A little boy having mass 15 kg climbing on a rope with one end fixed to the ceiling. If it wishes to go up with an acceleration of 1 m text/s2 and rope is 5m long, also monkey starts from roof then how much time will it take to reach the ceiling? (take g = 10 m text/s2 )

Question

A little boy having mass 15 kg climbing on a rope with one end fixed to the ceiling. If it wishes to go up with an acceleration of 1 m text/s2 and rope is 5m long, also monkey starts from roof then how much time will it take to reach the ceiling? (take g = 10 m text/s2 ) (A)  √5 s  (B)  √7 s  (C)  2√10 s  (D)  √10 s

Tardigrade Admin · Accepted Answer

Given, mass m = 15 kg acceleration, a=1 m/s2 and initial velocity, u=0 As the length of rope is 5 m. So, height attained by child in given as s=ut+(1/2)at2 ⇒ 5=0+(1/2)× 1× t2⇒ t2=5× 2 or, t=√10 s

Q. A little boy having mass 15 kg climbing on a rope with one end fixed to the ceiling. If it wishes to go up with an acceleration of $1 m / s^{2}$ and rope is 5m long, also monkey starts from roof then how much time will it take to reach the ceiling? (take g = 10 $m / s^{2}$ )

$5 s$

$7 s$

$210 s$

$10 s$

Given, mass m = 15 kg acceleration, $a = 1 m / s^{2}$ and initial velocity, $u = 0$ As the length of rope is 5 m. So, height attained by child in given as $s = u t + \frac{1}{2} a t^{2}$ $\Rightarrow$ $5 = 0 + \frac{1}{2} \times 1 \times t^{2} \Rightarrow t^{2} = 5 \times 2$ or, $t = 10 s$

Q. A little boy having mass 15 kg climbing on a rope with one end fixed to the ceiling. If it wishes to go up with an acceleration of 1m/s2 and rope is 5m long, also monkey starts from roof then how much time will it take to reach the ceiling? (take g = 10 m/s2 )

5​s

7​s

210​s

10​s