Question 503: Thermodynamics & KTG

Question

A simple pendulum made of a bob of mass m and a metallic wire of negligible mass has time period 2 s at T=0oC. If the temperature of the wire is increased and the corresponding change in its time period is
plotted against its temperature, the resulting graph is a line of slope S. If the
coefficient of linear expansion of metal is $$\alpha $$ then the value of S is :

Accepted Answer

Change of length of wire with temperature,
$$\Delta $$$$\ell $$   =   $$\alpha \ell \Delta 	heta $$
Time period of pendulum at temperature $$	heta $$,
T
$$	heta $$
= 2$$\pi $$$$\sqrt {{{\ell  + \Delta \ell } \over g}} $$
= 2$$\pi $$$$\sqrt {{{\ell \left( {1 + \alpha \Delta 	heta } ight)} \over g}} $$
= $$2\pi \sqrt {{\ell  \over g}} {\left( {1 + \alpha \Delta 	heta } ight)^{{1 \over 2}}}$$
= $$2\pi \sqrt {{\ell  \over g}} {\left( {1 + {{\Delta \ell } \over \ell }} ight)^{{1 \over 2}}}$$
$$ \simeq $$  T
0
$$\left( {1 + {{\Delta \ell } \over {2\ell }}} ight)$$
Here T
0
= time period at temperature 0
o
C.
$$ 	herefore $$   Change in time period,
$$\Delta $$T = T
$$	heta $$
$$-$$ T
0
= $${{{T_0}\Delta \ell } \over {2\ell }}$$
= $${{{T_0}\left( {\alpha \ell \Delta 	heta } ight)} \over {2\ell }}$$
$$ 	herefore $$   $${{\Delta T} \over {\Delta 	heta }}$$ = $${{{T_0}\alpha } \over 2}$$
Given that T
0
= 2,
$$ 	herefore $$   $${{\Delta T} \over {\Delta 	heta }}$$ = $${{2\alpha } \over 2}$$ = $$\alpha $$
$${{\Delta T} \over {\Delta 	heta }}$$ is the shape of $$\Delta $$T and $${\Delta 	heta }$$
curve = S (given)
$$ 	herefore $$   S = $$\alpha $$

Answer

$$\alpha $$

Answer

$${\alpha  \over 2}$$

Answer

2$$\alpha $$

Answer

$${1 \over \alpha }$$

Thermodynamics & KTG

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