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NEET Physics Solved Question Paper 2008

Multiple Choice Questions

For inelastic collision between two spherical rigid bodies

the total kinetic energy is conserved
the total mechanical energy is not conserved
the linear momentum is not conserved
the linear momentum is conserved

For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

2
$\frac{1}{2}$
$\frac{1}{\sqrt{2}}$
$\sqrt{2}$

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Taking g =10 m/s², work done against friction is

200J
100 J
zero
1000 J

A particle moves along a straight line OX. At a time t (in second) the distance x ( in metre ) of the particle from O is given by x = 40 + 12 t - t³ . How long would the particle travel before coming to rest?

24 m
40 m
56 m
16 m

A small disc of radius 2 cm is cut from a disc of radius 6 cm. If the distance between their centres is 3.2 cm, what is the shift in the centre of mass of the disc?

0.4 cm
2.4 cm
1.8 cm
1.2 cm

The displacement of particle is given by $x = α_{o} + \frac{α_{1} t}{2} - \frac{α_{2} t^{2}}{3}$ .What is its acceleration?

$\frac{2 α_{2}}{3}$
$- \frac{2 α_{2}}{3}$
$α_{2}$
Zero

The area of the acceleration-displacement curve of a body gives

impulse
change in momentum per unit mass
change in KE per unit mass
total change in energy

The speed of earth's rotation about its axis is o. Its speed is increased to x times to make the effective acceleration due to gravity equal to zero at the equator. Then x is

9.
A projectile is thrown in the upward direction making an angle of 60° with the horizontal direction with a velocity of 147 ms^-1 Then the time after which its inclination with the horizontal is 45°, is

15 s

10.98 s

5.49 s

2.745 s

5.49 s

At the two points of the trajectory during projection, the horizontal component of the velocity is the same.

The horizontal component of velocity at angle 60^o = Horizontal component of velocity at 45^o

⇒ u cos 60^o = v sin 45^o

$147 \times \frac{1}{2} = ν \times \frac{1}{\sqrt{2}} v = \frac{147}{\sqrt{2}} m / s$

Vertical component of u = u sin 60^o

$= \frac{147 \sqrt{3}}{2} m$

Vertical component of v = v sin 45^o

$\frac{147}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{147}{2} m$

but v_y = u_y + $α t$

$∴ \frac{147}{2} = \frac{147 \sqrt{3}}{2} - 9.8 t \Rightarrow 9.8 t = \frac{147}{2} (\sqrt{3} - 1)$

t = 5.49 s

10.

A motor cycle is going on an overbridge of radius R. The driver maintains a constant speed. As the motor cycle is ascending on the overbridge, the normal force on it