A pendulume clock loses $12\;s$ a day if the temperature is $40^oC$ and gains $4\;s$ a day if the temperature is $20^oC$. The temperature at which the clock will show correct time, and the coeffecient of linear expansion $(\alpha)$ of the metal of the pendulum shaft are respectively
  • A$30^o $ $C$ ,$\;\alpha $ $= 1.85 \times 10^{-3}/^o C$
  • B$55^o C$ ,$\;\alpha $ $= 1.85 \times 10^{-2}/^o C$
  • C$25^o C$ ,$\;\alpha $$ = 1.85 \times 10^{-5}/^o C$
  • D$60^o $ $C$ ,$\;\alpha $ = $1.85  \times10^{-4}/^o C$
JEE MAIN 2016, Diffcult
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
    The oscillation of a body on a smooth horizontal surface is represented by the equation $x= Acos$$\omega t$ 

    where $x=$ displacement at time $t$

    $\omega =$ frequency of oscillation

    Which one of the following graphs shows correctly the variation $a$ with $t$ ?

    Here $a=$ acceleration at time $t$

    $T=$ time period

    View Solution
  • 2
    A particle executes simple harmonic motion according to equation $4 \frac{d^2 x}{d t^2}+320 x=0$. Its time period of oscillation is .........
    View Solution
  • 3
    In the situation as shown in figure time period of vertical oscillation of block for small displacements will be 
    View Solution
  • 4
    Three simple harmonic motions of equal amplitudes $A$ and equal time periods in the same direction combine. The phase of the second motion is $60^o$ ahead of the first and the phase of the third motion is $60^o$ ahead of the second. Find the amplitude of the resultant motion
    View Solution
  • 5
    A particle of mass $m$ oscillates with simple harmonic motion between points ${x_1}$ and ${x_2}$, the equilibrium position being $O$. Its potential energy is plotted. It will be as given below in the graph
    View Solution
  • 6
    If the maximum velocity and maximum acceleration of a particle executing $SHM$ are equal in magnitude, the time period will be .... $\sec$
    View Solution
  • 7
    The amplitude of vibration of a particle is given by ${a_m} = ({a_0})/(a{\omega ^2} - b\omega + c);$ where ${a_0},a,b$ and $c$ are positive. The condition for a single resonant frequency is
    View Solution
  • 8
    A uniform rod of mass $m$ and length $I$ is suspended about its end, Time period of small angular oscillations is ..........
    View Solution
  • 9
    An object of mass $0.2 \mathrm{~kg}$ executes simple harmonic motion along $\mathrm{x}$ axis with frequency of $\left(\frac{25}{\pi}\right) \mathrm{Hz}$. At the position $\mathrm{x}=0.04 \mathrm{~m}$ the object has kinetic energy $0.5 \mathrm{~J}$ and potential energy $0.4 \mathrm{~J}$ The amplitude of oscillation is ............ cm.
    View Solution
  • 10
    A particle is executing Simple Harmonic Motion $(SHM)$. The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be
    View Solution