question_answer

A mass of 1 kg is suspended by a thread. It is

(i) lifted up with an acceleration \[4.9\text{ }m/{{s}^{2}}\].

(ii) lowered with an acceleration 4.9 m/s.

The ratio of the tensions is:         [AIPMT 1998]

A)             3 : 1

B) 1 : 3     

C) 1 : 2

D) 2 : 1

Correct Answer: A

Solution :

Key Idea: In a lift weight is the net force acting on the mass while going upwards or downwards.

(i) When mass is lifted upwards with an acceleration a, then apparent weight.

\[{{T}_{1}}-mg=ma\]

\[\Rightarrow \]   \[{{T}_{1}}=mg+ma\]

\[{{T}_{1}}=m(g+a)\]

Substituting the values, we obtain

\[\therefore \]      \[{{T}_{1}}=(1)\,(9.8+4.9)=14.7\,V\]

(ii) When mass is lowered downwards with an acceleration a, then

\[mg-{{T}_{2}}=ma\]

\[\Rightarrow \]   \[{{T}_{2}}=mg-ma=m(g-a)\]

Substituting the values, we have

\[{{T}_{2}}=(1)\,(9.8-4.9)\,=4.9\,N\]

Then, ratio of tensions

\[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{14.7}{4.9}=\frac{3}{1}\]

\[\Rightarrow \]   \[{{T}_{1}}:{{T}_{2}}=3:1\]