was d when the capacitor was unchanged). The force constant of the spring is approximately
(a) 4 0 2
3
ε AE
d (b) 2 0
2
ε AE d (c) 6 0 2
3
ε E
Ad (d) ε0 3
2 3
AE d
Detailed solutions of these questions are available on http://www.arihantbooks.com/Physics%20Spectrum.pdf
Answers
1. (c) 2. (b) 3. (a) 4. (b) 5. (d) 6. (b) 7. (d) 8. (a) 9. (c) 10. (a)
11. (a) 12. (a) 13. (b) 14. (a) 15. (c) 16. (b) 17. (d) 18. (a) 19. (c) 20. (c)
21. (c) 22. (b) 23. (d) 24. (b) 25. (d) 26. (b) 27. (b) 28. (b) 29. (d) 30. (a)
M B
A
D C
45°
2 cm
O 2cm
220v50v 20kΩ R1 Output Voltage +
− 5kΩ
10cm
v=10m/s V
r
dr r2
B r1
A
× B i 10 A
E
1. (c) Let the heat produced be H, the current through the wire be I, the resistance be R and the time be t. Since heat is a form of energy, its dimensional formula is [ML T2 −2].
Let us assume that the required equation is H=KI R ta b c,
where, K is a dimensional constant.
Writing dimensions of both sides,
[ML T ] 2 −2 =[Ia][ML I T ] [T2 2− −3b c]
= [M L Tb 2b − 3 + b c aI −2b] Equating the exponents, we get
b = 1 2b =2
−3b c+ = −2 a−2b=0
Solving these, we get, a = 2, b = 1 and c = 1 Thus, the required equation is H=KI Rt2
2. (b) Velocity of the centre = v0 and the angular velocity about the centre =v
r
0
2 . Thus v0> ω . The sphere slips forward and thus the0r friction by the plane on the sphere will act backward. As the friction is kinetic, its value is µN=µMg and the sphere will be
This friction will also have torque τ = fr about the centre. This torque is clockwise and in the direction of ω0. Hence, the angular acceleration about the centre will be
α =
and the clockwise angular velocity at time t will be ω( )t ω f Eliminating t from Eqs. (i) and (ii), we get
5
Translational velocity after the sphere starts pure rolling, or v t( ) =2× v = v
7 3 6
0 7 0
3. (a) The field is required at a point outside the sphere. Dividing the sphere in concentric shells, each shell can be replaced by a point particle at its centre having mass equal to the mass of the shell.
Thus, the whole sphere can be replaced by a point particle at its centre having mass equal to the mass of the given sphere. If the mass of the sphere is M, the gravitational field at the given point is
E GM
The mass M may be calculated as follows. Consider a concentric shell of radius r and thickness dr
Its volume is The mass of the whole sphere is
M ardr
Thus, by Eq. (i) by the gravitational field is
E G a
The delivered power output and the power losses in the field coils,
PFC=I RF2 f and the armature, Pa=I Ra2 a
The efficiency of power output/power input where the denominator includes the stray losses.
Thus, Poutput =V It a=120×48=5760 W Efficiency of the generator =
+ 5760+ + ×
5760 240 322 56 520 100 .
= 0 84. = 84%
5. (d) The situation is shown in figure. Suppose the detector is placed at a distance of 4 m from the source. The direct wave received from the sources travels a distance of
x metre. The wave reaching the detector after reflection from the wall has a travelled distance of =2 (2) +
The path difference between the two waves is
∆ = + minimum distance x for a maximum corresponds to
∆ = λ ...(i)
The wavelength is λ
=cν
According to equation (i), we get
2 22 4 2
Answer with Explanations
S
D h
or 4 6. (b) For concave mirror,
Using this relation 1 1 1 v+u=f
and use sign convention, also the magnification in case of concave mirror is given by m v
=−u Distance of image
1 1 1
v+u=f
Distance from focus is nf, so distance from pole nf+f1 from the formula, 1 1 1 Magnification of image
m v
7. (d) A particle accelerate along x-component and y-component, ax=(1.5 m/s )(cos37 )2 °
The velocity makes an angle θ with x-axis where,
tan . 8. (a) In case of charging of a capacitor through a resistance
q q e So, substituting the value of q
q0
However in circuit (b) as capacitor is connected directly to the battery initially it acts like short circuit and hence, it will change instantaneously, i.e. t = 0 s.
9. (c) It is based on the combination of two long charged wires will act as a capacitor. Let us give equal and opposite charges to two wires so that they would have linear charge densities as + λ and − λ
Electric field at point P
E= x+ a x
where, a is radius of wire.
C= θv =πεn
| | η
0
1
10. (a) Consider a small length dx of the rod at a distance x from the fixed end. The part below this small element has length (L− ).x The tension T of the rod at the element equals to the weight of the rod below it.
T L xW
=( − )L
Elongation in the element is given by Elongation = Original length × Stress/Y
=Tdx= −
Electric field at point p P
The total elongation = − 11. (a) The light of wavelength λ is strongly reflected if
2 1
µd=n+2 λ
...(i)
where, n is a non-negative integer.
2µd =2×1 50. ×0 5 10. × −6 m
Thus, within 400 nm to 700 nm, the integer n can have the values 2 and 3. Putting these values of n in (i), the wavelengths becomes
λ µ
= 4+ = 2 d1 600
n nm and 429 nm
Thus, light of wavelengths 429 nm and 600 nm are strongly reflected.
12. (a) It is based on the mirror formula i.e. 1 1 1
v+u= , then the signf convention can be decided accordingly.
Then, take the cases such as 1. u>0,v>0, 2. u < 0, v < 0 3. u > 0 and v < 0
For part (1) u > 0, v > 0, so object is virtual and real image.
For part (2) u < 0, v < 0, then it is real object and virtual image.
For part (3) u > 0, v < 0, so it is virtual object and real image.
13. (b) If pressure near any area A of mercury column is A, then force on the column will be pA. The force on the column will be perpendicular to the surface.
Force on the column are shown in above diagram. For the equilibrium, forces must be balanced.
In first stage equilibrium of mercury column
Similarly in second stage, p A0 +Mg=p A0
14. (a) Suppose initial current is i0 then i t i t
Amount of heat generated in a coil, H q t
15. (c) Let us take the zero of potential energy at the table. Consider a part dx of the chain at a depth x below the surface of the table.
The mass of this part is dx m ldx
= and hence its potential energy is
−
m
l dx gx.
The potential energy of the l
3 of the chain that overhangs is
U m
This is also the potential energy of the full chain in the initial position because the part lying on the table has zero potential energy. The potential energy of the chain when it completely ships off the table is
V m The loss in potential energy = −
16. (b) Figure shows the right angle triangle ABC, such that
AC = ×2 10−2 m and BC =10−2 m. The charges of qA=4 10× −6 C and qB=2×10−6 C are placed at the
vertices A and B respectively.
Let EA and EB be electric intensity at
If θ is angle between the direction of EA and EB, then
Suppose that the resultant electric intensity E makes an angle α with line AC. Then,
These coordinates should satisfy the equation of projectile, i.e.
−h=R − gR +
∴ The maximum range is Rmax=2 a a( +h)
18. (a) The induced emf in the rod causes a current of flow counter clockwise in the circuit because of this current in the rod, it experiences a force to the left due to the magnetic field. In order to the pull the rod to the right with constant speed, this force must be balanced by the puller.
The emf induced in the loop is
| | ( )
Mechanical power supplied to circuit = Rate at which electric work is done on charge
Fv q Equating the buoyant force to the weight of the man plus the weight of the life jacket
ρwaterg 14V Vj ρwater g Vj
15 + 667 0 25
= + ( . )
Volume of the life jacket, V
gV Cm = 4184 J/K Since, the heat capacities of the two masses are equal, the final temperature will be average of the initial temperature. Mass of remaining portion,
M2=π( )32−4=(9π−4) sq cm
∴ M2=(9π−4) ,σr2=? As centre of mass of circular plate is at O,
r = 0 i.e. At 0.233 cm from point 0 towards M.
22. (b) Let T1 and T2 be the time periods of the pendulum with lengths 1.0 m and 1.21 m, respectively.
T
For the two pendulums to swing together, required condition is V1−V2= or V1 1=V2+1
23. (d) For D-region, ν =55 10× 6 Hz
25. (d) Let the wire run in the X-direction and the magnetic field in the Z-direction. In the Hall effect, the Lorentz force, which is due to B2 and the induced electric field Ey vanishes
eEy=eV Bx z ...(i)
Allowing the conduction electrons to move down the wire at drift speed vn.
v I
Ane I
n= =l ne2 ...(ii)
and the Hall voltage is given by
E V
y= lH ...(iii)
Substituting (ii) and (iii) in Eq. (i), we get V IB
H=nel2 Transverse magnetic field BH
= ×
26. (b) The heat energy required is given by
∆H m c T dT
Inserting the given numerical values, we find that
∆H = ×
27. (b) In case of only the source is moving with respect to the medium. Let | |vT denote the speed of the train and ν0 denote the emitted frequency, the observer measure a frequency
νa ν
As the train approaches and a frequency νT ν
As the train recedes from him. Dividing Eq. (i) by Eq. (ii), we find that
Solving this equation for the speed of the train,
| | | |vT v a r .
28. (b) According to right-hand rule, the magnetic field due to the current carrying wire is directed down into the plane of the figure all along the wire segment AB.
The magnitude of the field is B i
=µ r π
0
2 Since v is perpendicular to B,
|v B× |=Bv
Lorentz force to be zero (as required for the equilibrium of a mobile charge carries in the wire segment)
There must be an electric field Eopp= − × v B
Eopp is directed from A towards B along the wire, so that the electric potential is higher at A than at B, VA>VB, specifically
VA VB E r dr
It is an exponentially increasing graph of potential energy | |U with x2. Therefore, U versus x graph will be as shown.
From the graph, it is clear that origin potential energy U is minimum (therefore, kinetic energy will be maximum) and force acting on the particle is also zero because
F dU
= −dx = − (slope of U- x graph) = 0
Therefore, origin is the stable equilibrium position. Hence, particle will oscillate simple harmonically about n = 0 for small
displacement.
30. (a) In equilibrium, electrostatic attraction between the plates = spring force
Force constant of the spring, k AE d