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NEET PHYSICSEasy

A body, under the action of a force F⃗=6i^−8j^+10k^\vec{F} = 6\hat{i} - 8\hat{j} + 10\hat{k}, acquires an acceleration of 1 m/s21 \text{ m/s}^2. The mass of this body must be:

A

210 kg2\sqrt{10} \text{ kg}

B

10 kg10 \text{ kg}

C

20 kg20 \text{ kg}

D

102 kg10\sqrt{2} \text{ kg}

Step-by-Step Solution

  1. Concept: According to Newton's Second Law of Motion, the force vector is the product of mass and the acceleration vector (F⃗=ma⃗\vec{F} = m\vec{a}). In terms of magnitude, ∣F⃗∣=m∣a⃗∣|\vec{F}| = m |\vec{a}| [NCERT Class 11, Physics Part I, Section 5.5].
  2. Calculate Magnitude of Force: The force is given by F⃗=6i^−8j^+10k^\vec{F} = 6\hat{i} - 8\hat{j} + 10\hat{k}. The magnitude is calculated as: ∣F⃗∣=Fx2+Fy2+Fz2|\vec{F}| = \sqrt{F_x^2 + F_y^2 + F_z^2} ∣F⃗∣=62+(−8)2+102=36+64+100|\vec{F}| = \sqrt{6^2 + (-8)^2 + 10^2} = \sqrt{36 + 64 + 100} ∣F⃗∣=200=100×2=102 N|\vec{F}| = \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2} \text{ N}
  3. Calculate Mass: Given the magnitude of acceleration ∣a⃗∣=1 m/s2|\vec{a}| = 1 \text{ m/s}^2. m=∣F⃗∣∣a⃗∣=1021=102 kgm = \frac{|\vec{F}|}{|\vec{a}|} = \frac{10\sqrt{2}}{1} = 10\sqrt{2} \text{ kg} (Reference: NCERT Class 11, Physics Part I, Chapter 4: Motion in a Plane, Section 4.4 for vector magnitude, and Chapter 5: Laws of Motion, Section 5.5).
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