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Work Power and Energy question

2025 · 22 Jan · Shift 2 · Q69
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Work Power and Energy question

2025 · 22 Jan · Shift 2 · Q69

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1
A force F→=2i^+bj^+k^\overrightarrow{\mathrm{F}}=2 \hat{i}+\mathrm{b} \hat{j}+\hat{k}F=2i^+bj^​+k^ is applied on a particle and it undergoes a displacement i^−2j^−k^\hat{i}-2 \hat{j}-\hat{k}i^−2j^​−k^ What will be the value of bbb, if work done on the particle is zero.
  1. A
    13\frac{1}{3}31​
  2. B
    12\frac{1}{2}21​
  3. C
    0
  4. D
    2
View written solutionFree

Correct answer: B

  1. Given vectors

    Force: F⃗=2i^+bj^+k^\vec F = 2\hat i + b\hat j + \hat kF=2i^+bj^​+k^

    Displacement: s⃗=i^−2j^−k^\vec s = \hat i - 2\hat j - \hat ks=i^−2j^​−k^

  2. Condition for zero work

    Work done is given by the dot product: W=F⃗⋅s⃗W = \vec F \cdot \vec sW=F⋅s

    Since work done is zero, F⃗⋅s⃗=0\vec F \cdot \vec s = 0F⋅s=0

  3. Compute the dot product

    F⃗⋅s⃗=(2)(1)+(b)(−2)+(1)(−1)\vec F \cdot \vec s = (2)(1) + (b)(-2) + (1)(-1)F⋅s=(2)(1)+(b)(−2)+(1)(−1)

    =2−2b−1= 2 - 2b - 1=2−2b−1

    =1−2b= 1 - 2b=1−2b

  4. Set equal to zero

    1−2b=01 - 2b = 01−2b=0

    2b=12b = 12b=1

    b=12b = \frac{1}{2}b=21​

  5. Match with options

    b=12b = \frac{1}{2}b=21​

    So the correct option is B.

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