NEET physics

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Question
CBSEENPH11020791

For angles of projection of a projectile at angles (45° - θ) and (45° + θ), the horizontal ranges described by the projectile are in the ratio of

  • 1:1

  • 2:3

  • 1:2

  • 2:1

Solution

A.

1:1

For complementary angles of projection, their horizontal ranges will be same.
We know that, horizontal ranges for complementary angles of projection will be same.
The projectiles are projected at angles left parenthesis 45 degree minus straight theta right parenthesis and left parenthesis 45 degree plus straight theta right parenthesis which are complementary to each other i.e., two angles add up to give 90 degree. Hence, horizontal ranges will be equal. Thus, the required ratio is 1:1.

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Question
CBSEENPH11020792

A body of mass 3 kg is under a constant force which causes a displacement s in metres in it, given by the relation straight s equals 1 third straight t squared, where t is in s. Work done by the force in 2 s is:

  • 5 over 19 straight J
  • 3 over 8 straight J
  • 8 over 3 straight J
  • 19 over 5 straight J

Solution

C.

8 over 3 straight J

If a constant force is applied on the object causing a displacement in it, then it is said that work has been done on the body to displace it.
  Work done by the force  = Force x Displacement
or          W = F x s                            ...(i)
But from Newton's 2nd law, we have
  Force  =  Mass x Acceleration
i.e.,       F = ma                                ...(ii)
Hence, from Eqs. (i) and (ii), we get
WiredFaculty
Hence, Eq (ii) becomes
                straight W space equals 2 over 3 ms space equals space 2 over 3 straight m cross times 1 third straight t squared
                   equals 2 over 9 mt squared
We have given
                   WiredFaculty
 

Question
CBSEENPH11020793

A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle from O is given by
        straight x equals 40 straight t space plus 12 straight t minus straight t cubed
How long would the particle travel before coming to rest?

  • 24m

  • 40m

  • 56m

  • 16m

Solution

C.

56m

Velocity is rate of change of distance or displacement.
Distance travelled by the particle is
                      straight x equals space 40 space plus 12 straight t minus straight t cubed
We know that, velocity is rate of change of distance i.e., straight v equals dx over dt.
WiredFaculty
but final velocity v = 0
WiredFaculty
Hence, distance travelled by the particle before coming to rest is given by
                   WiredFaculty
            

Question
CBSEENPH11020794

The velocity v of a particle at time t is given by straight v equals at plus fraction numerator straight b over denominator straight t plus straight c end fraction comma where a, b and c are constants, The dimensions of a, b and c are respectively:

  • open square brackets LT to the power of negative 2 end exponent close square brackets comma space open square brackets straight L close square brackets space and space open square brackets straight T close square brackets
  • open square brackets straight L squared close square brackets space open square brackets straight T close square brackets space and space open square brackets LT squared close square brackets
  • open square brackets LT squared close square brackets comma space open square brackets LT close square brackets space and space open square brackets straight L close square brackets
  • open square brackets straight L close square brackets comma space open square brackets LT close square brackets space and space open square brackets straight T squared close square brackets

Solution

A.

open square brackets LT to the power of negative 2 end exponent close square brackets comma space open square brackets straight L close square brackets space and space open square brackets straight T close square brackets

According to principle of homogeneity of dimensions, the dimensions of all the terms in a physical expression should be same.
   The given expression is 
                      straight v equals at plus fraction numerator straight b over denominator straight t plus straight c end fraction
From principle of homogeneity
     WiredFaculty

Question
CBSEENPH11020795

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

  • 200 J

  • 100 J

  • Zero

  • 1000 J

Solution

B.

100 J

Net work done in sliding a body up to a height h on inclined plane
    = Work done against gravitational force + Work done against frictional force
WiredFaculty