If space straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top space equals space straight B with rightwards arrow on top space straight x space straight A with rightwards arrow on top then the angle between A and B isπ
  • π

  • π/3

  • π/2

  • π/2


A.

π

open parentheses straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top close parentheses space equals space left parenthesis straight B with rightwards arrow on top space straight x straight A with rightwards arrow on top right parenthesis
rightwards double arrow space open parentheses straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top close parentheses minus space left parenthesis straight B with rightwards arrow on top space straight x straight A with rightwards arrow on top right parenthesis space equals space 0
rightwards double arrow space open parentheses straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top close parentheses plus space left parenthesis straight B with rightwards arrow on top space straight x straight A with rightwards arrow on top right parenthesis space equals space 0
open square brackets therefore space left parenthesis straight B with rightwards arrow on top space straight x space straight A with rightwards arrow on top right parenthesis space equals space minus space left parenthesis straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top right parenthesis close square brackets
2 space open parentheses straight A with rightwards arrow on top space straight x space straight B with rightwards arrow on top close parentheses space equals 0
rightwards double arrow space space 2 AB space sin space straight theta space equals space 0
sin space straight theta space equals space 0 space space left square bracket space because space vertical line straight A with rightwards arrow on top vertical line space equals space straight A space not equal to 0 comma space vertical line straight B with rightwards arrow on top vertical line space equals space straight B not equal to space 0 right square bracket
straight theta space equals space 0 space or space straight pi space
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A projectile can have the same range R for two angles of projection. If T1 and T2 be the time of flights in the two cases, then the product of the two time of flights is directly proportional to

  • 1/R2

  • 1/R

  • R

  • R


C.

R

We know in advance that range of projectile is same for complementary angles i.e. for θ and (900 - θ )

straight T subscript 1 space equals space fraction numerator 2 straight u space sin space straight theta over denominator straight g end fraction

straight T subscript 2 space equals space fraction numerator 2 straight u space sin space left parenthesis 90 to the power of 0 minus straight theta right parenthesis over denominator straight g end fraction space equals space fraction numerator 2 space straight u space cos space straight theta over denominator straight g end fraction
and space straight R space equals space fraction numerator straight u squared space sin space 2 straight theta over denominator straight g end fraction
Therefore comma space straight T subscript 1 space straight T subscript 2 space equals space fraction numerator 2 space straight u space sin space straight theta over denominator straight g end fraction space straight x space fraction numerator 2 space straight u space cos space straight theta over denominator straight g end fraction
space equals space fraction numerator 2 straight u squared space left parenthesis 2 space sin space straight theta space cos space straight theta right parenthesis over denominator straight g squared end fraction
space equals space fraction numerator 2 straight u squared space left parenthesis sin space 2 straight theta right parenthesis over denominator straight g squared end fraction
space equals space 2 straight R divided by straight g
space equals space straight T subscript 1 straight T subscript 2 space proportional to space straight R

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A body of mass m accelerates uniformly from rest to v1 in time t1. The instantaneous power delivered to the body as a function of time t is

  • fraction numerator mv subscript 1 straight t over denominator straight t subscript 1 end fraction
  • fraction numerator mv subscript 1 superscript 2 space straight t over denominator straight t subscript 1 superscript 2 end fraction
  • fraction numerator mv subscript 1 straight t squared over denominator straight t subscript 1 end fraction
  • fraction numerator mv subscript 1 straight t squared over denominator straight t subscript 1 end fraction

B.

fraction numerator mv subscript 1 superscript 2 space straight t over denominator straight t subscript 1 superscript 2 end fraction

Let the constant acceleration of body of mass m is a.
From equation of motion

v1 = 0 + at1
⇒ a = t2/t= ...... (i)
At an instant t, the velocity v of the body v = 0 + at

straight v space equals space straight v subscript 1 over straight t subscript 1 straight t space space.... space left parenthesis ii right parenthesis
therefore space instantaneous space power
straight p space equals Fv
space equals mav space space left parenthesis because space straight F space equals space ma right parenthesis
equals space straight m open parentheses straight v subscript 1 over straight t subscript 1 close parentheses space straight x open parentheses straight v subscript 1 over straight t subscript 1 space straight x space straight t close parentheses space left square bracket space From space equ space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis
space equals space fraction numerator mv subscript 1 superscript 2 straight t over denominator straight t subscript 1 superscript 2 end fraction

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Which of the following statements is false for a particle moving in a circle with a constant angular speed?

  • The velocity vector is tangent to the circle.

  • The acceleration vector is tangent to the circle.

  • The acceleration vector points to the centre of the circle.

  • The acceleration vector points to the centre of the circle.


B.

The acceleration vector is tangent to the circle.

For a particle moving in a circle with constant angular speed, the velocity vector is always tangent to the circle and the acceleration vector always points towards the centre of the circle or is always along the radius of the circle. Since, the tangential vector is perpendicular to radial vector, therefore, velocity vector will be perpendicular to the acceleration vector. But in no case acceleration vector is tangent to the circle.
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A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particle takes place in a plane. It follows that

  • its velocity is constant

  • its acceleration is constant

  • its kinetic energy is constant

  • its kinetic energy is constant


C.

its kinetic energy is constant

When a force of constant magnitude acts on the velocity of particle perpendicularly, then there is no change in the kinetic energy of the particle. Hence, kinetic energy remains constant.

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