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MCQMediumJEE 2023Uniform Circular Motion

JEE Physics 2023 Question with Solution

A small block of mass 100 g100 \, \text{g} is tied to a spring of spring constant 7.5 N/m7.5 \, \text{N/m} and length 20 cm20 \, \text{cm}. The other end of the spring is fixed at a point AA. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity 5 rad/s5 \, \text{rad/s} about point AA, the tension in the spring is:

  • A

    0.75 N0.75 \, \text{N}

  • B

    1.5 N1.5 \, \text{N}

  • C

    0.25 N0.25 \, \text{N}

  • D

    0.50 N0.50 \, \text{N}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Mass of block = 100 g=0.1 kg100 \, \text{g} = 0.1 \, \text{kg}, spring constant = 7.5 N/m7.5 \, \text{N/m}, natural length of spring = 20 cm=0.2 m20 \, \text{cm} = 0.2 \, \text{m}, angular velocity = 5 rad/s5 \, \text{rad/s}.

Find: The tension in the spring.

A spring fixed to a wall at one end and attached to a block labeled m on a horizontal surface, showing spring force kx and total stretched length l plus x.

The tension in the spring provides the centripetal force needed for circular motion.

Fc=mω2rF_c = m\omega^2 r

Also, by Hooke's law,

T=kxT = kx

If the spring extends by xx, then the radius of the circular path is

r=l+xr = l + x

So,

kx=mω2(l+x)kx = m\omega^2(l + x)

Substituting the given values,

7.5x=0.1×25×(0.2+x)7.5x = 0.1 \times 25 \times (0.2 + x)7.5x=2.5(0.2+x)7.5x = 2.5(0.2 + x)7.5x=0.5+2.5x7.5x = 0.5 + 2.5x5x=0.55x = 0.5x=0.1 mx = 0.1 \, \text{m}

Now the tension is

T=kx=7.5×0.1=0.75 NT = kx = 7.5 \times 0.1 = 0.75 \, \text{N}

Therefore, the tension in the spring is 0.75 N0.75 \, \text{N}. The correct option is A.

The same result can also be written in the compact form shown in the solution:

kx=mω2rkx = m\omega^2 r7.5x=0.1×25×(0.2+x)7.5x = 0.1 \times 25 \times (0.2 + x)7.5x=2.5(0.2+x)7.5x = 2.5(0.2 + x)3x=0.2+x3x = 0.2 + x2x=0.22x = 0.2x=0.1 mx = 0.1 \, \text{m}

Hence,

T=kx=7.5×0.1=0.75 NT = kx = 7.5 \times 0.1 = 0.75 \, \text{N}

Centripetal Force and Spring Relation

Given: The block moves with constant angular velocity on a smooth horizontal surface, so the only horizontal inward force is the spring tension.

Find: The value of spring tension.

Use the idea that for uniform circular motion, inward force equals centripetal force.

T=mω2rT = m\omega^2 r

Since the spring is stretched, the tension is also

T=kxT = kx

and the distance of the block from the fixed point is

r=l+xr = l + x

Thus,

kx=mω2(l+x)kx = m\omega^2(l + x)

Now substitute:

m=0.1 kg,ω=5 rad/s,l=0.2 m,k=7.5 N/mm = 0.1 \, \text{kg}, \quad \omega = 5 \, \text{rad/s}, \quad l = 0.2 \, \text{m}, \quad k = 7.5 \, \text{N/m}7.5x=0.1×(5)2×(0.2+x)7.5x = 0.1 \times (5)^2 \times (0.2 + x)7.5x=0.1×25×(0.2+x)7.5x = 0.1 \times 25 \times (0.2 + x)7.5x=2.5(0.2+x)7.5x = 2.5(0.2 + x)7.5x=0.5+2.5x7.5x = 0.5 + 2.5x5x=0.55x = 0.5x=0.1 mx = 0.1 \, \text{m}

Therefore,

T=kx=7.5×0.1=0.75 NT = kx = 7.5 \times 0.1 = 0.75 \, \text{N}

So, the tension in the spring is 0.75 N0.75 \, \text{N} and the correct option is A.

Common mistakes

  • Using the spring's natural length l=0.2 ml = 0.2 \, \text{m} directly as the radius is incorrect because the block is moving after the spring has stretched. The correct radius is r=l+xr = l + x.

  • Equating centripetal force to klkl instead of kxkx is wrong. Spring tension depends on extension, not on the natural length. Always use Hooke's law as T=kxT = kx.

  • Not converting 100 g100 \, \text{g} to 0.1 kg0.1 \, \text{kg} and 20 cm20 \, \text{cm} to 0.2 m0.2 \, \text{m} leads to inconsistent SI units. Convert all quantities before substitution.

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