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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Physics
Chapter
Rotational Motion

BITSAT 2023 · Paper 2 · Physics · Rotational Motion

The torque of a force 5 i +3 j -7 k about the origin is . If the...

The torque of a force 5i^+3j^−7k^5 \hat{i}+3 \hat{j}-7 \hat{k} about the origin is τ\tau. If the force acts on a particle whose position vector is 2i^+2j^+k^2 \hat{i}+2 \hat{j}+\hat{k}, then the value of τ\tau will be :

Options

  1. A11i^+19j^−4k^11 \hat{\mathrm{i}}+19 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}
  2. B−11i^+9j^−16k^-11 \hat{\mathbf{i}}+9 \hat{\mathbf{j}}-16 \hat{\mathrm{k}}
  3. C−17i^+19j^−4k^-17 \hat{\mathrm{i}}+19 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}
  4. D17i^+9j^+16k^17 \hat{i}+9 \hat{j}+16 \hat{k}

Correct answer

C. −17i^+19j^−4k^-17 \hat{\mathrm{i}}+19 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}

Step-by-step solution

Given that torque, τ⃗=5i^+3j^−7k^\vec{\tau}=5 \hat{i}+3 \hat{j}-7 \hat{k}
Position vector, r→=2i^++2j^+k^\overrightarrow{\mathrm{r}}=2 \hat{\mathrm{i}}++2 \hat{\mathrm{j}}+\hat{\mathrm{k}}
τ⃗=r→×F→=∣ijk22153−7∣=i^(−14−3)−j^(−14−5)+k^(6−10)=−17i^+19j^−4k^\begin{aligned} & \vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 2 & 2 & 1 \\ 5 & 3 & -7\end{array}\right| \\ & =\hat{\mathrm{i}}(-14-3)-\hat{\mathrm{j}}(-14-5)+\hat{\mathrm{k}}(6-10) \\ & =-17 \hat{\mathrm{i}}+19 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}\end{aligned}
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