As shown in the figure,an iron block A of vol | vedclass

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(B) The free body diagrams of the blocks are shown in the figure.At equilibrium,the net force on a body is zero.For block $A$ hanging vertically:$m_A

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$1.3$

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(B) The free body diagrams of the blocks are shown in the figure.At equilibrium,the net force on a body is zero.For block $A$ hanging vertically:$m_A g = k x$$\Rightarrow k = \frac{m_A g}{x}$For block $B$ on the inclined plane:$m_B g \sin 30^{\circ} = k x^{\prime}$$\Rightarrow x^{\prime} = \frac{m_B g \sin 30^{\circ}}{k} = \frac{m_B g \sin 30^{\circ}}{(m_A g / x)} = \frac{m_B}{m_A} x \sin 30^{\circ}$Since both blocks are made of the same material,mass is proportional to volume $(m = ho V)$:$\Rightarrow x^{\prime} = \frac{V_B}{V_A} x \sin 30^{\circ} = \frac{0.75}{0.25} 	imes 0.2 	imes \sin 30^{\circ}$$\Rightarrow x^{\prime} = 3 	imes 0.2 	imes 0.5 = 0.3 \ m$The total distance of the block from the top is the sum of the unstretched length $l$ and the extension $x^{\prime}$:$d = l + x^{\prime} = 1.0 \ m + 0.3 \ m = 1.3 \ m$.

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