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JIPMER 2009
Exam
JIPMER
Year
2009
Subject
Physics
Chapter
Rotational Motion

JIPMER 2009 · Physics · Rotational Motion

A force of -F k acts on O , the origin of the coordinate system....

A force of −Fk^-F \hat{\mathbf{k}} acts on OO, the origin of the coordinate system. The torque about the point (1,−1)(1,-1) is A force of -F k acts on O , the origin of the coordinate system. The torque about the point (1,-1) is

Options

  1. AF(i^−j^)F(\hat{\mathbf{i}}-\hat{\mathbf{j}})
  2. B−F(i^+j^)-F(\hat{\mathbf{i}}+\hat{\mathbf{j}})
  3. CF(i^+j^)F(\hat{\mathbf{i}}+\hat{\mathbf{j}})
  4. D−F(i^−j^)-F(\hat{\mathbf{i}}-\hat{\mathbf{j}})

Correct answer

B. −F(i^+j^)-F(\hat{\mathbf{i}}+\hat{\mathbf{j}})

Step-by-step solution

Position vector (\vec{r}=\mathrm{PO}=\overrightarrow{\mathrm{O}}-(1 \hat{i}-1 \hat{\mathrm{j}})=-\hat{i}+\hat{j}) Torque about point (P, \vec{\tau}=\vec{r} \times \vec{F}) ∴τ⃗=(−i^+j^)×(−Fk^)OR τ⃗=F[(i^×k^)−(j^×k^)]=F[−j^−i^]⇒τ⃗=−F[i^+j^]\begin{aligned} & \therefore \vec{\tau}=(-\hat{i}+\hat{j}) \times(-F \hat{k}) \\ & \text{OR } \vec{\tau}=F[(\hat{i} \times \hat{k})-(\hat{j} \times \hat{k})]=F[-\hat{j}-\hat{i}] \\ & \Rightarrow \vec{\tau}=-F[\hat{i}+\hat{j}] \end{aligned})$
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