Question

Answer it fast please

Answer 1

Question :

A Bob is suspended with the help of a thread whose breaking load is twice the weight of the Bob. Taking g=10m/s² , what is the minimum time in which the Bob can be raised by 10m?

Answer:

Let's start from the given information,

Mass of the Bob (m) = x kg

Breaking strength = 2x kg wt

=> 2x * g

=> 2x (10)

=> 20x

Here, g is the acceleration due to gravity.

Let the maximum acceleration be as 'a'

So the breaking strength = mg + ma

20x = x*g + x*a

20x = 10x + ax

10x = ax

a = 10m/s²

So maximum acceleration allowed is 10m/s²

Height (h) = 10m

Initial velocity (u) = 0

Time taken = t (say)

According to the formula

h = ut + 0.5at²

10 = 0 + 0.5(10)t²

t² = 2

t = √2

So, the correct option is (C)