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]\right]\)

    The dimension of distance, \(\Delta x = \left[M L T^{-2}\right]\)

    \(=\left[M L T^{-2}\right][L]=\left[M L^{2} T^{-2}\right]\)

    Torque = force \(\times\) distance

    \(=\left[M L^{2} T^{-2}\right]\)

    Option (B):

    Angular momentum \(=m v r\)

    \(=[M]\left[L T^{-1}\right][L]=\left[M L^{2} T^{-1}\right]\)

    Planck's constant \(=\frac{E}{\nu}\)

    \(=\frac{\left[M L^{2} T^{-2}\right]}{\left[T^{-1}\right]}=\left[M L^{2} T^{-1}\right]\)

    Option (D):

    Impulse \(=F \times \Delta t\)

    \(=\left[M L T^{-2}\right][T]=\left[M L T^{-1}\right]\)

    Linear momentum \(=\) mass \(\times\) velocity

    \(=[M]\left[L T^{-1}\right]=\left[M L T^{-1}\right]\)

    So, among the above pairs only tension and surface tension does not have same dimensional formula. They both sound similar but they both have different meaning and different applications.

  • Question 8
    4 / -1

    The first law of motion is also known as the law of:

    Solution

    Newton's first law of motion is also known as the law of inertia. Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. It may be seen as a statement about inertia, that objects will remain in their state of motion unless a force acts to change the motion.

  • Question 9
    4 / -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 \(10^{4} NC C ^{-1}\).

    Solution

    Given that:

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

    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 10
    4 / -1
    A capacitor and an inductor is connected in series with an ac source in a circuit. If the capacitance of the capacitor is \(18 \mu \mathrm{F}\) and the inductance of the inductor is \(8 H\) then find the resonance frequency of the circuit.
    Solution

    Given,

    Capacitance \((\mathrm{C})=18 \mu \mathrm{F}\)

    \(=18 \times 10^{-6} \mathrm{~F}\)

    Inductance \((\mathrm{L})=8 \mathrm{H}\)

    Now,

    Resonance frequency \((f)=\frac{1}{2 \pi \sqrt{L C}}\)

    \(=\frac{1}{2 \pi \sqrt{8 \times 18 \times 10^{-6}}}\)

    \(=\frac{1}{2 \pi \sqrt{144} \times 10^{-3}}\)

    \(=\frac{1000}{24 \pi}\)

    \(=\frac{125}{3 \pi}\)

    This resonance frequency \((f)=\frac{125}{3 \pi} \mathrm{Hz}\).

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