JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2021
- Medium
- Numerical
A person standing on a spring balance inside a stationary lift measures \(60 \,kg\). The weight of that person if the lift descends with uniform downward acceleration of \(1.8\, m / s ^{2}\) will be \(_{-} N\). \(\left[ g =10\, m / s ^{2}\right]\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(492\)
Step-by-step Solution
Detailed explanation
When lift is at rest
\(N = mg\)
\(\Rightarrow 60 \times 10=600 N\)
When lift moves with downward acceleration.
In frame of lift pseudo force will be in upward direction.
\(N^{\prime}=M(g-a)\)
\(\Rightarrow 60(10-1.8)\)
\(N^{\prime} \Rightarrow 492 N\)
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