NIMCET 2013 — Mathematics PYQ

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Question

NIMCET 2013 — Mathematics PYQ

NIMCET | Mathematics | 2013

Forces $3\hat{i} + 2\hat{j} + 5\hat{k}$ and $2\hat{i} + \hat{j} - 3\hat{k}$ are acting on a particle and displace it from the point $2\hat{i} - \hat{j} - 3\hat{k}$ to the point $4\hat{i} - 3\hat{j} + 7\hat{k}$ . The work done by the force is:

Choose the correct answer:

Explanation

Solution

1. Calculate the Resultant Force ( $\vec{F}$ ):

\vec{F} = \vec{F_1} + \vec{F_2}

\vec{F} = (3\hat{i} + 2\hat{j} + 5\hat{k}) + (2\hat{i} + \hat{j} - 3\hat{k})

\vec{F} = (3+2)\hat{i} + (2+1)\hat{j} + (5-3)\hat{k}

\vec{F} = 5\hat{i} + 3\hat{j} + 2\hat{k}

2. Calculate the Displacement Vector ( $\vec{d}$ ):

Let the initial position be $\vec{r_1} = 2\hat{i} - \hat{j} - 3\hat{k}$ and the final position be $\vec{r_2} = 4\hat{i} - 3\hat{j} + 7\hat{k}$ .

\vec{d} = \vec{r_2} - \vec{r_1}

\vec{d} = (4\hat{i} - 3\hat{j} + 7\hat{k}) - (2\hat{i} - \hat{j} - 3\hat{k})

\vec{d} = (4-2)\hat{i} + (-3+1)\hat{j} + (7+3)\hat{k}

\vec{d} = 2\hat{i} - 2\hat{j} + 10\hat{k}

3. Calculate Work Done ( $W$ ):

W = \vec{F} \cdot \vec{d}

W = (5\hat{i} + 3\hat{j} + 2\hat{k}) \cdot (2\hat{i} - 2\hat{j} + 10\hat{k})

W = (5 \times 2) + (3 \times -2) + (2 \times 10)

W = 10 - 6 + 20

W = 4 + 20

W = 24\text{ units}

Choose the correct answer:

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