Two wires of same material and length are stretched by the same force. Their masses are in the ratio 4:3 . The ratio of their elongation will be

Question

Two wires of same material and length are stretched by the same force. Their masses are in the ratio 4:3 . The ratio of their elongation will be (A)  16:9  (B)  9:16  (C)  4:3  (D)  3:4

Tardigrade Admin · Accepted Answer

Y=(Fl/AΔ l) Here y,l and F are constant So, Δ l propto (1/A) Also m=(V)ρ =(Al)ρ i.e. m propto A ∴ Δ l propto (1/m) ⇒ (Δ l1/Δ l2)=(m2/m1)=(3/4) (∵ m1:m2=4:3)

Q. Two wires of same material and length are stretched by the same force. Their masses are in the ratio $4 : 3$ . The ratio of their elongation will be

$16 : 9$

$9 : 16$

$4 : 3$

$3 : 4$

$Y = \frac{Fl}{A Δ l}$ Here $y, l$ and F are constant So, $Δ l \propto \frac{1}{A}$ Also $m = (V) ρ = (A l) ρ$ i.e. $m \propto A$ $∴$ $Δ l \propto \frac{1}{m}$ $\Rightarrow$ $\frac{Δ l _{1}}{Δ l _{2}} = \frac{m _{2}}{m _{1}} = \frac{3}{4}$ $(∵ m_{1} : m_{2} = 4 : 3)$

Q. Two wires of same material and length are stretched by the same force. Their masses are in the ratio 4:3 . The ratio of their elongation will be

16:9

9:16

4:3

3:4