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Laws of Motion question

2023 · 11 Apr · Shift 2 · Q55
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Laws of Motion question

2023 · 11 Apr · Shift 2 · Q55

JEE MainPhysicsLaws of MotionMCQ+4 / −1
A body of mass 500 g500 \mathrm{~g}500 g moves along x\mathrm{x}x-axis such that it's velocity varies with displacement x\mathrm{x}x according to the relation v=10x m/sv=10 \sqrt{x} \mathrm{~m} / \mathrm{s}v=10x​ m/s the force acting on the body is:-
  1. A
    166 N
  2. B
    5 N
  3. C
    25 N
  4. D
    125 N
View written solutionFree

Correct answer: C

  1. Given data
  • Mass of body: m=500 g=0.5 kgm = 500\,\text{g} = 0.5\,\text{kg}m=500g=0.5kg
  • Velocity as a function of displacement: v=10xv = 10\sqrt{x}v=10x​

We need to find the force acting on the body.

  1. Use the relation between acceleration and velocity

When velocity is given as a function of position, acceleration is: a=dvdt=dvdxdxdt=vdvdxa = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}a=dtdv​=dxdv​dtdx​=vdxdv​

  1. Differentiate vvv with respect to xxx

Given: v=10x1/2v = 10x^{1/2}v=10x1/2

So, dvdx=10⋅12x−1/2=5x\frac{dv}{dx} = 10 \cdot \frac{1}{2}x^{-1/2} = \frac{5}{\sqrt{x}}dxdv​=10⋅21​x−1/2=x​5​

  1. Compute acceleration

a=vdvdx=(10x)(5x)=50 m/s2a = v\frac{dv}{dx} = (10\sqrt{x})\left(\frac{5}{\sqrt{x}}\right) = 50\,\text{m/s}^2a=vdxdv​=(10x​)(x​5​)=50m/s2

  1. Find force using Newton's second law

F=ma=0.5×50=25 NF = ma = 0.5 \times 50 = 25\,\text{N}F=ma=0.5×50=25N

  1. Match with options
  • A: 166 N166\,\text{N}166N
  • B: 5 N5\,\text{N}5N
  • C: 25 N25\,\text{N}25N
  • D: 125 N125\,\text{N}125N

Hence, the correct option is: C: 25 N\boxed{\text{C: }25\,\text{N}}C: 25N​

  1. Comparison with stored correct answer

Stored correct answer: C

Our derived answer is also C, so they agree.

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