A monkey of mass $40 \mathrm{~kg}$ climbs up a rope which can stand a maximum tension of $600 \mathrm{~N}$. The rope will break, when the monkey

Mass of the monkey, $m=40 \mathrm{~kg}$ Maximum tension that the rope can bear, $T_{\max }=600 \mathrm{~N}$ In option (a), acceleration of monkey, $a=6 \mathrm{~m} / \mathrm{s}^{2}$ (upward) $\therefore$ By equation of motion, $$ \Rightarrow \quad \begin{array}{ll} T & -m g=m a \\ T & =m(g+a) \\ & =40(10+6)=640 \mathrm{~N} \end{array} $$ Since, $T>T_{\text {max }}$, hence in this case, rope will break. Therefore, option (a) is correct. In option (b), acceleration of monkey, $a=4 \mathrm{~m} / \mathrm{s}^{2}$ (downward) $\therefore$ By equation of motion, $$ T=m(g-a)=40(10-4)=240 \mathrm{~N} $$ Since, $T < T_{\max }$, then the rope will not break. In option (c), monkey climbs up with uniform speed, $v=5 \mathrm{~m} / \mathrm{s}$ $\therefore$ Acceleration, $a=0 \mathrm{~m} / \mathrm{s}^{2}$ Hence, by equation of motion, $T=m g$ $$ =40 \times 10=400 \mathrm{~N} $$ Since, $T < T_{\max }$, hence rope will not break. In option $(\mathrm{d})$, acceleration, $a=g$ $$ \therefore \quad T=m(g-a)=m(g-g)=0 \mathrm{~N} \text {. } $$ Since, $T < T_{\max }$, the rope will not break. Therefore, only option (a) is correct.