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Circular Motion question

2023 · 6 Apr · Shift 1 · Q49
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  5. /2023 · 6 Apr · Shift 1 · Q49

Circular Motion question

2023 · 6 Apr · Shift 1 · Q49

JEE MainPhysicsCircular MotionMCQ+4 / −1
A small block of mass 100 g100 \mathrm{~g}100 g is tied to a spring of spring constant 7.5 N/m7.5 \mathrm{~N} / \mathrm{m}7.5 N/m and length 20 cm20 \mathrm{~cm}20 cm. The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity 5 rad/s5 ~\mathrm{rad} / \mathrm{s}5 rad/s about point A\mathrm{A}A, then tension in the spring is -
  1. A
    0.50 N
  2. B
    1.5 N
  3. C
    0.75 N
  4. D
    0.25 N
View written solutionFree

Correct answer: C

  1. Given data
  • Mass of block: m=100 g=0.1 kgm = 100\text{ g} = 0.1\text{ kg}m=100 g=0.1 kg
  • Spring constant: k=7.5 N/mk = 7.5\text{ N/m}k=7.5 N/m
  • Natural length of spring: l0=20 cm=0.2 ml_0 = 20\text{ cm} = 0.2\text{ m}l0​=20 cm=0.2 m
  • Angular velocity: ω=5 rad/s\omega = 5\text{ rad/s}ω=5 rad/s
  1. Physics idea

The block moves in a horizontal circle about point AAA. The only horizontal force providing centripetal force is the tension in the spring.

So, T=mω2rT = m\omega^2 rT=mω2r where rrr is the radius of circular motion, i.e. the stretched length of the spring.

Also, by Hooke’s law, T=k(r−l0)T = k(r-l_0)T=k(r−l0​)

Since both expressions represent the same tension, equate them: k(r−l0)=mω2rk(r-l_0)=m\omega^2 rk(r−l0​)=mω2r

  1. Substitute values

7.5(r−0.2)=0.1×52×r7.5(r-0.2)=0.1\times 5^2 \times r7.5(r−0.2)=0.1×52×r

7.5(r−0.2)=0.1×25×r7.5(r-0.2)=0.1\times 25\times r7.5(r−0.2)=0.1×25×r

7.5(r−0.2)=2.5r7.5(r-0.2)=2.5r7.5(r−0.2)=2.5r

  1. Solve for rrr

7.5r−1.5=2.5r7.5r-1.5=2.5r7.5r−1.5=2.5r

5r=1.55r=1.55r=1.5

r=0.3 mr=0.3\text{ m}r=0.3 m

  1. Find tension

Using T=mω2rT=m\omega^2 rT=mω2r

T=0.1×25×0.3T=0.1\times 25\times 0.3T=0.1×25×0.3

T=0.75 NT=0.75\text{ N}T=0.75 N

  1. Check with options

0.75 N0.75\text{ N}0.75 N corresponds to Option C.

Final Answer

0.75 N\boxed{0.75\text{ N}}0.75 N​

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