JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 19)
The value of tension in a long thin metal wire has been changed from T1 to T2. The lengths of the metal wire at two different values of tension T1 and T2 are l1 and l2 respectively. The actual length of the metal wire is :
$${{{l_1} + {l_2}} \over 2}$$
$$\sqrt {{T_1}{T_2}{l_1}{l_2}} $$
$${{{T_1}{l_2} - {T_2}{l_1}} \over {{T_1} - {T_2}}}$$
$${{{T_1}{l_1} - {T_2}{l_2}} \over {{T_1} - {T_2}}}$$
Forklaring
Suppose, I0 be the actual length of metal wire and Y be its Young's modulus.
From Hooke's law,
$$Y = {{T{I_0}} \over {A\Delta I}}$$
where, $$\Delta I = I - {I_0}$$
$$ \Rightarrow Y = {{T{I_0}} \over {A(I - {I_0})}}$$ or $$I - I = {{T{I_0}} \over {AY}}$$
$$\therefore$$ $${{{I_1} - {I_0}} \over {{I_2} - {I_0}}} = {{{T_1}{I_0}} \over {AY}} \times {{AY} \over {{T_2}{I_0}}} = {{{T_1}} \over {{T_2}}} $$
$$\Rightarrow {I_1}{T_2} - {I_0}{T_2} = {I_2}{T_1} - {I_0}{T_1}$$
$$ \Rightarrow {I_0} = {{{I_1}{T_2} - {I_2}{T_1}} \over {{T_2} - {T_1}}} = {{{T_1}{I_2} - {T_2}{I_1}} \over {({T_1} - {T_2})}}$$
From Hooke's law,
$$Y = {{T{I_0}} \over {A\Delta I}}$$
where, $$\Delta I = I - {I_0}$$
$$ \Rightarrow Y = {{T{I_0}} \over {A(I - {I_0})}}$$ or $$I - I = {{T{I_0}} \over {AY}}$$
$$\therefore$$ $${{{I_1} - {I_0}} \over {{I_2} - {I_0}}} = {{{T_1}{I_0}} \over {AY}} \times {{AY} \over {{T_2}{I_0}}} = {{{T_1}} \over {{T_2}}} $$
$$\Rightarrow {I_1}{T_2} - {I_0}{T_2} = {I_2}{T_1} - {I_0}{T_1}$$
$$ \Rightarrow {I_0} = {{{I_1}{T_2} - {I_2}{T_1}} \over {{T_2} - {T_1}}} = {{{T_1}{I_2} - {T_2}{I_1}} \over {({T_1} - {T_2})}}$$