A brass cube of side a and density rho is flo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $a$ be the side of the cube,$\rho$ be the density of the cube,and $\sigma$ be the density of the mercury.The mass of the cube is $M = a^3 \rho

Vedclass · Accepted Answer

$2\pi \sqrt {\frac{{\rho \,a}}{{\sigma \,g}}} $

Vedclass · Answer

(B) Let $a$ be the side of the cube,$\rho$ be the density of the cube,and $\sigma$ be the density of the mercury.The mass of the cube is $M = a^3 \rho$.In equilibrium,the weight of the cube is balanced by the buoyant force: $Mg = a^2 h \sigma g$,where $h$ is the depth of immersion.Thus,$a^3 \rho g = a^2 h \sigma g$,which gives $h = \frac{a \rho}{\sigma}$.When the cube is pushed down by a small vertical displacement $y$,the additional buoyant force provides the restoring force:$F_{restoring} = - (a^2 y) \sigma g$.The equation of motion is $M \frac{d^2y}{dt^2} = - (a^2 \sigma g) y$.Substituting $M = a^3 \rho$:$a^3 \rho \frac{d^2y}{dt^2} = - a^2 \sigma g y \implies \frac{d^2y}{dt^2} = - \left( \frac{\sigma g}{a \rho} \right) y$.Comparing this with the standard $S.H.M.$ equation $\frac{d^2y}{dt^2} = - \omega^2 y$,we get $\omega^2 = \frac{\sigma g}{a \rho}$.The time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{a \rho}{\sigma g}}$.

$A$ brass cube of side $a$ and density $\rho$ is floating in mercury of density $\sigma$. If the cube is displaced a bit vertically,it executes $S.H.M.$ Its time period will be

Similar Questions

$A$ man measures the time period of a pendulum $(T)$ in a stationary lift. If the lift moves upward with an acceleration of $\frac{g}{4}$,then the new time period will be:

$A$ pendulum performs $S.H.M.$ with period $\sqrt{3} \ s$ in a stationary lift. If the lift moves up with acceleration $\frac{g}{3}$,the period of the pendulum is $[g=$ acceleration due to gravity $]$. (in $s$)

For a simple pendulum,the graph between $L$ and $T$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module