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NEET physics

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Question
CBSEENPH11020623
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A particle moves in a circle of radius 5 cm with constant speed and time period 0.2 πs. The acceleration of the particle is 

  • 25 m/s2

  • 36 m/s2

  • 5 m/s2

  • 15 m/s2

Solution

C.

5 m/s2

D.

15 m/s2

Given comma space straight r space equals space 5 space cm space equals space 5 space straight x space 10 to the power of negative 2 end exponent space straight m and space straight T space equals space 0.2 space πs We space know space that space acceleration straight a space equals space rω squared equals fraction numerator 4 straight pi squared over denominator straight T squared end fraction straight r equals space fraction numerator 4 space straight x space straight pi squared space straight x space 5 space straight x space 10 to the power of negative 2 end exponent over denominator left parenthesis space 0.2 space straight pi right parenthesis squared end fraction space equals space 5 space ms to the power of negative 2 end exponent

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Question
CBSEENPH11020624
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The instantaneous angular position of a point on a rotation wheel is given by the equation 

Q(t)  = 2t3 - 6t2

The torque on the wheel becomes zero at

  • t = 0.5 s 

  • t = 0.25 s

  • t = 2s

  • t = 1s

Solution

D.

t = 1s

According space to space question comma space torque times straight tau space equals space 0 space it space means space that comma space straight alpha space equals space fraction numerator straight d squared straight theta over denominator dt squared end fraction Given space straight theta left parenthesis straight t right parenthesis space equals space 2 straight t cubed space minus space 6 straight t squared so space dθ over dt space equals space 6 straight t squared space minus space 12 straight t fraction numerator straight d squared straight theta over denominator dt squared space end fraction space equals space 12 straight t minus 12 open parentheses therefore space straight alpha space equals space fraction numerator straight d squared straight theta over denominator dt squared end fraction space equals space 0 close parentheses 12 straight t minus 12 space equals space 0 straight t space equals 1 straight s

Question
CBSEENPH11020625
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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
CBSEENPH11020627
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The moment of inertia of a thin uniform rod of mass M  and length L about an axis passing through its mid - point and perpendicular to its length is Io. Its moment of inertia about an axis passing through one of its ends and perpendicular to its length is

  • Io + ML2/4

  • Io + 2ML2

  • Io + ML2

  • Io + ML2/2

Solution

A.

Io + ML2/4

The theorem of parallel axis for moment of inertia
I = ICM + Mh2
space straight I space equals space straight I subscript straight o space plus straight M space open parentheses straight L over 2 close parentheses squared straight I space equals thin space straight I subscript straight o space plus ML squared over 4

Question
CBSEENPH11020631
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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

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