SCRA Whole-Testpaper Contributed by Aravind updated on May 2019

SCRA General Ability Test SCRA Exam General Ability Physical Science

1. What is the lateral displacement of a ray of light passing through a parallel plate of glass of thickness t with angle of incidence and refraction respectively as µ and b?

(a) t sec (µ/b)

(b) t sin (µ-b) sec b (Ans)

(c) t sin (µ-b)

(d) t cos (µ-b) cosec b

Solution :  The lateral displacement of a ray

d = t sec b sin (µ-b)

2. A spaceship is launched into a circular orbit close to the Earth's surface. What additional velocity has to be imparted to the spaceship to overcome the gravitational pull ? (Radius of the Earth is R)

(a) gR

(b) √2gR

(c) (√2 - 1) √gR (Ans)

(d) (√2 - 1) gR

Solution :  The orbital velocity close to the

Earth's surface  = √gR

Escape velocity = √2 * orbital velocity

= √2gR

Additional velocity = escape velocity - orbital velocity

= √2gR - √gR = √gR (√2 - 1)

3.  What is the trajectory of a particle if it moves according to the law x = a sin (wt) and y = a cos (2wt) ?

(a) Circle

(b) Parabola  (Ans)

(c) Hyperbola

(d) Ellipse

Solution :  Given,     x = a sin wt

and  y = a cos 2wt

y = a (1 - 2sin2 wt)

y = a (1 - 2x2 / a2)        [? sin wt = x / a]

or            y = (a2 - 2x2) / a

or          ay = a2 - 2x2

or         2x2 = a2 - ay

or          x2 = a (a - y) / 2

This equation shows the equation of parabola. So, the trajectory of a particle will be parabola.

4. The velocity of a particle moving in the positive direction of x-axis varies as v = µ√x where µ is a positive constant. Assuming t = 0, the particle is at x = 0.  What is the time dependence on displacement ?

(a) t =  2x
µ

(b) t =  2√x     (Ans)
µ

(c) t = 2√ x
µ

(d) t = √2x
µ

Solution :  Velocity of particle, v = µ√x

x                            t
or    dx / dt = µ√x    or    ∫    dx / √x  =