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Physics Test - 23

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Physics Test - 23
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  • Question 1
    4 / -1

    A body is dropped from height 'h'. If the coefficient of restitution is 'e', calculate the height achieved (h1) after one bounce.

    Solution
    If a body falls from a height 'h' onto a hard floor and rebounds to height \(h_{1},\) the coefficient of the
    restitution is defined as the ratio of the velocity of approach to the velocity of separation.
    Velocity of approach
    \(v=\sqrt{2 g h}\)
    Velocity of separation
    \(v_{1}=\sqrt{2 g h_{1}}\)
    Coefficient of restitution
    \(e=\frac{v_{1}}{v}=\sqrt{\frac{2 g h_{1}}{2 g h}}\)
    Or \(e=\sqrt{\frac{h_{1}}{h}}\)
    Then,
    \(h_{1}=e^{2} h\)
  • Question 2
    4 / -1

    If the values of each of length, mass and time are doubled, then the work done becomes

    Solution
    \(W=F \cdot s=(m a) \cdot s\)
    \(\mathrm{w}=m \frac{v}{t} s=m \frac{s}{t^{2}} s=\mathrm{m} \frac{s^{2}}{t^{2}}\) (where \(\mathrm{m}\) is the mass, s is the displacement and t is the time)
    Now, mass, length and time are doubled.
    So, \(w^{\prime}=2 m \frac{4 s^{2}}{4 t^{2}}=2 m \frac{s^{2}}{t^{2}}=2 \mathrm{W}\)
  • Question 3
    4 / -1

    A body of mass m1 moving at a constant speed undergoes an elastic collision with a body of mass m2 initially at rest. The ratio of the kinetic energy of mass m1 after the collision to that before the collision is

    Solution
    Let \(u_{1}\) be the speed of mass \(m_{1}\) before the collision. Here \(u_{2}=0 .\) Therefore, the speeds of masses \(m_{1}\) and \(m_{2}\) after the collision respectively are
    \[
    v_{1}=\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right) u_{1}
    \]
    and \(\quad v_{2}=\left(\frac{2 m_{1}}{m_{1}+m_{2}}\right) u_{1}\)
    \(\therefore\) KE of \(m_{1}\) after collision \(=\frac{1}{2} m_{1} v_{1}^{2}=\)
    \(\frac{1}{2} m_{1}\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)^{2} u_{1}^{2} .\) KE of \(m_{1}\) before collision \(=\)
    \(\frac{1}{2} m_{1} u_{1}^{2} .\) The ratio of the two is \(\left(\frac{m_{1}-m_{2}}{m_{1}+m_{2}}\right)^{2}\)
  • Question 4
    4 / -1

    A thin uniform metallic triangular sheet of mass M has sides AB = BC = L as shown in the figure. What is its moment of inertia about axis AC lying in the plane of the sheet?

    Solution
    It is clear from the figure that the moment of inertia of triangular sheet \(A B C=\frac{1}{2} \times\) moment of inertia of a square sheet \(\mathrm{ABCD}\) about its diagonal \(\mathrm{AC}\) or \(\mathrm{I}_{t}=\frac{1}{2} \mathrm{I}_{\mathrm{S}}\)
    Now, mass of square sheet \(=\mathrm{M}+\mathrm{M}=2 \mathrm{M}\).
    Therefore, \(\mathrm{I}_{\mathrm{s}}=(2 \mathrm{M}) \frac{\mathrm{L}^{2}}{12}=\frac{\mathrm{ML}^{2}}{6}\)
    \(\therefore \quad I_{t}=\frac{I_{s}}{2}=\frac{M L^{2}}{12}\)
  • Question 5
    4 / -1

    A body floats in a liquid contained in a beaker. The whole system falls freely under gravity. The upthrust on the body due to the liquid is

    Solution

    For a body falling freely under gravity, the effective value of g is zero. Hence, the upthrust is zero.

  • Question 6
    4 / -1

    A body of densityρis dropped from rest at a height 'h' into a lake of density '', where>ρ. Neglecting all the dissipative forces, calculate the maximum depth to which the body sinks before returning to float on the surface.

    Solution
    Velocity with which the ball strikes the surface of the water:
    \(u=\sqrt{2 g h}\)
    In the water, the resultant force, \(m a=F_{t h}-m g\) \(\rho \mathrm{Va}=\sigma \mathrm{Vg}-\rho \mathrm{Vg} \Rightarrow \mathrm{a}=\left(\frac{\sigma}{\rho}-1\right) \mathrm{g}\)
    Let \(d\) be the depth to which the ball goes. Using, \(v^{2}-u^{2}=2 a s\)
    \(0-2 \mathrm{gh}=-2 \mathrm{d}\left(\frac{\mathrm{o}}{\rho}-1\right) \mathrm{g} \Rightarrow \mathrm{d}=\frac{\rho \mathrm{h}}{(\sigma-\rho)}\)
  • Question 7
    4 / -1

    If the Earth stops rotating, the value of g at the equator will

    Solution

    At the equator, the value of acceleration due to gravity is g' = g - ω2R.

    If the Earth stops rotating, ω = 0, g' = g

    So, the value of acceleration due to gravity will increase.

  • Question 8
    4 / -1
    If e, \(\epsilon_{0},\) h and c respectively are electronic charae. electric permittivity, Planck's constant and velocity of light in vacuum, then what are the dimensions of \(\frac{e^{2}}{4 \pi \varepsilon_{0} h c} ?\)
    Solution
    \(\left[e^{2}\right]=\left[Q^{2}\right]=\left[A^{2} T^{2}\right]\)
    \(\left[e_{0}\right]=\left[M^{-1} L-3 T^{4} A^{2}\right]\)
    \([h]=\left[M L^{2} T^{-1}\right]\)
    \([C]=\left[L T^{-1}\right]\)
    \(\left[\frac{e^{2}}{4 \pi^{2} e_{0} h c}\right]=\frac{\left[A^{2} T^{2}\right]}{\left[M^{-1} L^{-3} T^{4} A^{2}\right]\left[M L^{2} T^{-1}\right]\left[L T^{-1}\right]}\)
    \(=\left[M^{0} L^{0} T^{0}\right]\)
  • Question 9
    4 / -1

    Two particles of masses m and 4 m have kinetic energies in the ratio 2 : 1. What is the ratio of their linear momenta?

    Solution
    Given \((\mathrm{KE})_{1}=\frac{1}{2} m_{1} v_{1}^{2}=2 K\)
    and \((\mathrm{KE})_{2}=\frac{1}{2} m_{2} v_{2}^{2}=K,\) so that
    \(\begin{aligned} \frac{m_{1} v_{1}^{2}}{m_{2} v_{2}^{2}} &=2 \\ \text { or } \frac{m_{1}^{2} v_{1}^{2}}{m_{2}^{2} v_{2}^{2}} &=\frac{2 m_{1}}{m_{2}} \\ \text { or } \frac{p_{1}}{p_{2}} &=\sqrt{\frac{2 m_{1}}{m_{2}}}=\sqrt{\frac{2}{4}}=\frac{1}{\sqrt{2}} \quad\left(\because m_{2}=4 m_{1}\right) \end{aligned}\)
  • Question 10
    4 / -1

    Kepler's second law states that the straight line joining a planet to the sun sweeps out equal areas in equal time. This statement is equivalent to the statement that

    Solution

    Earth and all other planets move around the sun under the effect of gravitational force. This force always acts along the line joining the centre of the planet and the sun and is directed towards the sun. In other words, a planet moves around the sun under the effect of a purely radial force. Therefore, areal velocity of the planet must always remain constant.

    \(\therefore \quad \frac{\Delta \overrightarrow{\mathrm{A}}}{\Delta t}=\frac{\overrightarrow{\mathbf{L}}}{2 m}=\mathrm{a}\) constant vector
    Therefore, Kepler's \(2^{\text {nd }}\) law is the consequence of the principle of conservation of angular momentum
    ()
    \[
    \tau=\frac{d L}{d t}=0
    \]
    \(\mathrm{NoW}\)
    \(\tau=I \mathbf{c}\)
    \(\therefore \quad I \alpha=0 \quad\) or \(\quad \alpha=0\)
    \(a_{T}=r \alpha=0\)
    le, tangential acceleration is zero.
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