Laws of Motion

Question 525
CBSEENPH11020625

A person of mass 62 kg is inside of a lift of mass 940 kg and presses the button on control panel. The lift starts moving upwards with an acceleration 1.0 m/s2. If g = 10 m/s2 ,the tension in the supporting cable is

  • 9680 N

  • 11000 N

  • 1200 N 

  • 8600 N

Solution

B.

11000 N


Total mass = Mass of lift + Mass of person
= 940 +60 = 1000 kg
T - mg = ma
Hence, T - 1000 x 10 = 1000 x 1
 T = 11000 N
Question 526
CBSEENPH11020631

A planet moving along an elliptical orbit is closer to the sun at a distance r1 and farthest away at a distance of r2. if v1 and v2 are the linear velocities at these points respectively, then the ratio v1/v2 is

  • r2/r1

  • (r2/r1)2

  • r1 /r2

  • (r1/r2)2

Solution

A.

r2/r1

From the law of conservation of angular momentum
mr1v1 = mr2v2
r1v1 = r2v2
v1/v2 = r2/r1

Question 527
CBSEENPH11020639

Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12 m is

  • 21 J

  • 26 J 

  • 13 J

  • 18 J

Solution

C.

13 J

Work done = Area under (F - x) graph
= 2 x (7-3) + space equals space 2 space straight x space left parenthesis 7 minus 3 right parenthesis space plus 1 half space straight x space 2 space straight x space left parenthesis 12 minus 7 right parenthesis

equals space 8 space plus 1 half space straight x space 10

equals space 8 space plus space 5 space equals space 13 space straight J

Question 528
CBSEENPH11020640

Uniform magnetic field acting along AB. If the magnetic force on the arm BC Is F, the force on the arm AC is

  • -F
  • F
  • square root of 2 bold F
  • negative square root of 2 space bold F

Solution

A.

-F
FAB = 0
FAB + FBC + FCA = 0
FBC + FCA = 0
FBC + FCA = 0
FCA = - FBC = - F

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