Laws of Motion

Question 481
CBSEENPH11020311

A mass m hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass m and radius R.Assuming pulley to be a perfect uniform circular disc, the acceleration of the mass m, if the string does not slip on the pulley, is

  • g

  • 2/3g

  • g/3

  • 3/2g

Solution

B.

2/3g


mg - T = ma
TR space equals space fraction numerator mR squared straight alpha over denominator 2 end fraction
straight T space equals space mRα over 2 space equals space ma over 2
mg space minus ma over 2 space equals space ma
fraction numerator 3 space ma over denominator 2 end fraction space equals space mg
straight a space equals space fraction numerator 2 straight g over denominator 3 end fraction
Question 482
CBSEENPH11020312

Work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly. (Surface tension of soap solution = 0.03 Nm-1)

  • 4π mJ 

  • 0.2π mJ

  • 2π mJ

  • 0.4π mJ

Solution

D.

0.4π mJ

Work done = Change in surface energy
⇒ W = 2T x 4π (R22-R12)
 = 2 x 0.03 x 4π [ (5)2-(3)2] x 10-4
 = 0.4 π mJ

Question 483
CBSEENPH11020323

A pulley of radius 2m is rotated about its axis by a force F = (20t - 5t2) Newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg-m2, the number of rotations made by the pulley before its direction of motion if reversed is

  • more than 3 but less than 6

  • more than 6 but less than 9

  • more than 9

  • less than

Solution

A.

more than 3 but less than 6

To reverse the direction
integral τdθ space equals 0
straight tau space equals space left parenthesis 20 space straight t minus 5 straight t squared right parenthesis 2 space equals space 40 straight t minus 10 straight t squared
straight alpha space equals space straight tau over straight I space equals space fraction numerator 40 straight t minus 10 straight t squared over denominator 10 end fraction space equals space 4 straight t minus straight t squared
straight omega space equals space integral subscript 0 superscript straight t space αdt space equals space 2 straight t squared minus straight t cubed over 3
straight omega space is space zero space at
2 straight t squared minus straight t cubed over 3 space equals space 0
straight t cubed space equals space 6 straight t squared
straight t space equals space 6 space sec
straight theta space equals space integral ωdt
space equals space integral subscript 0 superscript 6 left parenthesis 2 straight t squared minus straight t cubed over 3 right parenthesis dt
open square brackets fraction numerator 2 straight t cubed over denominator 3 end fraction minus straight t to the power of 4 over 12 close square brackets subscript 0 superscript 6 space equals space 216 space open square brackets 2 over 3 minus 1 half close square brackets space equals space 36 space rad.
No space of space revolution space fraction numerator 36 over denominator 2 straight pi end fraction space Less space than space 6

Question 484
CBSEENPH11020326

STATEMENT – 1
Two particles moving in the same direction do not lose all their energy in a completely inelastic collision.
STATEMENT – 2
Principle of conservation of momentum holds true for all kinds of collisions.

  • The statement I is True, Statement II is False.

  • The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.

  • The statement I is True, Statement II is True; Statement II is not the correct explanation for Statement I.

  • Statement I is False, Statement II is

Solution

B.

The statement I is True, Statement II is True; Statement II is a correct explanation for Statement I.

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