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Laws of Motion

Question 537
CBSEENPH11020722
Wired Faculty App

The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration is 

  • 30 ms-2 downward

  • 4 ms-2  upwards

  •  4 ms-2  downwards

  • 14 ms-2  upwards

Solution

B.

4 ms-2  upwards

Apparent weight > actual weight, then the lift is accelerating upward.
Lift is accelerating upward at the rate of a 
Hence, equation of motion is written as
R - mg = ma
28000-20000 = 2000a
a = 8000/2000 = 4 ms-2 upwards.

Question 538
CBSEENPH11020736
Wired Faculty App

A body under the action of a force,straight F with rightwards arrow on top space equals space 6 space bold i with bold hat on top space minus space 8 space bold j with bold hat on top space plus 10 space bold k with hat on top acquires an acceleration of 1 ms-2. The mass of this body must be

  • 2 square root of 10 space kg

  • 20 kg

  • 10 kg

  • 10 square root of 2 space kg

Solution

D.

10 square root of 2 space kg

According to Newton's second law of motion, force = mass x acceleration.
Here, 
straight F with rightwards arrow on top equals space 6 space bold i with bold hat on top space minus 8 space bold j with bold hat on top space plus 10 space bold k with hat on top vertical line straight F vertical line space equals space square root of 36 plus 64 plus 100 end root space equals space 10 square root of 2 space straight N straight a space equals space 1 space ms to the power of negative 2 end exponent straight m space equals space fraction numerator 10 square root of 2 over denominator 1 end fraction space equals 10 space square root of 2 straight g

Question 539
CBSEENPH11020751
Wired Faculty App

A closed loop PQRS carrying a current is placed in a uniform magnetic field. If the magnetic forces on segments PS, SR and RQ are F1, F2 and F3 respectively and are in the plane of the paper and along the directions shown, the force on the segment of QP is

  • F3 - F1-F2

  • square root of left parenthesis straight F subscript 3 minus straight F subscript 2 right parenthesis squared space plus straight F subscript 2 squared end root
  • square root of left parenthesis straight F subscript 3 minus straight F subscript 1 right parenthesis squared minus straight F subscript 2 squared end root

  • F3 - F2 + F1

Solution

B.

square root of left parenthesis straight F subscript 3 minus straight F subscript 2 right parenthesis squared space plus straight F subscript 2 squared end root
F4 sin θ = F2
F4 cos θ = (F3-F1)
straight F subscript 4 space equals space square root of left parenthesis straight F subscript 3 minus straight F subscript 1 right parenthesis squared space plus straight F subscript 2 squared end root
Question 540
CBSEENPH11020752
Wired Faculty App

Three forces acting on a body are shown in the figure. To have the resultant force only along the y-direction, the magnitude of the minimum additional force needed is

  • 0.5 N

  • 1.5 N

  • fraction numerator square root of 3 over denominator 4 end fraction space straight N
  • square root of 3 space straight N

Solution

B.

1.5 N

Minimum additional force needed
F = - (Fresultant)x
straight F subscript resultant space equals space left square bracket left parenthesis 4 minus 2 right parenthesis left parenthesis cos space 30 space bold j with hat on top space minus space sin space 30 space bold i with hat on top right parenthesis space plus space left parenthesis cos space 60 space bold i with hat on top space plus sin space 60 space bold j with hat on top right parenthesis space equals space open square brackets 2 open parentheses fraction numerator square root of 3 over denominator 2 end fraction bold j with hat on top minus 1 half bold i with hat on top close parentheses space plus open parentheses 1 half bold i with hat on top space plus fraction numerator square root of 3 over denominator 2 end fraction bold j with hat on top close parentheses close square brackets equals open square brackets open parentheses square root of 3 space plus fraction numerator square root of 3 over denominator 2 end fraction close parentheses bold j with hat on top space plus open parentheses negative bold i with bold hat on top space plus fraction numerator bold i with bold hat on top over denominator 2 end fraction close parentheses close square brackets equals open square brackets negative 1 half bold i with bold hat on top space plus fraction numerator 3 square root of 3 over denominator 2 end fraction bold j with hat on top close square brackets equals fraction numerator bold i with hat on top over denominator 2 end fraction plus fraction numerator 3 square root of 3 over denominator 2 end fraction space bold j with hat on top therefore comma straight F space equals space minus space open parentheses fraction numerator bold i with hat on top over denominator 2 end fraction close parentheses space equals space 1 half bold i with bold hat on top Hence comma space space vertical line straight F vertical line space equals space 0.5 space straight N

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