SCRA Physical Science |   805

SCRA Physical Science

SCRA - Solved Paper 2009 Physical Science solved questions papers for learn and practice, SCRA upcoming written test-model questions for reference, SCRA Previous years solved question papers with detailed explanations

 SCRA-Previous years Physical science solved questions with detailed answers and explanations

1. A particle of mass m is at x = 0 and moving along the x-axis with velocity v0, at time t = 0. It is subjected to a frictional force - bvx, where b is a constant and vx is the velocity in the x-direction. At what position x on the x-axis will it come to rest ?

(a) x = bv0

(b) x = bv0/m

(c) x = mbv0

(d) x = mv0/b (Ans)

Solutions :  Force, F = m (vf - vi) / t

- bv=   m (0 - v0) / t

                                        (v= 0 as body comes to rest)

- bx/t= - mv0/ t

    x = mv0/b

 

2.  A large tank filled with water to a height h is to be emptied through a small hole at the bottom. What is the ratio of the time taken for the level to fall from h to h/2 and that taken for the level to fall from h/2 to 0 ?

(a) √2

(b) 1/√2

(c) √2 - 1  (Ans)

(d) 1/(√2 -1)

Solutions :  t =  A/A√2/g [√H - √H']

According to problem, the time taken for the level to fall from h to h/2,

t1 =  A/A0√2/g [√h - √h/2]

and similarly time taken for the level to fall from h/2 to zero,

t2 = A/A0√2/g [√h/2 - 0]
 

    t1 / t2 = 1 - 1/√2   = √2 - 1
                   1/√2 - 0

 

3.  A body kept on a smooth inclined plane having inclination 1 in l will remain stationary relatives to the inclined plane if the plane is given a horizontal accleration equal to

(a) g/√l2 - 1 (Ans)

(b) gl/√l2 - 1

(c) g/2√l2 - 1

(d) 2g/√l2 - 1

Solutions :  Here, sin θ = 1/l

 

Let required acceleration of inclined plane be a for the object to remain stationary relative to incline, we have

ma cos θ = mg sin θ

a = g  tan θ = 1/√l2 - 1

 

4.  A body of mass m falling vertically downward with speed v0 is given an upward impulse I = F?t >mv0. What is the maximum height h that it will reach from the location of the impulse source ?

(a)

(b)  (Ans)

(c)

(d)

Solutions :  From equation of motion v2 = u2 + 2as

Here, a = g

s = h

vnet = (I/m - v0)

Substituting the above values in Eq. (i),we get

 

 

5.  A body B of mass m moving forward with velocity v along the x-axis, collides elastically with a stationary object C of mass 2m at the origin. After the collision, body B moves backward along the x-axis. Given that the kinetic energy of the system is conserved, what is the speed of the object C after the collision ?

(a) 2v/3   (Ans)

(b) v/2

(c) v/3

(d) 3v/4

Solutions :  If in a collision, kinetic energy after collision is equal to kinetic energy before collision, the collision is said to be perfectly elastic.

v2 = [m2 - m1 / m1 + m2] u2 + 2m1 u1 / m1 + m2

v2 = (2m - m) / 3m * 0 + 2mv / 3m

= 2v / 3

 

6.  A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to

(a) t1/2

(b) t3/4

(c) t3/2

(d) t2   (Ans)

Solutions :  Power,      P = Force / Area

   =  m * acc / l2

P = m * x  / l2t2

   x  ∝   t2                [as P and m are constants]

 

7.  A body of mass kg is under a force which causes displacement  in it, given by s = t2 / 3 in metre, with time t in seco

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