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System of Particles and Rotational Motion · Question 48

JEE Main 2025 (Online) 28th January Morning ShiftMCQEasy📈 Topic in 10% papers+4−1
1

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 )
  2. ( 23a, b2 )
  3. ( 13a, b2 )
  4. ( a2, b2 )

Answer

(2) ( 23a, b2 )

Solution

Take a strip of width dx at distance x from the y-axis:

dm = b dx = 0 xab b dx = 0ax dx

M = 0a 0ax dx = 0 a2

xcm = 1M 0a x dm = 0a a33 0 a2 = 2a3

does not depend on y mass is uniform along y ycm = b2

Centre of mass = ( 23a, b2 )

Answer: (2)

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2

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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3

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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4

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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