Q 1 :

A vector has magnitude same as that of A→=3i^+4j^ and is parallel to B→=4i^+3j^. The x and y components of this vector in first quadrant are x and 3 respectively where x= _________ .                    [2024]



(4)    Given A→=3i+4j, B→=4i+3j,

         |A|=32+42=5, B^=4i+3j5,B→|B|=B^

         Now, as per question, C→=|A|B^, mag of A→ and direction of B

         =(5)(4i+3j)5=4i+3j,4i+3j,          ∴x=4

 



Q 2 :

The position vector of a moving body at any instant of time is given as r→=(5t2i^–5tj^) m. The magnitude and direction of velocity at t = 2 s is,           [2025]

  • 515 m/s, making an angle of tan–1 4 with –ve Y axis

     

  • 515 m/s, making an angle of tan–1 4 with +ve X axis

     

  • 517 m/s, making an angle of tan–1 4 with –ve Y axis

     

  • 517 m/s, making an angle of tan–1 4 with +ve X axis

     

(3)

r→=5t2i^–5j^

v→=10ti^–5j^

v→=20i^–5j^ at t=2 sec

|v→|=517 m/s

tanθ=205=4

θ=tan–1 4 with –ve Y-axis



Q 3 :

A light wave is propagating with plane wave fronts of the type x + y + z = constant. The angle made by the direction of wave propagation with the x-axis is:          [2025]

  • cos–1(13)

     

  • cos–1(23)

     

  • cos–1(13)

     

  • cos–1(23)

     

(1)

Direction of propagation A→=i^+j^+k^

The direction of propagation of light is perpendicular to the wave front and is symmetric about x, y and z-axis.

Direction cosine associated with x-axis,

cosα=(AxAx2+Ay2+Az2)

cos α=(11+1+1) ⇒ α=cos–1(13)



Q 4 :

A vector in x-y plane makes an angle of 30° with y-axis. The magnitude of y-component of vector is 23. The magnitude of x-component of the vector will be          [2023]

  • 13

     

  • 6

     

  • 3

     

  • 2

     

(4)

      Ay=A cos 30°=23

⇒ A32=23

⇒ A=4

Now, Ax=A sin 30°=4×12=2



Q 5 :

When the position vector r→=xi^+yj^+zk^  changes sign as -r→ which one of the following vector will not flip under sign change ?  [2026]

  • Velocity

     

  • Linear momentum

     

  • Angular momentum

     

  • Acceleration

     

(3)

r→=xi^+yj^+zk^

v→=dr→dt=vxi^+vyj^+vzk^

p→=mv→

L→=m(r→×v→)

    =(xi^+yj^+zk^)×m(vxi^+vyj^+vzk^)

When sign of r→ changes, L→ remains same.