JEE MAIN - Physics (2022 - 25th June Morning Shift - No. 27)
A force on an object of mass 100 g is $$\left( {10\widehat i + 5\widehat j} \right)$$ N. The position of that object at t = 2 s is $$\left( {a\widehat i + b\widehat j} \right)$$ m after starting from rest. The value of $${a \over b}$$ will be ___________.
Answer
2
Explanation
$$\overrightarrow F = m\overrightarrow a $$
$$ \Rightarrow \overrightarrow a = 100\widehat i + 50\widehat j$$
So, $$\overrightarrow S = {1 \over 2}\overrightarrow a {t^2}$$
$${1 \over 2}\left( {100\widehat i + 50\widehat j} \right){2^2}$$
$$ = 200\widehat i + 100\widehat j$$ m
so a = 200 m and b = 100 m
so $${a \over b} = 2$$
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