NEET physics

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Question
CBSEENPH11020765

The position x of a particle with respect to time t along x- axis is given by x = 9t2 -t3 where x is in meter and t in second. What will be the position of this particle when it achieves maximum speed along the +x direction? 

  • 32 m

  • 54 m

  • 81 m 

  • 24 m

Solution

B.

54 m

At the instant when speed is maximum, its acceleration is zero.
Given, the position x of particle with respect to time t along x- axis
x = 9t2-t3 ... (i)
differentiating Eq. (i) with respect to time, we get speed, ie,
straight v space equals space dx over dt space equals space straight d over dt left parenthesis 9 straight t squared minus straight t cubed space right parenthesis
straight v space equals space 18 space straight t space minus space 3 straight t squared space... space left parenthesis ii right parenthesis
Again differentiating Eq. (ii) with respect to time, we get acceleration, ie,
straight a space equals space dv over dt space equals straight d over dt left parenthesis 9 straight t squared minus straight t cubed right parenthesis
straight a space equals space 18 minus 6 straight t
Now, when speed of particle is maximum, its acceleration is zero, ie,
a= 0
18-6t = 0 or t = 3s
Putting in eq (i) We, obtain the position of a particle at that time. 
x = 9 (3)2 - (3)3 = 9 (9) -27 
= 81-27 = 54 m

Sponsor Area

Question
CBSEENPH11020766

Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at a temperature  to C, the power received by a unit surface (normal to the incident rays) at a distance R from the centre of the sun is :

  • fraction numerator 4 πr squared σt to the power of 4 over denominator straight R squared end fraction
  • fraction numerator straight r squared straight sigma left parenthesis straight t plus 273 right parenthesis to the power of 4 over denominator 4 πR squared end fraction
  • fraction numerator 16 space straight pi squared straight r squared σt to the power of 4 over denominator straight R squared end fraction
  • fraction numerator straight r squared straight sigma left parenthesis straight t plus 273 right parenthesis to the power of 4 over denominator straight R squared end fraction

Solution

D.

fraction numerator straight r squared straight sigma left parenthesis straight t plus 273 right parenthesis to the power of 4 over denominator straight R squared end fraction

From Stefan's law, the rate at which energy is radiated by sun at its surface si 
P = σ x 4 πr2T4

[Sun is a perfectly black body as it emits radiations of all wavelengths and so for it e =1]
The intensity of this power at earth's surface (under the assumption R >>ro) is
straight I space equals space fraction numerator straight P over denominator 4 space πR squared end fraction
space equals space fraction numerator straight sigma space straight x space 4 space πr squared straight T to the power of 4 over denominator 4 πR squared end fraction
space equals space fraction numerator σr squared straight T to the power of 4 over denominator straight R squared end fraction
equals space fraction numerator σr squared left parenthesis straight t space plus 273 right parenthesis squared over denominator straight R squared end fraction

Question
CBSEENPH11020767

A particle starting from the origin (0,0) moves in a straight line in the (x,y) plane. Its coordinates at a later time are  ( square root of 3 comma space 3). The path of the particle makes with the x -axis an angle of:

  • 30o

  • 45o C

  • 60o C

  • 00

Solution

C.

60o C

The slope of the path of the particle gives the measure of angle required.

Draw the situation as shown. OA represents the path of the particle starting from origin O (0,0). Draw a perpendicular from point A to X- axis. Let path pf the particle makes and angle θ with the x -axis, then 
tan θ = slope of line OA

AB over OB space equals space fraction numerator 3 over denominator square root of 3 end fraction space equals space square root of 3
theta space equals space 60 to the power of 0

Question
CBSEENPH11020768

A Wheel has an angular acceleration of 3.0 rad/s2 and an initial angular speed of 2.00 rad/s. In a time of 2s it has rotated thorough an angle (in radian) of:

  • 6

  • 10

  • 12

  • 4

Solution

B.

10

Angular acceleration is time derivative of angular speed and angular speed is time derivative of angular displacement.
By definition,
straight alpha space equals space dω over dt
ie comma space dω space equals space αdt
So comma space if space in space time space straight t space the space angular space speed space of space straight a space body space changes space from space straight omega subscript straight o space to space straight omega
integral subscript straight omega subscript straight o end subscript superscript straight omega space dω space equals space integral subscript 0 superscript straight r space straight alpha space dt
If space straight alpha space is space constant
straight omega space minus straight omega subscript straight o space equals space αt
straight omega space equals space straight omega subscript straight o space plus αt
Now comma space as space by space definition
straight omega space equals space dθ over dt
Eq space left parenthesis straight i right parenthesis space becomes
dθ over dt space equals space space straight omega subscript straight o space plus αt
dθ space equals space left parenthesis straight omega subscript straight o space plus αt right parenthesis dt
So space straight m space if space in space time space straight t space angular space displacement space is space straight theta
integral subscript 0 superscript straight theta dθ space equals space integral subscript 0 superscript straight t space left parenthesis straight omega subscript straight o space plus αt right parenthesis space dt
Given comma space straight alpha space equals space 3.0 space rad divided by straight s squared comma space space equals space 2.0 space rad divided by straight s comma space straight t space equals space 2 straight s
Hence comma space straight theta space equals space 2 space straight x space 2 space plus space 1 half space straight x space 3 space left parenthesis 2 right parenthesis squared
or space straight theta space equals space 4 plus 6 space equals space 10 space rad

Question
CBSEENPH11020769

A uniform rod AB of length l and mass m is free to rotate about point A. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about A  is ml2/3, the initial angular acceleration of the rod will be:

  • 2g/3l

  • mgl/2

  • 3gl/2

  • 3g/2l

Solution

D.

3g/2l

The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is 
l = ml3/3
where m is a mass of rod and l is the
straight tau space equals space mg straight iota over 2
Iα space equals space mg straight l over 2
ml squared over 3 straight alpha space equals space mg straight l over 2
straight alpha space equals space fraction numerator 3 straight g over denominator 2 straight l end fractionlength.
Torque acting on centre of gravity of rod is given by