NEET physics

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Question
CBSEENPH11020511

At what height from the surface of earth the gravitation potential and the value of g are -5.4 x 107 J kg-2 and 6.0 m/s2 respectively? (Take the radius of the earth as 6400 km)

  • 1600 km

  • 1400 km

  • 2000 km

  • 2600 km

Solution

D.

2600 km

Gravitational potential at some height h from the surface of the earth is given by,
V = fraction numerator negative GM over denominator straight R plus straight h end fraction    ... (i)
Acceleration due to gravity at some height h from the earth surface can be given as,

straight g apostrophe space equals space fraction numerator GM over denominator left parenthesis straight R plus straight h right parenthesis squared end fraction  ... (ii)
From equation (i) and (ii), we get

WiredFaculty    ... (iii)
because V = -54 x 107 J kg-2
g' = 6.0 m/s2
Radius of Earth, R = 6400 km


Putting the values in Eqn. (iii), we get
fraction numerator 5.4 space straight x space 10 to the power of 7 over denominator 6.0 end fraction = R+h
rightwards double arrow space 9 space straight x space 10 to the power of 6 space equals space straight R plus straight h
rightwards double arrowh = (9-6.4)x106 = 2.6 x 106 m
rightwards double arrow h = 2600 km

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Question
CBSEENPH11020513

A car is negotiating a curved road of radius R. The road is banked at angle straight theta. The coefficient of friction between the tyres of the car and the road is straight mu subscript straight s. The maximum safe velocity on this road is,

  • square root of gR open parentheses fraction numerator straight mu subscript straight s plus tan space straight theta over denominator 1 minus straight mu subscript straight s space tan space straight theta end fraction close parentheses end root
  • square root of straight g over straight R open parentheses fraction numerator straight mu subscript straight s plus tan space straight theta over denominator 1 minus straight mu subscript straight s space tanθ end fraction close parentheses end root
  • square root of straight g over straight R squared open parentheses fraction numerator mu subscript s plus tan theta over denominator 1 minus mu subscript s tan theta end fraction close parentheses end root
  • square root of g R squared open parentheses fraction numerator straight mu subscript straight s plus tanθ over denominator 1 minus straight mu subscript straight s tanθ end fraction close parentheses end root

Solution

A.

square root of gR open parentheses fraction numerator straight mu subscript straight s plus tan space straight theta over denominator 1 minus straight mu subscript straight s space tan space straight theta end fraction close parentheses end root

A car is negotiating a curved road of radius R. The road is banked at angle straight theta and the coefficient of friction between the tyres of car and the road is straight mu subscript straight s.
The given situation is illustrated as:
WiredFaculty
In the case of vertical equilibrium,
N cos straight theta = mg + f1 sin straight theta
rightwards double arrowmg = N cos straight theta space minus space straight f subscript 1 space sin space straight theta    ... (i)
In the case of horizontal equilibrium,
straight N space sin space straight theta space plus space straight f subscript 1 space cosθ space equals space mv squared over straight R  ... (ii)
Dividing Eqns. (i) and (ii), we get
WiredFaculty

Question
CBSEENPH11020514

A body of mass 1 kg begins to move under the action of a time dependent force F = left parenthesis 2 straight t space straight i with hat on top space plus space 3 straight t squared space straight j with hat on top right parenthesis N, where straight i with hat on top space a n d space j with hat on top are units vectors along X and Y axis. What power will be developed by the force at the time (t)?

  • (2t2 + 4t4) W

  • (2t3 + 3 t4) W

  • (2t3 + 3t5) W

  • (2t + 3t3)W

Solution

C.

(2t3 + 3t5) W

A body of mass 1 kg begins to move under the action of time dependent force,
F = (2t straight i with hat on top+3t2 straight j with hat on top) N,
where stack straight i space with hat on top a n d space j with hat on top are unit vectors along X and Y axis.
F = ma
rightwards double arrow space straight a space equals space straight F over straight m
WiredFaculty
Acceleration, a = dv over dt
rightwards double arrow space dv space equals space straight a. space dt space space space space space space... space left parenthesis straight i right parenthesis
Integrating both sides, we get
WiredFaculty
Power developed by the force at the time t will be given as,
P = F.v = (2t straight i with hat on top + 3t2 straight j with hat on top).(straight t squared space straight i with hat on top space plus space straight t cubed space straight j with hat on top)
   = (2t. t2 + 3t2.t3)
P = (2t3 + 3t5) W

Question
CBSEENPH11020515

From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc, about a perpendicular axis, passing through the centre?

  • 13 MR2/32

  • 11MR2/32

  • 9MR2/32

  • 15MR2/32

Solution

A.

13 MR2/32

Illustrating the above figure,
WiredFaculty
As given in the above fig.,
Moment of Inertia of the disc, I = Iremain + I(R/2)
rightwards double arrow space straight I subscript remain space equals space straight I space minus space straight I subscript straight R divided by 2 end subscript

Now, using the values, we get
WiredFaculty
That is,
Iremainfraction numerator 13 space MR squared over denominator 32 end fraction

Question
CBSEENPH11020518

If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is,

  • 90o

  • 45o

  • 180o

  • 0o

Solution

A.

90o

There are two vectors P and Q.
It is given that,
open vertical bar straight P plus straight Q close vertical bar space equals open vertical bar P minus Q close vertical bar
Let, angle between P and Q is straight ϕ,
straight P squared plus straight Q squared space plus space 2 PQ space cosϕ space equals space straight P squared space plus space straight Q squared space minus space 2 space PQ space cos space straight ϕ
WiredFaculty