As per the given figure,a weightless pulley P | vedclass

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(B) Let the mass $m_1 = 4 \, kg$ be on the $60^{\circ}$ incline and $m_2 = 1 \, kg$ be on the $30^{\circ}$ incline.For the $4 \, kg$ block,the equatio

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$4(\sqrt{3} + 1) \, N$

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(B) Let the mass $m_1 = 4 \, kg$ be on the $60^{\circ}$ incline and $m_2 = 1 \, kg$ be on the $30^{\circ}$ incline.For the $4 \, kg$ block,the equation of motion is: $m_1 g \sin 60^{\circ} - T = m_1 a \implies 4 	imes 10 	imes \frac{\sqrt{3}}{2} - T = 4a \implies 20\sqrt{3} - T = 4a \dots (1)$For the $1 \, kg$ block,the equation of motion is: $T - m_2 g \sin 30^{\circ} = m_2 a \implies T - 1 	imes 10 	imes \frac{1}{2} = 1a \implies T - 5 = a \dots (2)$From equation $(2)$,$a = T - 5$. Substituting this into equation $(1)$:$20\sqrt{3} - T = 4(T - 5)$$20\sqrt{3} - T = 4T - 20$$5T = 20\sqrt{3} + 20$$T = 4(\sqrt{3} + 1) \, N$.

As per the given figure,a weightless pulley $P$ is attached on a double inclined frictionless surface. The tension in the string (massless) will be (if $g = 10 \, m/s^2$):

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