Question

A stone of $10\,kg$ is suspended with a rope of breaking strength $30\,kg.wt$ The minimum time in which the stone can be raised through s height $10\,m$ starting from rest is: (take $g = 10\,NK{g^{ - 1}}$ )
A. $0.5s$
B. $1.0s$
C. $\sqrt {\dfrac{2}{3}} s$
D. $2.0s$

Answer 1

Hint:In order to solve this question, we will first calculate the maximum acceleration that can be produced by maximum given force breaking point of rope and then will calculate net acceleration of the stone against gravity and later using newton’s law of equation of motion we will calculate the time to reach the height of $10m$ starting from rest.

Formula used:
If $u, t$ is the initial velocity, time and $S, a$ be the distance and acceleration of a body then by equation of motion its written as $S = ut + \dfrac{1}{2}a{t^2}$.
$a = \dfrac{F}{m}$
where, $F, m$ be the force and mass of a body having acceleration $a$.

Complete step by step answer:
According to the given problem we have, the maximum force breaking point of rope is $F = 30 \times 10$ where $g = 10\,NK{g^{ - 1}}$ is used as acceleration for rope.
$m = 10\,kg$
So maximum acceleration can be found as
$a = \dfrac{F}{m}$
$\Rightarrow a = \dfrac{{30 \times 10}}{{10}}$
$ \Rightarrow a = 30m{s^{ - 2}}$
Now, for stone we have, initial velocity is zero because starting from rest
$u = 0$
Net acceleration for stone be ${a_{net}} = a - g$.
${a_{net}} = 20\,m{s^{ - 2}}$
and distance $S = 10\,m$ height to which stone is raised
so using,
$S = ut + \dfrac{1}{2}a{t^2}$ we get,
$\Rightarrow 10 = 0 + \dfrac{1}{2}(20){t^2}$
$\Rightarrow {t^2} = 1$
$ \therefore t = 1\sec $

Hence, the correct option is B.

Note:It should be remembered that, acceleration for stone to raise up is in upward direction but direction of acceleration due to gravity is in downward direction so net acceleration for stone is the difference between two accelerations. Also, the square root of $1$ can be $ \pm 1$ but we know, time has no meaning in negative magnitude so we always take positive value while calculating time.