JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Elasticity & SHM

A block of weight 100 N is suspended by copper and steel wires of same cross sectional area

0.5 cm^{2}

and, length

3 m

and 1 m, 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° and 60°, respectively. If elongation in copper wire is

(Δ ℓ_{C})

and elongation in steel wire is

(Δ ℓ_{S})

, then the ratio

\frac{Δ ℓ _{C}}{Δ ℓ _{S}}

is ____.
[Young's modulus for copper and steel are

1 \times 1 0^{11} N / m^{2}

and

2 \times 1 0^{11} N / m^{2}

respectively]

Answer: 2

Solution

Resolving forces horizontally:

T_{C} cos 30° = T_{S} cos 60°

T_{C} \times \frac{3}{2} = T_{S} \times \frac{1}{2}

T_{S} = 3 T_{C}

... (i)

Resolving vertically:

T_{C} sin 30° + T_{S} sin 60° = 100

\frac{T _{C}}{2} + \frac{T _{S} 3}{2} = 100

... (ii)

Solving (i) and (ii):

T_{C} = 50 N, T_{S} = 503 N

Elongation formula:

Δ L = \frac{F L}{A Y}

\frac{Δ L _{C}}{Δ L _{S}} = \frac{T _{C} L _{C}}{A _{C} Y _{C}} \times \frac{A _{S} Y _{S}}{T _{S} L _{S}}

Since

A_{C} = A_{S}

\frac{Δ L _{C}}{Δ L _{S}} = \frac{50 \times 3}{1 \times 1 0 ^{11}} \times \frac{2 \times 1 0 ^{11}}{50 3 \times 1} = 2

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Numerical answer

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Elasticity & SHM

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