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CBSE Questions for Class 11 Engineering Physics Work,Energy And Power Quiz 15 - MCQExams.com
CBSE
Class 11 Engineering Physics
Work,Energy And Power
Quiz 15
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0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct
A body is falling a height
h
. After it has fallen a height
h
2
. it will possess
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0%
only potential energy
0%
only kinetic energy
0%
half potential energy and half kinetic energy
0%
more kinetic and less potential energy
Explanation
(
c
)
half potential and half kinetic energy
Explanation: When the body is height
h
; its potential energy is at maximum and kinetic energy is zero. When the body hits the ground, its potential energy becomes zero and kinetic energy is at maximum. At mid-way i.e., half the height; its potential energy becomes half of the maximum potential energy and same happens to the kinetic energy.
Four alternatives are given to each of the following incomplete statements/questions. Choose the right answer:
Which of the following object has higher potential energy?
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0%
mass
=
10
k
g
,
g
=
9.8
m
s
−
2
,
h
=
10
m
0%
mass
=
5
k
g
,
g
=
9.8
m
s
−
2
,
h
=
12
m
0%
mass
=
8
k
g
,
g
=
9.8
m
s
−
2
,
h
=
100
m
0%
mass
=
6
k
g
,
g
=
9.8
m
s
−
2
,
h
=
20
m
Explanation
The potential energy is given as-
P
E
=
m
g
H
m
; mass of the object
g
=
9.8
/
m
s
−
2
h
; height gained from ground
For option A:
m
=
10
k
g
,
h
=
10
m
P
E
=
10
×
9.8
×
10
⇒
P
E
=
980
J
For option B:
m
=
5
k
g
,
h
=
12
m
P
E
=
5
×
9.8
×
12
⇒
P
E
=
588
J
For option C:
m
=
8
k
g
,
h
=
100
m
P
E
=
8
×
9.8
×
100
⇒
P
E
=
7840
J
For option D:
m
=
6
k
g
,
h
=
20
m
P
E
=
6
×
9.8
×
20
⇒
P
E
=
1176
J
For option C potential energy is highest so option C is the correct answer.
A force
F
=
K
(
y
i
+
x
j
)
( where
K
is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particles is
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0%
−
2
K
a
2
0%
2
K
a
2
0%
−
K
a
2
0%
K
a
2
Explanation
While moving from
(
0
,
0
)
t
o
(
a
,
0
)
Along positive x-axis
y
=
0
∴
→
F
=
−
k
x
ˆ
j
i.e. force is in negative y-direction while displacement is in positive x-direction.
∴
W
1
=
0
Because force is perpendicular to displacement
Then particle moves from
(
a
,
0
)
to
(
a
,
a
)
along a line parallel to y-axis
(
x
=
+
a
)
during this
→
F
=
−
k
(
y
ˆ
j
+
a
ˆ
J
)
The first component of force
−
k
y
ˆ
i
will not contribute any work because this component is along negative x-direction
(
−
ˆ
i
)
while displacement is in positive y- direction
(
a
,
0
)
to
(
a
,
a
)
The second component of force i.e.
−
k
a
ˆ
j
will perform negative work.
∴
W
2
=
(
−
k
a
ˆ
j
)
(
a
ˆ
j
)
=
(
−
k
a
)
(
a
)
=
−
k
a
2
So net work done on the particle
W
=
W
1
+
W
2
=
0
+
(
−
k
a
2
)
=
−
k
a
2
Is the work required to be done by an external force on an object on a frictionless, horizontal surface to accelerate it from a speed
v
to a speed
2
v
.
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0%
equal to the work required to accelerate the object from
v
=
0
to
v
,
0%
twice the work required to accelerate the object from
v
=
0
to
v
0%
three times the work required to accelerate the object from
v
=
0
to
v
,
0%
four times the work required to accelerate the object from
0
to
v
0%
not known without knowledge of the acceleration
Explanation
Answer (c). The net work needed to accelerate the object from
v
=
0
to
v
is
W
1
=
K
E
1
f
−
K
E
1
i
=
1
2
m
v
2
−
1
2
m
(
0
)
2
=
1
2
m
v
2
The work required to accelerate the object from speed
v
to speed
2
v
is
W
2
=
K
E
2
f
−
K
E
2
i
=
1
2
m
(
2
v
)
2
−
1
2
m
v
2
=
1
2
m
(
4
v
2
−
v
2
)
=
3
(
1
2
m
v
2
)
=
3
W
1
Bullet
2
has twice the mass of bullet
1
. Both are fired so that they have the same speed. If the kinetic energy of bullet
1
is
K
, is the kinetic energy of bullet
2
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0%
0.25
K
0%
0.5
K
0%
0.71
K
0%
K
0%
2
K
Explanation
Let
m
be the mass of bullet
1
.
According to the given condition, the mass of bullet
2
is
2
m
.
Kinetic energy of bullet
1
is
K
E
1
=
1
2
m
v
2
=
K
KE of bullet
2
is
K
E
2
=
1
2
(
2
m
)
v
2
=
2
(
1
2
m
v
2
)
=
2
K
The displacement of a particle of mass
m
=
2
k
g
on a smooth horizontal surface is a function of time given by
x
=
e
2
t
. Find net the work done by external agent from
t
=
l
n
1
to
t
=
l
n
2
.
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0%
20
J
0%
30
J
0%
50
J
0%
60
J
Explanation
Displacement of particle
x
=
e
2
t
∴
velocity at any instant
′
t
′
is given by
v
=
d
x
d
t
=
2
e
2
t
∴
acceleration at any instant
′
t
′
is given by
a
=
d
v
d
t
=
4
e
2
t
=
d
2
x
d
t
2
work done by the force
W
=
∫
F
.
d
x
=
∫
F
.
d
x
d
t
.
d
t
W
=
∫
l
n
2
l
n
1
m
a
(
d
x
d
t
)
d
t
m
=
2
k
g
=
∫
l
n
2
l
n
1
2
×
4
e
2
t
×
2
e
2
t
d
t
=
16
∫
l
n
2
l
n
1
e
4
t
d
t
=
16
4
[
e
4
t
]
l
n
2
l
n
1
=
16
4
[
e
4
l
n
2
−
e
4
l
n
1
]
=
4
[
e
l
n
(
2
4
)
−
e
l
n
(
1
4
)
]
=
4
[
2
4
−
1
4
]
=
4
[
15
]
=
60
J
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