JEE Mains · Physics · STD 11 - Newtons laws of motion
- JEE Mains 2022
- Medium
- Numerical
A force on an object of mass \(100\, g\) is \((10 \hat{i}+5 \hat{j}) N\). The position of that object at \(t=\) \(2 s\) is \((a \hat{i}+b \hat{j}) m\) after starting from rest. The value of \(\frac{a}{b}\) will be..........
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ F }=10 \hat{ i }+5 \hat{ j }\)
\(m =100 \,g =0.1 \,kg\)
\(\vec{a}=\frac{\overrightarrow{ F }}{ m }=100 \hat{ i }+50 \hat{ j }\)
\(\overrightarrow{ S }=\overrightarrow{ u t}+\frac{1}{2} \overrightarrow{ a }{ }^{2}=\frac{1}{2} \overrightarrow{ a } t ^{2}(\) as \(\overrightarrow{ u }=0)\)
\(=\frac{1}{2}(100 \hat{ i }+50 \hat{ j }) 2^{2}\)
\(=200 \hat{i}+100 \hat{j}\)
\(=a \hat{i}+b \hat{j}\)
\(a=200, b=100\)
\(\therefore \frac{a}{b}=2\)
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