Three bars having length $l, 2 l$ and $3 l$ and area of cross-section $A, 2 A$ and $3 A$ are joined rigidly and to end. Compound rod is subjected to a stretching force $F$ . The increase in length of rod is (Young's modulles of material is $Y$ and bars are massless)

\frac{13 F l}{2 A Y}

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\frac{3 F l}{A Y}

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\frac{9 F l}{A Y}

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\frac{13 F l}{A Y}

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Solution

The correct option is B $\frac{3 F l}{A Y}$
Changed in length of compound bar $= Δ l_{1} + Δ l_{2} + Δ l_{3}$
$Δ l = \frac{P L}{A E} = \frac{F L}{A E} = \frac{F L}{A Y}$ ( $E =$ Young's modulus)
$= \frac{F l}{A Y} + \frac{F (2 l)}{2 A Y} + \frac{F (3 l)}{3 A Y}$
$= \frac{3 F l}{A Y}$

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Three bars having length l,2l and 3l and area of cross-section A,2A and 3A are joined rigidly and to end. Compound rod is subjected to a stretching force F. The increase in length of rod is (Young's modulles of material is Y and bars are massless)

The correct option is B 3FlAYChanged in length of compound bar =Δl1+Δl2+Δl3Δl=PLAE=FLAE=FLAY (E=Young's modulus)=FlAY+F(2l)2AY+F(3l)3AY=3FlAY

Three bars having length $l, 2 l$ and $3 l$ and area of cross-section $A, 2 A$ and $3 A$ are joined rigidly and to end. Compound rod is subjected to a stretching force $F$ . The increase in length of rod is (Young's modulles of material is $Y$ and bars are massless)

The correct option is B $\frac{3 F l}{A Y}$
Changed in length of compound bar $= Δ l_{1} + Δ l_{2} + Δ l_{3}$
$Δ l = \frac{P L}{A E} = \frac{F L}{A E} = \frac{F L}{A Y}$ ( $E =$ Young's modulus)
$= \frac{F l}{A Y} + \frac{F (2 l)}{2 A Y} + \frac{F (3 l)}{3 A Y}$
$= \frac{3 F l}{A Y}$