Verified Solution • 2025–26 Session moderate

QJEEM21Ph3ca5: JEE Main 2021 Physics - Class 11 JEEM21Ph3ca5 Solution: Two wires of same length and radius are joined end to end and…

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 :

Multiple Choice Options

  • A Y=2Y1Y23(Y1+Y2)Y = {{2{Y_1}{Y_2}} \over {3({Y_1} + {Y_2})}}
  • B Y=2Y1Y2Y1+Y2Y = {{2{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}
  • C Y=Y1Y22(Y1+Y2)Y = {{{Y_1}{Y_2}} \over {2({Y_1} + {Y_2})}}
  • D Y=Y1Y2Y1+Y2Y = {{{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}

Step-by-Step Verified Solution

In series combination Δ\Delta l = l1 + l2

Y=F/AΔl/l⇒Δl=FlAYY = {{F/A} \over {\Delta l/l}} \Rightarrow \Delta l = {{Fl} \over {AY}}

⇒Δl∝lY\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

Δl=Δl1+Δl2\Delta l = \Delta {l_1} + \Delta {l_2}

leqYeq=lY1+lY2⇒2lY=lY1+lY2{{{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=2Y1Y2Y1+Y2Y = {{2{Y_1}{Y_2}} \over {{Y_1} + {Y_2}}}

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