A man having a wrist watch and a pendulum clock rises on a $TV$ tower. The wrist watch and pendulum clock per chance fall from the top of the tower. Then
Easy
Download our app for free and get startedPlay store
(d) Function of wrist watch depends upon spring action so it is not effected by gravity but pendulum clock has time period, $T = 2\pi \sqrt {\frac{l}{g}} $. During free fall effective acceleration becomes zero, so time period comes out to be infinity i.e. the clock stops.
art

Download our app
and get started for free

Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*

Similar Questions

  • 1
    Amplitude of a wave is represented by $A = \frac{c}{{a + b - c}}$ Then resonance will occur when
    View Solution
  • 2
    An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $M$. The piston and the cylinder have equal cross sectional area $A$. When the piston is in equilibrium, the volume of the gas is $V_0$ and its pressure is $P_ 0$. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency
    View Solution
  • 3
    $A$ particle of mass m is constrained to move on $x$ -axis. $A$ force $F$ acts on the particle. $F$ always points toward the position labeled $E$. For example, when the particle is to the left of $E, F$ points to the right. The magnitude of $F$ is a constant $F$ except at point $E$ where it is zero. The system is horizontal. $F$ is the net force acting on the particle. The particle is displaced a distance $A$ towards left from the equilibrium position $E$ and released from rest at $t = 0.$ Find minimum time it will take to reach from $x = - \frac{A}{2}$ to $0$.
    View Solution
  • 4
    A particle performs $SHM$ on $x-$ axis with time period of $0.5\,sec,$ such that it's velocity is zero at $x = -3\,cm$ and at $x = 9\,cm$. It was located at $x = 0$ and moving in negative $'x'$ at $t = 0$. The equation of $SHM$ of the particle is
    View Solution
  • 5
    A simple pendulum has time period $T_1$. The point of suspension is now moved upward according to equation $y = k{t^2}$ where $k = 1\,m/se{c^2}$. If new time period is $T_2$ then ratio $\frac{{T_1^2}}{{T_2^2}}$ will be
    View Solution
  • 6
    The kinetic energy and the potential energy of a particle executing $S.H.M.$ are equal. The ratio of its displacement and amplitude will be 
    View Solution
  • 7
    A particle free to move along the $x-$axis has potential energy given by $U(x) = k[1 - \exp {( - x)^2}]$ for $ - \infty \le x \le + \infty $, where k is a positive constant of appropriate dimensions. Then
    View Solution
  • 8
    Equations ${y_1} = A\sin \omega t$ and ${y_2} = \frac{A}{2}\sin \omega t + \frac{A}{2}\cos \omega t$ represent $S.H.M.$ The ratio of the amplitudes of the two motions is
    View Solution
  • 9
    The angular frequency of motion whose equation is $4\frac{{{d^2}y}}{{d{t^2}}} + 9y = 0$ is ($y =$ displacement and $t =$ time)
    View Solution
  • 10
    The position co-ordinates of a particle moving in a $3-D$ coordinates system is given by $x = a\,cos\,\omega t$ , $y = a\,sin\,\omega t$ and $z = a\omega t$ The speed of the  particle is
    View Solution