inspiration

Solution

1 D MOTION Q75 XI PHY (JULY2022) TEST

Class 11 - Physics
Section A
1. (c) 8.33 m/s
Explanation: 8.33 m/s
2. (c) 158.33 m/s
= 150 + 8.33 = 158.33 m/s
Explanation: 158.33 m/s
3. (d) 105 m/s
Explanation: 105 m/s
= 158.33 - 53.33 = 105 m/s
4. (a) 17.22 m/s
Explanation: 17.22 m/s
192 - 30 = 162 km/h changed to m/s
5. (c) 53.33 m/s
Explanation: 53.33 m/s
6. (d) P ≡ (+360, 0, 0); R ≡ (-120, 0, 0)
Explanation: The position coordinates of point P = (+360, 0, 0) and point R =(-120, 0, 0)
7. (d) All of these
Explanation: Displacement is a vector quantity, it can be positive, negative or zero.
8. (b) +360 m and -120 m
Explanation: +360 m and -120 m
9. (b) 0 m
Explanation: 0
10. (c) 720 m
Explanation: 720 m
Section B
11. (c) A is true but R is false.
Explanation: At the point of intersection, the two vehicles share the same position at that instant of time and then the positions
of the two vehicles change differently. So, the vehicles cross each other at that point and they cannot have same velocity at that
instant. So, the assertion is true and the reason is false.
12. (c) If the assertion is true but the reason is false.
Explanation: An object is said to be in uniform motion if it undergoes equal displacement in equal intervals of time.
s1 + s2 + s3 +……….
∴ Vav. =
t1 + t2 + t3 +………..

s+s+s………. ns
= =
t+t+t+……… nt


and V ins. =
s

t


Thus, in uniform motion, average and instantaneous velocities have same value.
13. (a) Both A and R are true and R is the correct explanation of A.
Explanation: According to statement of reason, as the graph is a straight line, P ∝ Q, or, P = constant × Q ∴ P

Q
= constant.
Equation of a straight line is y = mx + c
14. (c) A is true but R is false.
Explanation: When slope of a time displacement graph of a particle is zero means the graph is a straight line parallel to the
time axis. The assertion is true.
It means that the particle does not change its position w.rt. time. So the particle is at rest. So, the assertion is false.
15. (b) Both A and R are true but R is not the correct explanation of A.
Explanation: According to the definition, displacement = velocity × time. Since displacement is a vector quantity so its value



1 D MOTION Q75 XI PHY (JULY2022) TEST

, is equal to the signed sum of the area under the velocity-time graph.
When considering the signed sum areas above the X-axis are considering positive and areas below the X-axis are
considering negative.
Section C
16. 15
Explanation:
Time taken by first drop to reach the ground,
−
−−
t=√ 2 h

g

−−−−
2×20
∴ t=√ 10
= 2s
Time taken by second drop to reach the ground
= s = 1s
t

2


In this time, distance of second drop from the tap = 1

2
gt2 = 1

2
g(1)2 = 5 m
Its distance from the ground = 20 - 5 = 15 m

17. 8
Explanation:
The resultant of two forces of 10 N and 6 N lies between (10 + 6) N, i.e., 16 N and (10 - 6) N, i.e., 4 N. Hence, the possible value
of resultant is 8 N.


18. -8
Explanation:
x = 10t - 4t2
y = 5t
dx

dt
= 10 - 8t
dy


dt
=5
vx = 10 - 8t
vy = 5
dv x

dt
= -8
dv y


dt
=0
ax = -8 m/s2

ay = 0 m/s2
Acceleration of particle is given by,
a = ax + ay
= -8 + 0
a = -8 m/s2 in x direction

Section D
19. i. We know that, a
​ = dv

dt


adt = dv
Integrating we get
t v

∫ adt = ∫ dv
0 u


at = v - u
v = u+at
dv
ii. a = dt


Multiply and Divide by dx, we get
dv dx
a= ×
dt dx
dv
a= × v
dx

dx
adx = vdv (∵ = v)
dt





1 D MOTION Q75 XI PHY (JULY2022) TEST

, Integrating within the limits, we have
s v

a ∫ dx = ∫ vdv
0 u

2 2
υ ν
as = −
2 2
2 2
υ −ν
as =
2


v2 – u2 = 2as
20. Here u = 2 ms-1, g = -9.8 ms-2, t = 2 s
i. v = u + gt = 2 - 9.8 × 2 = -17.6 ms-1.
A negative sign shows that the velocity is directed vertically downwards.
ii. Distance covered by the food packet in 2 s,
s = ut + 1

2
at2 = 2 × 2 - 1

2
× 9.8 × 22
= 4 - 19.6 = - 15.6 m
Thus the food packet falls through a distance of 15.6 m in 2 s but in the meantime, the helicopter rises up through a distance
= 2 ms-1 × 2 s = 4 m
∴ After 2 s, the distance of the food packet from the helicopter

= 15.6 + 4 = 19.6 m
21. i. It is not correct because in the time interval between t1 and t2, a is not constant.
ii. This relation is also not correct for the same reason as in (a).
iii. This relation is correct.
iv. This relation is also correct.
v. This relation is not correct because average acceleration cannot be used in this relation.
vi. This relation is correct.
22. Distance travelled in nth second is given by, Sn = u + a(n - 1

2
).
Thus,
12 = u + a(2 - 1

2
) ⇒ 12 = u + 3

2
a ...(i)
20 = u + a(4 - 1

2
) ⇒ 20 = u + 7

2
a ...(ii)
Eq. (ii) - Eq.(i), we get,
7 3
20 - 12 = (u + a) - (u + 2 2
a)
7−3
⇒ 8= 2
a
⇒ 8 = 2a
⇒ a= 8

2
= 4 ms-2
Substituting value of a in Eq. (i),
12 = u + × 4 3

2


u = 12 - 6 = 6 ms-1
Then,
Δs = s − s = [ut2 + ] - [ut1 + ]
1 2 1 2
t2 t1 at at
2 2 2 1


Δ s = u(t2 - t1) + 1

2
a (t
2
2
− t )
2
1


Now, t2 = 9 and t1 = 5, an using above calculated values of u and a, we get,

a(92 - 52) = 6(9 - 5) + 4(92 - 52)
1 1
Δ s= u(9 - 5) + 2 2
×

= 6(4) + 2(56)
Δ s = 136
∴ Distance covered by a body in 4 seconds after the 5th second is 136 m.
23. Distance Covered in nth Second of motion
Snth = Sn - Sn-1
1 2
Sn = un + an
2
1 2
Sn−1 = u(n − 1) + a(n − 1)
2
1 2 1 2
Snth = un + an − u(n − 1) − a(n − 1)
2 2





1 D MOTION Q75 XI PHY (JULY2022) TEST