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System of Particles and Rotational Motion · Question 46

JEE Main 2025 (Online) 2nd April Evening ShiftNumericalEasy📈 Topic in 35% papers+4
1

[figure] A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by a force of 10 N as shown in figure. The established torque produces an angular acceleration of 2 rad/s2. Moment of inertia of the wheel is kg m2. (Acceleration due to gravity = 10 m/s2)

Answer

1

Solution

The string leaves the rim tangentially, so the force acts at a perpendicular distance r from the axis.

Torque: = Fr = 10 0.2 = 2 N m

= I I = = 22

I = 1 kg m2

Answer: 1

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The coordinates of a particle with respect to origin in a given reference frame is (1, 1, 1) meters. If a force of F = i - j + k acts on the particle, then the magnitude of torque (with respect to origin) in z-direction is .

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3

The center of mass of a thin rectangular plate (fig - x) with sides of length a and b, whose mass per unit area ( ) varies as = 0 xab (where 0 is a constant), would be . [figure]

  1. ( 23a, 23b )
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  3. ( 13a, b2 )
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Two iron solid discs of negligible thickness have radii R1 and R2 and moment of inertia I1 and I2, respectively. For R2 = 2R1, the ratio of I1 and I2 would be 1/x, where x = .

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The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is n times higher than the moment of inertia of the given ring. Here, n = . Consider all the bodies have equal masses.

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6

A uniform rod of mass 250 g having length 100 cm is balanced on a sharp edge at 40 cm mark. A mass of 400 g is suspended at 10 cm mark. To maintain the balance of the rod, the mass to be suspended at 90 cm mark, is

  1. 290 g
  2. 200 g
  3. 190 g
  4. 300 g
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