JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2023
- Medium
- Numerical
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 \(\left.g =10\,m s ^{-2}\right)\)
Pick an option to check your answer.
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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