• # question_answer                 A mass of 1 kg is suspended by a thread. It is                 (i) lifted up with an acceleration $4.9\,m/{{s}^{2}},$                 (ii) lowered with an acceleration $4.9\,m/{{s}^{2}}$.                 The ratio of the tensions is:                                                                         A)                            3 : 1  B)                 1 : 3        C)                                            1 : 2    D)                                            2 : 1

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                 $\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{14.7}{4.9}=\frac{3}{1}$                 $\Rightarrow$               ${{T}_{1}}:{{T}_{2}}=3:1$