DC Pandey Solutions: Laws of Motion- 3 | DC Pandey Solutions for NEET Physics PDF Download

 Page 1


10. Had q been 90°.
For every x displacement of wedge ( ) w the
vertical fall in mass would have been 2x as 
the string passes through pulley B.
i.e., v v
1 2
2 =
For the situation given the speed of mass
would thus be 
          v v
1 2
2 90 = ° - cos ( ) q
i.e., v v
1 2
2 = sin q
Option (c) is correct.
11. The cylinder will start rising up the
inclined plane if
ma mg cos sin q q >
i.e., a g > tan q
\ a g
min
tan = q [cylinder will be at the
point of rising up the inclined plane]
12. At the position of maximum deflection the
net acceleration of bob (towards its mean
position will be)
g a sin cos q q -
as explained in figure.
13. Mg T Ma - =
T mg ma - =
Þ   a
M m
M m
g =
-
+
\     T m g a = + ( )
       = +
-
+
é
ë
ê
ù
û
ú
m g
M m
M m
g
       = +
-
+
é
ë
ê
ù
û
ú
mg
M m
M m
1
       = + mg[ ] 1 1
{as m M << , M m M - » and 
M m M + » }
           = 2 mg
\ Tension in the string suspended from
ceiling = = 2 4 T mg
Option (a) is correct.
14. For 0° < £ f q  frictional force ( ) sin F mg = q
For q ³ f  frictional force ( ) cos F mg = m q
[f = angle of repose]
For example
Let m =
1
3
     \      tan f =
1
3
f = ° 30
q F (force of friction) Condition
10° mgsin10° = 0.174 mg
20° mgsin20° = 0.342 mg Increase but not
linearly only
30° mmgcos30° = 0.500 mg
40° mmgcos40° = 0.442 mg Decrease but not
linearly only
60° mmgcos60° = 0.287 mg
90° Zero
Option (b) is correct.
15.   a
mg
m
B
=
m
2
        =
mg
2
Laws of Motion 91
ma cos q
ma
(Pseudo
furce)
O
a cos q
a
g cos (90° – q)
= g sin q
90°– q
q
g
T
T T
T
mg
Mg
2T
2T
m
v
2
w
m
q
v
1
90° – q
v
1
B
O
90°
q
F
f
Page 2


10. Had q been 90°.
For every x displacement of wedge ( ) w the
vertical fall in mass would have been 2x as 
the string passes through pulley B.
i.e., v v
1 2
2 =
For the situation given the speed of mass
would thus be 
          v v
1 2
2 90 = ° - cos ( ) q
i.e., v v
1 2
2 = sin q
Option (c) is correct.
11. The cylinder will start rising up the
inclined plane if
ma mg cos sin q q >
i.e., a g > tan q
\ a g
min
tan = q [cylinder will be at the
point of rising up the inclined plane]
12. At the position of maximum deflection the
net acceleration of bob (towards its mean
position will be)
g a sin cos q q -
as explained in figure.
13. Mg T Ma - =
T mg ma - =
Þ   a
M m
M m
g =
-
+
\     T m g a = + ( )
       = +
-
+
é
ë
ê
ù
û
ú
m g
M m
M m
g
       = +
-
+
é
ë
ê
ù
û
ú
mg
M m
M m
1
       = + mg[ ] 1 1
{as m M << , M m M - » and 
M m M + » }
           = 2 mg
\ Tension in the string suspended from
ceiling = = 2 4 T mg
Option (a) is correct.
14. For 0° < £ f q  frictional force ( ) sin F mg = q
For q ³ f  frictional force ( ) cos F mg = m q
[f = angle of repose]
For example
Let m =
1
3
     \      tan f =
1
3
f = ° 30
q F (force of friction) Condition
10° mgsin10° = 0.174 mg
20° mgsin20° = 0.342 mg Increase but not
linearly only
30° mmgcos30° = 0.500 mg
40° mmgcos40° = 0.442 mg Decrease but not
linearly only
60° mmgcos60° = 0.287 mg
90° Zero
Option (b) is correct.
15.   a
mg
m
B
=
m
2
        =
mg
2
Laws of Motion 91
ma cos q
ma
(Pseudo
furce)
O
a cos q
a
g cos (90° – q)
= g sin q
90°– q
q
g
T
T T
T
mg
Mg
2T
2T
m
v
2
w
m
q
v
1
90° – q
v
1
B
O
90°
q
F
f
For no slipping
        a a
A B
=
   
F mg
m
g -
=
m m
2
i.e.,      F mg =
3
2
m
          m =
2
3
F
mg
Slipping will obviously be m
2
 there if m is
greater than above mentioned value
\ m
min
=
2
3
F
mg
For no slipping.
Option (c) is correct.
16. F
net
 (downward) = + mg ma sin cos q q
= + m g a ( sin cos ) q q
\      g g a
eff
= + sin cos q q
Time ( ) T required to cover 2L distance
along inclined would be
       T
L
g
=
2
eff
=
+
2L
g a ( sin cos ) q q
Option (c) is correct.
17. F
net
 on block along incline in the upward
direction
= - ma mg cos sin q q
= - m a g ( cos sin ) q q
\     g a g
eff
= - cos sin q q
\ Time ( ) t to move s distance would be
given by
s g t =
1
2
2
eff
i.e.,  t
s
g
s
a g
= =
-
2 2
eff
( cos sin ) q q
Substituting s = 1 m,
           q = ° 30 ,
a =10 3 m/s
2
 and g = 10 m/s
2
t =
´
´
æ
è
ç
ö
ø
÷
´
æ
è
ç
ö
ø
÷
2 1
10 3
3
2
10
1
2
_
       =
1
5
 s
Option (b) is correct.
18. N
w
w
w
+ ° = +
2
30
2
sin
               f = frictional force
       N
w
=
5
4
      f N
w
max
= = × m m
5
4
The block will remain stationary if
w w
2
30
5
4
cos ° £
m
or
w w
2
3
2
5
4
£ m
or 3 5 £ m
or
3
5
£ m
\    Block will move if m <
3
5
Option (d) is correct.
92 | Mechanics-1
2m
m
A
B
mmg
F
F
F
mmg
F F
q
a
ma
ma cos q
mg sin q
q
ma cos q
mg sin q
a
w
2
sin 30°
w/2
30°
w
2
cos 30°
w
f
w/2
N
Page 3


10. Had q been 90°.
For every x displacement of wedge ( ) w the
vertical fall in mass would have been 2x as 
the string passes through pulley B.
i.e., v v
1 2
2 =
For the situation given the speed of mass
would thus be 
          v v
1 2
2 90 = ° - cos ( ) q
i.e., v v
1 2
2 = sin q
Option (c) is correct.
11. The cylinder will start rising up the
inclined plane if
ma mg cos sin q q >
i.e., a g > tan q
\ a g
min
tan = q [cylinder will be at the
point of rising up the inclined plane]
12. At the position of maximum deflection the
net acceleration of bob (towards its mean
position will be)
g a sin cos q q -
as explained in figure.
13. Mg T Ma - =
T mg ma - =
Þ   a
M m
M m
g =
-
+
\     T m g a = + ( )
       = +
-
+
é
ë
ê
ù
û
ú
m g
M m
M m
g
       = +
-
+
é
ë
ê
ù
û
ú
mg
M m
M m
1
       = + mg[ ] 1 1
{as m M << , M m M - » and 
M m M + » }
           = 2 mg
\ Tension in the string suspended from
ceiling = = 2 4 T mg
Option (a) is correct.
14. For 0° < £ f q  frictional force ( ) sin F mg = q
For q ³ f  frictional force ( ) cos F mg = m q
[f = angle of repose]
For example
Let m =
1
3
     \      tan f =
1
3
f = ° 30
q F (force of friction) Condition
10° mgsin10° = 0.174 mg
20° mgsin20° = 0.342 mg Increase but not
linearly only
30° mmgcos30° = 0.500 mg
40° mmgcos40° = 0.442 mg Decrease but not
linearly only
60° mmgcos60° = 0.287 mg
90° Zero
Option (b) is correct.
15.   a
mg
m
B
=
m
2
        =
mg
2
Laws of Motion 91
ma cos q
ma
(Pseudo
furce)
O
a cos q
a
g cos (90° – q)
= g sin q
90°– q
q
g
T
T T
T
mg
Mg
2T
2T
m
v
2
w
m
q
v
1
90° – q
v
1
B
O
90°
q
F
f
For no slipping
        a a
A B
=
   
F mg
m
g -
=
m m
2
i.e.,      F mg =
3
2
m
          m =
2
3
F
mg
Slipping will obviously be m
2
 there if m is
greater than above mentioned value
\ m
min
=
2
3
F
mg
For no slipping.
Option (c) is correct.
16. F
net
 (downward) = + mg ma sin cos q q
= + m g a ( sin cos ) q q
\      g g a
eff
= + sin cos q q
Time ( ) T required to cover 2L distance
along inclined would be
       T
L
g
=
2
eff
=
+
2L
g a ( sin cos ) q q
Option (c) is correct.
17. F
net
 on block along incline in the upward
direction
= - ma mg cos sin q q
= - m a g ( cos sin ) q q
\     g a g
eff
= - cos sin q q
\ Time ( ) t to move s distance would be
given by
s g t =
1
2
2
eff
i.e.,  t
s
g
s
a g
= =
-
2 2
eff
( cos sin ) q q
Substituting s = 1 m,
           q = ° 30 ,
a =10 3 m/s
2
 and g = 10 m/s
2
t =
´
´
æ
è
ç
ö
ø
÷
´
æ
è
ç
ö
ø
÷
2 1
10 3
3
2
10
1
2
_
       =
1
5
 s
Option (b) is correct.
18. N
w
w
w
+ ° = +
2
30
2
sin
               f = frictional force
       N
w
=
5
4
      f N
w
max
= = × m m
5
4
The block will remain stationary if
w w
2
30
5
4
cos ° £
m
or
w w
2
3
2
5
4
£ m
or 3 5 £ m
or
3
5
£ m
\    Block will move if m <
3
5
Option (d) is correct.
92 | Mechanics-1
2m
m
A
B
mmg
F
F
F
mmg
F F
q
a
ma
ma cos q
mg sin q
q
ma cos q
mg sin q
a
w
2
sin 30°
w/2
30°
w
2
cos 30°
w
f
w/2
N
19.
At t = 2 s
F
1
4 = N
f
1
 and f
2
 are the frictional forces
( )
max
f
1
1 10 = ´ ´ 0.6
= 6 N
( )
max
f
2
2 10 = ´ ´ 0.5
  = 10 N
At t = 2 s
Net external force ( ) F
net
 on system
          = 15 N - 4 N 
= 11 N
As F f f
net
> + ( ) ( )
max max 1 2
, the system will
remain at rest and the values frictional
forces on the blocks will be given
T f = + 4
1
 and T f = - 15
2
4 15
1 2
+ = - f f …(i)
f f
1 2
11 + = N
Let direction being + ive for Eq. (i)
Option (a) f
1
4 = - N, f
2
5 = = N
Þ f f
1 2
1 + = N wrong
Option (b) f
1
2 = - N, f
2
5 = + N
Þ f f
1 2
3 + = N wrong
Option (c) f
1
0 = N,f
2
10 = +  N
Þ f f
1 2
10 + = N wrong
Option (d) f
1
1 = + N, f
2
10 = + N
Þ f f
1 2
11 + = N correct.
OR
As the likely movement would be towards
right f
2
 will be at its maximum.
\ f
2
10 = N
Þ f
1
1 = N
Option (d) is correct.
20. 2 2 mg T ma sin a - =
and T ma =
\ 2 3 mg ma sin a =
Þ a
g
=
° 2 30
3
sin
 =
g
3
\ T
mg
=
3
Force ( ) R applied by clamp on pulley
would be
| | | | T T
1 2
® ®
= = T
T
1
3
30
3
30
®
= ° + °
mg mg
(cos ) (sin )
^ ^
i j
= +
mg mg 3
6 6
i j
^ ^
| |
^
T
2
3
®
=
mg
j
Force by clamp on pulley P
  = +
® ®
T T
1 2
  = + +
mg mg mg 3
6 6 3
i j j
^ ^ ^
  = +
mg mg 3
6
3
6
$ $
i j  = +
mg
6
3 3 ( )
^ ^
i j
Option (b) is correct.
21. f
1
4 10 (max) = ´ ´ 0.3 = 12 N
f
1
 and f
2
 are frictional forces.
f
2
2 10 12 (max) = ´ ´ = 0.6 N
As, f f
1 2
16 (max) (max) + < N ( ) F
ext
Laws of Motion | 93
f
1 m = 0.6
1
m = 0.5
2
f
2
F = 15N F = 2t
1
T T
15N
4N T T
B A
a = 30°
m
T T
T
T
P
T
T
2m
60°
x
i
y
j
®
®
T
1
x
y
T
2
60°
f
1
m = 0.6
1
m = 0.3
2
f
2
16N 2 kg 4 kg
Page 4


10. Had q been 90°.
For every x displacement of wedge ( ) w the
vertical fall in mass would have been 2x as 
the string passes through pulley B.
i.e., v v
1 2
2 =
For the situation given the speed of mass
would thus be 
          v v
1 2
2 90 = ° - cos ( ) q
i.e., v v
1 2
2 = sin q
Option (c) is correct.
11. The cylinder will start rising up the
inclined plane if
ma mg cos sin q q >
i.e., a g > tan q
\ a g
min
tan = q [cylinder will be at the
point of rising up the inclined plane]
12. At the position of maximum deflection the
net acceleration of bob (towards its mean
position will be)
g a sin cos q q -
as explained in figure.
13. Mg T Ma - =
T mg ma - =
Þ   a
M m
M m
g =
-
+
\     T m g a = + ( )
       = +
-
+
é
ë
ê
ù
û
ú
m g
M m
M m
g
       = +
-
+
é
ë
ê
ù
û
ú
mg
M m
M m
1
       = + mg[ ] 1 1
{as m M << , M m M - » and 
M m M + » }
           = 2 mg
\ Tension in the string suspended from
ceiling = = 2 4 T mg
Option (a) is correct.
14. For 0° < £ f q  frictional force ( ) sin F mg = q
For q ³ f  frictional force ( ) cos F mg = m q
[f = angle of repose]
For example
Let m =
1
3
     \      tan f =
1
3
f = ° 30
q F (force of friction) Condition
10° mgsin10° = 0.174 mg
20° mgsin20° = 0.342 mg Increase but not
linearly only
30° mmgcos30° = 0.500 mg
40° mmgcos40° = 0.442 mg Decrease but not
linearly only
60° mmgcos60° = 0.287 mg
90° Zero
Option (b) is correct.
15.   a
mg
m
B
=
m
2
        =
mg
2
Laws of Motion 91
ma cos q
ma
(Pseudo
furce)
O
a cos q
a
g cos (90° – q)
= g sin q
90°– q
q
g
T
T T
T
mg
Mg
2T
2T
m
v
2
w
m
q
v
1
90° – q
v
1
B
O
90°
q
F
f
For no slipping
        a a
A B
=
   
F mg
m
g -
=
m m
2
i.e.,      F mg =
3
2
m
          m =
2
3
F
mg
Slipping will obviously be m
2
 there if m is
greater than above mentioned value
\ m
min
=
2
3
F
mg
For no slipping.
Option (c) is correct.
16. F
net
 (downward) = + mg ma sin cos q q
= + m g a ( sin cos ) q q
\      g g a
eff
= + sin cos q q
Time ( ) T required to cover 2L distance
along inclined would be
       T
L
g
=
2
eff
=
+
2L
g a ( sin cos ) q q
Option (c) is correct.
17. F
net
 on block along incline in the upward
direction
= - ma mg cos sin q q
= - m a g ( cos sin ) q q
\     g a g
eff
= - cos sin q q
\ Time ( ) t to move s distance would be
given by
s g t =
1
2
2
eff
i.e.,  t
s
g
s
a g
= =
-
2 2
eff
( cos sin ) q q
Substituting s = 1 m,
           q = ° 30 ,
a =10 3 m/s
2
 and g = 10 m/s
2
t =
´
´
æ
è
ç
ö
ø
÷
´
æ
è
ç
ö
ø
÷
2 1
10 3
3
2
10
1
2
_
       =
1
5
 s
Option (b) is correct.
18. N
w
w
w
+ ° = +
2
30
2
sin
               f = frictional force
       N
w
=
5
4
      f N
w
max
= = × m m
5
4
The block will remain stationary if
w w
2
30
5
4
cos ° £
m
or
w w
2
3
2
5
4
£ m
or 3 5 £ m
or
3
5
£ m
\    Block will move if m <
3
5
Option (d) is correct.
92 | Mechanics-1
2m
m
A
B
mmg
F
F
F
mmg
F F
q
a
ma
ma cos q
mg sin q
q
ma cos q
mg sin q
a
w
2
sin 30°
w/2
30°
w
2
cos 30°
w
f
w/2
N
19.
At t = 2 s
F
1
4 = N
f
1
 and f
2
 are the frictional forces
( )
max
f
1
1 10 = ´ ´ 0.6
= 6 N
( )
max
f
2
2 10 = ´ ´ 0.5
  = 10 N
At t = 2 s
Net external force ( ) F
net
 on system
          = 15 N - 4 N 
= 11 N
As F f f
net
> + ( ) ( )
max max 1 2
, the system will
remain at rest and the values frictional
forces on the blocks will be given
T f = + 4
1
 and T f = - 15
2
4 15
1 2
+ = - f f …(i)
f f
1 2
11 + = N
Let direction being + ive for Eq. (i)
Option (a) f
1
4 = - N, f
2
5 = = N
Þ f f
1 2
1 + = N wrong
Option (b) f
1
2 = - N, f
2
5 = + N
Þ f f
1 2
3 + = N wrong
Option (c) f
1
0 = N,f
2
10 = +  N
Þ f f
1 2
10 + = N wrong
Option (d) f
1
1 = + N, f
2
10 = + N
Þ f f
1 2
11 + = N correct.
OR
As the likely movement would be towards
right f
2
 will be at its maximum.
\ f
2
10 = N
Þ f
1
1 = N
Option (d) is correct.
20. 2 2 mg T ma sin a - =
and T ma =
\ 2 3 mg ma sin a =
Þ a
g
=
° 2 30
3
sin
 =
g
3
\ T
mg
=
3
Force ( ) R applied by clamp on pulley
would be
| | | | T T
1 2
® ®
= = T
T
1
3
30
3
30
®
= ° + °
mg mg
(cos ) (sin )
^ ^
i j
= +
mg mg 3
6 6
i j
^ ^
| |
^
T
2
3
®
=
mg
j
Force by clamp on pulley P
  = +
® ®
T T
1 2
  = + +
mg mg mg 3
6 6 3
i j j
^ ^ ^
  = +
mg mg 3
6
3
6
$ $
i j  = +
mg
6
3 3 ( )
^ ^
i j
Option (b) is correct.
21. f
1
4 10 (max) = ´ ´ 0.3 = 12 N
f
1
 and f
2
 are frictional forces.
f
2
2 10 12 (max) = ´ ´ = 0.6 N
As, f f
1 2
16 (max) (max) + < N ( ) F
ext
Laws of Motion | 93
f
1 m = 0.6
1
m = 0.5
2
f
2
F = 15N F = 2t
1
T T
15N
4N T T
B A
a = 30°
m
T T
T
T
P
T
T
2m
60°
x
i
y
j
®
®
T
1
x
y
T
2
60°
f
1
m = 0.6
1
m = 0.3
2
f
2
16N 2 kg 4 kg
The system will remain at rest.
For the equilibrium of 4 kg mass :
\ 16
1
= + T f …(i)
As f
1
 will be at its maximum value
f
1
12 = N
\    T = - 16 12
                       = 4 N [from Eq. (i)]
Further, for the equilibrium of 2 kg mass.
T f =
1
\ f
1
4 = N 
Option (c) is correct.
22. For the rotational equilibrium of rod
Taking moment about O.
R
l
s
l
´ =
2 2
cos sin q q
or mg ma cos sin q q =
Þ      a g = cotq
Option (d) is correct.
23. v t = 2
2
\ a
dv
dt
d
dt
t t = = = ( ) 2 4
2
At t = 1 s, a =
-
4
2
ms
As a g
s
= m
m
s
a
g
= = =
4
10
0.4
Option (c) is correct.
24. Just at the position of tipping off, R
1
 will
be zero.
\ Taking moment about point Q
( )( ) ( )( ) 10 4 80 g g x =
\   x =
1
2
 m
Option (a) is correct.
25. N ma mg = + sin cos q q       …(i) 
Now, as the block does not slide
ma mg cos sin q q =
i.e.,         a g = tan q
Substituting the found value of a in Eq. (i)
    N m g mg = + ( tan ) sin cos q q q
 = +
é
ë
ê
ù
û
ú
= mg mg
sin
cos
cos
2
q
q q q sec
When, the block stops a = 0, the value of
normal force will be 
N mg ¢ = cos q
\ 
N
N
mg
mg
¢
=
cos q
q sec
Option (c) is correct.
26. For the ro ta tional equi lib rium of the block
Taking moment about O.
Nx f
a
=
2
or ( cos ) ( sin ) mg x mg
a
q q =
2
or x
a
=
2
tan q
or
x
a /
tan
2
= q
or tan tan f = q
or f = q
94 | Mechanics-1
q
O
mg
ma
s (= ma)
R (= mg)
a
A
8 m
P
R
1
R
2
1 m
B A
1 m
x
10 g
mg cos q
a
ma
ma sin q
N
mg sin q
q
q = 45°
mg
N
mg sin q
B
O
f
mg cos q
x
A
f
f
1
f
2
16N
T T
4 kg 2 kg
Page 5


10. Had q been 90°.
For every x displacement of wedge ( ) w the
vertical fall in mass would have been 2x as 
the string passes through pulley B.
i.e., v v
1 2
2 =
For the situation given the speed of mass
would thus be 
          v v
1 2
2 90 = ° - cos ( ) q
i.e., v v
1 2
2 = sin q
Option (c) is correct.
11. The cylinder will start rising up the
inclined plane if
ma mg cos sin q q >
i.e., a g > tan q
\ a g
min
tan = q [cylinder will be at the
point of rising up the inclined plane]
12. At the position of maximum deflection the
net acceleration of bob (towards its mean
position will be)
g a sin cos q q -
as explained in figure.
13. Mg T Ma - =
T mg ma - =
Þ   a
M m
M m
g =
-
+
\     T m g a = + ( )
       = +
-
+
é
ë
ê
ù
û
ú
m g
M m
M m
g
       = +
-
+
é
ë
ê
ù
û
ú
mg
M m
M m
1
       = + mg[ ] 1 1
{as m M << , M m M - » and 
M m M + » }
           = 2 mg
\ Tension in the string suspended from
ceiling = = 2 4 T mg
Option (a) is correct.
14. For 0° < £ f q  frictional force ( ) sin F mg = q
For q ³ f  frictional force ( ) cos F mg = m q
[f = angle of repose]
For example
Let m =
1
3
     \      tan f =
1
3
f = ° 30
q F (force of friction) Condition
10° mgsin10° = 0.174 mg
20° mgsin20° = 0.342 mg Increase but not
linearly only
30° mmgcos30° = 0.500 mg
40° mmgcos40° = 0.442 mg Decrease but not
linearly only
60° mmgcos60° = 0.287 mg
90° Zero
Option (b) is correct.
15.   a
mg
m
B
=
m
2
        =
mg
2
Laws of Motion 91
ma cos q
ma
(Pseudo
furce)
O
a cos q
a
g cos (90° – q)
= g sin q
90°– q
q
g
T
T T
T
mg
Mg
2T
2T
m
v
2
w
m
q
v
1
90° – q
v
1
B
O
90°
q
F
f
For no slipping
        a a
A B
=
   
F mg
m
g -
=
m m
2
i.e.,      F mg =
3
2
m
          m =
2
3
F
mg
Slipping will obviously be m
2
 there if m is
greater than above mentioned value
\ m
min
=
2
3
F
mg
For no slipping.
Option (c) is correct.
16. F
net
 (downward) = + mg ma sin cos q q
= + m g a ( sin cos ) q q
\      g g a
eff
= + sin cos q q
Time ( ) T required to cover 2L distance
along inclined would be
       T
L
g
=
2
eff
=
+
2L
g a ( sin cos ) q q
Option (c) is correct.
17. F
net
 on block along incline in the upward
direction
= - ma mg cos sin q q
= - m a g ( cos sin ) q q
\     g a g
eff
= - cos sin q q
\ Time ( ) t to move s distance would be
given by
s g t =
1
2
2
eff
i.e.,  t
s
g
s
a g
= =
-
2 2
eff
( cos sin ) q q
Substituting s = 1 m,
           q = ° 30 ,
a =10 3 m/s
2
 and g = 10 m/s
2
t =
´
´
æ
è
ç
ö
ø
÷
´
æ
è
ç
ö
ø
÷
2 1
10 3
3
2
10
1
2
_
       =
1
5
 s
Option (b) is correct.
18. N
w
w
w
+ ° = +
2
30
2
sin
               f = frictional force
       N
w
=
5
4
      f N
w
max
= = × m m
5
4
The block will remain stationary if
w w
2
30
5
4
cos ° £
m
or
w w
2
3
2
5
4
£ m
or 3 5 £ m
or
3
5
£ m
\    Block will move if m <
3
5
Option (d) is correct.
92 | Mechanics-1
2m
m
A
B
mmg
F
F
F
mmg
F F
q
a
ma
ma cos q
mg sin q
q
ma cos q
mg sin q
a
w
2
sin 30°
w/2
30°
w
2
cos 30°
w
f
w/2
N
19.
At t = 2 s
F
1
4 = N
f
1
 and f
2
 are the frictional forces
( )
max
f
1
1 10 = ´ ´ 0.6
= 6 N
( )
max
f
2
2 10 = ´ ´ 0.5
  = 10 N
At t = 2 s
Net external force ( ) F
net
 on system
          = 15 N - 4 N 
= 11 N
As F f f
net
> + ( ) ( )
max max 1 2
, the system will
remain at rest and the values frictional
forces on the blocks will be given
T f = + 4
1
 and T f = - 15
2
4 15
1 2
+ = - f f …(i)
f f
1 2
11 + = N
Let direction being + ive for Eq. (i)
Option (a) f
1
4 = - N, f
2
5 = = N
Þ f f
1 2
1 + = N wrong
Option (b) f
1
2 = - N, f
2
5 = + N
Þ f f
1 2
3 + = N wrong
Option (c) f
1
0 = N,f
2
10 = +  N
Þ f f
1 2
10 + = N wrong
Option (d) f
1
1 = + N, f
2
10 = + N
Þ f f
1 2
11 + = N correct.
OR
As the likely movement would be towards
right f
2
 will be at its maximum.
\ f
2
10 = N
Þ f
1
1 = N
Option (d) is correct.
20. 2 2 mg T ma sin a - =
and T ma =
\ 2 3 mg ma sin a =
Þ a
g
=
° 2 30
3
sin
 =
g
3
\ T
mg
=
3
Force ( ) R applied by clamp on pulley
would be
| | | | T T
1 2
® ®
= = T
T
1
3
30
3
30
®
= ° + °
mg mg
(cos ) (sin )
^ ^
i j
= +
mg mg 3
6 6
i j
^ ^
| |
^
T
2
3
®
=
mg
j
Force by clamp on pulley P
  = +
® ®
T T
1 2
  = + +
mg mg mg 3
6 6 3
i j j
^ ^ ^
  = +
mg mg 3
6
3
6
$ $
i j  = +
mg
6
3 3 ( )
^ ^
i j
Option (b) is correct.
21. f
1
4 10 (max) = ´ ´ 0.3 = 12 N
f
1
 and f
2
 are frictional forces.
f
2
2 10 12 (max) = ´ ´ = 0.6 N
As, f f
1 2
16 (max) (max) + < N ( ) F
ext
Laws of Motion | 93
f
1 m = 0.6
1
m = 0.5
2
f
2
F = 15N F = 2t
1
T T
15N
4N T T
B A
a = 30°
m
T T
T
T
P
T
T
2m
60°
x
i
y
j
®
®
T
1
x
y
T
2
60°
f
1
m = 0.6
1
m = 0.3
2
f
2
16N 2 kg 4 kg
The system will remain at rest.
For the equilibrium of 4 kg mass :
\ 16
1
= + T f …(i)
As f
1
 will be at its maximum value
f
1
12 = N
\    T = - 16 12
                       = 4 N [from Eq. (i)]
Further, for the equilibrium of 2 kg mass.
T f =
1
\ f
1
4 = N 
Option (c) is correct.
22. For the rotational equilibrium of rod
Taking moment about O.
R
l
s
l
´ =
2 2
cos sin q q
or mg ma cos sin q q =
Þ      a g = cotq
Option (d) is correct.
23. v t = 2
2
\ a
dv
dt
d
dt
t t = = = ( ) 2 4
2
At t = 1 s, a =
-
4
2
ms
As a g
s
= m
m
s
a
g
= = =
4
10
0.4
Option (c) is correct.
24. Just at the position of tipping off, R
1
 will
be zero.
\ Taking moment about point Q
( )( ) ( )( ) 10 4 80 g g x =
\   x =
1
2
 m
Option (a) is correct.
25. N ma mg = + sin cos q q       …(i) 
Now, as the block does not slide
ma mg cos sin q q =
i.e.,         a g = tan q
Substituting the found value of a in Eq. (i)
    N m g mg = + ( tan ) sin cos q q q
 = +
é
ë
ê
ù
û
ú
= mg mg
sin
cos
cos
2
q
q q q sec
When, the block stops a = 0, the value of
normal force will be 
N mg ¢ = cos q
\ 
N
N
mg
mg
¢
=
cos q
q sec
Option (c) is correct.
26. For the ro ta tional equi lib rium of the block
Taking moment about O.
Nx f
a
=
2
or ( cos ) ( sin ) mg x mg
a
q q =
2
or x
a
=
2
tan q
or
x
a /
tan
2
= q
or tan tan f = q
or f = q
94 | Mechanics-1
q
O
mg
ma
s (= ma)
R (= mg)
a
A
8 m
P
R
1
R
2
1 m
B A
1 m
x
10 g
mg cos q
a
ma
ma sin q
N
mg sin q
q
q = 45°
mg
N
mg sin q
B
O
f
mg cos q
x
A
f
f
1
f
2
16N
T T
4 kg 2 kg
Thus, the normal force ( ) N will pass
through point A.
Option (a) is correct.
[Note : The cube will be just at the point of
tilting (about point A). The cube will tilt if q is
made greater than 45°].
27. For the rotational equilibrium of the cube
Moment of couple ( , ) N mg
= Moment of couple ( , ) F f
\   mgx Fa =
or mgx
mg
a = ×
3
i.e., x
a
=
3
Option (b) is correct.
28. Taking moment about point O.
N
l l
N
l l
1 2
2 4 2 6
-
æ
è
ç
ö
ø
÷
= -
æ
è
ç
ö
ø
÷
N
l
N
l
1 2
4 3
=
\ N N
1 2
4 3 : : =
Option (c) is correct.
29.
( / ) T ma 2 = ® Box
T
ma mg
+ = ®
2 2
 Pendulum with
respect to
box
\ 2
2 2
ma
ma mg
+ =
\                  a g = / 3
30.        N m a
B B
sinq=           …(i)
            a a
A B
= tanq   …(ii)
Þ      N m
a
B
A
sin
tan
q
q
=
N
m a
B A
=
sin tan q q
\   Force on rod by wedge
 N
m a
B A
cos
cos
sin tan
q
q
q q
=
       =
m a
B A
tan
2
q
       =
´
æ
è
ç
ö
ø
÷
=
10 9
3
4
160
2
 N
Option (c) is correct.
31. Net downward force on ring = - mg ma m
                      = - m g a ( ) m
\       g g a
eff
= -m
          t
L
g
=
2
eff
           =
-
2L
g a m
           =
´
- ´
2 1
10 4 ( ) 0.5
           = =
1
2
0.5 s
Laws of Motion | 95
l/6 l/4
N
1
N
2
mg
a
T
T
ma
mg mg
q = 37°
A
N sin q
N
N
B
–2
9 ms
q
a
B
a
A
®
®
a
a
A
A
B
B
=
=
®
®
| |
| |
a
a
L = 1m
–2
a = 4 ms
f = mN
  = mma
mg
T
N = ma
a
F (        )
=
mg
3
N (= mg)
mg
x
f = F
3
4
37°
5