Page 1
CBSE | Physics
Sample Paper 3 Solution
CBSE Board
Class XI Physics
Sample Paper-3 Solution
1.
Yes. Example is angle
2.
Motion of a body thrown vertically/obliquely under constant g
3.
- x-axis
4.
Work
5.
Since ? ? ? rF , larger arm means larger r which requires less F for same ? .
6.
3
7.
8.
No process is possible whose sole result is the absorption of heat from a reservoir and the
conversion of all of this heat into work.
Page 2
CBSE | Physics
Sample Paper 3 Solution
CBSE Board
Class XI Physics
Sample Paper-3 Solution
1.
Yes. Example is angle
2.
Motion of a body thrown vertically/obliquely under constant g
3.
- x-axis
4.
Work
5.
Since ? ? ? rF , larger arm means larger r which requires less F for same ? .
6.
3
7.
8.
No process is possible whose sole result is the absorption of heat from a reservoir and the
conversion of all of this heat into work.
CBSE | Physics
Sample Paper 3 Solution
9.
Systematic Errors Random Errors
1. Errors in which the deviation
from true value tends to have
fixed size and sign.
2. They can be attributed to a
fixed cause and can be
eliminated.
Deviation from true value is
irregular in size as well as sign.
Irregular pattern does not allow
them to be attributed to any fixed
cause and hence cannot be
eliminated, only minimized.
10.
For 3 kg:
F – F 32 = m 3a (i)
For 2 kg:
F 23 = m 2a (ii)
But from Newton’s third law
23 32
FF ??
Therefore, from (i) and (ii), F – m 2a = m 3a
F = (m 2 + m 3)a
2
5
a 1 m/s
5
? ? ?
Therefore, F 32 = m 2a = 2.1 = 2N (1/2)
Page 3
CBSE | Physics
Sample Paper 3 Solution
CBSE Board
Class XI Physics
Sample Paper-3 Solution
1.
Yes. Example is angle
2.
Motion of a body thrown vertically/obliquely under constant g
3.
- x-axis
4.
Work
5.
Since ? ? ? rF , larger arm means larger r which requires less F for same ? .
6.
3
7.
8.
No process is possible whose sole result is the absorption of heat from a reservoir and the
conversion of all of this heat into work.
CBSE | Physics
Sample Paper 3 Solution
9.
Systematic Errors Random Errors
1. Errors in which the deviation
from true value tends to have
fixed size and sign.
2. They can be attributed to a
fixed cause and can be
eliminated.
Deviation from true value is
irregular in size as well as sign.
Irregular pattern does not allow
them to be attributed to any fixed
cause and hence cannot be
eliminated, only minimized.
10.
For 3 kg:
F – F 32 = m 3a (i)
For 2 kg:
F 23 = m 2a (ii)
But from Newton’s third law
23 32
FF ??
Therefore, from (i) and (ii), F – m 2a = m 3a
F = (m 2 + m 3)a
2
5
a 1 m/s
5
? ? ?
Therefore, F 32 = m 2a = 2.1 = 2N (1/2)
CBSE | Physics
Sample Paper 3 Solution
OR
The two ways are:
1. Friction adjusts its direction to be always opposite to applied force.
2. Friction adjusts its magnitude up to a certain limit, to be equal to the applied force.
F ms = ?
s
N = ? ? ? ? ?
s
mg 0.2 2 10 4N
Since, applied force < F ms, the static friction acting = f s = 2 N.
11.
We know that
p 2mk ?
??
? ? ? ? ?
??
??
21 11 11
and p' 2mk' 2m k k 2mk p
100 10 10
Therefore,
11
pp
p1
10
p p 10
?
?
??
p
100 10%
p
?
??
12.
? ? ? rpsin mv
Direction of ? rp ? direction of
13.
Polar satellites - Their orbit is perpendicular to the orbit of geostationary satellites. These
are used for communication purpose. Also, the height above the Earth’s surface is lower.
Negative sign of total energy indicates attractive nature of force between the satellite and
the Earth.
Page 4
CBSE | Physics
Sample Paper 3 Solution
CBSE Board
Class XI Physics
Sample Paper-3 Solution
1.
Yes. Example is angle
2.
Motion of a body thrown vertically/obliquely under constant g
3.
- x-axis
4.
Work
5.
Since ? ? ? rF , larger arm means larger r which requires less F for same ? .
6.
3
7.
8.
No process is possible whose sole result is the absorption of heat from a reservoir and the
conversion of all of this heat into work.
CBSE | Physics
Sample Paper 3 Solution
9.
Systematic Errors Random Errors
1. Errors in which the deviation
from true value tends to have
fixed size and sign.
2. They can be attributed to a
fixed cause and can be
eliminated.
Deviation from true value is
irregular in size as well as sign.
Irregular pattern does not allow
them to be attributed to any fixed
cause and hence cannot be
eliminated, only minimized.
10.
For 3 kg:
F – F 32 = m 3a (i)
For 2 kg:
F 23 = m 2a (ii)
But from Newton’s third law
23 32
FF ??
Therefore, from (i) and (ii), F – m 2a = m 3a
F = (m 2 + m 3)a
2
5
a 1 m/s
5
? ? ?
Therefore, F 32 = m 2a = 2.1 = 2N (1/2)
CBSE | Physics
Sample Paper 3 Solution
OR
The two ways are:
1. Friction adjusts its direction to be always opposite to applied force.
2. Friction adjusts its magnitude up to a certain limit, to be equal to the applied force.
F ms = ?
s
N = ? ? ? ? ?
s
mg 0.2 2 10 4N
Since, applied force < F ms, the static friction acting = f s = 2 N.
11.
We know that
p 2mk ?
??
? ? ? ? ?
??
??
21 11 11
and p' 2mk' 2m k k 2mk p
100 10 10
Therefore,
11
pp
p1
10
p p 10
?
?
??
p
100 10%
p
?
??
12.
? ? ? rpsin mv
Direction of ? rp ? direction of
13.
Polar satellites - Their orbit is perpendicular to the orbit of geostationary satellites. These
are used for communication purpose. Also, the height above the Earth’s surface is lower.
Negative sign of total energy indicates attractive nature of force between the satellite and
the Earth.
CBSE | Physics
Sample Paper 3 Solution
14.
The stress required to fracture a material whether by compression, tension, or shear is
called breaking stress.
Yes, the wire is under stress as its own weight acts as load.
15.
For adiabatic expression PV
?
= const.
Therefore, PV
?
= '' PV
?
?
5/3 5/3 5 5/3
1600 '(8 ) 2 ' V P V PV ??
Or ??
a
1600
P' 50P
32
Therefore, fall in pressure = 1600 – 500 = 1550 P a
16.
(i) For isothermal expansion ?T = 0, hence ?U = 0
(ii) For adiabatic expansion ?U = ?Q – ?W = - ?W = -P ?V as ?Q = 0.
17.
?
? ? ? ?
?
A
36 1000
36 km/h 10 m/s
60 60
AC
54 km/h 15 m/s ? ? ? ? ?
BA B A
15 10 5 m/s ? ? ? ? ? ? ? ?
CA C A
15 ( 10) 25 m/s ? ? ? ? ? ? ? ? ?
Time taken by C to cover 1 km = ?
1000
40s
25
To avoid accident, B should cover 1 km is less than 40s.
??
2
1
s ut at
2
2
1
1000 5 40 a.(40)
2
? ? ? ?
200 800a ??
800a = 1000 – 200 = 800
? a = 1 m/s
2
Page 5
CBSE | Physics
Sample Paper 3 Solution
CBSE Board
Class XI Physics
Sample Paper-3 Solution
1.
Yes. Example is angle
2.
Motion of a body thrown vertically/obliquely under constant g
3.
- x-axis
4.
Work
5.
Since ? ? ? rF , larger arm means larger r which requires less F for same ? .
6.
3
7.
8.
No process is possible whose sole result is the absorption of heat from a reservoir and the
conversion of all of this heat into work.
CBSE | Physics
Sample Paper 3 Solution
9.
Systematic Errors Random Errors
1. Errors in which the deviation
from true value tends to have
fixed size and sign.
2. They can be attributed to a
fixed cause and can be
eliminated.
Deviation from true value is
irregular in size as well as sign.
Irregular pattern does not allow
them to be attributed to any fixed
cause and hence cannot be
eliminated, only minimized.
10.
For 3 kg:
F – F 32 = m 3a (i)
For 2 kg:
F 23 = m 2a (ii)
But from Newton’s third law
23 32
FF ??
Therefore, from (i) and (ii), F – m 2a = m 3a
F = (m 2 + m 3)a
2
5
a 1 m/s
5
? ? ?
Therefore, F 32 = m 2a = 2.1 = 2N (1/2)
CBSE | Physics
Sample Paper 3 Solution
OR
The two ways are:
1. Friction adjusts its direction to be always opposite to applied force.
2. Friction adjusts its magnitude up to a certain limit, to be equal to the applied force.
F ms = ?
s
N = ? ? ? ? ?
s
mg 0.2 2 10 4N
Since, applied force < F ms, the static friction acting = f s = 2 N.
11.
We know that
p 2mk ?
??
? ? ? ? ?
??
??
21 11 11
and p' 2mk' 2m k k 2mk p
100 10 10
Therefore,
11
pp
p1
10
p p 10
?
?
??
p
100 10%
p
?
??
12.
? ? ? rpsin mv
Direction of ? rp ? direction of
13.
Polar satellites - Their orbit is perpendicular to the orbit of geostationary satellites. These
are used for communication purpose. Also, the height above the Earth’s surface is lower.
Negative sign of total energy indicates attractive nature of force between the satellite and
the Earth.
CBSE | Physics
Sample Paper 3 Solution
14.
The stress required to fracture a material whether by compression, tension, or shear is
called breaking stress.
Yes, the wire is under stress as its own weight acts as load.
15.
For adiabatic expression PV
?
= const.
Therefore, PV
?
= '' PV
?
?
5/3 5/3 5 5/3
1600 '(8 ) 2 ' V P V PV ??
Or ??
a
1600
P' 50P
32
Therefore, fall in pressure = 1600 – 500 = 1550 P a
16.
(i) For isothermal expansion ?T = 0, hence ?U = 0
(ii) For adiabatic expansion ?U = ?Q – ?W = - ?W = -P ?V as ?Q = 0.
17.
?
? ? ? ?
?
A
36 1000
36 km/h 10 m/s
60 60
AC
54 km/h 15 m/s ? ? ? ? ?
BA B A
15 10 5 m/s ? ? ? ? ? ? ? ?
CA C A
15 ( 10) 25 m/s ? ? ? ? ? ? ? ? ?
Time taken by C to cover 1 km = ?
1000
40s
25
To avoid accident, B should cover 1 km is less than 40s.
??
2
1
s ut at
2
2
1
1000 5 40 a.(40)
2
? ? ? ?
200 800a ??
800a = 1000 – 200 = 800
? a = 1 m/s
2
CBSE | Physics
Sample Paper 3 Solution
18.
a = -kx
dv
a kx
dx
? ? ? ?
dv kx dx ? ? ?
Integrating both sides, we get
x
uo
dv kx dx
?
? ? ?
??
2 2 2
11
( u ) kx
22
? ? ? ?
or
2 2 2
11
m( u ) m kx
22
? ? ? ?
Therefore, loss in K.E. =
2
1
m kx
2
19.
(a) During free fall acceleration of thief = g = acceleration of load
So that load is unable to apply any force.
Let the force by load be N.
mg – N ? N = 0 = force applied by load on man
(b) Along horizontal direction, ?
? ext
F0 . Net linear momentum is conserved.
Before firing, the system is at rest.
Therefore, 0 =
b b g g
mm ? ? ?
b
gb
g
m
m
? ? ? ?
So, to conserve linear momentum, the gun recoils.
(c) The sand yields but the cemented floor doesn’t.
Hence, the time taken by man to come to rest increases in case of sand.
Since,
p
F
t
?
?
?
, force on man is less.
20.
We know that
12
KA(T T )t
Q
x
?
?
A = area of 6 faces = 6 × (3 × 10
-1
)
2
= 54 × 10
-3
m L =
12
KA(T T )t
x
?
or
12
KA(T T )t
m
xL
?
?
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