TS EAMCET · Physics · Motion In Two Dimensions
At a given instant of time the position vector of a particle moving in a circle with a velocity \(3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}\) is \(\hat{\mathbf{i}}+9 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}\). Its angular velocity at that time is
- A \(\frac{(13 \hat{\mathbf{i}}+29 \hat{\mathbf{j}}-31 \hat{\mathbf{k}})}{\sqrt{146}}\)
- B \(\frac{(13 \hat{\mathbf{i}}-29 \hat{\mathbf{j}}-31 \hat{\mathbf{k}})}{146}\)
- C \(\frac{(13 \hat{\mathbf{i}}+29 \hat{\mathbf{j}}-31 \hat{\mathbf{k}})}{\sqrt{146}}\)
- D \(\frac{(13 \hat{\mathbf{i}}+29 \hat{\mathbf{j}}+31 \hat{\mathbf{k}})}{146}\)
Answer & Solution
Correct Answer
(B) \(\frac{(13 \hat{\mathbf{i}}-29 \hat{\mathbf{j}}-31 \hat{\mathbf{k}})}{146}\)
Step-by-step Solution
Detailed explanation
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