The Young's modulus of brass and steel are respectively 1.0× 1011Nm-2 and 2.0× 1011Nm-2. .A brass wire and a steel wire of the same length are extended by 1 mm each under the same force. If radii of brass and steel wires are RB and RS respectively, then

Question

The Young's modulus of brass and steel are respectively 1.0× 1011Nm-2 and 2.0× 1011Nm-2. .A brass wire and a steel wire of the same length are extended by 1 mm each under the same force. If radii of brass and steel wires are RB and RS respectively, then (A) Rs = √2 RB (B) RS = (RB/√2)  (C) Rs = 4RB (D) Rs = (RB/2)

Tardigrade Admin · Accepted Answer

Increase in length, Δ L=(F L/Y ⋅ A)=(F L/Y ⋅ π R2) As F, L and Δ L are same hence, Y ⋅ R2= a constant ∴ 2.0 × 1011 RS2=1.0 × 1011 RB2 ⇒ RS=(RB/√2)

$R_{s} = 2 R_{B}$

$R_{S} = \frac{R _{B}}{2}$

$R_{s} = 4 R_{B}$

$R_{s} = \frac{R _{B}}{2}$

Increase in length,
$Δ L = \frac{F L}{Y \cdot A} = \frac{F L}{Y \cdot π R ^{2}}$
As $F, L$ and $Δ L$ are same hence,
$Y \cdot R^{2} = a$ constant
$∴ 2.0 \times 1 0^{11} R_{S}^{2} = 1.0 \times 1 0^{11} R_{B}^{2}$
$\Rightarrow R_{S} = \frac{R _{B}}{2}$

Q. The Young's modulus of brass and steel are respectively 1.0×1011Nm−2 and 2.0×1011Nm−2. .A brass wire and a steel wire of the same length are extended by 1mm each under the same force. If radii of brass and steel wires are RB​ and RS​ respectively, then

Rs​=2​RB​

RS​=2​RB​​

Rs​=4RB​

Rs​=2RB​​

Increase in length, ΔL=Y⋅AFL​=Y⋅πR2FL​ As F,L and ΔL are same hence, Y⋅R2=a constant ∴2.0×1011RS2​=1.0×1011RB2​ ⇒RS​=2​RB​​

$R_{s} = 2 R_{B}$

$R_{S} = \frac{R _{B}}{2}$

$R_{s} = 4 R_{B}$

$R_{s} = \frac{R _{B}}{2}$

Increase in length,
$Δ L = \frac{F L}{Y \cdot A} = \frac{F L}{Y \cdot π R ^{2}}$
As $F, L$ and $Δ L$ are same hence,
$Y \cdot R^{2} = a$ constant
$∴ 2.0 \times 1 0^{11} R_{S}^{2} = 1.0 \times 1 0^{11} R_{B}^{2}$
$\Rightarrow R_{S} = \frac{R _{B}}{2}$