JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 5)
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 :
$$\alpha $$
$${\alpha \over 2}$$
2$$\alpha $$
$${1 \over \alpha }$$
Explanation
Change of length of wire with temperature,
$$\Delta $$$$\ell $$ = $$\alpha \ell \Delta \theta $$
Time period of pendulum at temperature $$\theta $$,
T$$\theta $$ = 2$$\pi $$$$\sqrt {{{\ell + \Delta \ell } \over g}} $$
= 2$$\pi $$$$\sqrt {{{\ell \left( {1 + \alpha \Delta \theta } \right)} \over g}} $$
= $$2\pi \sqrt {{\ell \over g}} {\left( {1 + \alpha \Delta \theta } \right)^{{1 \over 2}}}$$
= $$2\pi \sqrt {{\ell \over g}} {\left( {1 + {{\Delta \ell } \over \ell }} \right)^{{1 \over 2}}}$$
$$ \simeq $$ T0 $$\left( {1 + {{\Delta \ell } \over {2\ell }}} \right)$$
Here T0 = time period at temperature 0oC.
$$ \therefore $$ Change in time period,
$$\Delta $$T = T$$\theta $$ $$-$$ T0
= $${{{T_0}\Delta \ell } \over {2\ell }}$$
= $${{{T_0}\left( {\alpha \ell \Delta \theta } \right)} \over {2\ell }}$$
$$ \therefore $$ $${{\Delta T} \over {\Delta \theta }}$$ = $${{{T_0}\alpha } \over 2}$$
Given that T0 = 2,
$$ \therefore $$ $${{\Delta T} \over {\Delta \theta }}$$ = $${{2\alpha } \over 2}$$ = $$\alpha $$
$${{\Delta T} \over {\Delta \theta }}$$ is the shape of $$\Delta $$T and $${\Delta \theta }$$
curve = S (given)
$$ \therefore $$ S = $$\alpha $$
$$\Delta $$$$\ell $$ = $$\alpha \ell \Delta \theta $$
Time period of pendulum at temperature $$\theta $$,
T$$\theta $$ = 2$$\pi $$$$\sqrt {{{\ell + \Delta \ell } \over g}} $$
= 2$$\pi $$$$\sqrt {{{\ell \left( {1 + \alpha \Delta \theta } \right)} \over g}} $$
= $$2\pi \sqrt {{\ell \over g}} {\left( {1 + \alpha \Delta \theta } \right)^{{1 \over 2}}}$$
= $$2\pi \sqrt {{\ell \over g}} {\left( {1 + {{\Delta \ell } \over \ell }} \right)^{{1 \over 2}}}$$
$$ \simeq $$ T0 $$\left( {1 + {{\Delta \ell } \over {2\ell }}} \right)$$
Here T0 = time period at temperature 0oC.
$$ \therefore $$ Change in time period,
$$\Delta $$T = T$$\theta $$ $$-$$ T0
= $${{{T_0}\Delta \ell } \over {2\ell }}$$
= $${{{T_0}\left( {\alpha \ell \Delta \theta } \right)} \over {2\ell }}$$
$$ \therefore $$ $${{\Delta T} \over {\Delta \theta }}$$ = $${{{T_0}\alpha } \over 2}$$
Given that T0 = 2,
$$ \therefore $$ $${{\Delta T} \over {\Delta \theta }}$$ = $${{2\alpha } \over 2}$$ = $$\alpha $$
$${{\Delta T} \over {\Delta \theta }}$$ is the shape of $$\Delta $$T and $${\Delta \theta }$$
curve = S (given)
$$ \therefore $$ S = $$\alpha $$
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