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