JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
As shown in figure, a \(70\,kg\) garden roller is pushed with a force of \(\overrightarrow{ F }=200\,N\) at an angle of \(30^{\circ}\) with horizontal. The normal reaction on the roller is \(.......\,N\) \(\left(\right.\) Given \(\left.g =10\,m s ^{-2}\right)\)
- A \(800 \sqrt{2}\)
- B \(600\)
- C \(800\)
- D \(200 \sqrt{3}\)
Answer & Solution
Correct Answer
(C) \(800\)
Step-by-step Solution
Detailed explanation
\(N = mg + F \sin 30^{\circ}\) \(=700+200 \times \frac{1}{2}=800 \text { newton }\)
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