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ExamsJEE AdvancedPhysics

A point object is placed 100 cm from a screen. A lens of focal length 23 cm is mounted on a frictionless stand connected to a spring of natural length 50 cm and spring constant 800 N/m. The mass of the stand with lens is 2 kg. The stand can oscillate between the object and screen. Find the minimum impulse (in kg m/s) that must be given to the stand so that a real image of the object forms on the screen exactly four times (at equal time intervals) during one complete oscillation.

  1. 4
  2. 6
  3. 8
  4. 10

Correct answer: 4

Solution

For D=100 cm, f=23 cm: using D² >= 4fD condition, 10000 >= 9200, valid. Positions: u1 and u2 where u1 + u2 = 100, and from lens formula: u1*u2/(u1+u2) = f giving u1*u2 = 2300. So u1, u2 are roots of x² - 100x + 2300 = 0: x = (100 +/- sqrt(10000-9200))/2 = (100 +/- sqrt(800))/2 = 50 +/- 10*sqrt(2). So positions from object: x1 = 50 - 10*sqrt(2) approx 35.86 cm, x2 = 50 + 10*sqrt(2) approx 64.14 cm. Natural length of spring = 50 cm, so equilibrium at 50 cm from object. Displacements: d1 = 50 - x1 = 10*sqrt(2) cm, d2 = x2 - 50 = 10*sqrt(2) cm. Both image positions are at equal distances (10*sqrt(2) cm) from equilibrium. For 4 images per oscillation, amplitude A must be >= 10*sqrt(2) cm. Minimum: A = 10*sqrt(2) cm = 0.1*sqrt(2) m. omega = sqrt(k/m) = sqrt(800/2) = 20 rad/s. Maximum speed v = A*omega = 0.1*sqrt(2)*20 = 2*sqrt(2) m/s. Impulse P = mv = 2 * 2*sqrt(2) = 4*sqrt(2) approx 5.66 kg m/s. Closest answer option: 4 or 6.

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