JEE Main 2026 — Apr 4, Shift 2 — Physics solution

JEE Main2026Apr 4, Shift 2Physics

A metal string A is suspended from a rigid support and its free end is attached to a block of mass M. Second block having mass 2M is suspended at the bottom of the first block using a string B. The area of cross sections of strings A and B are same. The ratio of lengths of strings of A to B is 2 and the ratio of their Young's moduli ( $Y_{A} / Y_{B}$ ) is 0.5. The ratio of elongations in A to B is ________.

Solution

Let string B have length $ℓ_{B} = ℓ$ and Young's modulus $Y_{B} = Y$ ; then string A has $ℓ_{A} = 2 ℓ$ and $Y_{A} = Y /2$ . Both have area $A$ .

Tensions: String B supports only the lower block ( $2 M$ ): $T_{B} = 2 M g$ . String A supports both blocks ( $3 M$ ): $T_{A} = 3 M g$ .

Elongations using $Δ ℓ = \frac{T ℓ}{Y A}$ :

$Δ ℓ_{A} = \frac{( 3 M g ) ( 2 ℓ )}{( Y /2 ) A} = \frac{12 M g ℓ}{Y A}$ .

$Δ ℓ_{B} = \frac{( 2 M g ) ( ℓ )}{Y A} = \frac{2 M g ℓ}{Y A}$ .

$∴ \frac{Δ ℓ _{A}}{Δ ℓ _{B}} = \frac{12}{2} = 6$ .

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