Question

A thin string is held at one end and oscillates vertically so that, y (x=0, t) = 8 sin 4t (cm). Neglect the gravitational force. The string passes through bath filled with 1 kg water. Due to friction, heat is transferred to the both. The heat transfer efficiency is 50%. Calculate how much time passes before the temperature of the bath rises by one degree kelvin.

• A. 3.1 day
• B. 4.2 day
• C. 2.5 day
• D. 1.1 day

Answer 1

The correct option is B 4.2 day

y(x = 0, t) = 8 sin 4t
$\Rightarrow A=8cm,\omega =4rad{s}^{-1}$
$v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{1}{0.2}}=2.236m{s}^{-1}$
The average power over one time period is $P=\frac{1}{2}\rho {\omega }^2{A}^{2}sv$
$=\frac{1}{2}{\omega }^2{A}^{2}v.\mu$
$(\because \mu =\rho s,where,s=cross-sectionarea)$
$\because P=\frac{1}{2}\left(4{)}^2\right(8\times {10}^{-2}{)}^2\left(2.236\right)\left(0.2\right)=2.29\times {10}^{-2}J{s}^{-1}$
Power transferred to the bath is
$P^{\prime }=0.5P=1.145\times {10}^{-2}J{s}^{-1}$
Now, $P^{\prime }t=mC\Delta \theta ,whereC=4.2\times {10}^{3}Jk{g}^{-1}{k}^{-1}$
$t=\frac{mC\Delta \theta }{p^{\prime }}=\frac{1\times 4.2\times {10}^3\times 1}{1.145\times {10}^{-2}}=3.6\times {10}^{5}S=4.2day$