The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from $10\, cm$ to $8\, cm$ in $40\, seconds$ . Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is $1.3$ . The time in which amplitude of this pendulum will reduce from $10\, cm$ to $5\, cm$ in carbon dioxide will be close to ..... $s$ $(ln\, 5 = 1.601,ln\, 2 = 0 .693)$
JEE MAIN 2014, Medium
Download our app for free and get started
As we know,
$x=x_{0} e^{-b t/ 2 m}$
From question,
$8=10 \mathrm{e}^{-\frac{40 \mathrm{b}}{2 \mathrm{m}}}$ $...(i)$
Similarly, $5=10 e^{-\frac{b t}{2 m}}$ $...(ii)$
Solving eqns $(i)$ and $(ii)$ we get
$t \cong 142 s$
Download our appand 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
- 1The resultant of two rectangular simple harmonic motions of the same frequency and equal amplitudes but differing in phase by $\frac{\pi }{2}$ isView Solution
- 2The length of the second pendulum on the surface of earth is $1\, m$. The length of seconds pendulum on the surface of moon, where g is 1/6th value of $g$ on the surface of earth, isView Solution
- 3$Assertion :$ In simple harmonic motion, the motion is to and fro and periodicView Solution
$Reason :$ Velocity of the particle $(v) = \omega \sqrt {k^2 - x^2}$ (where $x$ is the displacement). - 4A particle of mass $10\, gm$ moves in a field where potential energy per unit mas is given by expression $v = 8 \times 10^4\, x^2\, erg/gm$. If the total energy of the particle is $8 \times 10^7\, erg$ then the relation between $x$ and time $t$ isView Solution
- 5A plank with a small block on top of it is under going vertical $SHM.$ Its period is $2\, sec.$ The minimum amplitude at which the block will separate from plank is :View Solution
- 6A second's pendulum is placed in a space laboratory orbiting around the earth at a height $3R$, where $R$ is the radius of the earth. The time period of the pendulum isView Solution
- 7The maximum velocity of a body undergoing $S.H.M$. is $0.2\,m/s$ and its acceleration at $0.1\,m$ from the mean position is $0.4\,m/s^2$. The amplitude of the $S.H.M.$ is .... $m$View Solution
- 8A particle is executing $S.H.M.$ with total mechanical energy $90 \,J$ and amplitude $6 \,cm$. If its energy is somehow decreased to $40 \,J$ then its amplitude will become ........ $cm$View Solution
- 9The length of a seconds pendulum at a height $h=2 R$ from earth surface will be.(Given: $R =$ Radius of earth and acceleration due to gravity at the surface of earth $g =\pi^{2}\,m / s ^{-2}$ )View Solution
- 10Two masses ${m_1}$ and ${m_2}$ are suspended together by a massless spring of constant k. When the masses are in equilibrium, ${m_1}$ is removed without disturbing the system. Then the angular frequency of oscillation of ${m_2}$ isView Solution