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1)
The displacement of an object attached to a spring and executing simple harmonic motion is given
by
x = 2 X 10-2 cos πt metres. The time at which the maximum speed first occurs is t
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1 )
0.5 s
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2 )
0.75 s
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3 )
0.125 s
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4 )
0.25 s
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| see the answer see the solution |
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2)
In an a.c. circuit the voltage applied is E = E0 sin ωt.
The resulting current in the circuit is
I = I0 sin (ωt - π/2) . The power consumption in the circuit is given by
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1 )
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P =
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E0I0
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2 )
P = zero
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3 )
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4 )
P = E0I0
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| see the answer see the solution |
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3)
An electric charge 10-3 μ C is placed at the
origin (0, 0) of X - Y co-ordinate system.
Two points A and B are situated at
( , )
and (2, 0) respectively.
The potential difference between the points A and B will be
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1 )
9 volt
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2 )
zero
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3 )
2 volt
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4 )
4.5 volt
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| see the answer see the solution |
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4)
A battery is used to charge a parallel plate capacitor till the potential difference
between the plates becomes equal to the electromotive force of the battery.
The ratio of the energy stored in the capacitor and the work done by the battery will be
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1 )
1
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2 )
2
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3 )
1/4
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4 )
1/2
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| see the answer see the solution |
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5)
An ideal coil of 10 H is connected in series with a resistance of 5 Ω
and a battery of 5V.
2 seconds after the connection is made,
the current flowing in amperes in the circuit is
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1 )
( 1 - e )
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2 )
e
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3 )
e-1
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4 )
( 1 - e-1 )
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| see the answer see the solution |
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6)
A long straight wire of radius a carries a steady current i.
The current is uniformly distributed across its cross section.
The ratio of the magnetic field at a/2 and 2a is
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1 )
1/4
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2 )
4
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3 )
1
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4 )
1/2
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| see the answer see the solution |
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7)
A current I flows along the length of an infinitely long, straight, thin walled pipe. Then
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1 )
the magnetic field is zero only on the axis of the pipe
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2 )
the magnetic field is different at different points inside the pipe
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3 )
the magnetic field at any point inside the pipe is zero
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4 )
the magnetic field at all points inside the pipe is the same, but not zero
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| see the answer see the solution |
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8)
If M0 is the mass of an oxygen isotope 8O17 ,
MP and MN are the masses of a proton and a neutron respectively,
the nuclear binding energy of the isotope is
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1 )
(M0 - 8 MP) C2
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2 )
(M0 - 8 MP - 9 MN) C2
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3 )
M0 C2
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4 )
(M0 - 17 MN) C2
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| see the answer see the solution |
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9)
In gamma ray emission from a nucleus
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1 )
both the neutron number and the proton number change
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2 )
there is no change in the proton number and the neutron number
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3 )
only the neutron number changes
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4 )
only the proton number changes
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| see the answer see the solution |
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10)
If in a p-n junction diode, a square input signal of 10V is applied as
shown
Then the output signal across RL will be
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1 )
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2 )
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3 )
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4 )
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| see the answer see the solution |
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11)
Photon of frequency ν has a momentum associated with it. If c is the velocity of light, the
momentum is
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1 )
ν/c
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2 )
h ν c
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3 )
h ν /c2
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4 )
h ν /c
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| see the answer see the solution |
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12)
The velocity of a particle is v = v0 + gt + ft2 .
If its position is x = 0 at t = 0, then its displacement
after unit time (t = 1) is
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1 )
v0 + 2g + 3f
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2 )
v0 + g/2 + f/3
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3 )
v0 + g + f
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4 )
v0 + g/2 + f
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| see the answer see the solution |
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13)
For the given uniform square lamina ABCD, whose centre is O,
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1 )
IAC = IEF
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2 )
IAD = 3 IEF
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3 )
IAC = IEF
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4 )
IAC = IEF
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| see the answer see the solution |
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14)
A point mass oscillates along the x-axis according to the law x = x0 cos (ω t - π/4).
If the acceleration of the particle is written as a = A cos (ω t + δ ). , then
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1 )
A = x0 , δ = - π/4
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2 )
A = x0 ω2 , δ = π/4
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3 )
A = x0 ω2 , δ = - π/4
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4 )
A = x0 ω2 , δ = 3π/4
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| see the answer see the solution |
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15)
Charges are placed on the vertices of a square as shown. Let E be the
electric field and V the potential at the centre. If the charges on A and B
are interchanged with those on D and C respectively, then
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1 )
E remains unchanged, V changes
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2 )
both Both E and V change
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3 )
E and V remains unchanged
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4 )
E changes, V remains unchanged
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| see the answer see the solution |
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16)
The half-life period of a radio-active element X is same as the mean life time of another radioactive
element Y. Initially they have the same number of atoms. Then
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1 )
X will decay faster than Y
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2 )
Y will decay faster than X
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3 )
X and Y have same decay rate initially
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4 )
X and Y decay at same rate always
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| see the answer see the solution |
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17)
A Carnot engine, having an efficiency of η = 1/10 as heat engine, is used as a refrigerator. If the
work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower
temperature is
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1 )
99 J
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2 )
90 J
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3 )
1 J
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4 )
100 J
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| see the answer see the solution |
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18)
Carbon, silicon and germanium have four valence electrons each. At room temperature which
one of the following statements is most appropriate ?
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1 )
The number of free conduction electrons is significant in C but small in Si and Ge
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2 )
The number of free conduction electrons is negligible small in all the three
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3 )
The number of free electrons for conduction is significant in all the three
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4 )
The number of free electrons for conduction is significant only in Si and Ge but small in C
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| see the answer see the solution |
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19)
A charged particle with charge q enters a region of constant, uniform and mutually orthogonal
fields E and B with a velocity v perpendicular to both
E and B ,
and comes out without any change in magnitude or direction of v . Then
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1 )
v = E X B/B2
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2 )
v = B X E/B2
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3 )
v = E X B/E2
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4 )
v = B X E/E2
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| see the answer see the solution |
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20)
The potential at a point x (measured in μ m) due to some charges situated on the x-axis is given
by V(x) = 20/(x2 - 4) Volts. The electric field E at x = 4 μ m is given by
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1 )
5/3 Volt/μ m and in the -ve x direction
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2 )
5/3 Volt/μ m and in the +ve x direction
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3 )
10/9 Volt /μ m and in the -ve x direction
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4 )
10/9 Volt/μ m and in the +ve x direction
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| see the answer see the solution |
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21)
Which of the following transitions in hydrogen atoms emit photons of highest frequency ?
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1 )
n = 2 to n = 6
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2 )
n = 6 to n = 2
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3 )
n = 2 to n = 1
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4 )
n = 1 to n = 2
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| see the answer see the solution |
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22)
A block of mass 'm' is connected to another block of mass 'M' by a spring (massless) of spring
constant 'k'. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and
the spring is unstretched. Then a constant force 'F' starts acting on the block of mass 'M' to pull
it. Find the force on the block of mass 'm'
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1 )
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2 )
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3 )
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4 )
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| see the answer see the solution |
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23)
Two lenses of power - 15 D and + 5D are in contact with each other. The focal length of the
combination is
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1 )
- 20 cm
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2 )
- 10 cm
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3 )
+ 20 cm
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4 )
+ 10 cm
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| see the answer see the solution |
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24)
One end of a thermally insulated rod is kept at a
temperature T1 and the other at T2. The rod is
composed of two sections of lengths l1 and l2 and
thermal conductivities k1 and k2 respectively. The
temperature at the interface of the two sections is
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1 )
( k2 l2 T1 + k1 l1 T2 ) / ( k1 l1 + k2 l2 )
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2 )
( k2 l1 T1 + k1 l1 T2 ) / ( k2 l1 + k1 l2 )
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3 )
( k1 l2 T1 + k2 l1 T2 ) / ( k1 l2 + k2 l1 )
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4 )
( k1 l1 T1 + k2 l2 T2 ) / ( k1 l1 + k2 l2 )
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| see the answer see the solution |
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25)
A sound absorber attenuates the sound level by 20 dB. The intensity decreases by a factor of
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1 )
1000
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2 )
10000
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3 )
10
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4 )
100
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| see the answer see the solution |
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26)
If Cp and Cv denote the specific heats of nitrogen per unit mass at constant pressure and constant
volume respectively, then
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1 )
Cp - Cv = R/28
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2 )
Cp - Cv = R/14
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3 )
Cp - Cv = R
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4 )
Cp - Cv = 28 R
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| see the answer see the solution |
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27)
A charged particle moves through a magnetic field perpendicular to its direction. Then
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1 )
the momentum changes but the kinetic energy is constant
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2 )
both momentum and kinetic energy of the particle are not constant
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3 )
both, momentum and kinetic energy of the particle are constant
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4 )
kinetic energy changes but the momentum is constant
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| see the answer see the solution |
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28)
Two identical conducting wires AOB and COD are placed at right angles to each other. The wire
AOB carries an electric current I1 and COD carries a current I2.
The magnetic field on a point lying
at a distance 'd' from O, in a direction perpendicular to the plane of the wires AOB and COD, will
be given by
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1 )
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2 )
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3 )
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4 )
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| see the answer see the solution |
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29)
The resistance of a wire is 5 ohm at 50°C and 6 ohm at 100°C.
The resistance of the wire at 0°C will be
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1 )
2 ohm
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2 )
1 ohm
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3 )
4 ohm
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4 )
3 ohm
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| see the answer see the solution |
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30)
A parallel plate condenser with a dielectric of dielectric constant K between the plates has a
capacity C and is charged to a potential V volts. The dielectric slab is slowly removed from
between the plates and then reinserted. The net work done by the system in this process is
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1 )
1/2 (K - 1) C V2
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2 )
C V2(K - 1)K
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3 )
(K - 1)C V2
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4 )
zero
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| see the answer see the solution |
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31)
If gE and gm are the accelerations due to gravity on the surfaces
of the earth and the moon respectively and if Millikan's oil drop experiment could be
performed on the two surfaces, one will find the ratio
(electronic charge on the moon)/(electronic charge on the earth) to be
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1 )
1
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2 )
0
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3 )
gE/gm
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4 )
gm/gE
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| see the answer see the solution |
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32)
A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the
circumferences of the discs coincide. The centre of mass of the new disc is αR from the centre of
the bigger disc. The value of α is
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1 )
1/3
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2 )
1/2
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3 )
1/6
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4 )
1/4
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| see the answer see the solution |
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33)
A round uniform body of radius R, mass M and moment of inertia 'I' , rolls down (without slipping)
an inclined plane making an angle θ with the horizontal. Then its acceleration is
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1 )
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2 )
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3 )
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4 )
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| see the answer see the solution |
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34)
Angular momentum of the particle rotating with a central force is constant due to
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1 )
Constant Force
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2 )
Constant linear momentum
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3 )
Zero Torque
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4 )
Constant Torque
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| see the answer see the solution |
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35)
A 2 kg block slides on a horizontal floor with a speed of 4 m/s. It strikes a uncompressed spring,
and compresses it till the block is motionless. The kinetic friction force is 15 N and spring constant
is 10,000. N/m. The spring compresses by
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1 )
5.5 cm
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2 )
2.5 cm
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3 )
11.0 cm
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4 )
8.5 cm
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| see the answer see the solution |
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36)
A particle is projected at 60° to the horizontal with a kinetic energy K.
The kinetic energy at the highest point is
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1 )
K
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2 )
Zero
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3 )
K/2
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4 )
K/4
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| see the answer see the solution |
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37)
In a Young's double slit experiment the intensity at a point where the path difference is
λ/6 ( λ being the wavelength of the light used ) is I .
If I0 denotes the maximum intensity , I/I0 is equal to
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1 )
1/
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2 )
 /2
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3 )
1/2
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4 )
3/4
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| see the answer see the solution |
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38)
Two springs, of force constants k1 and k2 are
connected to a mass m as shown. The frequency of oscillation of the mass is f.
If both k1 and k2 are made four times their original values,
the frequency of oscillation becomes
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1 )
f/2
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2 )
f/4
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3 )
4f
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4 )
2f
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| see the answer see the solution |
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39)
When a system is taken from state i to state f along the path iaf, it
is found that Q = 50 cal and W = 20 cal. Along the path ibf
Q = 36 cal. W along the path ibf is
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1 )
6 cal
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2 )
16 cal
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3 )
66 cal
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4 )
14 cal
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| see the answer see the solution |
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40)
A particle of mass m executes simple harmonic motion with amplitude 'a' and frequency 'ν' . The
average kinetic energy during its motion from the position of equilibrium to the end is
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1 )
&pi2 m a2ν2
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2 )
(1/4) &pi2 m a2ν2
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3 )
4 &pi2 m a2ν2
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4 )
2 &pi2 m a2ν2
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| see the answer see the solution |
1 back to 1
x = 2 X 10-2 cos πt
Differentiating both side
=> dx/dt = -0.02πsinπt
=> v = -0.02π sinπt
=> v is maximum when, πt = π/2
=> t = 1/2 = 0.5 sec
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2 back to 2
In AC circuit average power is :
Pavg = VrmsIrms cosφ
In AC circuit voltage and current is represented as:
V = V0sin(ωt+φ)
I = I0sin(ωt)
So, φ = -π/2 = -90°
So, Pavg = VrmsIrms cos(-90°)
So, Pavg = 0
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3 back to 3
Distance between point A and origin(0,0) = 2
Distance between point B and origin(0,0) = 2
As the distance from the electric charge of A and B are same
so both A and B will have same potential.
So potential difference between point A an B will be = 0
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4 back to 4
Q = charge on the capacitor = CV
Work done by battery = Q.V = CV2
Energy stored in capacitor = (1/2)CV2
So, (Energy stored in capacitor)/(Work done by battery) = 1/2
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5 back to 5
I=I0[1-e-Rt/L]
I0 = 5/5 = 1
So, I=1-e-5t/10
=> I=1-e-t/2
As t=2, so
I=1-e-1
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6 back to 6
Magnetic Field around a wire (B1) when r is greater than the radius of the wire.
where
I = current
r = distance from wire
and r ≥ Radius of the wire
Magnetic Field around a wire (B2) when r is less than the radius of the wire.
where
I = current
R = radius of wire
r = distance from wire
and r ≤ Radius of the wire (R)
So, B2/B1 = 1
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7 back to 7
If electric current flowing through a hollow pipe,
it will induce a magnetic field outside the pipe.
The magnetic field inside the pipe will be
zero because a closed loop just inside the pipe
will not have any current flowing through it.
This situation is similar to a Faraday cage where
the electric field inside a hollow conducting shell is zero.
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8 back to 8
Nuclear binding energy = [mass of nucleus - mass of nucleons].C2
Nuclear binding energy = (M0 - 8 MP - 9 MN) C2
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9 back to 9
Gamma ray is aform electromagnetic radiation.
It is produced by sub-atomic particle interactions,
such as electron-positron annihilation or radioactive decay.
It does does not involve any change in
proton number or neutron number
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10 back to 10
In the first half cycle, the diode is in forward biased.
In the next half-cycle, the diode is in reverse biased
Diode is forward biased in first half-cycle.
Voltage applied: 10 V (it difference the peak ie. +5V and -5V)
So the amplitude of signal: 5V.
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11 back to 11
Relation between energy and frequency
E = hν
where
E = Energy
h = Planck's constant
ν = frequency
As per de Broglie equation
where
p = momentum
λ = wavelength
h = Planck's constant
v = velocity
By solving above two equation
p= hν/c
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12 back to 12
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13 back to 13
Using perpendicular Axis Theorem:
IAC = IEF
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14 back to 14
x = x0 cos (ω t - π/4)
dx/dt = v = -x0ωsin(ωt-π/4)
dv/dt = a = -x0ω2cos(ωt-π/4)
a = x0ω2cos(ωt+3π/4)
By comparing it with , a = A cos (ω t + δ )
A = x0ω2
δ = 3π/4
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15 back to 15
Electric field (E) is vector quantity.
Electric Potential (V) is scalar quantity.
By changing the charge Electric Potential (V) do not change, but the direction
of Electric field (E) changes.
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16 back to 16
Half life of radioactive element
Average life of radioactive element
So,
So,
1.4λx = λy
λx < λy
So element Y will decay faster than element X
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17 back to 17
Efficiency of Carnot cycle
W = η.Q where Q=total energy put into system
=> 10=(1/10)Q
=> Q = 100
=> So energy absorbed by the system = (total energy put into system)-(Work done)
=> energy absorbed by the system = 100-10 = 90
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18 back to 18
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19 back to 19
Magnetic force acting on charge = Fm = q v x B
Electric force acting on charge = Fe = q E
So, q v x B = q E
=> q v x B = q E
=> v =(E x B)/B2
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20 back to 20
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E = -
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dV
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=
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20
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(2x-0) =
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160
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=
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10
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(+ve)
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dX
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(x2-4)2
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144
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9
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21 back to 21
Emitting Photons(Rydberg Formula)
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Ephoton = E0(
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1
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-
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1
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)
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n12
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n22
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where
n1 < n2
E0 = 13.6 eV
By using above formula E is maximum when n=2 to n=1
As, E=hν
So ν is maximum if E is maximum.
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22 back to 22
Assuming acceleration of both blocks are: a
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So force acting on m = Fm =
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Fm
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M+m
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23 back to 23
P=P1+P2 = -15+5=-10
The focal length of the combination = 1/P = -1/10 m = -10cm
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24 back to 24
Assume that the temperature at the interface = T
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(T1-T)k1
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=
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(T-T2)k2
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l1
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l2
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By solving above equation, T =
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T1k1l2+T2k2l1
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k1l2+k2l1
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25 back to 25
Sound intensity is sound power Pac per unit area A
Sound intensity I = 10log10(I/I0)
B1 = 10log10(I1/I0)
B2 = 10log10(I2/I0)
B1-B1=20
20 = 10log10(I1/I2)
=> 2 = log10(I1/I2)
=> I1/I2 = 100
=> I1 = 100I2
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26 back to 26
cp - cv = R [for one mole of gas]
=> cp - cv = R [for one 28 gm of nitrogen gas]
=> cp - cv = R/28 [for one 1 gm of nitrogen gas]
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27 back to 27
Force acting on the particle = F = q v x B
The force acts perpendicular to the velocity so there is no work done on particle.
So kinetic energy will not change.
As force acts on the particle so there is change in its momentum
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28 back to 28
|
Force acting at a distance d from O, due to wire AOB = F1 =
|
μ0I1
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|
2 π d
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Force acting at a distance d from O, due to wire COD = F2 =
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μ0I2
|
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2 π d
|
F1 and F2 are perpendicular to each other, So net force will be
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29 back to 29
5=R0(1+50α) ..................... (i)
6=R0(1+100α) ..................... (ii)
By solving (i) and (ii)
α = 1/200
R0 = 4Ω
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30 back to 30
Net work done by the system is zero because there is no change in energy of the system.
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31 back to 31
Net work done by the system is zero because there is no change in energy of the system.
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32 back to 32
Let mass of the circular disc = M
So mass of the removed disc = M/4
So mass of the remaining desc = 3M/4
Assume that centre of mass of remaining disc is at a distance x from centre
So, (3M/4).x = (M/4).R
=> x = R/3 = αR
=> α = 1/3
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33 back to 33
Assume that acceleration = a
So angular acceleration = a/R
mgsinθ - f = ma .............. (i)
fR = I(a/R) ....................... (ii)
By solving (i) and (ii)
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34 back to 34
The direction of the force that acts on the rotating particle
passes thru the center. So there is no torque acting on the particle.
As no torque acts on the particle so angular momentum is constant.
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35 back to 35
Assume that spring is compressed by distance=x
Kinetic energy of the block = (1/2)×2×42 = 16
While compressing the spring, energy loss due to friction = 10x
Energy transferred to spring due to block = (1/2)×10000×x2
(1/2)×10000×x2 + 10x = 16 ...................(i)
By solving equation (i), x = 5.5 (approx.)
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36 back to 36
Kinetic energy of the particle at the beginning : K = (1/2)mv2
Horizontal velocity of the particle = vcos60°
Kinetic energy of the particle at the beginning : K = (1/2)mv2
Kinetic energy of the particle at the hightest point = (1/2)m(vcos60°)2
Kinetic energy of the particle at the hightest point = (1/2)mv2cos260° = K/4
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37 back to 37
Now, I/I0 = cos2&phi/2 = cos2(2π/λ)(λ/12) =
cos230°
I/I0 = 3/4
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38 back to 38
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39 back to 39
As per the first law of thermodynamics :
ΔQ = ΔU + ΔW
For the route iaf, 50 = 20 + ΔU
ΔU = 30
For the route ibf, 36 = ΔW + 30
ΔW = 6
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40 back to 40
Kinetic Energy = (1/2)mω2a2sin2ωt
and ω = 2πν
Kinetic Energy = π2ma2ν2
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