Self Studies

Physics Test - 12

Result Self Studies

Physics Test - 12
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    3 / -1

    A sonometer wire, with a suspended mass of \(M=1 \mathrm{~kg}\), is in resonance with a given tuning fork. The apparatus is taken to the moon where the acceleration due to gravity is \(\frac{1}{6}\) that on earth. To obtain resonance on moon, the value of \(M\) should be

    Solution

    The suspended mass \(M\) on the sonometer wire is \(1 \mathrm{~kg}\).

    \(M=1 \mathrm{~kg}\)

    The resonance tension \(T\) of the sonometer wire is balanced by the weight \(M g\) of the suspended mass.

    \(T=M g\)

    Substitute \(M g\) for \(T\) in equation (1).

    \(f=\frac{1}{2 l} \sqrt{\frac{M g}{m}}\)

    Rewrite the above equation for the resonance frequency \(f^{\prime}\) on the moon.

    \(f^{\prime}=\frac{1}{2 l} \sqrt{\frac{M^{\prime} g^{\prime}}{m}}\)

    Here, \(g^{\prime}\) is the acceleration due to gravity on the moon.

    The acceleration due to gravity \(g^{\prime}\) on the moon is \(\frac{1}{6}\) that on the earth.

    \(g^{\prime}=\frac{1}{6} g\)

    We want to obtain the same resonance frequency on the earth and the moon.

    \(f=f^{\prime}\)

    Substitute \(\frac{1}{2 l} \sqrt{\frac{M g}{m}}\) for \(f\) and \(\frac{1}{2 l} \sqrt{\frac{M^{\prime} g^{\prime}}{m}}\) for \(f^{\prime}\) in the above equation.

    \(\frac{1}{2 l} \sqrt{\frac{M g}{m}}=\frac{1}{2 l} \sqrt{\frac{M^{\prime} g^{\prime}}{m}}\)

    \(\Rightarrow \sqrt{M g}=\sqrt{M^{\prime} g^{\prime}}\)

    \(\Rightarrow M g=M^{\prime} g^{\prime}\)

    Rearrange the above equation for \(M^{\prime}\).

    \(M^{\prime}=\frac{M g}{g^{\prime}}\)

    Substitute \(\frac{1}{6} g\) for \(g^{\prime}\) and \(1 \mathrm{~kg}\) for \(M\) in the above equation.

    \(M^{\prime}=\frac{(1 \mathrm{~kg}) g}{\frac{1}{6} g}\)

    \(\Rightarrow M^{\prime}=6 \mathrm{~kg}\)

    Therefore, the mass \(M\) on the moon should be \(6 \mathrm{~kg}\). 

  • Question 2
    3 / -1

    The moment of inertia of a circular ring of radius ' \(\mathrm{R}\)' and mass '\(\mathrm{M}\)' about diameter is:

    Solution

    We know that moment of inertia of a ring through its center perpendicular to the ring plane is \(=\mathrm{MR}^{2}\)

    \(I_{z}=M R^{2}\)

    By perpendicular axis theorem

    \(\mathrm{I}_{\mathrm{z}}=\mathrm{I}_{\mathrm{x}}+\mathrm{I}_{\mathrm{y}}\)

    \(M R^{2}=I_{x}+I_{y}\)

    By symmetry \(I_{x}=I_{y}=I\)

    \(\mathrm{MR}^{2}=I+I\)

    \(\mathrm{I}=\frac{\mathrm{MR}^{2}}2\)

    The moment of inertia of a circular ring of radius 'R' and mass 'M' about diameter is \(\frac{\mathrm{MR}^{2}}2\).

  • Question 3
    3 / -1

    The work function of a metal surface is \(3.5 \mathrm{eV}\). If this surface is illuminated by monochromatic light of frequency \(6.4 \times 10^{14} \mathrm{~Hz}\), then find the maximum kinetic energy of the photoelectrons.

    Solution

    Given

    \(\phi_{a}=3.5 \mathrm{eV}\), and \(v=6.4 \times 10^{14} \mathrm{~Hz}\)

    We know that the energy of a photon is given as,

    \(E=h v\)....(1)

    Where

    \(h=\) Planck's constant \(=6.63 \times 10^{-34} \mathrm{~J}\)-sec

    \(\mathrm{v}=\) Frequency

    \(\phi_{a}=\) Work function

    By equation (1) the energy of the photon is given as,

    \(E=6.63 \times 10^{-34} \times 6.4 \times 10^{14}\)

    \(E=6.63 \times 6.4 \times 10^{-20} \mathrm{~J}\)

    \(E=\frac{6.63 \times 6.4 \times 10^{-20} \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}\)

    \(E=2.652 \mathrm{eV}\)...(2)

    According to Einstein's equation of photoelectric effect:

    \(KE_{\max}=E-\phi_{a}\)

    \(KE_{\max }=2.652-3.5\)

    \(KE_{\max }=-0.8 ~\mathrm{eV}\)

    Since the maximum kinetic energy is negative, therefore we can say that no photoelectron will emit.

  • Question 4
    3 / -1

    Action and reaction forces are:

    Solution

    Newton's third law: This law states that for every action there is an equal and opposite reaction. Action and reaction forces act on two different bodies. Action and reaction forces are not balanced because they act on two different bodies. If a pair of equal and opposite forces act on the same body then it is not an action-reaction pair.

  • Question 5
    3 / -1

    Two stones are thrown up simultaneously from the edge of a cliff \(240~m\) high with initial speed \(10~m/s\) and \(40~m/s\) respectively. Which of the following graph best represents the time variation of the relative position of the second stone with respect to the first?

    (Assume stones do not rebound after hitting the ground and neglect air resistance, take \(g=10~m/s^{2}\))

    Solution

    The concept of relative motion can be applied to predict the nature of the motion of one particle with respect to the other. Consider the stones thrown up simultaneously as shown in the diagram below.

    Considering the motion of the second particle with respect to the first we have relative acceleration.
    \(\left|a_{21}\right|=\left|a_{2}-a_{1}\right|=g-g=0\)
    Thus, motion of first particle is straight line with respect to second particle till the first particle strikes ground at a time given by
    \(-240=10 t-\frac{1}{2} \times 10 \times t^{2}\) or \(t^{2}-2 t-48=0\) or \(t^{2}-8 t+6 t-48=0\) or \(t=8-6\) (not possible)
    Thus, distance covered by second particle with respect to first particle in \(8~s\) is \(S_{12}=\left(v_{21}\right) t =(40-10)(8~s) =30 \times 8 =240 ~m\)
    Similarly, time taken by second particle to strike the ground is given by, 
    \(-240 =40 t-\frac{1}{2} \times 10 \times t^{2}\) or \(240=40 t-5 t^{2}\) or \(5t^{2}-40 t-240=0\) or \(t^{2}-8 t-48=0\)
    \(t^{2}-12 t+4 t-48=0\) or \(t(t-12)+4(t-12)=0\) or \(t=12,-4 \quad\)(not possible)
    Thus, after 8 s, the magnitude of relative velocity will increase up to \(12~s\) when the second particle strikes the ground.

  • Question 6
    3 / -1

    A capillary tube of radius \(0.5 \mathrm{~mm}\) is dipped vertically in a liquid of surface tension \(0.04 \mathrm{~N} / \mathrm{m}\) and density \(0.8 \mathrm{~g} / \mathrm{cc}\). Calculate the height of the capillary rise, if the angle of contact is \(10^{\circ} .\left(\cos 10^{\circ}\right.\) \(=0.98, g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) )

    Solution

    Given,

    Radius of capillary tube \(=r=0.5 \mathrm{~mm}=0.5 \times 10^{-3} \mathrm{~m}\)

    Surface tension \(=T=0.04 \mathrm{~N} / \mathrm{m}\)

    Density \(=\rho=0.8 \mathrm{~g} / \mathrm{Cc}=0.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}, \cos 10^{\circ}=0.98\)

    \(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\)

    \(\mathrm{h}=\frac{2 \mathrm{~T} \mathrm{} \cos \theta}{\mathrm{r} \rho \mathrm{g}} \)

    \(\mathrm{h}=\frac{2 \times 0.04 \times 0.98}{0.5 \times 10^{-3} \times 0.8 \times 10^{3} \times 9.8}=0.02 \mathrm{~m}=2 \mathrm{~cm} \)

    \(\mathrm{~h}=2 \mathrm{~cm} \)

  • Question 7
    3 / -1

    Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of \(30^{\circ}\) with each other. When suspended in a liquid of density \(0.8 \mathrm{gcm}^{-3}\), the angle remains the same. If density of the material of the sphere is \(1.6 \mathrm{gcm}^{-3}\), the dielectric constant of the liquid is:

    Solution
    Initially, the forces acting on each ball are:
    (i) Tension \(\mathrm{T}\)
    (ii) Weight mg
    (iii) Electrostatic force of repulsion F
    For its equilibrium along vertical,
    \(\operatorname{Tcos} \theta=\mathrm{mg} \ldots\)...(i)
    and along horizontal, \(\operatorname{Tsin} \theta=\mathbf{F}_{\ldots}\) (ii)
    Dividing equation (ii) by (i), we get,
    \(\tan \theta=\mathrm{F} / \mathrm{mg} \quad \ldots\)...(III)
    When the balls are suspended in a liquid of density \(\sigma\) and dielectric constant \(\mathrm{K}\), the electrostatic force will become \((\frac{1}{\mathrm{K}})\) times, i.e., \(\mathrm{F}^{\prime}=\) \((\frac{\mathrm{F}}{ \mathrm{K}})\) while weight,
    \(m g^{\prime}=m g-\) Upthrust
    \(=m g-V \sigma g \quad\) [As Upthrust = \(V \sigma g]\)
    \(m g^{\prime}=m g\left[1-\frac{\sigma}{\rho}\right] \quad\left[\right.\) As \(\left.V=\frac{m}{\rho}\right]\)
    For equilibrium of balls,
    \(\tan \theta^{\prime}=\frac{F^{\prime}}{m g^{\prime}}=\frac{F}{K m g[1-(\frac{\sigma}{\rho})]}\)\(\quad\)......(iv)
    According to given problem, \(\theta^{\prime}=\theta\)
    From equations (iv) and (iii), we get,
    \(K=\frac{1}{\left(1-\frac{\sigma}{\rho}\right)}\)
    \(K=\frac{\rho}{(\rho-\sigma)}\)
    \(=\frac{1.6}{(1.6-0.8)}\)
    \(=2\)
     
  • Question 8
    3 / -1

    A box is floating in the river of speed \(5 {~m}/{s}\). The position of the block is shown in the figure at \({t}=0 .\) A stone is thrown from point \(O\) at time \(t=0\) with a velocity \(v=\left(v_{1} \hat{\imath}+v_{2} \hat{\jmath}+40 \hat{k}\right) m/ s\). Find the value of \(v_{1}, v_{2}\) such that the stone hits the box.

    Solution

    Position vector of box and stone are

    \(r_{B}=(5 t) \hat{\imath}+60 \hat{\jmath}\)

    \(r_{S}=v_{1} \hat{\imath}+v_{2} \hat{\jmath}+\left(40 t-\frac{1}{2} g t^{2}\right) \hat{k}\)

    When stone hits the block,

    \(5 t=v_{1} t \rightleftarrows ; \quad 60=v_{2} t ; \quad\left(40 t-\frac{1}{2} g t^{2}\right)=0\)

    \(v_{1}=5 \mathrm{~m} / \mathrm{s}, \mathrm{v}_{2}=\frac{60}{t} ; \quad \mathrm{t}=8\)

    \(v_{1}=5 m / s, v_{2}=7.5 \mathrm{~m} / \mathrm{s}\)

  • Question 9
    3 / -1

    The value of alternating emf E in the given circuit will be:

    Solution

    Given:

    Voltage across the Resistor \(\mathrm{R}\left(V_{\mathrm{R}}\right)=80 \mathrm{~V}\)

    Voltage across the Inductor \(L\left(V_{L}\right)=40 \mathrm{~V}\)

    Voltage across the Capacitor \(\mathrm{C}\left(\mathrm{V}_{\mathrm{C}}\right)=100 \mathrm{~V}\)

    For a series LCR circuit, the total potential difference of the circuit is given by:

    \(V=\sqrt{V_{R}^{2}+\left(V_{L}-V_{C}\right)^{2}}\)

    Putting all the above values in the equation,

    \(V=\sqrt{80^{2}+(100-40)^{2}} \)

    \(=\sqrt{6400+3600} \)

    \(=\sqrt{10000} \)

    \(=100 \mathrm{~V}\)

  • Question 10
    3 / -1

    What happens to the normal force on a boy if he throws a ball vertically upwards?

    Solution

    According to Newton's Third Law of Motion for every action force, there is equal and opposite reaction force. This Action-Reaction pair acts on two different bodies.

    When the boy throws the ball upwards he applies force to it. Simultaneously the ball applies the same amount of force on the boy in a downward direction. Now this force is added to the weight of the boy (force applied by the boy on the ground/floor). Considering the second system now the normal force (force applied by the floor/ground on the boy) is increased.

Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now