& 3Q $ ! %R
& 2Q $ ! %R
3. Let the body is acted upon by a force at an
angle
9 with horizontal. FBD :
Charge b is negative and charge a is positive 10. (A) p (B) q,s (C) p (D) q,s (A) Speed of point P changes with time (B) Accele ration of poin t P is equal to ;2x (; = angular speed of disc and x = OP). The acceleration is directed from P towards O. (C) The angle betwe en accelera tion of P (constant in magnitude) and velocity of P changes with time. Therefore, tangential acceleration of P changes with time. (D) The acceleration of lowest point is directed towards centre of disc and remains constant with time
F cos
9 = > (mg – F sin 9)
3F=
> mg cos 9 + > sin 9 . For min. force
(cos 9 + > sin 9) should be max.
3 – sin 9 + > cos 9 = 0 3 tan 9 = >. or 9 = tan 1 (1/ –
DPP NO. - 22 1. Q AB =
H U AB + W AB
3
` 2
` 2
nR H T
(H PV)
H UAB =
;
3 ) = 30 ° Substituting
Fmin = 12.5 kg f
4. Change in momentum = Impulse "
HP ( Jx ˆi + Jy ˆj + Jzkˆ
W AB = 0
H UAB =
b
5 (H PV) 2
= 30(0.1) ˆi +
1 (80) (0.1) ˆj + (–50) × (0.1) kˆ 2
= 3ˆi + 4ˆj ' 5kˆ " m | HP | = 5 2 kg sec .
6. Plotting velocity v against time t, we get
Q AB = 2.5 P 0 V0 Process BC QBC = H U BC + W BC QBC = 0 + 2P 0 V0 !n 2 = 1.4 P 0 V0 Q net = Q AB + Q BC = 3.9 P 0 V0
Area under the v Distance =
–t
curve gives distance.
1 1 ×2×2+ × 2 × 2 = 4m 2 2
DPPS FILE # 186
c=
7. (C)
T ; and as the re is no change in length
10. By angular momentum conservation ;
1
3Q=
L=
T
Q' ( Q
T'
Q Q'
T' =
3
R + mvR = 2mR 2; 2
3 mvR = 2mR 2; 2
T
3
N ; 3 mv
;= T
3v 4R
2
T ’ = (2) T = 4T. Hence (C).
9
Position 2
8.
9
u
v
O
mg cos9
Principle axis
mg
N2
u
v
N1
Also at the time of contact ;
Position 1
mgcos 9
N1 u N 2 v For first & second position = , = u O v O v2
3
u2 v u
3 O
N2
=
=
N
N1 N2
( 2.2
–
8 N = mg cos 9 – = 4.84
and v + u = 96
3
v = 66 , u = 30
8 mgcos 9 is increasing and v u
( 2.2 ( 11 3 5
Focal length of lens f =
A is True
uv u+v
(
66 7 30 = 20.63 66 + 30
3
C is False Distance of lens from shorter image = u = 30 cm
9.
D is True
Q' =
332 ' 32 V ' Vs = = 0.3 m 1000 f
( V + V0 ) 332 + 64 f' = f V ' V = 1000 × = 1320 Hz 332 ' 32 s
Q'' =
mv 2 R
9 decreases so cos 9 increases
when it ascends and v decreases.
8 we can say N
Distance between two position of lens = v - u = 36 cm 3 B is True
3
mv 2 R
N=
V ' V0 = 0.2 m. f'
mv 2 is decreasing R
increases as wheel ascends.
DPP NO. - 23 1.
Q = 2 ! = 3m Equation of standing wave y = 2A sin kx cos ;t y = A as amplitude is 2A. A = 2A sin kx 2<
x =
Q
3 and
3
< 6
1 x1 = m 4 2<
Q
.x=
5< 6
x 2 = 1.25 m
3 x 2 – x 1 = 1m
DPPS FILE # 187
A
1 , gcos9 = a t 2
2. tan9 =
20
10 ×
10 2 + 20 2
=
200 100 + 400
=
20 5
normal i 45°
T=
2uy g
where u x = cos 9 and u y = u sin 9
normal
incident ray
P
°
45 4 5
30 °
m/s2.
3. Since time of flight depends only on vertical component of velocity and acceleration . Hence time of flight is
60 °
B
°
C Q R
5. One can create a mental view of distribution of charge i.e. how much charge is nearer and how much is comparatively farther away.
8 In horizontal2(x) direction d = uxt + ½ gt = u cos
=
& 2u sin 9 # 9 $$ g !! % "
+
1 & 2u sin 9 # ! g$ 2 $% g !"
2
2u2 (sin 9 cos 9 + sin 29) g
We want to maximise f( 9) = cos 9 sin 9 + sin 29 3 f ’(9) = – sin 29 + cos 29 + 2 sin 9 cos 9 = 0 3 cos2 9 + sin2 9 = 0 3 tan2 9 = –1 or 2 9 =
3< or 4
9=
3< = 67.5 ° 8
Alternate : As shown in figure, the net acceleration of projectile makes on angle 45 ° with horizontal. For maximum range on horizontal plane, the angle of projection should be along angle bisector of horizontal and opposite direction of net acceleration of projectile.
6.
As a rod AB moves, the point ‘P’ will always lie on the circle. 8 its velocity will be along the circle as shown by ‘VP’ in the figure. If the point P has to lie on the rod ‘AB’ also then it should have component in‘x’ direction as ‘V’. 8 VP sin 9 = V 3 VP = V cosec 9 3R 3 x 1 = . = 5 5 R R
here cos9 =
8
sin9 =
8
VP =
4 5
5 V 4
5 4
...Ans.
VP 5V = R 4R
;=
Sol. (b)
8 cosec 9 =
ALTERNATIVE SOLUTION : Sol. (a) Let ‘P’ have coordinate (x, y) x = R cos 9, y = R sin 9.
8 9=
135 5 = 67.5 ° 2
VX =
–
d9
3 4. Let angle of incidence i for which deviation due to first prism is minimum, then sin i = n sin30 ° or i = 45°. The net deviation shall be minimum if deviation due to each prism is minimum. From the ray diagram in figure, it is clear that angle between AC and PQ for net deviation to be minimum is 90 °.
dx = dt
R sin
9
d9 =V dt
'V
dt = R sin 9
VY = R cos
9
8
Vx2
VP =
and
d9 = R cos dt
+ Vy2
=
& V # 9 $$ ' R sin 9 !! % " V2
=
–
V cot
9
+ V 2 cot 2 9
= V cosec 9 ...Ans. Sol. (b) ; =
VP 5V = R 4R DPPS FILE # 188
7. (A) Electric field at r = R
10. (A) p (B) q (C) p,r (D) q,s (A) If velocity of block A is zero, from conservat ion of momentum, speed of b lock B is 2u. Then K.E. 1 m(2u) 2 2
of block B =
E=
= 2mu 2 is greater than net
mechanical energy of system. Since this is not possible , velocity of A can never be zero. (B) Since initial velocity of B is zero, it shall be zero for many other instants of time. (C) Since momentum of system is non-zero, K.E. of system cannot be zero. Also KE of system is minimum at maximum extension of spring. (D) The potential energy of spring shall be zero whenever it comes to natural length. Also P.E. of
KQ
R2 where Q = Total charge within the nucleus = Ze
So E = KZe R2 So electric field is independent of a
spring is maximum at maximum extension of spring.
DPP NO. - 24 8. Q =
I @r 4< r 2dr d R
for a = 0,
@r (
8
d (R ' r ) R
R
or,
Q=
1. (B) To an observer who starts falling freely under gravity from rest at the instant stones are projected, the motion of stone A and B is seen as
@ ( r R 'r
d
IR
(R
–
r) 4 < r 2 dr
=
=
I
I
=
4
< dR 3
8
3 Q = Ze =
< dR 3 3
or
d=
3Ze
9. From the formula of uniformly (volume) charged solid sphere E=
.......(1)
d! = u dt
.......(2)
8
0
R R / 4
dx = u dt
@r 3 \0
2.
x = ! and d BOA = 60°
& Vob # !! ` = `0 $$1 + V sound " % Vob ` (straight line) 3 ` ( 1+ V 0 sound Vob when V =0 ; sound and as
Vob Vsound
U1
= 1.
` 3 ` U 0
2
3. Distance travelled by A = s = 10 × 2 + 1 × 10 × 8 = 60 2 Distance travelled by B = s = 10 × 3 +
For E A r, @ should be constant throughout the volume of nucleus This will be possible only when a = R.
` `0
;
1 × 10 × 4 = 50 2
Average sp eed of A =
60 = 7.5 8
Average speed of B =
50 = 7.1 7 DPPS FILE # 189
4. Taking C as srcin and x & y – axes as shown in figure. Due to symmetry about y – axis x cm = 0
7. T cos30º + N sin30º = mg
3
3 T + N = 2 mg
..............(i) mv 2
T sin30º – N cos30º =
3
T sin30º – 3N = 4mv 2
= 4mv2
ycm =
=
=
3T
–
3N
..............(ii)
2y2 $%& m1my11 '' m m2 !#"
&$ <(6<)2 #! & 4(6<) # 2 $% 2 !" $% 3< !" – [ <(2) (8)]
by (i),)(ii)
<(6<)2 ' <(2)2 2
(m
( 3 / 2)
3
A Area) 8(18 <2
' 4)
(18<2 ' 4)
= 8 cm.
T=
N=
2mg ' 4mv 2 4
;
6mg ' 4mv 2 4 3
for N > 0
3
v < 5 m/s
at v = 2 ij
T=
38
N
;
N=2N.
3
5. The slab does not contribute to deviation. For minimum deviation by prism, r1 = r 2 = 30 ° as shown in figure.
3 sini = 2 sin 30 ° or i = 45 ° 8 Minimum deviation = 2i – A = 90°
–
8. 8 For the given case Ray diagram will be as given here
60° = 30°
6. Amplitude phasor diagram : Here say ER will be seen by observer which appear to becoming fr om point A. To find x, the distance of object from A we reverse the light rays by considering ER as incident ray & find the position of image after two refractions. For I refraction we use 4 1' 43 1 ' 3 = v
e
v = – 3R For II refraction we use
e
4
3
x
8
'R
2R
3
'
1
'R
4
=
3
'1
R
3 resultant amplitude =
6 2 .
x = – 2R Hence OA = 2 × 4 = 8 cm. DPPS FILE # 190
9. Let the cube dips further by y cm and water level rises by 2 mm.
2. The magnitude of the electric field is maximum where the equipotentials are close together. The direction of the field is from high potential to low potential. 4. From given graphs : 3 t and a y = 4
ax =
3 vx =
3 2 t +C 8
At t = 0 : v x = 3C=–3
8 vx =
Then equating the volumes (/// volume = \\\ volume in figure) 3 volume of water raised = volume of extra depth of wood
3
100 y = (1500 – 100)
3 ' &$ t + 1#! %4 "
–
3
3 t2 – 3 8
&3 2 # 3 dx = $ 8 t ' 3 ! dt % "
.... (1)
Similarly
2 2 = 1400 × 10 10
& 3 # $ ' t 2 ' t + 4 ! dt % 8 "
dy =
= 280 8 y = 2.8 cm 8 Extra upthrust @water × (2.8 + 0.2) × 100 g = mg 3 m = 300 gm. m = 300 gm. ....Ans.
.... (2)
As dw = F. ds = F.( dx ˆi + dy ˆj )
8
W
4
0
0
2
/ 2
I dw ( I 01 34 t i ' &$% 34 t + 1#!" j-. . 01&$% 38 t ˆ
ˆ
2
/ 3 ' 3 #! ˆi + &$ ' t 2 ' t + 4 #! ˆj - dt " % 8 ".
8
W = 10 J Alternate Solution : Area of the graph ;
10. A – p,r,t RF = 0 So, linear momentum conservation and centre of mass will not move. B – q,s So, linear momentum will not be conserved and centre of mass will accelerate W ext = HE. C – p,s,t D – p,s,t
Ia
x
dt = 6 = V( x ) f
I a dt =
and
y
– 10
= – 6. Now work done =
' ('3) 3 V(x)f = 3. = V( y ) f ' ( 4) 3 V(y)f H KE = 10 J
DPP NO. - 25 5. 1. U =
1 2
a0
E2 =
1 2
a 0K 2Q 2 r4
In diagram angle of emergence = 60° 8 d BOC = 60° 8 r + r = 60 ° 3 r = 30°
KQ V= r
U V2
2 1 2 Q 2 a 0K r 4 = K 2Q2
=
1 a0 2 r2
8 >=
sin 60 5 sin 305
3 >=
3
r2 because
U V2
A
1 r2
6.
V0 V0 nV0
b (a)
so the correct option is B.
DPPS FILE # 191
8. Using equation of continuity A1v1 = A2v2 (12 cm2)vA = (6 cm 2) (8.0 m/s) vA = 4.0 m/s
V0 V0
nV0
b (b)
If V0 be the f low velocity of the river, then velocity of boat relative to water = nV 0. If the boat has to adopt the shortest path, then direction of velocity of boat relative to water should make an angle greater than 90° with the flow direction of river Resultant velocity of boat =
(nV0 ) 2
–
9. Applying Bernoulli's principle between point A and C that are at same horizontal level 1 @.VA 2 2
+ pA
=
1 @.VC 2 2
+ p atm
3
pA = (1.01 × 105 N/m2) +
=
4.27 × 105 N/m2
1 × 13,600 (82 – 42) 2
V02 . This
velocity can have a real value only when n > 1. If n = 10. By applying Bernoulli's equation between point B and C and using equation of continuity 1, then resultant velocity = 0. So the boat will follow vB = 8.57 m/s the shortest path only when n > 1. So options (b) and and pB = 3.70 × 104 Pascal (c) are correct. @ gh = 3.70 × 104 If boat is moved normal to flow direction, then it will b cross the river in a time nV , where b is the width of 0
h=
3.70 7 10 4 = 10 7 13,600
the river. If n f 0, the boat will cross the river. So option (d) is also correct.
7.
>N = 2
, N=8,a =
P=
nRT V
the
force F is reversed a1 = 9 m/s 2 (g) and distance covered by the block before it stops 16 7 7 = s1 = 27 9
DPP NO. - 26 1. V = KT + C
16 – 2 = 7 ( U) 2
v2 = 2 (7) 8 U ...........(i) when the direction of horizontal component of
272mm
3
nRT KT + C
P=
3
dP dT
nRC
( (KT + C)2
As C < 0 by diagram
3
dP < 0 for all T dT
3
P continuously decreases.
2. The free body diagram of the capillary tube is as shown in the figure. Net force F required to hold tube is F = force due to surface tension at cross-section (S 1 + S 2) + weight of tube. F
Again
s =8+s =8+ 2
1
anda 2 = 7 (g) v2 = 0 +2 (a 2) (s 2) = 2 (7)
3
v=
16 7 m/s 3
16 7 7
S 1 T ×2< R
279
S 2 T×2< R
& 8 + 16 7 7 # $ ! 279 " %
mg Free body daigram of capillary tube
= (2
DPPS FILE # 192
H h5 = h4 – h2 = 7, H h 6 = h4 – h3 = 5
3. From conservation of energy mgh =
1 mv 2 2
mgh =
1 mv 2 2
+
1 2
+
1& 2 # $ mr 2 ! ;2 2%5 "
gh =
7 10
(;2 r2)
10gh 7
=
;2 r2
KE =
Iw
2
which matches with the given set of beat frequencies. Hence (D).
# = wr
1&2 # $ mr 2 ! ;2 2%5 "
=
7. 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 a long upward because tendency of slipping is downwards.
2 mgh 7
Sol. 8 to 10 The time taken to reach maximum height and
4. Let ‘ v’ be the initial velocity. Tangential velocity remains same during collision and equal to v cos60° = v/2 Let vJ be the normal component of velocity after impact. v/2 In H OAB : tan 60° = vJ
3 vJ =
maximum height are t=
u sin 9 g
For remaining half, the time of flight is
v 2 3
u 2 sin2 9 2g
and H =
u2 sin 2 9
2H = (2g)
t' =
=
2g 2
t 2
8 Total time of flight is t + t' =
T=
& %
t $$1 +
1
# !!
2"
# u sin 9 & $1 + 1 !! g $% 2"
Also horizontal range is = u cos
9×T
# u sin 29 & $1 + 1 !! 2g %$ 2" 2
Then : e =
vJ v cos 305
=
(v / 2 3 ) ( 3 v / 2)
=
1 3
6. As no. of beats = Hh For option (A) : The frequencies are : h 1 = 550 Hz, h 2 = 552 Hz, h 3 = 553 Hz, h4 = 560 Hz. The beats produced will be : H h1 = h2 – h1 = 2, H h2 = h3 – h1 = 3, H h3 = h4 – h1 = 10, H h4 = h3 – h2 = 1 H h5 = h4 – h2 = 8, H h6 = h4 – h3 = 7 Which doesnot ma tches with the given set of beat frequencies. Hence (A) is not possible. Similarly (B) and (C) are also not p ossible. For option (D); frequencies w ere; h1 = 550, h2 = 551, h3 = 553, h4 = 558 H h1 = h2 – h1 = 1, H h2 = h3 – h1 = 3, H h3 = h4 – h1 = 8, H h4 = h3 – h2 = 2
=
Let u y and vy be initial and final vertical components of veloc ity. 8 uy2 = 2gH and v y2 = 4gH
8
vy =
2 uy
Angle ( ?) final velocity makes with horizontal is tan ? =
vy ux
(
2
uy ux
=
2 tan 9
DPP NO. - 27 1. If M 0 is molecular mass of the gas then for initial condition PV =
M . RT M0
...(1)
After 2M mass has been added PJ .
3M V T = M .R. 3 3 0
...(2)
By dividing (2) by (1) PJ = 3P DPPS FILE # 193
2. Snell’s law = >0 sin (90°
–
& x# 9) = >0 $1 – d ! sin90° % "
;
& x# 3 $1 – d ! = cos 9 % " 3 x = d(1 – cos 9)
& $ %
;= 4. All energy is transfered to other particles. 5. (A) Absolute velocity of ball = 40 m/s (upwards) hmax = hi = f f ( 40) 2 2 7 10
(30) 2 (B) Maximum height from left = = 45 m 2 7 10 (C) The ball unless meet the elevator again when displacement of ball = displacement of lift –
2
2# + MR !! ; 4 "
= mv.
R 3 = MR 2 ; 2 2
v 3R
7. As speed of ball is variable, so motion is non uniform circular motion. 8. At the highest position of ball, net tangential force is zero, hence tangential acceleration of ball is zero, 9. Tension in the string is minimum when ball is at the highest position. By conservation of energy
h = 90 m
40 t
2& 2 MR 2 # MR 2 / !+ 0$$ MR + - ; 4 !" 2 10% .
= $ MR
3. The electric field at P shall be zero if q = Q.
= 10 +
=
1 × 10 × t2 = 10 × t 2
3
–
10 = 40 × to –
v2 = 16 g! where v is the velocity of ball at the highest point. mv 2
So T + mg =
!
t = 6s.
(D) Let t0 be the total time taken by the ball to reach the ground then
1 1 mv2 + mg (2!) = m(20 g !) 2 2
1 × 10 × to2 2
3 8
t0 = 8.24 s. time taken by the ball for each the ground after crossing the elevator = t0 – t = 2.24 s.
T=
m 16 g! !
–
mg = 15 mg
10. (A) p,r (B) q,s (C) p,r (D) q,s (A) The fundamen tal frequ ency in the string, f0 =
T/> 2!
(
102 .4 17 10 '3
7
1 Hz = 320 Hz. 2 7 0 .5
Other possible resonance frequencies are f A and f 0 = 320 Hz, 640 Hz, 960 Hz. (B) The fundamen tal frequ ency in the string.
6.
f0 =
Let velocity of COM after collision is v velocity is ;. conserving linear momentum mv = 2mv J
3
vJ =
v 2
J & angular
............(1)
conserving angular momentum about COM mv.
R = 2
N:;
= (NRing COM +
T/> 4!
(
320 = 160 Hz. 4 7 0 .5
Other possible resonance frequencies are f B = 160 Hz, 480 Hz, 800 Hz. (C) The fundamen tal freque ncy in both ends ope n organ pipe is f0 =
v 2!
(
320 = 320 Hz. 2 7 0. 5
Other possible resonance frequencies are f c = 320 Hz, 640 Hz, 960 Hz (D) The fundamental frequency in one end open organ pipe is v ( 320 = 160 Hz. 4! 4 7 0. 5 Other possible resonance frequencies are f D = 160 Hz, 480 Hz, 800 Hz.
f0 =
Nmass )
DPPS FILE # 194
DPP NO. - 28 5. From graph (1) : vy = 0 1. for object O1 O1 1 v
+
1 20
(
1 f
'
1 40
1 sec. 2
i.e., time taken to reach maximum height H is .... (1) t=
uy g
for object O2 O2 1 v
att =
1 2
=
3
( '1
uy = 5 m/s . ..Ans.(i) from graph (2) : vy = 0 at x = 2m
.... (2)
f
i.e., when the particle is at maximum height, its displacement along horizontal x = 2 m x = ux × t
from equation 1 and 2 we get f = 80/3
3
ux = 4 m/s
....Ans (ii)
y2 + 4 x 2)1/2 dE dx
Force = p
{y is not changing since p is directed along x axis} = p i [4 y 2 ˆi + 8xy ˆj ] i = p4y | [ y ˆi + 2x ˆj ] |
3. Equation of process P2
3 @
2 = ux ×
6 . 4 py
2. As the reflector approaches OS, the beat frequency will decrease to zero. After this, the reflector moves away from OS thereby increasing the beat frequency but after a long time the beat frequency will to become constant. Hence the correct option is (D).
1 2
3
2
= constant = C
Equation of State
P
(
@
.... (1) R T M
.... (2)
From 1 and 2 PT = constant 3 C is false, D is true. As @-changes to
3
P changes to
3
A is false.
= 4py y Ans. 4 py
2
+ 42x
y + 4 x2)1/2
Sol.7 to 9. FBD of rod and cylinder is as shown.
@ 2 P 2
Hence T changes to
from equation (1)
2T .
3
B is true.
4. w.r.t.
Net torque on rod about hinge 'O' = 0
8 N1 × L
= mg ×
L 2
DPPS FILE # 195
or
N1 (
mg 2
Net torque on cylinder about its centre C is zero.
2. (B ) The line of impact for duration of collision is parallel to x-axis. The situation of striker and coin just before the collision is given as
8 f1R = f2 R or
f1 ( f2
striker line of impact
coin
Net torque on cylinder about hinge O is zero. 8 N2 × L = N1 × L + mgL or N2 =
u 3
3 mg 2
Figure (A) before collision
10. (A) p (B) q,s (C) q,s (D) q,s (A) Work done by an ideal gas during free expansion is zero. (B) The angle between normal reaction on block and velocity of block is acute (whether the block moves up or down the incline). Hence work done by this force is non-zero and positive. (C) Net electrostatic potential energy
= S1 + S2 + M 12 =
=
Q2 4< \0 a
'
Q2 8< \0 a
Q2 4< \0 b
+
Q2 8< \0 a
'
coin u 3
rest Figure (B) after collision
Q2 4< \0 b
= non-zero and positive.
(# b > > a) (D) The kinetic energy of cylinder is increasing and work is done on cylinder by only force of friction. Therefore work done by force of friction on cylinder is non-zero and positive.
striker
Because masses of coin and striker are same, their components of velocities along line of im pact shall exchange. Hence the striker comes to rest and the x-y component of velocities of and 3 m/s as shown in figure. coin
y
u
P 9
3
DPP NO. - 29 1. The free body diagram of cylinder is as shown. Since net acceleration of cylinder is horizontal, NAB cos30° = mg
coin are u
6 O
x
4
For coin to enter hole, its velocity must be along PO
8
tan
9=
3 6 = u 4
u = 2 m/s
or Ans. (2, 0)
3. For no ray to emerge out of side PR or NAB =
2 3
mg
....(1)
and NBC – NAB sin30° = ma or NBC = ma + N AB sin 30° .... (2) Hence NAB remains constant and NBC increases with increase in a.
A > 2C
3 sin
A > sinC 2
A 3 > 2 2 A > 120 °
3 sin or
DPPS FILE # 196
4.
@gh
5. F =
f2
DPP NO. - 30 = 10
–6
m 1. Flux will be maximum when maximum length of ring is inside the sphere.
+ (mg )2 2 f (f = m ;r)
=
This will occur when the chord AB is maximum. Now maximum length of chord AB = diameter of sphere. In this case the arc of ring inside the sphere sub-
F mg
Now when the angular speed of the rod is increasing at const. rate the resultant force
tends an angle of
"
will be more inclined towards Hence the angle between decreases so as with the rod.
"
K dq = r
r U[ r
I (
r U[
1
I
= (const.)
(R r
r
R
2
F and horizontal plane
K (C / r 3 ) 4< r 2 dr r
dr
1 %&$ R !
= (const.)
"
a0
R(x
E × 4 < x2
E × 4 < x2
3
=
=
E=
charge on this arc =
8 ?=
R< Q 3
\0
=
R< .Q 3
R
2. The change in length of rod due to increase in temperature in absence of walls is H! = ! = H T= 1000 × 10 –4 × 20 mm = 2 mm But the rod can expand upto 1001 m m only.
8
qin
"
8
At that temperature its natural length is = 1002 mm. 8 compression = 1mm
7. Using gauss theorem
I E. ds =
at the centre of ring.
f .
6. Consider a spherical shell of radius r(r > R) and thickness dr. Then potential at centre due to it, dV =
< 3
I @ dv ( (I
r R
C r3
a0
mechanical stress = Y
H! !
= 10 111 ×
1 1000
= 108 N/m 2 4 < r dr 2
a0
( C) &x# !n$ ! a0 x 2 % R " (C 4 < ) &x# !n$ ! a0 %R "
3. v1 = 4 cos 53 º ˆi + 4 sin 53 º ˆj = v2 = 3 cos 37 º ˆi + 3 sin 37 º ˆj =
v12 =
12 16 ˆ ˆi + j 5 5 12 9 ˆ ˆi + j 5 5
7 ˆ j = 1.4 ˆj 5
Relative velocity in horizontal direction is
zero.
4. 0.3 m (Range)1 = (Range)2
8.
Electric field will be radially outwards. Electric potential decreases as we move in the direction of electric field.
2g(! – 0.1)
2 7 0.1 g
(
2g( ! – 0.2)
2 7 0. 2 g
! = 0.3
DPPS FILE # 197
5. (a) 4.5 m/s (b) 1.5 m/s (c) 3.75 cm
7. Since integral number of waves shall cross a point is 5 seconds, therefore power transmitted in 5 seconds is = × 5 = 2 <2 f 2 A2 > v × 5 = 2 × <2 × (50) 2 × (2 × 10 –3)2 × (0.01) × 200 × 5 =
By conservation of angular momentu m 0.5 × V × 0.4 = 0.5 × 4 × 1.2 v = 3u
<
and A = 2
solving ?0 = –30° 8 y = 2 sin &$ < x ' 100
%
( 1 7 100 7 (0.3)2 + 1 7 0.5 7 u 2 2
5
8. The equation of waves is y = A sin(kx – ;t + ?0) 8 where K = 2Q< ( 2< , ; = 2
also by energy conservation 1 7 0 .5 7 v 2 2
<2
"
2
2
DPP NO. - 31
3
v2 4
9 2
(
2 +u 4
1. The velocity of profile of each elementary section of the pulse is shown in figure 1 and figure 2.
4 8u2 4
3
(
u=
pulse moving towards right elementary section
9 2
9
(3
4
2
velocity vector of elementary section
Velocity profile
So v = 3u = 3 × 1.5
& a ( K h ( 100 7 0.3 ( 60m / s 2 # $ n ! m 0.5 % "
3
r=
3
(1.5) 2 2.25 r= = ( 60) 60
( 0.0375 m
N2 cos 60 ° N1 cos 30 ° N2
N1 2.
= 3.75 cm
(Figure-2)
When both the pulses completely overlaps, the velocity profiles of both the pulses in overlap region are identical. By superposition, velocity of each elementary section doubles. Therefore K.E. of each section becomes four time s. Hence the K.E. in the complete width of overlap becomes four times, i.e., 4k.
u2 r
u2 an
Velocity profile
(Figure-1)
= 4.5 (c) an =
pulse moving towards left
N1 sin 30 °
30° 60° N2 sin 60 °
6.
Q = 4m and f = 50 Hz. 8 V = f Q = 200 m/s # V=
T
>
8 T = > v2 = (0.1) = 400 N
× (200) 2
60°
2
1
30°
N1 sin 30° = N2 sin 60° N1 cos 30° + N2 cos 60° = mg Solving above equation mg 10 7 10 ( ( 50 N2 = 2 2 DPPS FILE # 198
3. (i) (a) The charge on the outer most surface will be (qa + qb) and it will be uniformly distributed K ( q a + qb ) 8 v= Ans. ; r K ( q a + qb ) E= Ans. r2 1 where K = 4< \ 0 (b) At a point inside the cavity of radius ‘b’ the potential will be due to q b, – q b induced on its inner surface and due to (qa + qb) on the outer surface of the sphere. v = Kqb – Kq b + K( qa + qb ) b r R
Kq b r
(ii)
R
(iii) 0
d2 y dt 2
2 x dx a 2 dt 2vx a2
Ans.
and E at that point will be only due to qb (which is placed at B) E=
aN =
+ 2y2 dy ( 0 dt
b
0 + b2 y2 &$% dy dt #!" (
2 v dx a 2 dt
+
2 b
2
2 dy 2y & d2 y # $& !# + 2 $$ 2 !! ( 0 % dt " b % dt "
Ans.
2
[# v = const. along x-axis only
qa
=
+ qb
4
,
a
=
' qa 4
,
b
=
' qb 4
Ans.
2v 2 a
Ans.
4. (a) Parabola y = ax 2 is shown. It is clear from diagram that at x = 0 velocity is along x-axis and constant a N is along y-axis. So,
2
& 2 # ( ' 2(b2 ) $$ d 2y !! b % dt " 'v
R=
2
=
aN
3
dy = 0] dt
aN =
'
bv 2 a2
a2 b
7. The difference in K.E. at positions A and B is
B 2
VB L
O
2
VA L
gsin
A
aN =
gsin
d2 y dt 2
K A – KB =
= 2mgLcos 9
dy dx = 2a × = 2aVx dt dt d2 y dt 2
= 2av
dx = 2av2 dt
(#
d2 x dt 2
(0)
aN = 2av2 R=
(b)
v2
1 mv 2A 2
2av 2 2
2
x y $& !# + $& 2 !# ( 1 %a" %b "
Here again at x = 0 particle is at (0,± b) moving along positive or negative x-axis hence a N is along y-axis only.
1 mv B2 2
.... (1)
TA =
mv 2A L
TB =
mv B2 – mg cos L
= mg (2L cos 9) Ans.
+ mg cos 9
8 TA – TB = 1 = . 2a
'
mv 2A
9
' mv B2 L
+ 2 mg cos
9
.... (2)
from equation (1) and (2) T A – TB = 6 mg cos 9 Ans. The comp onent of accele rations of ball at A and B are as shown in figure. "
| aA
=
"
+ aB |(
(2g sin 9)2
4g sin 2 9 + 16g2 cos 2 9
& v 2 v 2 #2 + $$ A ' B !! %L L" = g 4 + 12 cos2 9 Ans. DPPS FILE # 199
8. (A) p,q (B) p,q (C) q,r (D) q,r 5. Let the charge on intermediate shell be q (after Sol. In all cases speed of balls after collision will be same. earthing) In case of elastic collision speed of both balls after Potential of the intermediate shell = 0 collision will be u, otherwise it will be less than u. K 2Q KQ Kq 3 2R + 2R - 3R = 0 DPP NO. - 32 Q q + 2 2
1. At the instant 3m is about to slip, tension in all the strings are as shown
T q = 60 °
3 mg
8
8 3 >mg
= T cos 60 ° .... (1) and mg = T sin 60 ° .... (2)
2.
& 2Q Q # $ ' !2 % 3 2"
3 mv 2
M
!
O
O
V=
2
|( H 3 |H | HP net | ( 3 mv Pnet
Px
2
+H
Py
=
&9 3# $ + ! % 4 4 " mv
& Since, dm ( A (v dt ) @ # $ ! $$ 3 dm ( A @ v !! dt % " & dm # . v ! % dt "
3$
@1gh – @2gh +
h=
V2
2 GM a
&$ $%1 –
=
>
=
dx F = dt Kx
2
!
I
x dx =
O
F K
t
I dt o
4! 3 . K = 9 f 8M! = 9F
4! . 2m 9F ! 2 8 7 45 7 1.5 = 2. 9 7 15
3 @ A v 2 Ans.
2T = P0 r
GMm 1 mh2 = 2a 2
#! ! 2"
1
F
!
Sol. 7 to 9.
4.
–
2M
3
t=
=
r (@ – @1) gh 2 2
KE = V1
dM dx
I dM = I Kx dx and K =
mv 3 + mv = mv | HP y | = 2 2
3T=
& 4Q ' 3Q # 2 $ ! % 6 "
No charge will flow.
> = Kx =
3 3
| HP x | = mv sin 60 ° =
3. P0 +
=
6. 2
1
3 | HFnet | (
2Q =0 3
Q = 3 , which is same as that before earthing
mg
8 >=
–
–
& – GMm # $ ! % a "
The FBD of A and B are Applying Newton's second law to block A and B along normal to inclined surface NB – mg cos 53° = ma sin 53° mg cos 37° – NA = ma sin 37° m m Solving NA = (4g – 3a) and NB = (3g + 4a) 5 5 For NA to be non zero 4g – 3a > 0 or
a<
4g 3
DPPS FILE # 200
5. (A)
DPP NO. - 33 1. The distribution of charge on the outer surface, depends only on the charges outside, and it distributes itself such that the net, electric field inside the outer surface due to the charge on outer surface and all the outer charges is zero. Similarly the distribution of charge on the inner surface, depends only on the charges inside the inner surface, and it distributes itself such that the net, electric field outside the inner surface due to the charge on inner surface and all the inner charges is zero. Also the force on charge inside the cavity is due to the charge on the inner surface. Hence answer is
When the ball is just released, the net force on ball is W eff (= mg – buoyant force) The terminal velocity ‘vf’ of the ball is attained when net force on the ball is zero. $ Viscous force 6%5r vf = W eff 2 When the ball acquires rd of its maximum velocity v f 3
option (A). 2. The distance between the orbiting stars is
the viscous force is =
d = 2rcos30° = 3 r. The net inward force on orbiting stars is Gm
2
cos 30 " #
d2
GMm r2
#
Gm
2
d2
cos 30 " !
+m ( 4 % 2r 3 # M& ! $ G) T2 3 * ' r
or
Hence net force is W eff
2
r
2 1 W = W 3 eff 3 eff a 3
6. If we complete the trapezium as shown It becomes an equilateral triangle 2 A = 60°
+ A # 7 min ( * 2 &' ! 6
A
sin )
A Smn 2
d M d
–
$ required acceleration is =
T =2 % G 1/ M # m ., / , 30
m
3.
3
mv r
2 W eff. 3
5
5
10
d
+ 60 # 7 min ( & 2 * '! 2 , 60
sin )
sin
1 1 2 k x 0 + Mgh = k(x 0+h)2 + 0 2 2
7min = 30°
2
2Mg – 2x 0 k Maximum downward displacement
2h=
=[
2Mg k
–
2x0 ]
u23 4. H = 2a 3 a3 is same for all the three cases. HA =
(u sin 4 )2 u2 , HB = 2a 3 2a 3
(u cos 4 )2 and H C = 2a 3
7.
3 Clearly, PM = 2 cm 37º > sin 3 > 5
1
–
1 n0 # a(3 / 2)
1 n0 #
3a 2
$ HB = HA + HC DPPS FILE # 201
9a >5 2
3n0 +
N=40
F1=20N
20kg
N=40
10kg
f=20N
f=20N
9a >1 2
F2=60N
FBD of both blocks
Hence magnitude of friction force on both blocks is 20 N and is directed to right for both blocks. Normal reaction exerted by 20 kg block on 10 kg block has magnitude 40 N and is directed towards right. Net force on system of both blocks is zero.
2 a> 9
DPP NO. - 34
8. 2. U = 3x + 4y
6!
2.5 = 0.1 g / cm = 10 25
2
–
ay =
Kg/m
Ist overtone
ax =
8 s = 25 cm = 0.25 m fs =
1
T
8s
6
2
Fy m Fy m
=
=
9 :
9 :
=
=
–
–
3
4
"
a = 5 m/s2
Let at time 't' particle crosses y-axis
pipe in fundamental freq
then
–
6=
1 (– 3) t2 2
2 t = 2 sec. Along y-direction :
=y =
2 particle crosses y-axis at y = – 4
8 p = 160 cm = 1.6 m
At (6, 4) : U = 34 & KE = 0 At (0, – 4) : U = – 16 2 KE = 50
V
fp = 8 p ! by decreasing the tension , beat freq is decreased $ f s > fp 2 fs –fp = 8
1
1 (– 4) (2)2 = – 8 2
T
or,
1 mv2 = 50 2
2 v = 10 m/s while crossing y-axis 3.
320
9 = 8 2 0.25 10 9 2 1.6
2
T = 27.04 N
9. (A) p,s (B) p,s (C) q,s (D) r The minimum horizontal force required to push the two block system towards left = 0.2 × 20 × 10 + 0.2 × 10 × 10 = 60. Hence the two block system is at rest. The FBD of both of blocks is as shown. The friction force f and normal reaction N for each block is as shown.
F1=20N
Tension in elementary section of width dx is T = 8 xg (8 = mass / length)
$ extension of length x (= BC) of wire is x
=x =
fmax=
... (1)
2 extension in total length of wire #(=AB) is 2=x
F2=60N 2=x =
fmax=40N
2 ( 8 x g) dx = 8 x YA 2YA 0
>
8 #2 g 2 YA
... (2)
DPPS FILE # 202
$ from equation (1) and (2)
7. (A) Charge distri bution on syste m is shown below
#
x=
2 AC #9x = =( 2 PC x
Ans.
–
1)
4. 8i = wavelength of the incident sound u 19u 2 = 2f f fi = frequency of the incident sound =
10u 9
10u 9 u
So electric lines of forces are
18 f = = f = frequency of the r 19 f 10u 9 u 2 reflected sound 8r = wavelength of the reflected sound
=
=
10u # u 11u 11? 19 u = × 19 = . fr 18 f 18 f
8i 19 u 18 f 9 ? 8r = 2f 11? 19u = 11 Ans. 5.
as shown below
Since number of lines of force are proportional to charge so no. of lines of forces emerging from inner sphere should be equal to the no. of lines of fo rces emerging from outer shell.
"
rP = ( ˆi + ˆj ) t "
rQ = (2 ˆi + ˆj ) + ( – ˆi + 2 ˆj )t "
"
"
rQP = rQ – rP = 2 ˆi + ˆj + ( –2 ˆi + t ˆj )
+ 6 ? 10 99 3 ? 10 99 3 ? 10 99 ( 9 # & = 200 V 92 92 18 ? 10 18 ? 10 92 '& *) 27 ? 10
8. VB = K )
"
rQP = (2 – 2t) ˆ + (1 + t) ˆj
i 1+t x = 2 – 2t y = 2 x = 2 – 2 (y – 1) x + 2y = 4 Ans.
+ 6 ? 10 99 3 ? 10 99 3 ? 10 99 ( 9 # & 92 92 18 ? 10 6 ? 10 92 &' )* 27 ? 10
VA = K )
6. Force on cone while it is penetrating the sand is shown in F.B.D. below Applying work energy theorem to the cone as x changes from 0 to d = KE = work done by mg + work done by resistive force R 9.
= 200 V – 150 V + 450 V = 500V VC = V B (as shell is conducting) Therefore, VC – V B = 0 VB – VA = – 300 V Potential at point B and point C increases by same value, keeping their difference unchanged.
DPP NO. - 35 d
KFinal – KInitial = mgd
–
> kx
2
dx
1.
F.B.D. of man and plank are -
0 d
0 – mgh = mgd –
> kx
2
dx
0
$ 2
kd 3 = (mgd + gh) 3 3mg (h # d) k= d3
DPPS FILE # 203
For plank be at rest, applying Newtons second law to plank along the incline
4. At position A balloon drops first particle So, uA = 0, a A = – g, t = 3.5 sec.
Mg sin 4= f ...............(1) and applying Newton ’s second law to man along
11 2. gt , 02 -
SA = /
the incline. mg sin 4 + f = ma
1 0
a = g sin 4 /1 #
Balloon is going upward from A to B in 2 sec.so distance travelled by balloon in 2 second.
...............(2)
M. , down the incline m-
1 1 . / SB ! a B t 2 , 2 0 -
aB = 0.4 m/s
Alternate Solution : If the friction force is taken up the incline on man, then application of Newton ’s second law to man
..........(ii)
2
,
t = 2 sec.
S1 = BC = (SB + SA)
...........(iii)
Distance travell by second stone which is droped from balloon at B B > u = u = a t = 0.4 × 2 = 0.8 m/s
and plank along incline yields. f + Mg sin 4 = 0 mg sin 4 – f = ma Solving (1) and (2)
...........(i)
..........(1) ..........(2)
2
B
B
t = 1.5 sec.
SB >
1 1 . / S 2 ! u 2 t 9 gt 2 , 2 0 -
1 M. a = g sin 4 /1 # , down the incline 0 m-
>
...........(iv)
SA
Distance between two stone =S = S1 – S 2 .
Alternate Solution : Application of Newton ’s seconds law to system of
A
C
>
man + plank along the incline yields
1 0
a = g sin 4 /1 #
F
F
mg sin 4 + Mg sin 4 = ma M. , down the incline m-
2R
R
5.
a
f 2. As ON = MN = OM = r So it is equilateral t riangle : $ Potential at N due to two dipoles ; V = V1 + V2
=ma f ...(i) F2R – FR – fR = I4 ...(ii) a=Ra ...(iii) FR – fR = I .
N r
@ @ O
60° r
3.
% - 60°
@@
Kp cos 60" r2
x x rel ! 1 6 d2 x rel dt 2
!6
1 0
M@ P (r,0,0)
+
Kp cos ( % 9 60")
x rel = 6 x d2 x dt 2
F = /m #
r2
=0
a=
a R
Ia
F – ma =
r
P (0,0,0)
=
1r r 3 / , , / 2 2 ,0 , 0
R2
I .
, a.
R2 -
F m # I/R2
mF f = ma = m # I R2
mF f = m# I R2
arel = 6 g f=
10 N 3
DPPS FILE # 204
6. a =
F m#
4=
=
I
5
m# 2 3 ! 2% 2# (C) T = 2 % mgd = 2% # 3g mg 2
3
C
R2
a 5 = 2 6
v=0+
5 5 ?3 = 6 2
B = Bo + 4 t = 0 + 1
m D / 2 A# (D) T = 2 % DAg = 2 % DAg 5 3
?3 = 5
1 2 IB 2 1 1 5 5 = ?2?5?5 # ? 4? ? 2 2 2 2
KE =
DPP NO. - 36
mv2 +
2
25 75 = 25 + = J 2 2
1. (C) Work done against friction mus t equal the initial kinetic energy. E
$
1 mv 2 = 2
7. F2R – FR = I4
1
1
1 v2 = A g 2 dx ; x 2 1
>
E
v2= 2gA
1 FR . B = 0 + / I ,t 0 1
>
E
6 mg dx ;
v2 + 1( = Ag )9 & 2 * x '1
FR 4= I
KE =
#
= 2 % 2g
2 v = 2g A
2. F.B.D. for minimum speed (w.r.t. automobile) Iw
2
2
1 F 2 R 2 .,t 2 2 , I -
= 2 ? I // 0 =
100 ? 1? 3 ? 3 25 ? 9 F 2R 2 = = = 112.5 J. 2? 4 2 2I
8. (A) p (B) q (C) p (D) s (A) In frame of lift effective acceleration due to g 3g gravity is g # ! downwards 2 2
$ T = 2%
2# 3g
:
Ff y' = N – mg cos G – Ff x' =
k g ! m L constant acceleration of lift has no effect in time
2
period of oscillation.
sin G)
(B) K # = m g
$
# m $ T = 2% = 2% g k
mv 2 sin G = 0. R
mv 2 cos G + 6N R
–
mg sin G = 0
mv 2 mv 2 cos G + 6(mg cos G + R R –
2 v2 =
mg sin G = 0 (6Rg cos G 9 Rg sin G) (cos G # 6 sin G)
for G = 45 º and 6 = 1 : vmin =
Rg 9 Rg = 0 1# 1
DPPS FILE # 205
5. In elastic collision the velocities are exchanged if masses are same. $ after the collision ; VC = 0 VA = v Now the maximum compression will occure when both the m asses A and B move with same velocity. $ mv = (m + m) V (for system of A – B and spring)
$V =
v 2
F.B.D for maxim um speed (w.r.t. automobile)
Ff =
mv 2
x'
cos G
–
$ KE of the A
mg sin G – 6(mg cos G
R
–
B system =
1v. 1 × 2m / , 2 02-
2
mv 2 4 And at the time of maximum compress ion ;
mv 2 sin G) = 0 R for G = 45 º and 6 = 1 vmax = E (infinite)
=
+
2
1v. 1 1 1 mv 2 = × 2m / , + K X2max 2 2 2 02m 2K
$ X max = v
3.
Taking cylindrical element of radius r and thickness dr M
dm =
%(R 22 9 R12 ) # × (2%r # dr) R2
>
>
2 CAB = dC e# = dm r =
> (R
R1
2M 2 2
9 R12 )
.r 3 dr
1 m (R 22 # R12 ) 2 Using parallel axis theorem =
1 2 2 IXY = m (R 2 # R1 ) + MR22 2
6.
E=
40 9 10 = 100 V/m 0 .3
(near the plate the electric field has to be uniform $ it is almost due to the plate). For conducting plate
I
4. Let m be minimum mass of ball. Let mass A moves downwards by x. From conservation of energy, mgx =
1 kx 2 2
1 2mg .
x= / k , 0 For mass M to leave contact with ground, kx = Mg
1 2mg . , = Mg 0 k -
K/
M m= . 2
E= H 0
2 I = H 0E
Therefore %, I = 8.85 × 10 = 8.85 × 10 10 C/m 2
–
12
× 100
–
7. Direction of E.F. at B is towards A that will exert force in this direction only, causing the positive "
charge to move. [ E is perpendicular to equipotential surface and its direction is from high potential to low potential.] 8. W = q.dV = –1 × 10 6[20 – (–20)] = – 4 × 10 5 J. –
–
DPPS FILE # 206
DPP NO. - 37
4. Kx = V(2000) (10) – V (1000) (10)
1. For first collision v = 10 m/s. t1 =
=
10 [ 1000 × 10] 2000
% (5 ) 10
= % /2 s ec. velocity of sep = e. velocity of opp. v2 – v1 =
1 (10) 2
–
v2 v1 = 5 m/s for second collision
$
t2 =
2 % (5 ) 5
Kx =50N
...(b)
=2%
$ total time t = t + t = % /2 + 2 % t = 2.5 %
Ustored
1 50 . 1 , × (100) / 2 0 100 -
=
2
2
=
1 2500 ? 2 100
= 12.5 J
5. At x = 5 cm,
< Ve < x = slope of figure 1
at x = 5 = +ve 2. So F x = The friction force will reduce v0 , hence translational K.E. The friction force will increase B There is no torque about the line of contact, angular momentum will remain constant The frictional force will decrease the mechanical energy.
–
< Ve q < x = – ve
at y = 15 cm,
< Vg = slope of figure 2
– ve < Vg So F y = – m = +ve
at y = 15
=
So particle will try to move towards and +y direction.
–
x direction
6. at (25, 35),
Fx x ! 25 =
3. Equation of the component waves are : y = A sin( B t – kx) and y = A sin ( B t + kx) where; B t – kx = constant or B t + kx = cosntant Diffeentaitin g w.r.t. 't' ;
B – k dx= 0 dt
2 v=
dx B = dt k
and B + k dx = 0 dt andv =
–
Hence (B)
Fy
< Ve q
x ! 25
1 2 ? 10 4 . , 9// 9 6 0.1 ,- × (20 × 10 ) = 4 N 0 –
y ! 35
! 9m
< Vg
1 10 93 , = – 2N = – (200) // , 0 0.10
B k
i.e.; the speed of component waves is
=
–
Fnet = 4 ˆi 9 2 ˆj = (200) a
1 B. / ,. 0k-
4 ˆ 2 ˆ i9 j =a 200 200
a = (2 ˆi 9 ˆj ) × 10 2 m/sec 2 –
DPPS FILE # 207
7. W (5, 15) @ (25, 35) = U (25, 35) = (0 + (200) (
–
1.5 × 10
–
–
3))
U (5, 15)
–
[(20 × 10
–
6)
(1/2 × 10 4) + (200) ( –1.5 / 10 3)] = – 0.1 J –
8. (A ) p, q, r, s ,t ; (B ) p , q , r , s ,t ; (C ) p , s, t; (D) p, q, r, s,t (p, s) since there is net impulse, translations motion will occurs for all cases. (r,q) only in C, impulse is passing through centre of mass. Hence rotation will occur and angular momentum will increase in all cases except (C). (t) About all the points on the line of action of the impulse, torque is zero. Hence angular momentum will conserve for many points.
DPP NO. - 38
3. Both blocks loose contact immediately after the release. T P = 2%
m , 4K
T Q = 2%
m K T Q = 2T P
2 $
Q comes at lowest position at time
2
TQ
, i.e. time period of P (T 2 come back to srcinal position
P)
the block P
$ The distance between P and Q is F L
1. T =
...........(i)
dv F = 5A dx
V J 5L2 = 5LV V L
F ...........(ii) LV From (i) (ii),(i) (ii)
4. At t =
0
2.
–
#
both the blocks are at extreme position 2 and their velocity is zero. [Soln. of SSI Sir] $ VP = V Q = 0
m
m #
B2
F# Y= A =#
2
2
x 2
F# =# = Ay
m B2 x 2 dx =# = # 2 AY
=# =
–
m B2 # 3 # 6 AY
6y
=# = B B2 = 2B1
–
V =P = =V
–
1.5 ? 140 ? 10 3
9 0.2 ? 10 93
=
2 m 21 1 1. m 1L. 1 m . L2 L / # , / , / , = + 4 4 12 4 2 0 12 4 0 0 -
=
m L2 12
$ Moment of i nertia of frame = ML 2/3. 9. (Fo r th e abo ve tw o qu es ti ons ) Newton's law applied on C.M. gives mg sin G – f = ma .... (1) Writing K ! C4 about C.M., we have 2
L
.... (2) 2 = mL 3 4 from the condition of rolling, we have f.
DB2 # 3 2
=
7. Moment of inertia of one rod about the axis of frame
a=
=# =
=P
=V / V = 1.05 × 109 Pa .
0
2 T= 2
6. B =
Ans. 1.05 × 109 Pa.
> =T ! > # dxB x T
2mg K
TQ
5=
+F ( ) & J [V]. *5'
travelling
2mg downwards. K
a distance
In time
TQ
L 2
4
.... (3)
from (1), (2) and (3) f=
2 mg sin G 5
and a =
3 g sin G 5
DPPS FILE # 208
DPP NO. - 39 1. From the free body diagram of the sphere : FV = 4 mg – 2 mg – FB 2 FV = 2 mg – FB
m = 1kg Q
4.
µ = 1.5
S
P
R
P
R
4 31I 2 6 % 5r V = 3 %r / 2 9 D, g 0
(since 4m =
2
2 V = 9r
2
M = 11kg
4 3 %r ? I ) 3
( I 9 2 D) g
5
If the point P has an acceleration a upwards then the acceleration of point R will be a downwards. R
M = 11kg
The point R has an acceleration a downwards so the block will also have an acceleration a downwards.
S
P
M = 11kg
2. In the figure shown
The point P has an acceleration aupwards, the block has an acceleration a downwards so the acceleration of S will be 3a downwards. (because "
!
=OOL P & =CCL P are similar
OO L
C CL
!
OP C P ..............(1)
"
a S # aP " ! ablock ). 2 The point Q will also have an acceleration 3atowards right.
2T
also ! = OOL C & = CCL C are similar OO L
OC
C CL = C C
.............(2)
2
a
The F.B.D. of 11kg block
By equation (1) and (2) OC OP = CC CP
T
110 N
OP ! CP Ans. OC CC
3a 3. As wave has been reflected from a rarer medium, therefore there is no change in phase. Hence equation for the opposite direction can be written as y = 0.5A sin (–kx – Bt + G) = – 0.5A sin (kx + Bt – G)
The F.B.D. of 1kg block
15N
T
Using FBD of 11 kg block, which will have acceleration a downwards.
DPPS FILE # 209
110 – 3T = 11a ........ (1) (in downwards direction) For 1 kg block, which will have acceleration 3a, T – 15 = 3a (in horizontal direction) or 3T – 45 = 9a ............. (2) on adding equation (1) & (2) we get 20a = 65 2 4a = 13 m/s 2 5. 1.15 × 108 =
9. (i) Cons. li near mo mentum
900(10 # a)
1 %d2 . / , / 4 , 0 -
–
2m.v + 2v.m = 0 = MV
cm
Vcm = 0 (ii) As ball sticks to Rod Conserving angular momentum about C
6
2 d=
2v.m. 2a + 2mva = CB
0.06
10 % cm =
10 %
6 ? 10 92
m=
m
10 % 6 ? 10 92
Ans.
10 %
m
6. The situation is shown in figure. (a) From figure h = # (cos G – cos G0) and M2 = 2gh = 2g# (cos G – cos G0) ....... (1) Again T – mg cos G = mM2 / # ....... (2) Substitting the value of M2 from eq. (1) in eq. (2) we get
=
6mv.a = 30 ma 2.B
2
B=
(iii) KE =
v 5a
1
CB2 =
2 =
G GN T
1 8m. 36a 2 . / # 2m. a 2 # m. 4a 2 ,, / 12 0 -
1
. 30 ma 2 ×
v2 2
25a
2
3mv 2 . 5
#
h
DPP NO. - 40
M
mg
1. Upward force by capillary tube on top surface of liquid is
mg cos G = m {2g# (cos G – cos G0) / # } or T = mg cos G + 2mg (cos G – cos G0) or T = mg (3 cos G – 2 cos G0) or T = 40g (3 cos G – 2 cos G0) newton T
–
Ans.
fup = 4 Ia cos
h then we use
4Ia cos G = ha 2 Dg or h =
4I cos G
T = 40 (3 cos G – 2 cos G0) kg f.
(b) Let G0 be the maximum amplitude. The maximum tension T will be at mean position where G = 0. $ Tmax = 40 (3 – 2 cos G0) But Tmax = 80 Solving we get G0 = 60° Ans. G0 = 60°
G
If liquid is raised to a height
Ans.
aDg
2. =# =
F#
%r2y 2
=# 4
#
r2
Only option 'radius 3mm, length 2m' is
satisfying
the above relation.
DPPS FILE # 210
0 .5 3. Velocity gradient = 2.5 ? 10 9 2
6. µ =
2
as force on the plate due to viscocity is from upper as well as lower portion of the oil, equal from each part, Then,
=
F=2 5A
#=
0.5
f=
1.25 ? 10 92
2 5 = 2.5 × 10 2 kg –
40 cm
=
3.2 ? 10 93 40 ? 10 9 2
–
2
sec/m 2
=
3 .2 32 = kg/m 40 4000
8 2
2 8 = 2#
dv dz
2 × 5 × (0.5)
3.2 gm
v
8
...........(1)
1 2#
=
T
µ
1 1000 64 = 2 ? 40 ? 10 9 2
+1000 ( ? 2 ? 40 ? 10 92 & 2 ) * 64 '
T 32 / 4000 2
32 =T 4000
1000 32 × =T 64 4000
4.
2 T=
5 sin i = 1 sin e 2 =1×
2
4 cm
5 2
2 cm 4cm2 # h 2
10 N 8
10 / 8 10 9 6 107 vc y = .05 ? 10 9 2 = 8 40 ? 10 92
40 (.05)
2 h = 4 cm = 109 N/m2.
16cm2 # h2
[ Ans. 1
109 N/m2 ]
7. Torque of f riction a bout A is zero. 8. Angular momentum conservation about Lin = mv 0r – mk 2B0 Lfin = 0 Lfin = Lin 2 v0 = B0k 2/r.
5.
From conservation of energy mgh =
1 mv 2 2
2 mg # sin G = 2 2g sin G =
point A.
9. acm = –6g 02 = v02 – 26gs
v 02
2 S = 26g
1 mv 2 2 v2 #
DPP NO. - 41 = aC
g cos G = at Total acceleration a =
ac2 # a 2t
= g cos 2 G # (2 sin G)2 = g 3 sin2 G # 1
1. (B) VB > VD = In a Wheatstone's bridge circuit shown if PS = QR, VB = VD . No current flows between B and D. If PS < QR , VB > VD current flows from B to D. If PS > QR , VB < VD current flows from D to B. DPPS FILE # 211
VB = VC VD – VC = i3S = positive 2 VD > VC 2 VD > VB If resistance R is zero and all other resistances are non–zero, then PS > QR and similarly, we get VD > VB Hence if PS < QR, VB > VD .
Alternate :
B Q
P A
C
G R
S D
Let resistance P = 0 and all other resistances Q,R,S,G are non–zero then PS < QR condition is satisfied.
0 P=
0 P=
2. G
J
i1
VA = VB
1 × 2 mR 2
2
B02 +
1 m(2R) 2B0 2 + 2
2 VA – i1R = VD 2 VA – VD = i1R = positive (or current flows from A to
K.E. system =
D through G, then VA > VD) 2 VA > VD 2 VB > VD Let resistance S = 0 and all other resistances P,Q,R,G are non–zero then PS < QR condition is also satisfied.
1 1 m( 2 R)2 Br2 + 2m( 2 R)2 B02 2 2
B
B Q
P
G
A
C
R
G
A
3. a = C
1 × 2m ( 2
0 S=
R
D
t=
D
2 0 = u – at
u um = a 2T#
Total time T = 2t = VD = VC
2 VB – i2Q = V C = V D 2 VB – VD = i2Q = positive 2 VB > VD
B
G
Q=0
P
C S
D
13 v
1# . 4 / # e, 02 -
=
7v
1# . 2 / # 2e , 02 -
B Q=0
P
R
um . T#
4. According to given condition,
Suppose resistance Q is zero and all other resistances P,R,S,G are non–zero then PS > QR.
A
2 2 2 2 R) B = 2mv 0 .
2T# m
v = u + at
J
0 S=
K.E. 2m =
i2 Q
P
= 6 mv 02
#
24
C
A
J
e=
G R D
i3
S
So, r =
r=
10e 6
5# 72 DPPS FILE # 212
5. For TIR at B, the angle of incidence i O c
8. The energy of any geostationary satellite is the sum of kinetic energy of satellite, interaction energy of satellite and its own planet and interaction energy of satellite and star. Both planets have same mass and same length of day. Geostationary satellite - planet system will have same interaction energy in either planet. Also kinetic energy of both satellites will be same. But the satellite-sun system will account for the energy difference.
& r + i = 90 2 i = 90 – r by snell’s law at pt A, sin 45° = n sin r = n cos i Now ! i > c 2 sin i > sin c
–
Uf =
–
GMm 0 + Usatellite 2r GMm 0 2 ( 4r )
Emin = Uf
1 n
2 cos r >
Ui =
–
+ Usatellite
Ui =
–
planet
–
planet
3 GMm 0 8r
DPP NO. - 42
2 n>
1 cos r
1. In a binary star system
2 n>
B1 = B2
1 1 9 sin2 r
2. Strain (P) = 1
2 n>
19
2 n>
= 2 × 10
=# #
= Q =T = (10 5) (200) –
3
–
1 2
Stress = Y (strain)
2n
Stress = 1011 × 2 × 10 3 = 2 × 108 N/m2
2n
2 Required force = stress × Area = (2 × 108) (2 × 10 6) = 4 × 102 = 400 N
–
–
2n 2 9 1
2 2n2 – 1 > 2 2 n >
3 2
6. The time in which the planet rotates about its axis
$ Mass to be attached =
400 = 40 kg g
3. [Ans. v 1 =30, v 2 = – 25, v 3 = -35/3, v 4 = -25 cm] 4. Ans.
is not given for either planet.
C=
64
p2 Dv
7. For geostationary satellite, time period = 1 planet day (by def.) Let T = 1 planet day
2 p2 C2 D1 v 1 = 2 × C D v 1
T 0 = 1 planet year Now
4% 2 3 1 m . 4%2 3 r / , rG = T = Gm 0 M Gm 2
=
p1
9 1.5 400 ? ? 16 3 1200 C1
2
4% 3 r ! T03 2 T = T 0 = GM
2
6 C = 2
6 ? 32 3
2
=
9 3 = 16 ? 3 ? 2 32
= 64
DPPS FILE # 213
5. By energy conservation, mg.
1 m(3# )2 2 3# .B = . 2 2 3 g
B=
J
#
and velocity of centre of mass of rod BC
$ Ratio = 3. g
Vcm =
#
.2#
2. The only force acting on the body is the viscous force
Vcm = 2 g#
Here, m
If time taken by centre of mass of rod BC frombreaking position to line PQ is t.
= – rv
1 2 8# = ? g ? t 2
2
vdv
0
G = B.t
=
mv . r
4radian. 4. a1 =
R (58.3)
where
2 x=
0
3. The image of a point closer to the focuswill be farther. As the transverse magnification of B will be more than A, the image of AB will be inclined to the optical axis.
#
& iX =
x
> mdv ! > 9 rdx v
t= 4 g
6. iR =
= –6%5rv
dx
It is same in both cases
R (68.5)
R is the potential gradient of the
GM F = 2 m r
$
a1 =1 a2
potentiometer
$
58.3 R = 68.5 X
58.3 10 = 2 X = 11.75 68.5 X
5. Loudness S = 10 log 10
$ S 2 – S 1 = 10log 10 C=
!
7. The maxim um P.D. which we can measure by this potentiometer is V 8. Any change can be done which assures p.d. across R or X less then or equal to V
DPP NO. - 43 1. Originally VA = VD = VE
$
C2 C1
&
P 4%r T
r12 C2 = 2 C1 r2
$ (S + 20)
–
S = 10 log 10
r2 r22
r = 20 log 10 r 2
2 After connecting C & B. The equivalent circuit will be [Now VA = VD =VE and VC = VB ]
C C0
r r2
2 r 2 = 0.1r
= 10
$ shift = r
–
0.1 r = 0.9 r.
9r Ans. 10
DPPS FILE # 214
3. a = 3t 2 + 1
Sol. 6 to 8 The angular speed of rod =
$B=
As given v B = 0
B= 2u #
u 9 vB #/2 Ans.
The time after which centre of rod reaches the highest point is t
0
u g
=
:
v = t3
2
1
> dv ! > (3 t 2 # 1) dt
2
0
0
# t ;0 = 2 m/s. 1
v = t3 + t s
1
3 $ > ds ! > ( t # t ) dt 0
The angular acceleration of rod is zero and in the given time to the rod undergoes angular displacement
0
2
% 2. $ from G = Bt
S=
1
#
4
1
= 0.75
2
4. 8000 Conserving angular momentum m.(V
u 2 = # ? g or u = 2
%
v
dv = 3 t2 + 1 dt
2u
%gL 4
1
cos60 °).
V2 ; = 2. V1
4R = m.V 2.R
Conserving energ y of the system
DPP NO. - 44 1. By energy conservation between A & B
2
2R MgR 1 Mg +0= + MV2 5 5 2
m.(V 1 cos60 °). 4R = m.V 2.R
9
A R–R cos53 =2R/5
1 2 1 2 3 GM V – V = 2 2 2 1 4 R 2
=
1
O 37º B 37º g
R–R cos37= R/5 Reference line g cos37
V2 = 2. V1
G Mm 1 GMm 1 # mV12 = 9 # mV22 4R 2 R 2
or V R 53º
;
V1 =
1 GM 2
1 2
R
64 ? 10 6 =
8000
m/s Ans. 8000
2
7. On the positive side of x axis, potential is zero at distance x 1 (it is between both charges), then
V=
k.6e k.10e ! r 8 9r
2gR 5
For the left side
Now, radius of curvature r =
V32 2gR / 5 R ! ! ar g cos 37 2
r=
ar
! 2gR / 5 ! R g cos 37
2
2. Relative displacement of glass window w.r.t. cyclist is 20 cm time taken = 1 sec. So, relative velocity of glass window w.r.t. cyclist = 20 cm/sec. = 0.2 m/sec. 1
y
k.6e k.10e ! x2 8 # x2
2 x 2 = 12nm x1 # x 2
R=
V32
2 r = 3nm
R x2
=7.5nm
xc x x1
2 |X c| = X 2 – R = 4.5 nm Vc =
k.6e 4.5 ? 10 99
–
k.10e 12.5 ? 10 99
= 9×10 9×10 9×1.6×10
–
19
+4 4( 8 ) 3 9 5 & = 1.44 × 15 * '
= 0.77V DPPS FILE # 215
8. (A) p,r (B) q, r (C) q, r (D) q, r Initially the image is formed at infinity. (A) As m is increased the focal length decreases. Hence the object is at a distance larger than focal length. Therefore final image is real. Also final image becomes smaller is size in comparision to size of image before the change was made. (B) If the radius of curvature is doubled, the focal length decreases. Hence the object is at a distance lesser than focal length. Therefore final image is virtual. Also final image becomes smaller is size in comparision to size of image before the change was made. (C) Due to insertion of slab the effe ctive object for lens shifts right wards. Hence final image is virtual. Also final image becomes smaller is size in comparision to size of image before the change was made. (D) The object comes to centre of curvature of right spherical surface as a result. Hence the final image is virtual. Also final image becomes smaller is size in comparision to size of image before the change was made.
DPP NO. - 45 1. The linear relationship between V and x is V = – mx + C where m and C are positive constants. $ Acceleration a=
V
dV dx
=
–
v 20 g cos G
3. R1 =
( v 0 cos G) 2 g
R2 =
$
g
R1 1 !8 = R2 (cos G)3
Ans. 8
1 334 9 9 . 325 ,700 = × 700 = 650 Hz. 350 0 334 # 16 -
f' = /
5. Consider two small elements of ring having charges +dq and – dq symme trically located about y-axis. The potential due to this pair at any point on yaxis is zero. The sum of potential due to all such possible pairs is zero at all points on y-axis. R
Hence potential y at P(0, +dq + ++ +
m( – mx + C)
+ ´
+ +
+ +
+
dG
+ ++
–
) is zero.
2
– –
dq
––
–
–
G G
–
dG
– – –
–
+ + + +
–
–8
+8 + + + +
+ ++
–
++
–
–
–
–
–
x
– –
–
y´
$ a = m 2x – mC Hence the graph relating a to x
is.
2. Relative to lift initial velocity and acceleration of coin are 0 m/s and 1 m/s2 downward
1 (1) t2 or 2
g
4.
x
$ 2=
v0cos
v0
t = 2 second
6. Since all charge lies in x-y plane, hence direction of electric field at point P should be in x-y plane Also y-axis is an equipotential (zero potential) line. Hence direction of electric field at all point on yaxis should be normal to y-axis. $ The direction of e lectric field at P should be in x-y plane and normal to y-axis. Hence directio n of electric field i s along positive-x directi on. 7. Consider two small elements of r ing having charge +dq and –dq as shown in figure. The pair constitutes a dipole of dipole moment. dp = dq 2R = ( 8 Rd G) 2R The net dipole moment of system is vector sum of dipole moments of all such pairs of elementary charges. DPPS FILE # 216
By symmetry the resultant dipole moment is
y ++ ++
–
–
–
+ +
x
–
+
+8
–
dG
+ + + + + + +
´
dq
––
–
++ +
– – –
G
–
–
DPP NO. - 46 1. The electric field intensity due to each uniformly charged infinite plane is uniform. The elec tric field intensity at points A, B, C and D due to plane 1, plane 2 and both planes are given by E 1, E2 and E as shown in figure 1. Hence the electric lines of forces are as given in figure 2. z
along negative x-direction.
E1
x
B
– –
+ + + +
+dq
+ ++
–
++
–
–
–
–
–
–
–8
I
–
E
E2
E2
A
E2
E2
x +I D
E
E
1 C
y´
E1
E1
E
2 (figure 1)
E1
$ net dipole moment #% / 2
=
–
z
#% / 2
> (dp cos G) i ! 9 > (28 R ˆ
9% / 2
2
ˆ
cos G dG) i
9% / 2
x
= – 4R2 8 ˆi
. 1 1 n# ! // 9 1,, ns 8. (A) f 0( %'% &9
(figure 2)
1 1 . // 9 1 ,, R2 1 0(R% '%&
Aliter : Electric lines of forces srcinate from positively charged plane and terminate at negatively charged plane. Hence the correct representation
#
f = – ve P = 1 = –ve f
. 1 1 n# ! // 9 1,, ns (B) f 0( %'% &#
of ELOF is as shown figure 2.
q,t
1 1 . // 9 1 ,, R2 1 0(R% '%&
2.
#
f = +ve 1 P= = f
+ve
p,r,s
Force on q = Eq =
I 2P 0
(1 9 cos Go ) q = f
. 1 1 1 1 n# 1 . ! // 9 1,, // 9 ,, f n R R (C) 1 0( %s'% &- 0(% '%2&9
f=+,
(D) f =
P=
P=
9
1= +ve f
p,r,s
R 2 1 = –ve f
Consider a ring of radius y and thickness dy. Flux through this ring dR =(E cosG)2%ydy G0
r,q,t
$
Total
flux =
1
> 4%P 0
q 0
r2
2%ydy cos G
DPPS FILE # 217
7. by energy conservation
q = 2P (1 9 cos G0 ) 0 So, R !
2
1 1 mR v2 1 ? 20 ? 0.3 ? 0.3 ! ? ? 2 # mv 2 2 2 2 2 R mv2 = 1.2 J
f
I Translation K.E. =
mv 2 = 0.6 J 2
8. (A) – p,q,s, (B) – r, (C) – r,t ; (D) – p,q,s,t. (p) Parallel beam can be obtained from concave mirror and convex lens when point object is at focus. (q) Real image for real object is for concave mirror
3. (Moderate) a = v
and convexand lens. (r) Virtual diminished images are obtained for convex mirror and concave lens. (s) Real and magnified image is obtained for concave mirror and convex lens. (t) The direction of motion of image is in the same direction as motion of object in lens and opposite in mirror.
dv = cx + d dx
Let at x = 0 v = u v
$
x
> v dv ! > (cx # d) dx u
or
v2
0
=
c x2
+ 2dx + u
DPP NO. - 47
2
v shall be linear function of x if cx perfect square
2
2
+ 2dx + u is
1. As seen from the figure
4. (C) Velocity of approach of P and O is –
dx = v cos 60° = 5 m/s dt
( AF ) 2 # (FD ) 2
the displacement is It can be seen that velocity of approach is always constant. 100 $ P reaches O after = = 20 sec. 5
= 7 2m 2. Max. frictional force fmax = 6N = 6(mg + F sin53°) = 0.2 (20 × 10 + 30 ×
5. a = (kx – f)/m f × R = mR 2 / 2 × a/R
N
6. by energy conservation 2
1 1 mR v2 1 ? 20 ? 0.3 ? 0.3 ! ? ? 2 # mv 2 2 2 2 2 R
mv2 = 1.2 J Rotational K.E. =
4 ) 5
mv 2 = 0.3 J 4
6N
53° mg
Fcos53°
F
Fsin53° = 44.8 N As applied horizontal force is Fcos53 ° = 18N < fmax, friction force will also be 18 N.
DPPS FILE # 218
3. Potential gradient = 1.6 v/m E.M.F. = potential gradient × balancing length 1.2 = 1.6 × #
$#=
1 .2 3 = = 0.75 m = 75cm 1.6 4
g R
4. mg = m B2 R , B =
DPP NO. - 49 1. Four lines, perpendicular to lines of electric field and passing through A, B, C and D are drawn. These are equipotential lines. As potential decreases in the direction of electric field, therefore VA > VB > V D > VC y B
6.
KQ x
2
K 4Q
=
C
(Since r >> R)
(3 r 9 x ) 2
A
30°
x
E D
2 (3r – x) = 2r 2x=r 7. Once the charge reaches the neutral point it will be accelerated towards center of ring 2, will cross it, be reatarded, come to rest and then return towards it.
2. For A,
1 ds = VA = dt 3
Thus the motion is oscillatory, but not SHM. For B, KQ q 4 KQ q 8. Uin = + 3r R
3
VA 1 = . VB 3
Its energy when it reaches the center of ring 1 Ufin. =
ds = VB = dt
4 KQ q KQ q + + K.E. = Uin 3r R
2 K.E. is positive 3.
DPP NO. - 48 If pot. drop between A and B is also 2V, then no currrent will pass through the gelvanomter.
1
1.
2
3
0
2Î
I
2Î
+
4
I 0
Electric field due to both the plates will be cancelled out for all the points. So the net electric field at the points will be governed only by the sphere. Farther the point from the sphere, lesser the magnitude of electric field. Therefore E 3 = E4 > E2 > E1 2. Since the block slides down the incline with uniform velocity, net force on it must be zero. Hence mg sin G must balance the frictional force ‘f’ on the block. Therefore f = mg sinG = 5 ? 10 ? ½ = 25 N.
1 12 ,R = 2 0R # 5
Pot. drop across R = / 12 R = 2R + 10 R=1U
4. Let the angular speed of disc when the balls reach the end be B. From conservation of angular momentum 1 1 m 2 m 2 2 2 2 mR B0= 2 mR B + 2 R B + 2 R B or B =
B0 3
5. In s econd case due to psuedo force acting on the block its acceleration will be more as compared to the first case. Hence t1 > t2 Ans. t1 > t2 DPPS FILE # 219
6. V = E + ir (during charging) = 14 V.
5. From given data "
VM ! 5 2 ˆi # 5 2 ˆj
7. P = I 2 r (Due to internal resistance) = 502 × 4 × 10 2 = 100 W
velocity of man Velocity of wind
–
"
8. Rate of charging = E.I. = 12 V. 50 A = 600 W
VW ! 5 2 ˆi The flag will flutter in the direction in which wind is blowing with respect to the man holding the flag.
DPP NO. - 50
"
Vv + 2 VA = 6 ...(2) 2 Solving (1) & (2) Vv = 4, VA = 2 . after closing S2 : – Vv = 0 VA = 6 So that value after closing S2 is 3/2 times the value after closing S1 . 2. [ Ans: V 1 =
10 V 5V , V2 = ,] 3 3
V RT 3. The speed of sound in air is v =
M
V
of H2 is least, hence speed of sound in H 2 shall be M maximum.
"
2 VWM = VW
1. Initially :– Vv + VA = 6 ..(1) ; VV & VA being the potential across voltmeter & ammeter respectively after closing S1 .
–
"
VM
"
VWM = ( 5 2 ˆi ) – (5 2 ˆi # 5 2 ˆj ) "
VWM =
–
5 2 ˆj = 5 2 ( 9 ˆj )
This implies direction of wind with respect to man in south. Flag will flutter in south direction. Ans.
1
6.
C
20V A
X
P
0V B
Let reference potential of B be zero. No current shall flow through g alvanometer. –
If VC Vp = 16 volts. Now V p = 2 volts. $ Vc should be 18 volts. Now
VA 9 VC VC 9 VB = 1 X
Solving X = 9 U. 4.
1
Decrease in PE = Gain in rotational K.E. Mg
L + MgL = 2
2
2 3 MgL = .ML 2 . 3 2
2
9g = B2 4L =
X
P
0V B
CR3 ! CR1 # CR2 7. Balance point is independent of r. It can be seen for balance point at P, V C – V P = E in absence of
1 4 ML2 . ML 2. B2 = ML 2 3 2 3
[ Ans.:
20V A
C
B2 =
2 B= 9g ] 4#
cell, jockey and galvanomete r. 2
9g 4L
4ML 3
8. For balance point at P. VC – V P = E = 12 ! VC = 18 , V P should be 6 volts. Therefore VA 9 VP #
!
VP 9 0 or # = 70 cm. 100 9 #
DPPS FILE # 220
DPP NO. - 51
1. E – ir1 = 0 2 i =
V Vmax
2E E and i = r # r # R r1 1 2
4.
=2 U
x = x0 is the point where potential is maximum . So, if the impulse is sufficient enough and point charge crosses the maximum PE barrier than point charge will move to infinity otherwise it will perform oscillatory motion and for very small impulse the motion may be SHM.
50V q=2 00/36 C i
1 2.
2
26F
26F
q= f 100 6C
46F
x
x=x0
Therefore R = r1 – r2
Sol.(69-72) Resitance of wire AB is -
q1=100/36 C
S
q2=50/36C
46F
q= f 506 C
q=2 00/36 C i
1 ] Ddx ZW WR \ ! Y A W W[ 0 X
1 D0# . # 24 % , = = 12U 2% 0 2 - A
>
RAB = /
15 = 1A 12 # 3 when switch is open, null point at C (AC = x)
Current in wire AB is C = Initial and final charges are marked on 4 6f
2
and 26f capacitors as shown. Hence charge passing through segment 1 and 2 are
q !
100
1
6C
q ! 2
3
50
when switch closed null point at D (AD = x)
6C
3
$ charge through switch = q1 + q2 = 50 6C.
q= f 100 6C
46F 1 2
26F
1 D 0 x . 1 x . D 0 x 2 24% 1 , / ,= 2 = 6U = 0 2 - 0 A - 2A 2% =V battery = 6 × 1 =V cSVjh = 6 × 1 RAD = /
8 r =6 r#3 r=1U
8–
50V q=2 00/36 C i
1 D 0 x . 1 x . D0 x 2 24% , / ,= 3 = 8U = 0 2 - 0 A - 2A 2% EMF E = 1 × 8 = 8 V RAC = /
26F
q1=100/36 C
DPP NO. - 52 S
q2=50/36C
q=2 00/36 C i
q= f 506 C
2. (C) For pipe A, second resonant frequency is third
46F
harmonic thus f =
3V 4L A
For pipe B, second resonant frequency is second 3. The equation of pressure variation due to sound is p=
–
B
ds = dx
–
= B ks0 sin (2Bt
B –
d [s sin2 (Bt dx 0
–
kx)]
harmonic thus f =
Equating,
2kx)
2 LB =
2V 2L B
3V 2V 4L A = 2L B 4 4 L = .(1.5) 3 A 3
= 2m.
DPPS FILE # 221
P
5. [ Ans: V AB =
:1 # 35 # 5 ; 2
= 10V ]
The distribution of charges is shown in fig. In closed loop (1)
P–
q C1
–
( q – q1) !0 C2
In closed loop (2)
...(i) q1 C1
–
–
q1 q – q1 # C2 C2 = 0
6. Torque equation KHinge = CHinge Q mg # = 3g = 7#
1 m(4#)2 . 2, / / 12 # m# , Q 0 -
Q
Tangential acceleration = Q# =
3g 7#
Radial acceleration = B2 # = 0
1 2C1 # C2 . / , or q = 0/ C1 -, q1
Ans.
7. mg
–
N1 =
3g 7
1 3g . N1 = m / 7 , 0 4mg 7
N2 = 0 8. Energy conservation
P–
From Eq. (i),
P#
P C2C12 C12 # 3C1C2 # C22
$ R A – RB !
q1 q ! 1 C2 C2
–
P C12 C12
mg # =
6g = 7#
1 2C # C . 1 C # C . q1 ! // 1 2 ,, q1 // 1 2 ,, C2 0 C1 - 0 C1C2 -
$ q1 =
=
–
q q # 1 C2 C2 = 0
1C #C . q P # C12 ! q /0 C1 1C2 2 ,-
or
or
q1 C1
# 3C1C2 # C22
7 1 . m #2 B2 3 2
B DPP NO. - 53
1. Moment of inertia of semicircular portion s about x and y axes are same. But moment of inertia of straight portions about x-axis is zero.
$ Cx < Cy
C
x or C ^ 1 y
2. As voltage applied across capacitor is same i.e. 10V in both case. Therefore in both case Ed = 10 2 E =
P =
C C2 1# 3 2 # 2 C1 C12
10 , as d is constant . Therefore d
electric field remians the same as
10 V/m
3. after collision
P = 1 # 35 # 52
= 10V
1 C2 . //! ! 5 ! 2 ,, 0 C1 By momentum conservation in horizontal direction DPPS FILE # 222
V = V1 + V2 and V2
e=
V1
–
V
.............(i)
P= 1 ! 2
By (i) and (ii)
1 3V . , 0 4 -
P=
and loss in K.E. 3 mV 2 16
2
4. Ans. 3000 Time period is minimum for the satellites with minimum radius of the orbit i.e. equal to the radius of the planet. Therefore.
R
2
!
1
& S 2 are closed,
Pp 2r # Pp P = r # 2r . # . 2 = 3
=m /
GMm
............... (2)
Alternate Solution : If in second case both S
3V 4
So impulse on B
=
1 2. . /R. , 0 3-
From (1) & (2) R = r. Ans.
.............(ii)
V2=
Pp R#r
Pp
2rx 2 . . 2rx 2r # x 3 r# 2r # x
Pp 3
4P p x
=
2r # x # 2 x
.
Sol.(57 to 59)
mv 2 R Final potential of spheres will be same
GM R
2V =
T
= min
=
So, K
K ( 9Q 9 x 9 y ) x Ky = = 3R R 2R
y = 2x and 3x =
2%R
Q–x
–
–
y
$ 6x = –Q
GM R
x= 9
2% R R
Q 6
y= 9
Q 3
Charge on sphere of radius 3R is 9
GM
Q 2
Change in potential energy of sphere of radius ‘R’ is 4 using M = 3
T min. =
p R 3.
D
3% GD
Using values T
min
Pp Pp $ P = 3r . r = 3 Pp
35KQ 2 . 72R
1. At highest point
............. (1)
when S 2 is closed (Let resistance of w R#r
=U =
DPP NO. - 54
P p = 3P pr 2r # r
i=
KQ 2 K( 9Q / 6)2 9 2R 2R
= 3000 s
5. When S 1 is closed i=
=U =
2
be R)
mu =
2m v– 3
mu +
2u 2 = v 3 3
2 v=
2u
m 3
5u 2
total horizontal distance =
5R R 7R + = 2 2 2 4
DPPS FILE # 223
V v 2. (i)
Vcm smooth
B
8. Ex =
–
Ez =
–
=4
3
E 2x # E 2z = 64 = 8 N/kg
|E | =
Ans. (A)
Vcm + BR = V Vcm = V – BR
B depends on value of friction between plank & cylinder, hence Vcm is undetermined. 2v V (ii) B = 2R = R Vcm = 0
2V V (iii) B = 2R = R
3V – V 2V = R R
(iv) BA/C =
2 B=
2V R
3. ! wheat stone bridge is in balanced condition 100 x = 100 # x
100
So
#1
DPP NO. - 55
1. Ei =
#2
KQ 2 KQ 2 KQ # # .Q 2a 4a 2a =
KQ 2 KQ 2 # a 4a
=
5 KQ 2 4a
100U 100 U x
Ef =
E #1
#1
!
#2
#2
=2
–
K (2Q )2 KQ 2 = 4a a
E= H =
KQ 2
i f 4a 2. In potentiometer wire potential difference is directly proportional to length
2 x = 100 U 5. As gravitational force provides centripetal force GMm mv 2 = r3 r i.e.,
v2 =
So that T =
GM r2 2%r r2 = 2%r v GM
$
T Qr
Ans.
4
2
4
1 × 1× (2)2 = – 3 2 VA = – 5 Ans. (C)
6. VA +
7.
–
GM R
Ans. (C)
Let potential drop unit length a potentiometer wire be K. For zero deflection the current will flow independently in two circles C R = K × 10 .... (1) C R + C X = K × 30 .... (2) (2) – (1) 2 C X = k × 20 .... (3) (1)/(2) =
R 1 ! X 2
DPPS FILE # 224
3. Let I be the charge density of conducting plate and V be the volume of either dielectric
$
11 k / 1 H0 U1 02 = U2 11 k H / 2 0 02
. 2. E2 , V E12 , V
2
1 I . ,, - = k2 2 k1 . / I , 0/ k 2 H0 -,
/ k 1 /0 k 1 H0 = k2 1
4. 2 PE = –4 MJ TE = –2MJ The additional energy required to make the satellite escape = +2MJ. 5. Applying conservation of m ech. energy.
6. (A)
v
5 m/s
A
from linear momentum conservation M AV = m b5 2 v =
4?5 = 0.5 m/s Ans. 40
7. (C) m A 0.5 + m b 5 = (M A + m b) V 1 V1 =
40 10 40 ? 0.5 # 4 ? 5 = = m/s Ans. 44 11 44
8. (C) after throught the ball velocity of man A is 0.5 m/s For man B 4 × 5 = 40 v 2 – 4 × 5 2 v2 = 1 m/s velocity B is 1 m/s after through the ball after through the ball second time, velocity of man A is 4 × 5 + 40 × 0.5 = 40 × v 3 – 4 × 5 v3 = 1.5 m/s similarly for man B v 4 = 2 m/s after 5 round trip and man A h old the ball velocity of man B is 5 m/s velocity of man A 4.5 × 40 + 4 × 5 = (40 + 4)v 5 v5 =
50 m/s 11
Ans.
9. (A) When man through the ball 6 times it velocity is greater than 5 m/s and velocity of B is 5 m/s therefor maximum number of times man A can through the ball is 6 . decrease in P.E. = increase in rotational K.E.
1 3R . 1 2 (2m) g 2 / 4 , = _C system ` B2 0 2 1 3 mgR = [C + Cmass ] B2 2 disc.
1 + mR 2 mR 2 25mR 2 ( 2 = 2 ) 4 # 16 # 16 & B * '
2
16 g = 5 R
44
B
40
(A+ball) X cm =
3 mgR
3g 1 + 30 ( 2 ! ) &B R 2 * 16 '
10. Fext = 0 , Centre of mass of system cannot move Initial position of centre of mass from A.
d
B
40 d 10 = d 44 # 40 21
DPP NO. - 56 1. Potential drop. @ V = i (R + R A) V = R + R A = R measured . i
15 . Vparticle = / R , B ! 5gR 04 -
DPPS FILE # 225
4. When charge on plate is constant electric field
3. Switches open :
Q 2AP 0
remains constant E =
R1
A
In case when potential difference is constant E = V d
R R2
Electric field increases when 'd' decreases and hence chances of breakdown increases. 5. C = k H0 A /d Af = 4A i df = 2 d i Because, AL L linear dimensions are doubled capacitance become doubled.
R3 E
E
C = R1 # R 2 # R 3 A
so
Switches closed : There will be no current through R Current through E and R 2
6. x = 3 sin 100 t + 8 cos2 50 t = 3 sin 100 t +
CL =
8 [1 # cos 100 t] 2
x = 4 + 3 sin 100 t + 4 cos 100 t
Amplitude = 5 units Maximum displacement = 9 units.
R CL RE CL A = R # R = (R # R ) R # R R 1 1 2 1
=
7. C = k H0 A /d formula suggest that it depends on area, separation and surrounding medium. 9. Most Approp riate capacitor is a capacitors of high capacitance & high dielectric strength. By dielectric strength C>B>A>D By capacitance C =
kP 0 A
E R1 # R 2 #
As CA = CL A R=
R1 R 2 R3
? 100 =
100 ? 300 = 600 50
U
t
DPP NO. - 57
T
=
4 . 4 t ime s
X = X0 +
=T
R 2R 1 R
A1 = A3 = 8 (area) A2 = A4 = 9 Position of the particle at any time t is given by
d
C=D>B>A So C is best.
2.
E R R1 R # R1
R2 #
Current through the ammeter 4Z ] \tan R ! Y 3X [
(x – 4) = 5 sin (100t + R)
3.
> Vdt 0
X0 = Initial position
1 =# ? 100 is not valid as =# is not 2 #
small.
T1 ! 2%
#
g
T2 ! 2%
2# g
% change =
T2 9 T1 × T1
100 = ( 2 – 1) ? 100 = 41.04. DPPS FILE # 226
t
>
Vdt = Area under the curve
0
Nowat t = 0 X =X 0 = – 1 at t=2 X=X 0 + A1 = – 1 + 8 = 7 at t=5 X=X 0 + A1 – A2 = – 1 + 8 – 9 =–2 t=7 X=X 0 + A1 – A2 + A3 = – 1 + 8 – 9 + 8 =6 t = 10 X = X 0 + A1 – A2 + A3 – A4 = – 1 + 8 – 9 + 8 – 7=3 As during 10 seconds four times the position of the particle changed in sign. Particles passes 4 times the srcin
4. 5 and Y are properties of material. These coefficients are independent of geometry of body. 5. Relative to liquid, the velocity of sphere is 2v 0 upwards. $ viscous force on sphere = 6 % 5 r 2v0 downward = 12 % 5 r v0 downward
6.
J
5. Once the switch is closed, the capacitor is charged through resistance R1 by the battery's e.m.f. Time constant is R1 C. 6. Using Vc = E (1 – e 2 Vc = 12 (1 – e 2) = 10.4 V
–
t/RC
)
J
–
7. At any moment in the circuit
–
Vc + VR1 = 12 V
C1 =
P0 A / 2
, C2 =
P0 A / 2 d/ 2 d # k 2
13 P 0 A 10 d
Ans.
d
2 VR1 = 12 V – 10.4 V = 1.6 V
= C1 + C2 =
8. If loop law is applied to the left hand loop in clockwise direction E – VR 2 = 0
i.e. VR2 does not change during the charging process.
DPP NO. - 58
13 H0 A 10 d
8. Potential difference across galvanometer = Potential difference across S. 2 i g . G = ( C – i g) . S 2 10 × 10 3 a 10 = (1 – 10 × 10 3) a S
2 RS =
3. Speed of block is m aximum at m ean position. A t mean position upper spring is extended and lower spring is compressed.
10 91 1 9 10 9 2
–
=
10 99
U
DPP NO. - 59
D
2. In(A) x f – x i 0 – x = – x = – ve So average velocity is – ve.
1 1000
of total current, the S << G.
–
1. By symmetry CAB = CBC & CAD = CDC $ No current in BO and OD $ T =T =T O
4 P0 A C 5d
7. The current through the galvanometer is ~
VR 2 = E = 12 V
B
=
3. (A) Moment of inertia of the rod w.r.t. the axis through centre of the disc is : (by parallel axis theorum).
C=
mL2 # mR 2 12
DPPS FILE # 227
DPP NO. - 60 2. The velocity of the body a time t is given by
b!
$ At &
t=0,
K.E. of rod w.r.t. disc
!
+ 2 1 2 = 2 mB )R *)
#
L2 ( & 12 '&
5. Since, Heat energy radiated per se c = Ae I T 4 where, e is emissivity of the surface which depends upon its nature and A is its area. T is its own temperature (independent of surrounding temperature) Hence, (A) and (C) are correct. 8. Let V A = 0 volts. ! Net current entering node C = 0 0 9 Vc 2
$ VC = Also
#
92 9 Vc 6 .5
#
1 9 Vc 2
!0
1 volt. = p.d across wire AC. 6
VC 9 VB VC # 2 1 VB # 2 ! = = 5 1 .5 6 .5 3
$ VC – VB =
5 = p.d. across wire BC 3
$ VC > V B > V A Hence potential gradient across BC = 2V
A
-2v
VC C
OV
1.5v
5
VB
5 / 3 10 ! V/m 1/ 2 3
also potential gradient across 1/ 6 5 ! V/m 1/ 5 6
!
= 1ms -1, Now,,
1 2
m!
2
m!
2
9
1 2
mu
2
!
1 2
m!
2
90
1 2
1
! ? 6 ? (1) 2 2
= 3J,
Hence the correct choice is (d). 3. From conservation of energy, the kinetic energy of ball at lowest portion is (v c = speed of centre of ball) 1 2 1 2 mv c2 + ? mv c = mgR 2 5 2
or
7 mv 2c = mgR 10
Since net tangential force on sphere at lowest point is zero, net force on sphere at lowest position is =
mv c2 R
!
10 mg upwards. 7
R . RV 6. R A = R # R < R V 7. R B = R + R G > R 8. % error in case A.
1V 1v
AC
= u = 0 and t = 2 s,
Ans.
!
$
. t ,! , 2 -
work done = increase in KE
1 2 CB 2
=
!
dx d 1 t2 ! / dt dt /0 4
RA 9 R × 100 = R
1 RV . // 9 1,, × 100 0 R # RV -
9R = R # R × 100 c - 1% V % error in case B
R RB 9 R × 100 = G × 100 c 10% R R Hence percentage error in circuit B is m that in A.
ore than
DPPS FILE # 228
DPP NO. - 61 2. C1B1 = C2B2 Since, men move towards middle of turn table C2 decreases hence B2 increases.
$ =K =
=
=
+ B2 ( 1 C 1B12 )1 9 B & < 0 2 1' *
2
1 k(x 02 – x2) = 6mgx 2
2
1 1 × 200(22 – x2) = × 60 × 10x 2 2
2 x = 1m
1 1 C B 2 – C2B22 2 1 1 2
+ C 2 B2 2 ( 1 C1B12 )1 9 C . 2 & & 1 B1 ' 2 *)
5. By W net = =K.E = 0
Also at this moment fmax > kx
] B2 Z \ O 1Y [ B1 X
So kinetic energy increases.
So, block will not move so total distance travelled = 2 + 1 = 3m.
6. In steady state, before the switch S is closed, potential difference across capacitor is 40 volts. Just after switch S is closed, charge and hence potential difference across the capacitor does not change appreciably. So, the potential difference acr oss R 2 is 40 –10 = 30 volt. The curent through R 2 is 3 ampere.
For steady state
3. (A,B)
7. The current through resi stors when t he capacitor is in steady state with switch S closed.
1 dQ . 1 dQ . , / , = / 0 dt -in 0 dt - out
40 9 10 I = R # R ! 1amp . Therefore potential difference
(V) (i55) = 45(T – 20)
1
(500) (4.5) = 45(T – 20) T55 = 70ºC. Resistance at 20ºC is R = v = 500 5 i
(C,D)
2
across R 2 is 10 × 1 = 10 volts. Hence the potential difference across th e capacitor i s 10 + I R 2= 20 volts. So, the charge on capacitor q = CV = 200 6C.
R20 = 100 U Resistance at 70ºC is R =
500 v = 11 U ~ 111 4 .5 – i
8. At the given instant, p.d across capacitor is 20 Volts. Hence the current through R 1 at the required instant of time is I =
40 9 20 ! 1amp R1
Rf = R0(1 + 4=T) 111 = 100(1 + 4(50))
DPP NO. - 62
4=
0.11 ~ 2.2 × 10 3 /ºC. 50 – –
4. For painter ; R+T
–
............(1)
T2
i dx =
–
2T (m + M)g = (m + M)a 2T = (m + M) (g + a) ..............(2) where ; m = 100 kg
–
A4
2
150 ? 15 2
=
1125N
2 i=
T dT
>
0
T1
2 i # = – A4
M = 50 kg
and ;
#
>
For the system ;
$ T=
dT dx
idx = – kA dT
mg = ma
R + T = m(g + a)
a = 5 m/sec
1. Heat current : i = – k A
(T22 9 T12 ) 2
A 4 (T12 9 T22 ) 2#
R = 375 N DPPS FILE # 229
2
2. i =
2
2
10 # R 2
10 ? x= 10 # R 10
VAB =
2 10 # R
× 10
8. By mechanical energy conservation for m, just after collision 1 mv 2 = mgL sin G 2
2 xmax = 0.2 V/m.
m L
"
"
"
6 0 2v ? r 3. B ! 4% r 3 B!
3m
6 0 qv sin G 4% r 2
= 10–7 ×
2 ? 100 ? sin 30" (2)2
= 10–7 ×
1 2 ? 100 ? 2 22
= 25 = 2.5
=M
CM L0
2
2
2 L = 2gvsin G ! 24 ? 5 = 36 meter.. 20 ? 4 Alternate : since there is no energy loss, center of mass of and M rises to the same initial position. 3mL 0 = m(L – L0) 2 4mL 0 = mL 2 L = 4L 0 = 36 meter.
× 10–7 T
m
× 10–6 T
DPP NO. - 63
= 2.5 6T
1. (B) Initially effective resistance = 2R. In parallel effective resistance = 5.
R . It has reduced by a factor 2
of 1/4 so rate of heat transfer would be increased by a factor of 4, keeping other parameters
same.
V0 = 5 × 2t = 10 t S = 1500 =
1 1 V0 .3t = 10t.3t 2 2
2 t = 10 sec. $ total time = 3t = 30 sec. 6. Just before collision velocity of M and m=
2gL 0 sin G = 12 m/s
Since collision is elastic, let velocity of m just after collision is v then by relative velocity of separ ation = relative velocity of approach v = 12 + 12 = 24 m/s Ans.
4.
F1L F1 # F2 F1
–
F2L L = F1 # F2 4 5
2 F = K 3 4 [ Ans.: 3: 5 ] 5. (A) Distance travelled = 24t0 + 21 t0
7. By momentum conservation during collision of m and M. 12 M – 12 m = 24 m m:M=1:3
DPPS FILE # 230
= 45 t0
2. (B) Point A shall record zero magnetic field (due to 4particle) when the 4-particle is at position P and Q as shown in figure. The time taken by4-particle to go from P to Q is
2 45t0 = 180 m 2 t0 = 4 seconds $ Distance translted by A is 24t0 = 24 × 4 = 96 m Sol(57,58,59) (57). Initial velocity of com u=
20 m # 0 m # m = 10 m/s d
acceleration of com 2
= g = 10 m/s e initial height = 10 m S = ut +
1 2 at 2
t=
1 (10) t2 2
10 = – 10t +
1 2% 3 B
or B =
3. Angular velocity
w
=
5t – 10 t – 10 = 0 t2 – 2t – 2 = 0 2
t=
(59).
3
u2 Hmax. = 10 + = 10 + 5 = 15 m 2g
=t =
2 ? 20 a
2 ? 20 =4 g
–
–
4. Potential difference across wire AB = 5 V $ p.d. across 40 cm of this wire =
dQ KA =T ! dt 2#
New rate
=
) dQ dt
=T 2# KA
=
2 = 2sec.
× 40 = 2 volt.
$ p.d. across wire CD =
2 20
of wire
× 80 = 8 volt.
p.d. across 2 U resistor = 2 × 2 = 4 volt $ Emf of the cell = 12 volt.
10 J /sec. 120
2
2
1
> v ds = > ds dt ds = 2t = 8 × 4 + 2 × 10 × 2
5.
=T
=
5 100
$ Potential difference across 20 cm CD = 2 volt.
DPP NO. - 64
1.
20 9 10 = 20 rad/sec. 0 .5
2# 4#8 2
t=1+ (58).
2% 3t
t = 21 s t – 20 = 1 s Ans.
#
2 KA So l . =
40 J /sec. ; 120
(1to3) mA × 0.8 = mA × 0.2 + mB × 1.0 mA × 0.6 mB × 1.0 mB = 0.6 m A – e = 1 0 .2 = 1 0 .8
=1.5
Cd = 6 × 0.5 – 6 × 0 = 3N – 5 = 10 × {0.8 – 0.5} = 10 × 0.3 = 3 NS So time taken is t =
20 × 120 sec. 40
= 60 sec.
=U =
1 1 × 10 × (0.8)2 – × 10 × (0.5)2 2 2
= 5 × 0.64 8 × 0.25 = 3.2 – 2.0 = 1.2 J DPPS FILE # 231
DPP NO. - 65 5.
1=
/ a , ma 2 ** + m -12 . 2+
6. mgh =
=
2
=
7ma2 = 7. 12
1 1 1 2 mv 2 + mv 2 + . mv 2 2 2 2 5
/ 2v , - * . r +
2
8 18 9 mv 2 1 1 mv 2 [1 + 1 + ] = mv 2 = 5 5 2 2 5
1.
& v=
5 gh 9
7. KE of the ball = M (x + L) = 0 3
Mx +
4M x= 3
–
x=
L 4
8. X = 2vt = 2v
1 qr E = 4" $0 r 3 !
and
= 2.
2
=
13 18
!
!
% B = #0 $0 ( v ! E) =
c
2h g
5 2h 2 10 gh h = 9 g 3
DPP NO. - 66
v !E
!
!
!
–
!
!
#0 v ! r B = 4" q r 3 !
/ 2v , - * . r +
mgh
ML 3 !
2.
1 1 2 mv 2 + mv 2 2 2 5
2
1. Initially the centre of mass is at L
50 1 3. v = V = 100 e 2
4 distance from the vertical rod.
2GM R
Applying energy conservation
& )
GMm 1 ( mv2 = R 2
v2 =
2 GM R
&
/1 1 2GM . = 2GM .R 4 R 1
)
)
GMm (R ( h)
1 / , - As, x 0 m ( 2 ) ( m (0) 0 L * cm m(m 4* . +
2 GM R(h
)
1 , * R(h+
h
& 4R 0 R(R ( h) h = 4h & & R 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.
centre of mass does not move in x-direction as
2F x = 0. After they lie on the floor, the pin jo int should be at L/4 distance from the srcin shown inorder to ke ep the centre of mass at rest.
% Finally x-displacement of the pin is
L and 4
y-displacement of the pin is obviously L. Hence net displacement =
L2
(
L2 16
0
17 L 4
DPPS FILE # 232
2. H =
–
dT dx
kA
&
Now as k increases,
dT )H = dx kA
dT becomes less ( dx
So slope becomes less (
–
–
)ve
6, 7 & 8. Let 5 be the angular acceleration of rod and a be acceleration of block just after its r elease. % mg – T = ma ..... (1) T"
)ve
So curve will be
–
"
mg
2
and a =
=
m"2 3
5
.... (2)
"5
.... (3)
3. Since !
!
#0 v ! r ! ! B = 4" q r 3 , v ! r must be same !
!
where v = velocity of charge with respect to observer Let A and B are the observers !
then or
(v C !
(v A
!
!
!
!
!
) v A ) ! r = (v C ) vB ) ! r
!
5 mg a nd 8
5=
3g 8"
Now from free body diagram of rod, let R be the reaction by hinge on rod
!
R+T
!
!
Solving we get
) vB ) ! r 0 0
(v A
T=
!
) v B ) || r 4. 1131 = 1131 (Angular momentum i s conserved) As 12 decreases. 32 increases. or
Solving we get
R=
–
mg = m a cm = m
1 5 2
9 mg 16
Thus T = 2" i.e. T decreases.
3
Therefore the earth is completing each circle around its own axis in lesser time. K.E. =
1 2
Therefore K.E. of rotation increases. Duration of the year is dependent upon time tak en to complete one revoluti on around the sun. 5. Using
DPP NO. - 67
1 32
4 axis theorem
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 axis AB mass distribution is farthest .Hence A
3
x B
"x
= "y
21x = 1.6 1x = .8 Ma 2 1AB = 1x + M(2a) 2 = 4.8 Ma 2 Ans.: 4.8 Ma 2
5
cm y
C
IBC< IAC< IAB & I P > IB > IH I C = ICM + my2 = IB1 – mx 2 + my2 = IB1 + m (y 2–x2) = IP + IB + m (y 2 – x2) > IP + I B > IP Here IB1 is moment of inertia of the plate about an axis perpendicular to it and passing through B. % IC > I P > I B > I H DPPS FILE # 233
2. (C) Taking the srcin at the centre of
the plank.
Qv 2"R
1= 40 kg
The magnetic field at point P is
60 kg
A
B
B= 40 kg
smooth
60 cm
2"1R 2
µ0
4"
(x 2
=
( R 2 )3 / 2
QvR
µ0
4 " (2R 2 )3 / 2
1 E c2 = µ $ v = B 0 0 v
%
5. The orbital velocity, GM r
x
v0 =
Its velocity is increased by m 16 x 1 + m 2 6 x 2 + m 36 x 3 = 0 ( # 6x CM = 0) (Assuming the centres of the two m en 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 t he righ t end of the p lank. 3. The slope of temperature variation is more in inner dQ KA .6T = dt "
6T =
0
2GM = escape velocity r
The path is parabolic in case of escape velocity. 8.
6L =
7 8 dt
=
7 x Fdt
= x6P
1 2
& m "2 3 = xmv &
3 12x 0 2 v
3 v
"
3
&
v
0
from equation 1 and 2 : x cannot be larger than
""
2
6d
9
"
2
.... (1)
revolution by th e
2" .... (2) d
Larger the conductivity, smaller is the slope. 4. Let Q be the charge on the ring. The electric field at point P is
x is
is obtained from the given
information that rod makes one time centre reaches the dot. 3t = 2 " and vt = d
1 K
(because
essentially constant during the quick blow) since, the rod starts at rest, the final values therefore satisfy L = xP.
Another expression for
" . dQ KA dt
Slope 5
GM r
2
v=
2 times, new velocity
&
"" 3
12x "
2
0
2" d
%x=
"" 2 6d
"
2
9d
DPP NO. - 68
/ D, -1 ( * f + = 1. MP = . Qx
1
% E = 4" $ 0
(x2
( R 2 )3 / 2
QR 1 = 4 " $0 (2R 2 )3 / 2
The rotating charged (Q) ring is equivalent to a ring in which current 1 flows, such that
/ 25 , -1 ( * 5 + =6 .
2. Heat radiated (at temp same temp) : A & Q : 4"R2 and Q' : (4"R2 + 2 × "R2)
&
Q' Q
0
6"R 2 4"R 2
= 1.5
Here "R is extra surface area of plane surface of one of the hemisphere. 2
DPPS FILE # 234
7. If m is pole strength , then m =m 0
M l
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,
3.
!
Megnetic moment M = !
% !8 = M
!
× B =
M' 0 m ! 2r
!
" r2 i ˆi & B = 3 i ( 4 j
" r 2 (3 ˆi
ˆ
–
ˆ
4 ˆi )
!
along the direction shown . 8 will be Hence , the point about which the loop up will be : (3, 4)
will be lift
4.
Radius of Curvature =
=
( 2v ) 2 v2 /R
( velocity ) 2 Normal Accelerati on
l
!
2l
"
0
2M
"
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 statement 1.
V0 2 By mechanical energy conservation for (bob + string + cart + earth)
E1 , r1
v=
1 1 mV02 + 0 + 0 = (2m)v2 + mgh + 0 2 2 2 1 1 V mV02 – (2m) 0 = mgh 2 2 4 Solving it,
V02 . 4g
When twothe drops of radius r each form a big drop, radius of big drop willcombine be giventoby 4 4" 2 r "R 3 = 3 3
(
or R3 = 2r3
or R = 21/ 3 r
2 1 /R, 0 - * = 23 0 43 .r
% VR= 5 × 4 1/3 cm/s
DPP NO. - 69 3. Equal area means equal power output. A 3 area pertains to highest wavelength range, thus photons with minimum range of fr equency. Thus maximum number of photons are required from this segment to keep the power same.
5. The torque of s ystem = T orque on loop [AFGH + BCPE + ABEF] = 1 SB ( ) ˆi ) + 1 SB ( ˆi )+ 1 SB kˆ ( I = current, S = area of loop, B = m agnetic field.
4" 3 r 3
2
E2, r2
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 a long upward because tendency of slipping is downwards.
6.
VR Vr
M
= 4R
5. By linear momentum conservation in horizontal direction = for (bob + string + cart) mV0 = (m + m)v
h=
0
Now
=
1 S B kˆ
= 1 × 1 × 2 kˆ = 2 kˆ units [Ans: 2 Kˆ ]
DPPS FILE # 235
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 MB (ii) Here <1 = 0° and <2 = 90° % W = M B (cos0° – cos 90°) = M B = [1– 0)] = MB
A
B R
R
1
2
E
R
E
3
Fig-1
7. If velocity of m2 is zero then by momentum conservation m1v= = m2 v m2 v v= = m 1 Now kinetic energy of m 1 1 1 = m1 v=2 = m 2 2 1
=
1 2
Therefore statement 1 is true. Statement 2 is obviously false.
2
/ m2 , 2 -- ** v . m1 +
/ , / , -- m 2 ** m v2 = -- m 2 ** 1 m v2 = m2 m1 . m1 + 2 . m1 + 2 2
DPP NO. - 70 × initial
1. In normal adjustment
Kinetic energy Kinetic energy of m1 > initial mechanical energy of system m1 Hence proved
f0 m= f e 100 so 50 =
/ 1 , * = 30° 8. C = sin . 2 / 1+ –
(# eyepiece is concave lens) and L = f 0 – fe = 100 – 2 = 98 cm
for i=37 T 1R so, > = " – 2 (37°) = 104° i = 25, Refraction
><
" 2
–
C
i = 45°, T1R so,
/ ", - * = 90° .4+
By applying snells law for prism : i = 90 r1 = 30 r2 = 30 e = 45 > = 90 + 45
2.
?m T = const. " n ? m + " nT = C d? m
> ="–2
fe
& fe = 2 cm
1
?m Now
(
dT T
00
d? m
?m =
–
d? m dT % ? = ) T m 1% =
)
1 ( –ve sign indicates 100
decrease) dT = 1 (given) % T = 100 K. –
60 = 75°
3. Emissive power = @ T4 = 6 × 10 8 × 100 4 W/m 2 –
9. The points Aand B are at same potential, then under given conditions points A and B on the circuit can be connected by a conducting wire. Hence the circuit can be redrawn as shown in figure 2.
DPPS FILE # 236
7. w.r.t. the wedge 4.
to reach < = 270 º, it has to cross the potential energy barrier at < = 180º and to cross < = 180º angular velocity at < = 180º should be 0+ ki + U i = k f + U y 1/3 , - MR 2 * 32 + ( –Mi AB cos 0º) = 0 + ( 2.2 +
cos180º)
3 = 80 A 9 rad/sec.
–
NiAB
As maximum height = 125 m & block want by a height 20m over the wedge & (v sin53º)2 = 2.g.20 v2
5. Equation for linear motion mgsin < – f = ma for rotary motion
f. R =
1.
a R
& f=
a
mgsin< = ma +
a=
2g sin < g = 3 3
using S = ut +
1=0+
6. Here Now =
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 = 105m.
.
MR2 2.R 2
.a =
3 ma 2
1 2 at for linear motion. 2
1 g 2 . .t 2 3
6 g sec.
t=
1 R2
16 = 400 25
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
> = sin
1
–
1
–
G nd D E n sin iB F r C
G nd D E sin iB F nr C
–
i
it is non-linear function and graph is
Ans.
BH = 0.22 T ; BV = 0.38 T B
0 BH 2 ( B V 2
(0.22)2
After i > C for D to R
T. 1HR. will occur and graph is straight line
( (0.38 )2 0 0.1928 = 0.44 T DPPS FILE # 237
3. Due to the motion of the loop, there will be an induced curre nt flowing in the circuit, resulting in a force acting on each element of the loop equally & radially. Therefore the net force on the loop is zero. Hence (D). Similarly for light going from R to D 5. Decrease in PE =
nr = i – sin–1 n sin i d
Increase in rotation K.E
and for prism graph is drawn from i = i min to i = e
that is graph(s)
9. Statement-2 is wrong as in this case A
=
B E E
1
2
2mg. 2 – mg. 2 = 2 . 2 2m m. 2 4 4 2 1
mg 2
r
= A is at high potential and B is at low potential and there is no current from A to B. It also justifies Statement-1.
1
=
2
4g
3m 2
.
.
4
=
and v = r =
3
[ Ans.: (a) V =
3m 2 8
2
4g
2
3
g / 3 ,
=
=
g 3
4 g / 3 ]
DPP NO. - 71 1. By right hand thumb rule, the field by both the segments are out of the plane i.e.along +ve z-axis .
6.
2. Let us compute the magnetic fi eld due to any one segment :
mg cos
+ T1 =
mv12
for leaving circle T1 = 0 mv 12 = mg B=
=
0i 4(d sin )
0i 4(d sin )
(cos 00
cos(180 ))
1 cos
=
0i 4d
tan
2
Resultan t field will be : B net = 2B =
0i tan 2d 2
k =
0i 2d
cos
...(i)
and by energy conservation 1 1 2 0 + m ( 3g )2 = mv+1mg 2 2 1 2
m (3g ) =
3mg 2
=
1 2
mv 12 + mg (1 + cos)
mg cos 2
( cos )
+ mg + mg cos
DPPS FILE # 238
Y
(by eqation (i)) mg
3
=
2
a
1
8 3
=
9
g cos
=
B
Z
1
v 12
X
a
3
sin =
D
O
1
cos =
ac =
upto
mg cos
2
A
parallel to 'y' axis
upto
= g cos 2.
at = g sin then
ac at
g cos
=
1/ 3
=
g sin
8 /3
=
1 8
=
1 2 2
The magnetic flux must remain constant
1
=
y 2
m = B0ab =
so y = 2 Ans. y = 2
dx = ak Ans. dt 3. (A) Let 'F' be the magnitude of force exerted on the rod due to the collision. Then : F = ma
So l. ( 7 to 9) Applying bernoulli's equation 1
v= 1 2
2
× 2 × V 2 = P 0 + 2 g ×
h 2
v=
+ gh and
F.
=
m 2
4
2gh
h
a =
3
2
.... (1) 1 2 at 2
Using ; S = ut + h g
S = ut + S=
R=v×t
2h
Applying continuity equati on
6 ×
.
12
(about 'O')
× g × t2 =
t=
bx
where x is as shown x = a(1 + kt) or
P0 +
B0 1 kt
2gh =
1 2 at 2
6 s
=
a
A = 2cm 2 5.
mv 2
1 2 t 2
1
= 0 t + 2 t2
and 6 =
S = 2
3gh × A
DPP NO. - 72
1 2 at 2
and = 0 t +
3
1 2 t 2
(from (1)) )
Ans.
= N – mg sin
R 1. BOD = 0 BOB = 0 BAB =
=
0 4 a 2
0 8a
[cos 45 ( ˆi ) cos 45 kˆ ]
( ˆi kˆ )
DPPS FILE # 239
mv 2
N=
R
Velocity of separation after collision = e (v elocity + mg sin
of approach before collision). From centre of m ass frame in a head-on collision,
By energy conservation, mgR sin = mv 2 R
1 2
if u1 and v 1 be velocity of a ball before and after
collision v 1 eu1 . Since, v
mv 2
cm
= 0 from ground
frame, ground frame and centre ofmass frame carry same meaning.
= 2mg sin
DPP NO. - 73
N = 3mg sin Ratio =
mv 2 RN
x=
=
2
1. For disc, from t orque equation (constant)
2
3 mg R – TR = mR .... (1) 2
3
2 . 3
By application of Newton's second law on block we get,
6. Consider a block of ice having volume V and density i. Let the volume of ice submerged in water (of density w) be V
T–mg=ma
....(2)
a= R
where
..... (3)
T
M,R
a T
F=3mg
Since the ice block is in equilibrium
i V g = w Vg or V = i V ... (1) w Let V volume of ice melt in to V of water. Then i V = w V i V or V = ... (2) w from (1) and (2) V = V
mg
4g
solving a =
3
3. (A) From continuity equation, velocity at crosssection (1) is more than that at cross-section (2). Hence ; P 1 < P 2 Hence (A)
Hence when V volume of ice melts it occupies V V volume of water. 4.
N mg
Hence the level of water doesnot change on melting of ice.
N = mr 2 N = mg
.............(i) .............(ii)
From (i) & (ii),
(mr2) = mg 9. The velocity of centre of mass is always zero . At maximum deformation during hea d on coll ision, velocity of each sphere is equal to velocity of centre of mass and hence ze ro. Therefore at maximum deformation K.E. of system is also zero.
g
2 = r = =
5 6
10 0.8 18
rad/s. DPPS FILE # 240
5. By newton's law : mg
DPP NO. - 74
dv
ilB=m
dt
(1)
1. When the key is at position (B) for a long time ; the energy stored in the inductor is :
UB =
By k v l
q
Blv=iR+
c
(2)
mg
1 2
2
L.E.2 E = 2 2R 2 R2
.L.
source is connected.
di i Bl =R + dt dt c dv
dv
2
Li 02 =
This whole energy will be dissipated in the f orm of heat when the inductor is connected t o R 1 and no
differentiate (2) w.r .t. time
Eliminate
1
(3)
by (1) & (3)
dt
m
di
i l B = B R
dt
Hence (A).
c i
2. P = AeT 4 2 = 2 x 10-6 x0.9x 5.6 x 10 -8 x T4
m g B l i B 2 l2 = m R di + m i dt c
i will be maxi mum when
(4)
di = 0. dt
Use this in (4)
m g B l c = i (B 2
i max =
l2
c + m)
1014 0.9 x 5.6
T = 2110 k
3. f =
1 p
1 2
metre
f = 0.5 m this is positiv e so lense is convex lense.
m gB c
Ans.
m B 2 2 c
6. Angle of dip can not be calculated by tangent galvanometer. 2
4.
5
7. K = 2 10 7 2 10 4 10 200 8. i = K tan
T4 =
= 0.005
using values ( = 60°) i = 0.005 ×
3
=
3 200
Condition for pure rolling V – R = V ' ...........(i)
DPPS FILE # 241
Momentum conservation
DPP NO. - 75
m'V' = mV
V' =
1. Rate of heat transfer is same through all walls
mV
...........(ii)
m'
K 1.A.(10) 40 cm
From (i) and (ii)
' m1V
V=
m'–m
Torque of friction about point P is zero Angular
mR
mR
2
K 2 A (20) 20 cm
= K 2 = 4K 3
40 cm
2
=
=
K 2 A ( 40) 10 cm
K1 = 4K2 = 16 K 3 .
K 2 A (20) 20 cm
=
K 2 A ( 40) 10 cm
K
1 mVR
2
Solving this we get
m'– m
' =
3m'– m
.
5. M1 is very large as compared to M 2. Hence for collision between M1 and M2, M1 can be considered equivalent to a wall and M 2 as a small block. T hus the velocity of M 2 will be 2v o after collision with M
1
. Similarly after collision betwee n M 2 and M 3, the velocity of M
4
K 1.A.(10)
momentum will remain c onserved about this point 2
K1
=
3
will be 2(2v o). In sequence, the
velocity of M4 shall be 2(2(2vo)) = 8 vo after collision with M 3. 6. The component of velocity of charged particle along the magnetic field does not change. The component of v elocity of charged par ticle normal to magnetic field only changes in direction but always remains normal to magnetic f ield. Hence angle betwee n velocit y and magnetic f ield remains same.
1 4 = K 2 = 4K 3
K1 = 4K2 = 16 K 3 .
2. For the elastic collision, total kinetic energy is constant. If we consider both the colliding bodies as system, then net work done on the system is zero (By work energy theorem) K1 + K 2 = K ' 1 + K'2 .... (1) K is kinetic energy of object before the collision and K' is kinetic energy of object af ter the collision In general masses of both the objects are different, So speed of each object beco mes different after collision K 1 K' 1 and K 2 K' 2 By work energy theorem W 1 = K' 1 – K 1 .... (2) W 2 = K' 2 – K 2 .... (3) From W(1), 0, and W 2 (3) (2) 0 1 W1 + W 2 = 0 W1 = – W 2 i.e. the work done by the first object on the second object is ex actly the opposite of t he work done by the second ob ject on the fi rst object. 3. A, B, D
7. f =
=
qB 2m
3
=
10 19
3800 2
4. The F.B.D. of cylinder is as shown. In equilibrium T = Mg = tension in string In equilibrium, net force on cylinder is zero Torque is same about any axis. Torque on cylinder about anypoint is zero. A = Mg 2R – mg sin R = 0
10 4
38
8. Pitch = T.V ||
=
=
1 V.B . f |B|
250
38.400 10 4. 3800
T
=
4 10 3
m
f
m
A mgsin
DPPS FILE # 242
M=
m sin
8. If R denotes radius of curvature of curved surface, then from above figure 3 × (2R – 3) = 4 × 4
Hence for only one value of M
2
cylinder can remain in equilibrium. A is true, B is false
or
When the cylinder rolls up the incline, sense of rotation of cylinder about center of mass is clockwise. Hence T > f. C is false. When the cylinder rolls down the incline, sense of rotation of cylinder about center of mass is
25 mm 6
R=
From the formulae of focal length for plano-convex lens
anticlockwise. Hence T < f. D is True.
4 3
5. When S 2 is closed current in inductor
2R-3 4
3 1 R 25 25 1 6 3 1 = 6 3 1 × 3 1
f=
remains, i =
V
1
R
2R
=
V
1
+
=
2R
2 V 1 3
2R
25 12
( 3
1) cm
DPP NO. - 76
Potential difference (V) =
2
=
3
And
L
di dt
Ans. 3
2
=
1.
di dt
3
2
=+
3L
Ans.
6. From ray diagram it is clear that ray emerges out of lens parallel to itself. Hence the angle of deviation caused by the lens is 0°. 21 2 2
i=
R1 R 2
=
R1 R 2
60° r r
Where
60°
3 s inr
sin(i r ) cos r
3
dt
is the net emf in the circuit.
or r = 30°
Since the emergent ray is parallel to initial incident ray, the portion of lens used for refraction can be assumed as slab Hence lateral displacement = t
d
R 2 – R1 V1 – V2 = ( – iR1) – ( – iR2) = R R 1 2
7. From snells law at first interface sin 60 =
=
sin(60 30 ) cos 30
=
3 mm.
2. Mirror formula : 1 v 1 v
1
280 1 20
1 20
1 280
DPPS FILE # 243
1 v
electric force is opposite to that of magnetic f orce. Direction of electric field between the plates is opposite to that of direction of force on the negative (electron) charge.
14 1 280 280
v=
15 2
v .v om u
v = –
2
280 .15 v = – 15 280
v =
15 15 15
v =
1 m/s 15
CM
6.
L/2 Hinge
Ans.
L/2 Ma
3.
A
10m/s
B
A
v1
CM
v2
B
Mg
m × 10 = mv 1 + mv 2
10 = v1 + v 2 and
1 2
...(i)
× 10 = v 2 – v 1
From the cart's frame W all = KE2 – KE1
...(ii)
From and v1 =
5 2
m /s;
v2 =
15 2
L L Ma 2 + Mg 2 = 0 – 0
m/s
a=g
Distance between the two blocks S = (– v 1 + v2) . t
5 15 × 5 = 25 m 2 2
=
N1
Temperature gradient
Hinge
N2
4. Heat obviously flows from higher temperature to lower temperature in steady state. A is true. 1
cross sec tion area in
7.
Ma
steady state. B is false. Thermal current through each cross section area is same. C is true. Temperature decreases along the length of the rod from higher temperature end to lower temperature end. D is false. 5. A,B,C
Using – e ( V B ) for the region outside the plates, direction of magnetic field can be found. Inside the plates, net force on the electron is zero hence
Mg
Initially rod is at rest So, N1 = Mg N1 = Mg ]
Torque =
ML2 3
L Ma 2
DPPS FILE # 244
3 a= 4
L 2
3
L
4
g=
3. The direction of forces on the two elements taken symmetrical on two sides of the y–axis are shown. Clearly the net force will be on negative x-axis.
2
Ma + N2 = MaCM
Mg + N2 = N2 = –
3 g 4
M
Mg 4
4. For a ball rolling without slipping on a fixed rough surface, no work is done by friction. 17 Mg 4
N12 N22 =
N=
5.
8. Equation of motion for the cart –
Mg 4
+ f = 2Ma
f = 2Ma + f=
p - Implse = Ft = 3mgt
6. Let the angular speed of disc when the balls reach the end be . From conservation of angular momentum
Mg
1
4
2
9Mg
=
4
mR2 0=
1
mR2 +
2
m 2
R2
+
m 2
R2
or
0 3
7. The angular speed of the disc just after the balls leave
DPP NO. - 77 the disc is 1. It is clearly visible from all graphs that as xincreases. Velocity changes sign. Since this is not possible, no graph represents the possible motion. Because of increase in magnetic field with time, electric field is induced in the circular region and represented by lines of forces as shown in figure. The signs of minimum work done by external agent in taking unit positive charge from A to C via path APC, AOC and AQC are
0 3
Let the speed of each ball just after they leave the disc be v. From conservation of energy 1
2.
=
2
1 1 1 m 2 1 mR 2 02 = 2 2 mR 2 + v 2 2 2 2 +
1 m 2 v 2 2
solving we get v=
2R0 3
NOTE : v = ;
( R ) 2
v r2
v = radial velocity of the ball r
8. Workdone by all forces equal Kf – Ki =
1m
2 2
v2 =
mR 2 02 W APC =–ve,
W
= 0,
AOC
W AQC = + ve
9
(C) is the correct choice. DPPS FILE # 245
DPP NO. - 78
5. Since angular velocity is constant, acceleration of centre of mass of disc is zero. Hence the magnitude of acceleration of point S is 2x where is angular speed o f di sc and x is the distance of S from c entre. Therefore the graph is
1.
For the just completing the circular motion, minimum velocity at bottom in vB =
5gR
Energy conservation b/w point A and B 1
MgH + 0 = 0 +
MgH =
H=
mv B2
m (5gR)
FR =
5R
1 2
MR 2 .... (3)
2
3.
1 2
2
Sol. 14 to 16 The free body diagram of plank and disc is Applying Newton's second law F – f = Ma1 .... (1) f = Ma 2 .... (2)
E . dr = –
d dt
and take the sign of f lux according to right hand curl rule get.
E . d r = – (– (– A) – (–A) + (–A)) = –A 4.
a2 =
F = F B
from equation 2 and 3
= AD = R ( ˆi ˆj)
R
2 From constraint a 1 = a2 + R .... (4) a 1 = 3a 2
B = B 0 ( ˆi ˆj kˆ )
Solvi ng we get a 1 =
F = RB 0 (ˆi ˆj) × (ˆi ˆj kˆ ) = RB 0 ˆi
ˆj
kˆ
1 0 1 1 1 1
and
=
F 2MR
If sphere moves by x the plank moves by L + x. The from equation (4) L + x = 3x or
= RB0 ( ˆi ˆj 2kˆ )
3F 4M
x=
L 2
F = RB0 6 Aliter :
B
B0 ( ˆi ˆj kˆ )
= R (ˆi ˆj)
B
=0
Angle = 90° F = B = 3 B0 2 R = 6 B0 R DPPS FILE # 246
Method II
DPP NO. - 79 1. FBD for sphere & block
k
m
m
fr
a1
m
fr
fr
a1 =
m fr
a2 =
a1
x1
force equation for first block;
mg
2
m
23k (x 1 + x 2 ) = –m d x21 dt
g ˆi
a2
Put x 1 = x 2
a1 a 2 2g ˆi arel = 2g. a rel
1 2
d2 x 1
2. The electron ejected with maximum speed v max are stopped by electric field E = 4N/C after travelling a distance d = 1m
x2
mx 1 = mx 2 x 1 = x 2
g ˆi
m
/////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
m
=
m
k
m
a2
mg
=
m
/////////////////////////////////////////////////////////////////////////////////////
2 =
4k × x1 = 0 3m
4k 3m
T = 2
mVmax2 = eE d = 4eV
The energy of incident photon =
+
dt 2
3m 4K
1240 = 6.2 eV 200
From equation of photo electric effect M
5.
1 mv 2 = h – 0 max 2
= ( 4 / 3)R 3 ( 4 / 3 )(R / 2)3
0 = 6.2 – 4 = 2.2 eV. 3.
M1 =
10 In steady state current from battery = = 5A 2
In parallel inductors L1I1 = L2I2 all the time
i1 =
=
L2 3 L1 L 2 i = 3 2 × 5 = 3A
K(2K )
K 2K Time period T = 2
=
;
=
T = 2
1 7
M
2 2 R 2 R M1R2 – M2 M2 5 2 5 2
2
=
57 MR 2 140
3 P0
= mm1mm2 1 2 6.
Here
M , M2 =
2K
where
K eq
7
4. Both the spring are in series
Keq =
8
H 2
m V
2
m 3 2 2K
= 2
3m 4K
V=
P0
2gh =
H = 2
2g
gH
DPPS FILE # 247
3 1 . mv 2 2 2
= 7.
K.E. of plank
2mv 2
8
K.E. of sphere = 3 2 = . 3 mv 4
2.
Ix
= Iy = I z =
2 5
mR2
3. The spring is never compressed. Hence spring shall Let h be height of water column just after putting cylinder,
exert least force on the block when the block is at topmost position.
A H h´ A A 3 2
h´ =
V´ =
8.
3 4
H
3
2gh´ =
2
gH
A H A = h A 3 2
Fleast = kx0 – kA = mg – m 2A = mg – 4
3
2
2gh =
3
gH
DPP NO. - 80 1. Let velocity of c .m. of sphere be v. The velocity of the plank = 2v. 1 2
Kinetic energy of cylinder =
4. KEmax = (5 – ) eV when these electrons are accelerated through 5V, they will reach the anode with maximum energy = (5 – + 5)eV 10 – = 8 = 2eV Ans. Current is less than saturation current because if slowest electron also reached the plateit would have 5eV energy at the anode, but there it is given that the
× m × (2v) 1 2
2
= 2mv 2
5. By principal of energy conservation. P B = P R + PL Near the starting of the circuit
mv 2 P R = i 2R
=
1 2 1 2
mA
minimum energy is 6eV.
Kinetic energy of plank =
+
T2
H
h =
v =
2
1 2 2 + mR 2
mv 2 1
1
2
As
di dt
di andP L = L i dt .
has greater value at the starting of the
circuit, P L > P R
DPPS FILE # 248
Sol. 1 to 3. : When connected with the DC source R=
12 4
=3
When connected to ac source
12
2.4 =
32
Using P = =
2 Vrms
Z
V
=
Z
Net torque on rod about hinge 'O' = 0
L = 0.08 H
2L2
N1 × L = mg ×
rms Vrms cos
cos
or
2 Vrms R = = 24 W 1 2 2 R ( L – ) C
2
mg 2
Net torque on cylinder about its centre C is zero.
f1R = f2 R
or
f1
f2
Net torque on cylinder about hinge O is zero. N2 × L = N1 × L + mgL
DPP NO. - 81 2.
N1
L
The potential difference across theinductor ise = E– iR. Hence the plot of e versus i is a straight line with negative slope.
3. Equation can be written as i = 2 si n 100
or
N2 =
3 mg 2
8. The magnetic force on bob does not produce any restoring torque on bob about the hinge. Hence this force has no effect on time period of oscillation. Therefore both statements are correct and statement2 is the correct explanation.
t + 2 sin
(100 t + 120º) so phase difference = 120º =
A 12
A 22 2A 1A 2 cos
=
4 4 2 2 2 –
1 = 2 so effective value will 2
rms. value = 2 / 2 = 4. Angular momentum =
(mvr) = n.
h 2
=
h 2
2A
nh 2
=
h 2
( n = 1)
(n = 1)
Sol.5 to 7. FBD of rod and cylinder is as shown. DPPS FILE # 249
DPP NO. - 82
8.
1
i dt 3.
=
1
0
sin t 4 dt =
= 2 2
1
0
4
4. The hydrogen atom is in n = 5 state. Max. no of possible photons = 4 To emit photon in ultra violetregion, it must jump to n = 1, because only Lyman series lies in u.v. region. Once it jumps to n = 1 photon, it reaches to its ground state and no more photons can be emitted. So only
B d
along any closed path within a uniform
magnetic field is always zero. Hence the closed path can be chos en of any size, even v ery small size enclosing a very smal l area. Hence we can prove that net current through each area of inf init esimall y sm all size wi thin region of uniform magnetic field is zero. Hence we cansay no current (rather than no net current) flows through region of uniform magnetic field. Hence statement -2 is correct explanation of statement-1.
one photon in u. v. range can be emitted. If H atom emits a photon and then another photon of Balmer series, option D will be correct. 7. The FBD of cy linder is as shown Resolvi ng forces along x and y axis
x y fB= NB NB
NA
fA= NA
NA + f B =
mg
mg
.... (1)
2
f B = NB NB – f A =
mg
.... (2)
2
f A = NA solving we get fA =
Mg(1 ) and 2 (1 2 )
fB =
Mg(1 ) 2 (1 2 )
Angular acceleration =
(fA
fB ) R
MR 2 2
=
2 2 g R(1 2 )
DPPS FILE # 250
