Answer
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Hint: The maximum tension of the rope will be two third of the weight of the boy so the acceleration with which the boy should climb down will be the minimum acceleration of the boy as the rope may break and the boy falls from the rope.
Formula used: The formula of the force is given by $F = ma$ where F is the force m is the mass of the body and a is the acceleration of the body.
Step by step solution:
It is given in the problem that the maximum tension in the rope is equal to the two third of the weight of the boy.
The weight of the boy is $w = mg$ so the maximum tension in the rope will be${T_{\max .}} = \dfrac{2}{3}mg$. As the boy is climbing down the rope so we need to find the value of the minimum acceleration here.
The relation for the motion is given by,
$ \Rightarrow mg - {T_{\max .}} = m{a_{\min .}}$
Replace the value of the maximum tension in the rope which is equal to ${T_{\max .}} = \dfrac{2}{3}mg$.
$ \Rightarrow mg - {T_{\max .}} = m{a_{\min .}}$
$ \Rightarrow mg - \left( {\dfrac{2}{3}mg} \right) = m{a_{\min .}}$
$ \Rightarrow \dfrac{{\left( {3mg - 2mg} \right)}}{3} = m{a_{\min .}}$
$ \Rightarrow \dfrac{{mg}}{3} = m{a_{\min .}}$
$ \Rightarrow {a_{\min .}} = \dfrac{g}{3}$
The minimum acceleration of the boy for maximum tension of the rope equals to two third the weight of the boy is equal to${a_{\min .}} = \dfrac{g}{3}$.
The correct answer for this problem is option A.
Note:It should be observed that we have subtracted the value of maximum tension from the weight of the body, it is done as the boy is climbing down the rope and with a minimum acceleration. Students advisable to observe the options and solve the problem accordingly as in this problem the value of acceleration due to gravity is not used but the term is only used.
Formula used: The formula of the force is given by $F = ma$ where F is the force m is the mass of the body and a is the acceleration of the body.
Step by step solution:
It is given in the problem that the maximum tension in the rope is equal to the two third of the weight of the boy.
The weight of the boy is $w = mg$ so the maximum tension in the rope will be${T_{\max .}} = \dfrac{2}{3}mg$. As the boy is climbing down the rope so we need to find the value of the minimum acceleration here.
The relation for the motion is given by,
$ \Rightarrow mg - {T_{\max .}} = m{a_{\min .}}$
Replace the value of the maximum tension in the rope which is equal to ${T_{\max .}} = \dfrac{2}{3}mg$.
$ \Rightarrow mg - {T_{\max .}} = m{a_{\min .}}$
$ \Rightarrow mg - \left( {\dfrac{2}{3}mg} \right) = m{a_{\min .}}$
$ \Rightarrow \dfrac{{\left( {3mg - 2mg} \right)}}{3} = m{a_{\min .}}$
$ \Rightarrow \dfrac{{mg}}{3} = m{a_{\min .}}$
$ \Rightarrow {a_{\min .}} = \dfrac{g}{3}$
The minimum acceleration of the boy for maximum tension of the rope equals to two third the weight of the boy is equal to${a_{\min .}} = \dfrac{g}{3}$.
The correct answer for this problem is option A.
Note:It should be observed that we have subtracted the value of maximum tension from the weight of the body, it is done as the boy is climbing down the rope and with a minimum acceleration. Students advisable to observe the options and solve the problem accordingly as in this problem the value of acceleration due to gravity is not used but the term is only used.
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