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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 \(_{-} N\). \(\left[ g =10\, m / s ^{2}\right]\)

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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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