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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 ms−21 \text{ ms}^{-2}. The mass of this body must be:

A

2010 kg20\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 magnitude of the net force acting on a body is equal to the product of its mass and the magnitude of its acceleration (∣F⃗∣=m∣a⃗∣|\vec{F}| = m|\vec{a}|) [NCERT Class 11, Physics Part I, Sec 5.5].
  2. Calculate Force Magnitude: The force is given as a vector F⃗=6i^−8j^+10k^\vec{F} = 6\hat{i} - 8\hat{j} + 10\hat{k}. Its magnitude is calculated as: ∣F⃗∣=Fx2+Fy2+Fz2|\vec{F}| = \sqrt{F_x^2 + F_y^2 + F_z^2} ∣F⃗∣=(6)2+(−8)2+(10)2|\vec{F}| = \sqrt{(6)^2 + (-8)^2 + (10)^2} ∣F⃗∣=36+64+100=200=102 N|\vec{F}| = \sqrt{36 + 64 + 100} = \sqrt{200} = 10\sqrt{2} \text{ N}
  3. Calculate Mass: Given acceleration magnitude a=1 ms−2a = 1 \text{ ms}^{-2}. m=∣F⃗∣a=1021=102 kgm = \frac{|\vec{F}|}{a} = \frac{10\sqrt{2}}{1} = 10\sqrt{2} \text{ kg}
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