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(B) According to the question, the system is in equilibrium. Let $T$ be the tension in the string connected to mass $m$, and $T_1$ be the tension in t

Vedclass · Accepted Answer

$0.5$

Vedclass · Answer

(B) According to the question, the system is in equilibrium. Let $T$ be the tension in the string connected to mass $m$, and $T_1$ be the tension in the string connected to the $1 \,kg$ mass.For the $1 \,kg$ mass, the tension $T_1 = 1g$.For the $2 \,kg$ mass on the inclined plane, the forces acting along the incline are the component of its weight $2g \sin 30^{\circ}$ and the tension $T_1$ acting upwards. The net force along the incline is $(2g - T_1) \sin 30^{\circ}$ (assuming the $2 \,kg$ mass is pulled by the tension $T$ up the incline).For the system to be in equilibrium, the tension $T$ must balance the net force along the incline:$T = (2g - T_1) \sin 30^{\circ}$Substituting $T_1 = 1g$:$T = (2g - 1g) \sin 30^{\circ} = g \sin 30^{\circ} = g 	imes 0.5 = 0.5g$For the mass $m$, the tension $T$ must balance its weight:$T = mg$Equating the two expressions for $T$:$mg = 0.5g$$m = 0.5 \,kg$

$A$ system containing masses and pulleys connected on an inclined plane is shown in the figure. If the system is in equilibrium, then the value of $m$ is (in $\,kg$)

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