A simple pendulum has a time period T in vacuum. Its time period when it is completely immersed in a liquid of density one-eight of the density one-eight of density of material of bob is :

Answers (2)

Given,
Density of liquid / density of bob =\frac{d_{L}}{d_{B}}=\frac{1}{8} 

Time period of simple pendulum, T \propto \frac{1}{\sqrt{g}}$ 

 

 Weight in liquid = weight in vaccume – buoyancy force.

(dB?V)a=(dB?V)g(dL?V)g      (Where, V= volume of bob) 

Due to buoyancy, downward acceleration become,a=g\left[1-\frac{d_{L}}{d_{B}}\right]$

Time period in vacuume / time period in liquid =\frac{T}{T_{2}}$

\begin{aligned} \frac{T}{T_{2}} &=\sqrt{\frac{g\left[1-\frac{d_{L}}{d_{B}}\right]}{g}} \\ \frac{T}{T_{2}} &=\sqrt{1-\frac{d_{B}}{8 . d_{B}}}=\sqrt{\frac{7}{8}} \\ T_{2} &=\sqrt{\frac{8}{7} T} \end{aligned}

Hence, time period in liquid is T_{2}=\sqrt{\frac{8}{7}} T$

Given,
Density of liquid / density of bob =\frac{d_{L}}{d_{B}}=\frac{1}{8} 

Time period of simple pendulum, T \propto \frac{1}{\sqrt{g}}$ 

 

 Weight in liquid = weight in vaccume – buoyancy force.

(dB?V)a=(dB?V)g(dL?V)g      (Where, V= volume of bob) 

Due to buoyancy, downward acceleration become,a=g\left[1-\frac{d_{L}}{d_{B}}\right]$

Time period in vacuume / time period in liquid =\frac{T}{T_{2}}$

\begin{aligned} \frac{T}{T_{2}} &=\sqrt{\frac{g\left[1-\frac{d_{L}}{d_{B}}\right]}{g}} \\ \frac{T}{T_{2}} &=\sqrt{1-\frac{d_{B}}{8 . d_{B}}}=\sqrt{\frac{7}{8}} \\ T_{2} &=\sqrt{\frac{8}{7} T} \end{aligned}

Hence, time period in liquid is T_{2}=\sqrt{\frac{8}{7}} T$

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