$(a)\;Y \to 2Z\qquad(b)\;W \to X+Z \qquad(c)\;W \to 2Y\qquad(d)\;X \to Y+Z$

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Answer : (c) $\;W\to2Y$

Explanation : If it mass A

$Y \to 2Z$

Reactant :

$R=60 \times 8.5 =10 MeV$

Product :

$P=2 \times30 \times5=300 MeV$

$\bigtriangleup \varepsilon =-210 MeV$

Therefore ,Endothermic

if it was B

$W \to X+Z$

$R=120 \times 7.5 =900 MeV$

$P=90 \times 8 +30 \times5 =870 MeV$

$\bigtriangleup \varepsilon =-30 MeV$

Therefore ,Endothermic

it was C

$W \to 2Y$

$R=120 \times 7.5=900 MeV$

$P=2\times60\times8.5=1020 MeV$

$\bigtriangleup \varepsilon =120 MeV$

Therefore ,Exothermic

if it was D

$X \to Y+Z$

$R=90 \times 8= 720MeV $

$P=60 \times 8.5 +30 \times 5 =660 MeV$

$\bigtriangleup \varepsilon =-60 MeV$

Therefore , Endothemic

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