JEE MAIN - Physics (2021 - 25th July Morning Shift - No. 10)
Two wires of same length and radius are joined end to end and loaded. The Young's modulii of the materials of the two wires are Y1 and Y2. The combination behaves as a single wire then its Young's modulus is :
$$Y = {{2{Y_1}{Y_2}} \over {3({Y_1} + {Y_2})}}$$
$$Y = {{2{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}$$
$$Y = {{{Y_1}{Y_2}} \over {2({Y_1} + {Y_2})}}$$
$$Y = {{{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}$$
Explanation
In series combination $$\Delta$$l = l1 + l2
$$Y = {{F/A} \over {\Delta l/l}} \Rightarrow \Delta l = {{Fl} \over {AY}}$$
$$ \Rightarrow \Delta l \propto {l \over Y}$$
Equivalent length of rod after joining is = 2l
As, lengths are same and force is also same in series
$$\Delta l = \Delta {l_1} + \Delta {l_2}$$
$${{{l_{eq}}} \over {{Y_{eq}}}} = {l \over {{Y_1}}} + {l \over {{Y_2}}} \Rightarrow {{2l} \over Y} = {l \over {{Y_1}}} + {l \over {{Y_2}}}$$
$$\therefore$$ $$Y = {{2{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}$$
$$Y = {{F/A} \over {\Delta l/l}} \Rightarrow \Delta l = {{Fl} \over {AY}}$$
$$ \Rightarrow \Delta l \propto {l \over Y}$$
Equivalent length of rod after joining is = 2l
As, lengths are same and force is also same in series
$$\Delta l = \Delta {l_1} + \Delta {l_2}$$
$${{{l_{eq}}} \over {{Y_{eq}}}} = {l \over {{Y_1}}} + {l \over {{Y_2}}} \Rightarrow {{2l} \over Y} = {l \over {{Y_1}}} + {l \over {{Y_2}}}$$
$$\therefore$$ $$Y = {{2{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}$$
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