question_answer

The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration is                           [AIPMT (S) 2009]

A) \[a=0,b=-1,c=-2\] downwards

B) \[_{Z}^{A}Z\xrightarrow[{}]{{}}{{\,}_{Z+}}_{1}^{4}Y\xrightarrow[{}]{{}}\,_{Z-1}^{A-4}{{B}^{*}}\xrightarrow[{}]{{}}\,_{Z-1}^{A-4}B,\] upwards

C) \[\beta ,\alpha ,\gamma \] downwards

D) \[\gamma ,\beta ,\alpha \] upwards

Correct Answer: C

Solution :

Key Idea Apparent weight > actual weight, then the lift is accelerating upward.

Here, lift is accelerating upward at the rate of a

Hence, equation of motion is written as \[R-mg=ma\]

\[28000-20000=2000a\]

\[\Rightarrow \]   \[{{V}_{C}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{\sigma 4\pi {{a}^{2}}}{c}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\sigma 4\pi {{b}^{2}}}{c}\]upwards