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

space space space space space fraction numerator open vertical bar straight V close vertical bar over denominator straight g apostrophe end fraction equals fraction numerator G M over denominator R plus h end fraction space x space fraction numerator left parenthesis R plus h right parenthesis squared over denominator G M end fraction

rightwards double arrow space fraction numerator open vertical bar straight V close vertical bar over denominator straight g apostrophe end fraction space equals space R plus h    ... (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

Sponsor Area

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:

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
straight v squared over Rg equals fraction numerator sin space theta space plus space mu subscript s space cos theta over denominator cos space theta space minus space mu subscript s space sin space theta end fraction space open square brackets f subscript 1 proportional to mu subscript s close square brackets

rightwards double arrow v space equals square root of R g open parentheses fraction numerator tan space theta space plus mu subscript s over denominator cos space theta space minus space mu subscript s space sin space theta space end fraction close parentheses end root
rightwards double arrow space v space equals space square root of R g open parentheses fraction numerator tan space theta space plus space mu subscript s over denominator 1 minus mu subscript s space tan space theta end fraction close parentheses end root

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
rightwards double arrow space straight a space equals space fraction numerator 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 over denominator 1 end fraction space space left square bracket space straight m space equals 1 space kg right square bracket
rightwards double arrow space straight a space equals space 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 space straight m divided by straight s squared
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
integral d v space equals space integral a. space d t
space space space space space space space space space equals space integral left parenthesis 2 t space i with hat on top space plus space 3 t squared space j with hat on top right parenthesis space d t
v space equals space t squared space space i with hat on top space plus straight t cubed space straight j with hat on top
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,

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
equals space MR squared over 2 space minus space open square brackets fraction numerator begin display style straight M over 4 end style open parentheses begin display style straight R over 2 end style close parentheses squared over denominator 2 end fraction plus straight M over 4 open parentheses straight R over 2 close parentheses squared close square brackets
equals space MR squared over 2 space minus open square brackets MR squared over 32 plus MR squared over 16 close square brackets
equals MR squared over 2 minus open square brackets fraction numerator MR squared space plus space 2 MR squared over denominator 32 end fraction close square brackets
equals space MR squared over 2 minus fraction numerator 3 MR squared over denominator 32 end fraction
equals space fraction numerator 16 space MR squared minus 3 MR squared over denominator 32 end fraction
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
rightwards double arrow space 4 PQ space cos space straight ϕ space equals space 0

rightwards double arrow space cos space straight ϕ space equals space 0 space left square bracket because space straight P comma space straight Q space not equal to space 0 right square bracket

rightwards double arrow space straight ϕ space equals space straight pi over 2 equals space 90 to the power of straight o