question_answer

A man stands on a weighing machine kept inside a lift. Initially the lift is ascending with the acceleration 'a' due to which the reading is W. Now the lift decends with the same acceleration and reading is 10% of initial. Find the acceleration of lift?

A) \[\frac{g}{19}\,m/{{\sec }^{2}}\]

B) \[\frac{9g}{11}\,m/{{\sec }^{2}}\]

C) \[0\,/{{\sec }^{2}}\]

D) \[g\,m/{{\sec }^{2}}\]

Correct Answer: B

Solution :

[b]

\[N=m\,(g+a)\]

\[N'=m\,(g-a)\]

\[N'=\frac{10}{100}\times N\]

\[m(g-a)=\frac{m(g+a)}{10}\]

\[10g-10a=g+a\]

\[9g=11a\,\,\Rightarrow \,\,a=\frac{9g}{11}\]