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Vector Algebra question

2023 · 8 Apr · Shift 1 · Q56
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Vector Algebra question

2023 · 8 Apr · Shift 1 · Q56

JEE MainPhysicsVector AlgebraMCQ+4 / −1
Two forces having magnitude AAA and A2\frac{A}{2}2A​ are perpendicular to each other. The magnitude of their resultant is:
  1. A
    5A2\frac{5 A}{2}25A​
  2. B
    5A4\frac{\sqrt{5} A}{4}45​A​
  3. C
    5A2\frac{\sqrt{5} A}{2}25​A​
  4. D
    5A22\frac{\sqrt{5} A^{2}}{2}25​A2​
View written solutionFree

Correct answer: C

  1. Let the two perpendicular forces be: F1=A,F2=A2F_1 = A, \qquad F_2 = \frac{A}{2}F1​=A,F2​=2A​

  2. Since they are perpendicular, the magnitude of the resultant is found using Pythagoras theorem: R=F12+F22R = \sqrt{F_1^2 + F_2^2}R=F12​+F22​​

  3. Substitute the values: R=A2+(A2)2R = \sqrt{A^2 + \left(\frac{A}{2}\right)^2}R=A2+(2A​)2​

  4. Simplify: R=A2+A24=5A24R = \sqrt{A^2 + \frac{A^2}{4}} = \sqrt{\frac{5A^2}{4}}R=A2+4A2​​=45A2​​

  5. Therefore, R=5A2R = \frac{\sqrt{5}A}{2}R=25​A​

  6. Now check the options:

    • A: 5A2\frac{5A}{2}25A​ ❌
    • B: 5A4\frac{\sqrt{5}A}{4}45​A​ ❌
    • C: 5A2\frac{\sqrt{5}A}{2}25​A​ ✅
    • D: 5A22\frac{\sqrt{5}A^2}{2}25​A2​ ❌ (wrong dimension)

Hence, the correct option is C.

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