One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is

Question

One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is (A) 0.25 (B) 0.50 (C) 2.00 (D) 4.00

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/6d5903f60ecf62c519a52d4239336590-.png" /> Both the rods are made up of same kind of material, i.e. their youngâs modulus is same. K1 =(π4R2Y/2L), k2 =(π R2Y/L) Force constants of two rods are, ∴ F=k1x=k2Y ⇒ (y/x)=(k1/k2)=2

$0.25$

$0.50$

$2.00$

$4.00$

Both the rods are made up of same kind of material, i.e. their young’s modulus is same.
$K_{1}$ $= \frac{π 4 R ^{2} Y}{2 L}$ , $k_{2}$ $= \frac{π R ^{2} Y}{L}$
Force constants of two rods are,
$∴$ $F = k_{1} x$ $= k_{2} Y$
$\Rightarrow$ $\frac{y}{x} = \frac{k _{1}}{k _{2}} = 2$

Q. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is

0.25

0.50

2.00

4.00

Both the rods are made up of same kind of material, i.e. their young’s modulus is same. K1​ =2Lπ4R2Y​, k2​ =LπR2Y​ Force constants of two rods are, ∴ F=k1​x=k2​Y ⇒ xy​=k2​k1​​=2

$0.25$

$0.50$

$2.00$

$4.00$

Both the rods are made up of same kind of material, i.e. their young’s modulus is same.
$K_{1}$ $= \frac{π 4 R ^{2} Y}{2 L}$ , $k_{2}$ $= \frac{π R ^{2} Y}{L}$
Force constants of two rods are,
$∴$ $F = k_{1} x$ $= k_{2} Y$
$\Rightarrow$ $\frac{y}{x} = \frac{k _{1}}{k _{2}} = 2$