A block of weight 100N is suspended by copper and steel wires of same cross-sectional area 0.5c m ^ 2 and, length 3 m and 1m, respectively. Their other ends are fixed on a ceiling as shown in figure. The angles

Correct answer: 0 2.00 Resolving in components ⇒ ∑ F _ x = 0 ⇒ T _ S 60 = T _ C ⋅ 30 ⇒ T _ S ⋅ (1)/(2) = T _ C ⋅ 3 2 ⇒ T _ S = 3 ⋅ T _ C Δ l _ C Δ l _ S = ( T _ C ⋅ l _ C T _ S ⋅ l _ S Y _ S ) ⇒ Δ l _ C Δ l _ S = ( T _ C T _ S )( l _ C l _ S )( Y _ S Y _ C ) = ( 1 3 )( 3 1 )( 2 × 10 ^ 11 1 × 10 ^ 11 ) = 2

JEE Advanced 2019Paper 1

Exam: JEE Advanced
Year: 2019
Paper: Paper 1
Subject: Physics
Chapter: Laws of Motion

JEE Advanced 2019 · Paper 1 · Physics · Laws of Motion

A block of weight 100N is suspended by copper and steel wires of...

A block of weight $100N$ is suspended by copper and steel wires of same cross-sectional area $0.5c{m}^{2}$ and, length $\sqrt{3}m$ and 1m, respectively. Their other ends are fixed on a ceiling as shown in figure. The angles subtended by copper and steel wires with ceiling are ${30}^{o} and {60}^{o},$ respectively. If elongation in copper wire is \left(\Delta{l}_{C}\right) $and elongation in steel wire is \left(\Delta{l}_{S}\right),$ then the ratio $\dfrac{\Delta{l}_{C}}{\Delta{l}_{S}}$ is _____.
[Young's modulus for copper and steel are $1 \times {10}^{11}N / {m}^{2} and 2 \times {10}^{11}N / {m}^{2}$ respectively]
A block of weight 100N is suspended by copper and steel wires of same cross-sectional area 0.5c m ^ 2 and, length 3 m

Correct answer

Step-by-step solution

2.00
A block of weight 100N is suspended by copper and steel wires of same cross-sectional area 0.5c m ^ 2 and, length 3 m
Resolving in components
$⇒ ∑ {F}_{x} = 0$
$⇒ {T}_{S}\cos 60 = {T}_{C} ⋅ \cos 30$
$⇒ {T}_{S} ⋅ \dfrac{1}{2} = {T}_{C} ⋅ \dfrac{\sqrt{3}}{2}$
$⇒ {T}_{S} = \sqrt{3} ⋅ {T}_{C}$
$\dfrac{\Delta{l}_{C}}{\Delta{l}_{S}} =$ \left(\dfrac{{T}_{C} ⋅ {l}_{C}}{\dfrac{{T}_{S} ⋅ {l}_{S}}{{Y}_{S}}}\right) $<br >$ ⇒ \dfrac{\Delta{l}_{C}}{\Delta{l}_{S}} = \left(\dfrac{{T}_{C}}{{T}_{S}}\right)\left(\dfrac{{l}_{C}}{{l}_{S}}\right)\left(\dfrac{{Y}_{S}}{{Y}_{C}}\right) = \left(\dfrac{1}{\sqrt{3}}\right)\left(\dfrac{\sqrt{3}}{1}\right)\left(\dfrac{2 \times {10}^{11}}{1 \times {10}^{11}}\right) = 2

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