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# Motion in Straight Line

#### Multiple Choice Questions

111.

The given graph shows the variation of velocity with displacement. Which one of the graphs given below correctly represents the variation of acceleration with displacement.

A.

Given line has positive intercept but negative slope. So, its equation can be written as

v = − mx + V0

By differentiating w.r.t. to time, we get

Now, substituting the value of v from Eq. (i), we get

i.e., the graph between a and x should have positive slope but negative intercept on a-axis. So, graph (a) is correct.

112.

Assertion: If  $\mathrm{\alpha }$$-$2t for a particle moving in a straight line starting with initial velocity 4 m/s from the origin, then distance travelled by it in 2 s is same as displacement.

Reason: Velocity changes direction after 2 s only.

• If both assertion and reason are true and reason is the correct explanation of assertion.

• If both assertion and reason are true but reason is not the correct explanation of assertion.

• If assertion is true but reason is false.

• If both assertion and reason are false.

B.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Acceleration is given by

$\mathrm{\alpha }$ = $-$ 2t

2t

⇒    ${\int }_{4}^{\mathrm{v}}$dv = t dt

⇒     ${\left[\mathrm{v}\right]}_{4}^{\mathrm{v}}$ =

⇒     v $-$ 4 = $-$ t2

⇒           v = 4 $-$ t2

⇒           v > 0   for   t < 2

v = 0   for   t = 2

v  < 0   for  t > 2

The body reverses its direction of motion after 2 s.

113.

A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field $\stackrel{\to }{E}$ . Due to the force q $\stackrel{\to }{E}$, its velocity increases from 0 to 6 m/s in one-second duration. At that instant, the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively

• 2 m/s, 4 m/s

• 1 m/s, 3 m/s

• 1.5 m/s, 3 m/s

• 1 m/s, 3.5 m/s

B.

1 m/s, 3 m/s

v = - 6ms-1

Acceleration,

114.

In figure shown below, a particle is travelling counter clockwise in circle of radius 10 m. The acceleration vector indicated at a specific time. Find the value of 'v' at this time.

• 10 m/s

• 15 m/s

• 20 m/s

• 7 m/s

C.

20 m/s