Young's moduli of two wires A and B are in the ratio 7: 4 . Wire A is 2m long and has radius R . Wire B is 1.5m long and has radius 2mm . If the two wires stretch by the same length for a given load, the value of R is

Correct answer: D. 1.7mm Strain = ∆ l / l Stress = F / A Young’s modulus = (Stress)/(Strain) = (F / A)/((∆ l / l)) = (F)/(A)((l)/(∆ l)) Y _ A Y _ B = ( l _ A A _ A )( A _ B l _ B ) = (7)/(4) ( 2 π r _ A ^ 2 )( π r _ B ^ 2 1.5 ) = (7)/(4) r _ A ^ 2 = (2)/(1.5 × 7) × (2mm) ^ 2 r _ A = 1.7mm

JEE Main 20198 AprilEvening Shift

Exam: JEE Main
Year: 2019
Date: 8 April
Shift: Evening Shift
Subject: Physics
Chapter: Mechanical Properties of Solids

JEE Main 2019 · 8 April · Evening Shift · Physics · Mechanical Properties of Solids

Young's moduli of two wires A and B are in the ratio 7: 4 . Wire A...

Young's moduli of two wires

A

and

B

are in the ratio

7: 4

. Wire

A

2m

long and has radius

R .

Wire

B

1.5m

long and has radius

2mm .

If the two wires stretch by the same length for a given load, the value of

R

is close to:

Options

A $1.3mm$
B $1.9mm$
C $1.5mm$
D $1.7mm$

Correct answer

D. $1.7mm$

Step-by-step solution

Strain

= ∆ l / l

Stress

= F / A

Young’s modulus

= \dfrac{Stress}{Strain} = \dfrac{F / A}{\left(∆ l / l\right)} = \dfrac{F}{A}\left(\dfrac{l}{∆ l}\right)

\dfrac{{Y}_{A}}{{Y}_{B}} = \left(\dfrac{{l}_{A}}{{A}_{A}}\right)\left(\dfrac{{A}_{B}}{{l}_{B}}\right) = \dfrac{7}{4}

\left(\dfrac{2}{\pi{r}_{A}^{2}}\right)\left(\dfrac{\pi{r}_{B}^{2}}{1.5}\right) = \dfrac{7}{4}

<br/>

{r}_{A}^{2} = \dfrac{2}{1.5 \times 7} \times {\left(2mm\right)}^{2}

<br/>

{r}_{A} = 1.7mm

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