1. DPP NO. - 65
1.
Mx +
3
M
(x + L) = 0
3
M
4
x = –
3
ML
x = –
4
L
2. B

= 3
0
r
r
v
q
4





and E

= 3
0 r
r
q
4
1



 B

= 0
0 )
E
v
(


 = 2
c
E
v



3. v =
100
50
Ve =
R
GM
2
2
1
Applying energy conservation

2
v
m
2
1
R
GMm

 =
)
h
R
(
GMm


v2 =
h
R
GM
2
R
GM
2



R
GM
2
.
4
1
= 







h
R
1
R
1
GM
2

)
h
R
(
R
h
R
4
1


 R + h = 4h
 h = R/3
4. Keq.
is same in all three cases. All other parameter
being same, rate of energy conduction is same in all
three cases.
Simlarly temperature difference across any material
in any wall is also same.
5.  =
12
ma2
+
2
2
a
m 







=
12
ma
7 2
= 7.
6. mgh =
2
1
mv2
+
2
1
mv2
+
2
1
.
5
2
mv2
2
r
v
2






=
2
1
mv2
[1 + 1 +
5
8
] =
2
1
mv2
5
18
=
5
mv
9 2
 v = gh
9
5
7. KE of the ball =
2
1
mv2
+
2
1
5
2
mv2
2
r
v
2






=
18
13
mgh
8. X = 2vt = 2v g
h
2
= 2. g
h
2
gh
9
5
= h
3
10
2
DPP NO. - 66
1. Initially the centre of mass is at
4
L
distance from the vertical rod.












4
L
m
m
)
0
(
m
)
(
m
x
,
As 2
1
cm
centre of mass does not move in x-direction as
Fx = 0.
After they lie on the floor, the pin joint should be at
L/4 distance from the origin shown inorder to keep
the centre of mass at rest.
 Finally x-displacement of the pin is
4
L
and
y-displacement of the pin is obviously L.
Hence net displacement =
4
L
17
16
L
L
2
2



2. 2. H = – kA
dx
dT

dx
dT
=
kA
H

Now as k increases,
dx
dT
becomes less (–)ve
So slope becomes less (–)ve
So curve will be
3. Since
B

= 3
0
r
r
v
q
4





, r
v


 must be same
where v

= velocity of charge with respect to
observer
Let A and B are the observers
then r
)
v
v
( A
C




 = r
)
v
v
( B
C





or 0
r
)
v
v
( B
A 





or r
|
|
)
v
v
( B
A




4. 11 = 11 (Angular momentum is conserved)
As 2 decreases. 2 increases.
Thus T =


2
i.e. T decreases.
Therefore the earth is completing each circle around
its own axis in lesser time.
K.E. =
2
1
2
Therefore K.E. of rotation increases.
Duration of the year is dependent upon time taken
to complete one revolution around the sun.
5. Using  axis theorem
lx
= ly
2x
= 1.6
x
= .8 Ma2
AB
= x
+ M(2a)2
= 4.8 Ma2
Ans.: 4.8 Ma2
6, 7 & 8.
Let  be the angular acceleration of rod and a be
acceleration of block just after its release.
 mg – T = ma ..... (1)
T – mg
2

= 
3
m 2

.... (2)
and a =  .... (3)
Solving we get
T =
8
mg
5
and  =

8
g
3
Now from free body diagram of rod, let R be the
reaction by hinge on rod
R + T – mg = m acm = m
2
1

Solving we get
R =
16
mg
9
DPP NO. - 67
1. Moment of inertia is more when mass is farther from
the axis. In case of axis BC, mass distribution is
closest to it and in case of axisAB mass distribution
is farthest .Hence
cm
x y
C
5
A
B
3
IBC
< IAC
< IAB
 I P
> IB
> IH
IC
= ICM
+ my2
= IB
1
– mx2
+ my2
= IB
1
+ m (y2
–x2
) = IP
+ IB
+ m (y2
– x2
)
> IP
+ IB
> IP
Here IB
1
is moment of inertia of the plate about an
axis perpendicular to it and passing through B.
 IC
> IP
> IB
> IH

3. 2. (C) Taking the origin at the centre of the plank.
A B
40 kg 60 kg
40 kg
60 cm
smooth
x
m1x1 + m2 x2 + m3x3 = 0
( xCM = 0)
(Assuming the centres of the two men are exactly
at the axis shown.)
60(0) + 40(60) + 40 (–x) = 0 , x is the displacement
of the block.
 x = 60 cm
i.e. A & B meet at the right end of the plank.
3. The slope of temperature variation is more in inner
dt
dQ
= T
.
KA


T =
dt
dQ
.
KA

Slope 
K
1
Larger the conductivity, smaller is the slope.
4. Let Q be the charge on the ring. The electric field at
point P is
 E =
0
4
1

 2
/
3
2
2
)
R
x
(
Qx

=
0
4
1

 2
/
3
2
)
R
2
(
QR
The rotating charged (Q) ring is equivalent to
a ring in which current  flows, such that
 =
R
2
Qv

The magnetic field at point P is
B =

4
µ0
2
/
3
2
2
2
)
R
x
(
R
2



=

4
µ0
2
/
3
2
)
R
2
(
QvR

B
E
=
v
µ
1
0
0 
=
v
c2
5. The orbital velocity,
v0 =
r
GM
Its velocity is increased by 2 times, new velocity
v =
r
GM
2
r
GM
2  = escape velocity
The path is parabolic in case of escape velocity.
8. L = dt =  Fdt
x = xP (because x is
essentially constant during the quick blow)
since, the rod starts at rest, the final values
therefore satisfy L = xP.

2
1
m2
 = xmv  2
x
12
v 


.... (1)
Another expression for
v

is obtained from the given
information that rod makes one revolution by the
time centre reaches the dot.
t = 2 and vt = d

d
2
v



.... (2)
from equation 1 and 2 :
d
2
x
12
2



 x =
d
6
2


x cannot be larger than
2

2
d
6
2




 d
3


DPP NO. - 68
1. MP = 






f
D
1 = 






5
25
1 = 6
2. Heat radiated (at temp same temp)  A
 Q  4R2
and Q'  (4R2
+ 2 × R2
)
 2
2
R
4
R
6
Q
'
Q


 = 1.5
Here R2
is extra surface area of plane surface of one
of the hemisphere.

4. 3.
Megnetic moment M

= r2 i î &  B

= ĵ
4
î
3 
 

= M

× B

= r2 (3 î – 4 î )


will be along the direction shown .
Hence , the point about which the loop will be lift
up will be : (3, 4)
4.
Radius of Curvature =
on
Accelerati
Normal
)
velocity
( 2
=
R
/
v
)
v
2
(
2
2
= 4R
5. By linear momentum conservation in horizontal di-
rection = for (bob + string + cart)
mV0 = (m + m)v
v =
2
V0
By mechanical energy conservation for
(bob + string + cart + earth)
2
1
mV0
2
+ 0 + 0 =
2
1
(2m)v2
+ mgh + 0
2
1
mV0
2
–
2
1
(2m)
4
V
2
0
= mgh
Solving it,
h =
g
4
V2
0
.
6. When two drops of radius r each combine to form a
big drop, the radius of big drop will be given by
3
R
3
4
 =
3
2
r
3
4
r
3
4 


or R3 = 2r3 or R = r
2 3
/
1 Now
2
r
R
r
R
V
V






 = 3
1
3
2
4
2 
 VR
= 5 × 41/3
cm/s
7. If m is pole strength , then
m =
l
M
m 
When the wire is bent into a semicircular arc, the
separation between the two poles changes from l to
2l, where new magnetic moment of the steel wire,







M
2
2
M
r
2
m
'
M
l
l
8. (A) Real image of a real object is formed by concave
mirror and convex lens.
(B) Virtual image of a real object is formed by all four.
(C) Real image of a virtual object may be formed by
all four.
(D) Virtual image of a virtual object may be formed by
convex mirror and concave lens.
(A) p,r (B) p,q,r,s (C) p,q,r,s (D) q,s
9. Let E1
< E2
and a current i flows through the circuit.
Then the potential difference across cell of emf E1
is
E1
+ ir1
which is positive, hence potential difference
across this cell cannot be zero. Hence statement 1
is correct.
For current in the circuit to be zero, emf of both the
cells should be equal. But E1
 E2
. Hence statement
2 is correct but it is not a correct explanation of state-
ment 1.
E , r
1 1 E , r
2 2
DPP NO. - 69
3. Equal area means equal power output. A3
area
pertains to highest wavelength range, thus photons
with minimum range of frequency. Thus maximum
number of photons are required from this segment
to keep the power same.
4. Work done by kinetic friction may be positive when it
acts along motion of the body.
Friction on rigid body rolling on inclined plane is along
upward because tendency of slipping is downwards.
5. The torque of system = Torque on loop
[AFGH + BCPE + ABEF]
= SB )
î
( + SB( î )+ SB k̂ (I = current,
S = area of loop, B = magnetic field.
=  S B k̂
= 1 × 1 × 2 k̂ = 2 k̂ units
[Ans: K̂
2 ]

5. 6. The work done to rotate a bar magnet from its initial
position  = 1
to the final position  = 2
is given by
W = M B (cos 1
– cos2
),
(i) Here 1
= 0° and 2
= 180°
 W = M B (cos 1
– cos180°) = M B = [1–(–1)] = 2
M B
(ii) Here 1
= 0° and 2
= 90°
 W = M B (cos0° – cos 90°) = M B = [1– 0)]
= M B
7. If velocity of m2
is zero then
by momentum conservation
m1
v = m2
v
v =
1
2
m
v
m
Now kinetic energy of m1
=
2
1
m1
v2
=
2
1
m1
2
2
1
2
v
m
m








=
2
1








1
2
m
m
m2
v2
= 







1
2
m
m
2
1
m2
v2
=
1
2
m
m
× initial
Kinetic energy
Kinetic energy of m1
> initial mechanical energy of
system m1
Hence proved
8. C = sin–1 





1
/
2
1
= 30°
for i = 37 TR
so,  =  – 2 (37°) = 104°
i = 25, Refraction  <
2

– C
i = 45°, TR
so,  =  – 2 




 
4
= 90°
By applying snells law for prism :
i = 90
r1
= 30
r2
= 30
e = 45
 = 90 + 45 – 60 = 75°
9. The pointsAand B are at same potential, then under
given conditions pointsAand B on the circuit can be
connected by a conducting wire. Hence the circuit
can be redrawn as shown in figure 2.
E
B
A
E
R1 R2
R3
Fig-1
Therefore statement 1 is true. Statement 2 is obvi-
ously false.
DPP NO. - 70
1. In normal adjustment
m =
e
0
f
f
so 50 =
e
f
100
 fe
= 2 cm
( eyepiece is concave lens)
and L = f0
– fe
= 100 – 2 = 98 cm
2. m
T = const.
ln m
+ lnT = C
0
T
dT
d
m
m





m
m
d


=
T
dT

Now
m
m
d


= – 1 % =
100
1
 (–ve sign indicates
decrease)
dT = 1 (given)
 T = 100 K.
3. Emissive power =  T4
= 6 × 10–8
× 1004
W/m2

6. 4.
to reach  = 270º, it has to cross the potential
energy barrier at  = 180º and to cross  = 180º
angular velocity at  = 180º should be 0+
ki
+ Ui
= kf
+ Uy
2
2
MR
2
3
2
1







+ (–Mi AB cos 0º) = 0 + (–NiAB
cos180º)
 = 9
80  rad/sec.
5. Equation for linear motion
mgsin – f = ma
for rotary motion
f. R =  .
R
a
 f = 2
R

.
a mgsin = ma + 2
2
R
.
2
MR
.a =
2
3
ma
a =
3
sin
g
2 
=
3
g
using S = ut +
2
1
at2
for linear motion.
1 = 0 +
2
1
.
3
g
. t2
t = g
6
sec. Ans.
6. Here BH
= 0.22 T ; BV
= 0.38 T
Now 2
V
2
H B
B
B 

= 1928
.
0
)
38
.
0
(
)
22
.
0
( 2
2

 = 0.44 T
7. w.r.t. the wedge
As maximum height = 125 m
 block want by a height 20m over the wedge
 (v sin53º)2
= 2.g.20
v2
25
16
= 400
v2
= 25 × 25
v = 25 m/sec.
 block left the wedge with a relative velocity
25 m/sec.
Now, time of flight = 2 + 5 = 7 sec.
horizontal range w.r.t. wedge
= vx
× T
= 25 cos 53 × 7
= 105 m.
8. For slab no deviation so  = 0 for any i
for slab for light from D to R
 = r – i ....(i)
nd sin i = nr sin r
 r = sin–1 





i
sin
n
n
r
d
 = sin–1 





i
sin
n
n
r
d
– i
it is non-linear function and graph is
After i > C T.R. will occur and graph is straight line
for D to R