WBJEE · Physics · Laws of Motion
A body of mass \(2 \mathrm{~kg}\) moves in a horizontal circular path of radius \(5 \mathrm{~m}\). At an instant, its speed is \(2 \sqrt{5} \mathrm{~m} / \mathrm{s}\) and is increasing at the rate of \(3 \mathrm{~m} / \mathrm{s}^2\). The magnitude of force acting on the body at the instant is,
- A \(6 \mathrm{~N}\)
- B \(8 \mathrm{~N}\)
- C \(14 \mathrm{~N}\)
- D \(10 \mathrm{~N}\)
Answer & Solution
Correct Answer
(D) \(10 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
Hint: \(F=\) ma \(\begin{aligned} & =m \sqrt{a_c^2+a_T^2} \\ & =m \sqrt{\left(\frac{20}{5}\right)^2+9} \\ & =2 \sqrt{16+9}=2 \times 5=10 \mathrm{~N} \end{aligned}\)
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