Question

A mass 1 kg is suspended by a thread. it is
(i) lifted up with an acceleration 4.9 m/s24.9 m/s2
The ratio of the tension is:

• A. 3:1
• B. 1:3
• C. 1:2
• D. 2:1

Answer 1

Answer: (A)

3:1

The magnitude of tension is equal to the net force acting on the body suspended by the string and tension and the weight always act in the opposite direction so tension will be directed upward along the length of the string.

Tension developed in the string when the mass lifted up with an acceleration of 4.9m/s^2 will be$T_1=mg+ma........\left(i\right)$

Tension developed in the string when the mass lowered with an acceleration of 4.9m/s^2 will be$T_1=mg-ma........\left(ii\right)$

The ratio of the tension will be,

$\frac{T_1}{T_2}=\frac{(g+a)}{(g-a)}$

$\frac{T_1}{T_2}=\frac{(9.8+4.9)}{(9.8-4.9)}$

$\frac{T_1}{T_2}=\frac{3}{1}$