NEET physics

Sponsor Area

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.

Sponsor Area

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
straight W space equals space mas space equals space straight m open parentheses fraction numerator straight d squared straight s over denominator dt squared end fraction close parentheses straight s space space space space space space... left parenthesis iii right parenthesis space space space open parentheses because space straight a space equals space fraction numerator straight d squared straight s over denominator dt squared end fraction close parentheses
Now comma space we space have comma space space straight s space equals space 1 third straight t squared
therefore space space space space space space space space space fraction numerator straight d squared straight s over denominator dt squared end fraction space equals space straight d over dt open square brackets straight d over dt open parentheses 1 third straight t squared close parentheses close square brackets
space space space space space space space space space space space space space space space space space space space equals straight d over dt cross times open parentheses 2 over 3 straight t close parentheses
space space space space space space space space space space space space space space space space space space equals 2 over 3 dt over dt space space space space space space
space space space space space space space space space space space space space space space space space space equals 2 over 3 space space space space space space space
space space space space space space space space space space space space space space space
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
                   straight m equals 3 space kg comma space straight t space equals space 2 straight s
therefore space space space space straight w space equals space 2 over 9 cross times 3 cross times left parenthesis 2 right parenthesis squared space equals space 8 over 3 straight J
 

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.
therefore space space space straight v space equals space straight d over dt left parenthesis 40 plus 12 straight t minus straight t cubed right parenthesis
space space space space space space equals space 0 plus 12 minus 3 straight t squared
but final velocity v = 0
therefore space space 12 minus 3 straight t squared space equals space 0
or space space space space space space space space straight t squared space equals space 12 over 3 space equals 4
or space space space space space space straight t space equals space 2 straight s
Hence, distance travelled by the particle before coming to rest is given by
                   straight x equals 40 plus 12 left parenthesis 2 right parenthesis minus left parenthesis 2 right parenthesis cubed
space space equals 40 plus 24 minus 8 space equals space 64 minus 8
space space equals space 56 space straight m
            

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
     open square brackets straight a close square brackets space open square brackets straight t close square brackets space equals space open square brackets straight v close square brackets
space space space space space open square brackets straight a close square brackets space equals space fraction numerator open square brackets straight v close square brackets over denominator open square brackets straight t close square brackets end fraction space equals space fraction numerator open square brackets LT to the power of negative 1 end exponent close square brackets over denominator open square brackets straight T close square brackets end fraction space equals space open square brackets LT to the power of negative 2 end exponent close square brackets
Similarly comma space space open square brackets straight c close square brackets space equals space open square brackets straight t close square brackets space equals space open square brackets straight T close square brackets
Further comma space space space space space fraction numerator open square brackets straight b close square brackets over denominator open square brackets straight t plus straight c close square brackets end fraction space equals open square brackets straight v close square brackets

or space space space space space space space space space space space space open square brackets straight b close square brackets space equals space open square brackets straight v close square brackets space open square brackets straight t plus straight c close square brackets
or space space space space space space space space space space space space open square brackets straight b close square brackets space equals space open square brackets LT to the power of negative 1 end exponent close square brackets space open square brackets straight T close square brackets space equals space open square brackets straight L close square brackets

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
rightwards double arrow space space space space space straight W space equals space straight W subscript straight g plus straight W subscript straight f space space space space space space space space space space space... left parenthesis straight i right parenthesis
But space space space space space straight W space equals space 300 space straight J
space space space space space space straight W subscript straight g space equals space mgh space equals space 2 space cross times 10 cross times 10 space equals space 200 space straight J
putting space in space Eq. space left parenthesis straight i right parenthesis comma space we space get
space space space space space space space space space 300 space equals space 200 plus straight W subscript straight f
rightwards double arrow space space space space space space space space space space space space space straight W subscript straight f space equals space 300 space minus space 200 space equals space 100 space straight J