Physics Private Tuitions
COMBINATION – 1 - SOLUTION
UNITS & MEASUREMENTS, SC & VC, M1D, MIP, LOM, WPE, CIRCULAR MOTION,
ROTATIONAL MOTION, GRAVITATION
1. 4) L x B x H = 5 mm x 10 mm x 5 mm
L. C. = 0.01 cm = 0.1 mm
L B H
%V= x 100
L B H
0.1 0.1 0.1
%V= x 100
5 10 5
0.2 0.1
= 5 10 x 100
0.4 0.1
= 10 x 100
% V =- 0.05 x 100 = 5 %
2. 1) F Energy J
V Velocity m/s
T Time s
F N J
S.T. =
L m m2
i.e. From options;
E J J
1 E1V–2T–2
2 2 2
V T m 2 m2
(s)
s
3. 3) Both bodies are dropped simultaneously; (u = 0)
They will fall through same distance, in 2 s & relative distance will remain same between the 2 bodies.
1
4. 4)
uy = 0
1 2
sy = uy + uyt
2
1 2
h= gt
2
2h
This is an example of horizontal projection & we know that for H.P. t = g
5. 4)
VR
VBR
5 1 KM
Distance 1 Km
Speed = VBR
time 1
15x hr
60
VBR = 4 Km / hr
& By Pythagoras,
VR = (5)2 (4)2 25 16
VR = 3 km/hr
6. 3) Here, H1 = H2
u12 sin2 1 u22 sin2 2
2g 2g
u12 sin2 2 sin2 (60)
u22 sin2 1 sin2 (45)
3
= 4 3
1 2
2
7. 4) At maximum height speed is; u cos
u cos = u/2 = 600
u2 sin2 u2 sin2 (60)
H=
2g 2g
2
42 x (3 / 4) 3u2
=
2g 8g
8. 1)
Here; T cos 60 = 40
T = 80 N
V1sin
V1 m
u1= V u2 = 0
m m
9. 1)
V1cos
V2
By LOCOE;
K.Ei = K.Ef m
1 1 1 1
m1u12 m2u22 m1v12 mv22
2 2 2 2
1 1 1
m1v2 0 mv12 mv22
2 2 2
v2 = v12 v22
v22 v2 v12
v2 = v12 – v2
mv2
10. 4) C.P.F. =
r
2
5
2000x 90 x
= 18
=
2000x 625
= 200 x 25
250 250
C.P.F. = 5000 N
3
11. 4)
P2
M
P1
P3
By LOCOM;
Pi = Pf
P1 P2 P3 0
P3 – (P1 P2 )
P3 – (– 3P i 2P j )
|P3 | |P1 P2 |
P3 ( 3p)2 (2P)2
= 13 p
12. 1) m1 = m2 = m, u1 = V, u2 = 0
By LOCOM; P = mv1 = mv
m1u1 + m2u2 = m1v1 + m2v2
P = m (v1 + v2)
P
V1 + v2 =
m
P
V1= – v2
m
Now, B gives an impulse of J to A meaning;
Impulse = PB
J = m (v2 – u2) = mv2
V2 = J/m
J P
( v2 )
v2 v1 m m
Now, e =
u1 u2 u1
J P J 2J P
= m m m= m – m
u1 P P
m m
4
2J
e= –1
p
13. 1) By LOCOM;
m1u1 + m2u2 = m1v1 + m2v2
mu + 0 = 2 mv2
u1 = 2v2 v = 2v2
v v1
e= 2
u1 u2
v2 1
e=
2v2 2
= 0.5
14. 1) We know that
dp
dp x v
K = dv dv
v k
Now change in pressure for both the sphere will be same assuming they are at same depth in the ocean.
1
dv
k
Kbrass < ksteel dvB > dvs
m(v u)
15. 1) F = ma =
t
3(3.5 2)
F=
25
3 x 1.5 3 x 15
F=
25 25 x 10
9
F= = 0.18 in the direction
5 x10
16. 2) A = (2m, 3m, 4m)
B = (1m, 2m, 3m)
Displacement = ( i 2 j 3 k) (2 i 3 j 4 k)
x – i j –k
w = F x
= (2 i 3 j 4 k) ( i j k)
= –2–3–4=–9J
5
17. 4)
18. 3) w uniform
dw
i.e. =0
dt
=I =0
2 cos
19. 4) w= and T = 2
T g
2 g
w= and cos = h
cos cos
2
g
g
w=
h
20. 2)
21. 4)
I2 I1 I3
ML2 ML2
Here I2 = I3 = and I1 =
3 12
I2 = I3 > I1
5
22. 1) T.E. of particle performing vertical C.M. remains constant & T.E. = mgr
2
23. 3) to its plane = 2MR2 & M.I. of disc = g – 2Rcos2 ()
g = g – R cos
2 2
g – g = R2cos2
6
5MR2
24. 1) M.I. ring about tangent to its palne = 2MR2 & M.I. of disc about tangent into the plane =
4
I1 2MR2 8
2
I2 5MR 5
4
2
25. 3) w= = 2n
T
100 10
w = 2 x
60 3
w = 10 x 60 = 600
26. 4)
mv2
27. 2) N at highest point is given by; N = mg
r
If body loses contact N = 0
mv2
mg w2r = g
r
g 10
w
r 3
28. 1) At top of loop;
mv2
N= mg
r
50 x 250 x 250
= – 500
5000
= 600 – 500
N = 125 N
29. 1)
A
D
h
r
C
7
To complete the track BCD, Cart should have the critical speed at B = 5rg
T . EA = r . F.B
1
mgh = mv2
2
5rg 5x3
gh = h= = 7.5 M
2 2
30. 4)
By LOCOE;
B.FH = K.FL
1 2
mg mv
2
v= 2g
and
mv2 m(2g ) mg
TL = mg
r
T = 3 mg
31. 3) T r3
3 3
r 3r
T =
r r
T 2T T 3 3 T
GMm
32. 2) T.E. = –
2(R h)
6.67 x 1011 x 6 x 1024 x 103
=
2(6400 3600)x 103
6.67 x 6 x 1024 x 1011
=
2(10000)
= – 6.67 x 3 x 109
= – 20.01 x 109
T.E. = – 2.001 x 1010 J
8
GM
33. 3) g=
R2
M
Gx
GMp
gp = 2 7
2
Rp R
2
4
gp = g
7
4
g
wp mgp 7
we mg g
4
wp = x 3.5 = 2 kg wt.
7
Wp = 20 N
34. 2)
Here; IA = 1 x 0 = 0, IB = 0
Ic = 3 x (1)2, ID = 4 x (1)2
IT
Radius of Gyration =
MT
7
= m
10
35. 2) K20 K2cm R2
K2cm K20 – R2
402 – 412 – 402
= 168 – 1600
K= 81
K = 9 cm
9
1
36. 3) K.E. = I2
2
1
K.E. = I (22 12 )
2
1
500 = Ix [(2n2 )2 (2n1 )2 ]
2
1 240
2
60
2
1000 = I x 4 2
2 60 60
1000
I x 16 1
4 2
250 50
I= = kg m2
15 2
3 2
0 0 30
37. 1) = 0.05
T 600
I
= 1 x 0.05 = – 0.05
Now;
1
= 30 x 600 – x 0.05 x (600)2
2
= 18000 – 9000
= 9000
work = = 9000 x 0.05
= 450 J
38. 2) Lwax = Lsystem
Mvr = Is
ML2 L
2
Mvr = M
12 r
L 1 1
vx L2
4 12 16
v 28
L
4 12 x 16
7
v = L x
12
12 V
=
7L
10
39. 1) M = 10 kg, = 10 Nm
MR2 10 x (0.2)2
Idisc =
2 2
Idisc = 0.2
10
I = 50
0.2
0 t = 50 x 6 = 300 rad/sec
vi
40. 2) Here, vf =
27
4 4
R13 R13 x 27
3 3
R1 = 3R2
2I 1
L = I = ,L
T T
No external torque applied L is constant
I1 I
2
T1 T2
2 2 2 2
MR1 MR2
5 5
T1 T2
2
1 24 8
T2 = x T1
3 9 3
41. 2) When beads are at the end. The M.I. of system is given by;
2
m 2
I= 2mb
12 2
m 2 mb 2
I=
12 2
By LOCOAM;
I1 1 + I2 2
m 2 m 2 mb 2
x 2
12 12 2
60 x 2 x 40 60 x 2 190 x 2
x 2
12 12 2
200 2 = 100 2 x 2
2 = 2 rad/s
11
= 40 – 2 = 38 rad/sec
42. 2) By LOCOAM;
I1 1 + I2 2
2n1 2n2
I1 = I2
60 60
I1 x 90 = (75% I1) n2
3
90 = n2 n2 = 120 RPM
4
43. 4) Initially angular velocities of two wheels are 1 & 2
Angular momentum are in opposite direction
L = – L1 + L2 = – I 1 + I 2
After coupling; L 2I
By LOCOM; L = L
– I 1 + I 2 = 2I 2 = 2 1
2 x 2 n = 2 (n2 – n1)
240 120
2n = = 60 rpm
2
44. 1) I1 = M.I. disc
I2 = M.I. of Blob
I2 = I1 + MR2
By LOCOAM;
I11 I22
I1 (2n1 ) (I2 MR2 )2n2
I1n1 = M1n2 + MR2n2
I1(n1 – n2) = 1.9 x 10–3 x (25)2 x 10–4 x 2
1.9x 625 x 10 4 x 10 3 x 2
I1 = = 2.375 x 10–7
(3 2)
I1 = 2.375 x 10–4 kg m2
45. 1)
At a given instant of time both front wheel and rear wheel will travel same distance;
distance, v x t Vr = Vf
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