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GATE Physics Mock Test Series - 3

60 Questions MCQ Test GATE Physics Mock Test Series | GATE Physics Mock Test Series - 3

Description

Attempt GATE Physics Mock Test Series - 3 | 60 questions in 180 minutes | Mock test for GATE preparation | Free important questions MCQ to study GATE Physics Mock Test Series for GATE Exam | Download free PDF with solutions

QUESTION: 1

At what speed the volume of an object will shrinks to half its rest volume?

A.
B.
C.
V = C
D.

Solution:

Let the volume (initial) be ‘V ’

QUESTION: 2

Calculate the energy released by 1 g of natural uranium assuming 200 MeV is released in each fission event and that the fissionable isotope ²³⁵U has an abundance of 0.7% by weight in natural uranium.

A.
5.7 x 10⁸ J
B.
2.56 x 10⁸ J
C.
5.12 x 10⁸ J
D.
2.8 x 10⁸ J

Solution:

235 gm uranium contains 6.02 x 10²³ nuclei
So, in 1 gm no. of nuclei

So, total energy released = 2.56 x 10²¹ x 200 MeV = 5.12 x 10²³MeV
According to energy released = 5.12 x 10²³ x 1.6 x 10^-19 x 10⁶x Joule
= 5.7 x 10⁸ J

QUESTION: 3

A charged particle of mass'm',charge 'q' and constant velocity enters a uniform magnetic field at an angle θ to the direction of magnetic field.The angle'θ'(if,one revolution of the helical motion,the particle advances along the direction of the magnetic field at a distance equal to the radius of the helical path) is:-

A.
tan^-1π
B.
tan^-12π
C.
cot^-12π
D.

Solution:

Radius of helical path=r =
time taken for 1 revolution (T)=

QUESTION: 4

A free particle is moving in positive direction with linear momentum (p) The wave function of the particle normalized in a length (L) is:-
Applying normalization, we get

Solution:

Applying Schrodinger’s equation :-

Solution o f this equation is
Applying boundary condition ψ(O) = 0
⇒ A sinO + B cos 0 = 0 ⇒ B = 0
Hence wave function is
Applying normalization, we get

QUESTION: 5

The solution of the differential equation with the boundary condition that y = 1, at x=1 is:—

A.
y = 5x⁴-4
B.
C.
D.

Solution:

We have

solution is y .x =
but y = 1 at x = 1
solution is yx =

QUESTION: 6

A particle of mass 2kg is moving such that at time t second, its position in meter is given by The angular momentum of the particle at t = 2 sec about the origin in kg m²/sec is

Solution:

Given that

at time t = 2 sec

velocity vector

at time t = 2 sec

momentum vector
Angular momentum

QUESTION: 7

If z = f(x, y), where x = g(t), y = h(t), g(3) = 2, g'(3) = 5, h(3) = 7, h'(3) = -4, f_x(2, 7) = 6, f_y(2,7) = -8, then the value of at t = 3 is

A.
62
B.
60
C.
72
D.
52

Solution:

z = f(x, y)

At t = 3, x = g(3) = 2, y = h(3) = 7

= f_x(2, 7).g'(3) + f_y(2, 7)h'(3) 6 x 5 + (-8 ) x ( - 4) = 62

QUESTION: 8

If are non coplanar unit vectors such that then the angle between is

A.
B.
C.
D.
π

Solution:

We have

then

QUESTION: 9

A system of four particles is in x-y plane of these two particle of masses m are located at (1,1) and (-1,1). The remaining two particles each of mass 2m are located at (1,-1) and (-1,-1). The xy component of moment of inertia tensor of the system of particle is

A.
10m
B.
-10m
C.
0m
D.
-2m

Solution:

The xy component of inertia tensor is

I_xy = -[(m x1x1) + (m x-1x1) + (2m x1x-1) + 2m(-1x-1)]
= - [ m - m - 2m + 2m ]
= 0

QUESTION: 10

Consider circular orbits in a central force potential V(r) = , where k > 0 and o < n < 2. If the time period of a circular orbit of radius R is T₁ and that of radius 3R is T₃, then is,

A.
B.
C.
D.
3ⁿ

Solution:

Veffective =

QUESTION: 11

After a perfectly elastic collision of two identical balls, one of which was initially at rest, the velocities of both the balls are non zero. The angle θ between the final.
Velocities (in the lab frame) is

A.
B.
θ = π
C.
D.

Solution:

Angle between two particle θ₁ + θ₂ = 0
Applying the conservation of linear momentum

Conservation of Kinetic energy

from (i) and (ii)

QUESTION: 12

The escape velocity of the earth is v₀. For a planet having radius 4 times and density three times that of the earth, the escape velocity will be

Solution:

Escape velocity

If p becomes three times and radius becomes 4 times then Escape velocity v =

QUESTION: 13

A pump is used to push water of density p through a tube of constant cross- section area ‘A’ shown below. What is the value of ΔP?

A.
B.
0
C.
pgH
D.
-pgH

Solution:

Since, cross-section area at A and B is same. So, velocity of water at A and B will remain same. Let us apply Bernoulli equation at A & B.

Both equations are equal
ΔP = pgH

QUESTION: 14

Nearest neighbour distance in Na crystal is 1.83 A°. Density of electron in Na crystal is:-

A.
2.12 x 10^-1m^-3
B.
5.3 x 10²⁹m^-3
C.
3.3 x lO⁺²⁹m^-3
D.
6.3 x I0²⁹m^-3

Solution:

‘Na’ has bcc structure
Nearest Neighbour distance =

Density of electrons =

QUESTION: 15

The electric potential due to a linear quadrupole varies inversely with distance ‘r’ as :-

A.
r
B.
r²
C.
r³
D.
none of these

Solution:

The potential at the point 'P ' is v =

Q = 2qa² → Quardrupole moment of charge distribution

thus v is inversely proportional to r³ & not r^-3

QUESTION: 16

Calculate the collector voltage (V_c) of the transistor circuit shown in the fig be- low :-

Given

A.
3.78 V
B.
3.82 V
C.
4.72 V
D.
9.7 V

Solution:

since V_BE« V_cc
we can neglect V_BE in comparison to V_cc.

QUESTION: 17

The logic circuit shown in the following figure yields the given truth table.

The gate ‘x’ in the diagram is:-

A.
NAND
B.
XOR
C.
AND
D.
NOR

Solution:

Truth table for AND is given as:-

NOT of AND i.e. NAND is given as:-

than the given gate “x” is NAND gate

QUESTION: 18

A metal with body centered cubic (bcc) structure shows the first (i.e. smallest angle) diffraction peak at a Bragg angle of θ = 30°. The wavelength of X-ray used is 2.1 . The volume of the primitive unit cell of the metal is (for first peak, h + k + 1 should be even)

A.
26.2
B.
13.1
C.
9.4
D.
4.6

Solution:

λ = 2.1Aº for bcc →h+k+l should be even for first peak,
h = 1, k=1, l=0

QUESTION: 19

Consider the surface corresponding to the equation 4x² + y² + z = 0 .A possible unit tangent to this surface at the point (1,2,-8) is

Solution:

Unit normal for the surface =

Unit normal =
So, the possible unit tangent to this surface will be
Where is tangent
According to option

QUESTION: 20

In the laboratory frame, two observers A and B are moving along the sides of an equilateral triangle with equal speeds C/2, as shown in the figure. The speed of B as measured by A will be ______ .

Solution:

According to figure

The speed of B as measured by A

Resultant velocity u =

QUESTION: 21

The instantaneous electric and magnetic fields created at a distance r by a point source at the origin are given by

Where ω, A, B are constants and the unit vectors from an orthonormal set. The time averaged power radiated by the source is .

A.
B.
C.
C²AB
D.

Solution:

Given that

Pointing vector (Energy per unit time per unit area)

We know that

=> The time averaged power radiated by the source is = <S>. A = AB C²

QUESTION: 22

Electric field component of an electromagnetic radiation varies with time as E = , where a is a constant and the values of the ω and ω₀ are 2 x 10¹⁵ sec^-1 and 1 x 10¹⁶ sec^-1 respectively. This radiation falls on a metal of work function 4.4eV. The maximum kinetic energy (in eV) of photoelectrons is

A.
0.64eV
B.
1.30 eV
C.
1.95 eV
D.
3.475 eV

Solution:

The maximum frequency of photon =

QUESTION: 23

If E₁is the energy of the electron in the six fold degenerative energy state (1,2,3) in a cubical potential box of side a and E₂ is the energy of a particle of mass twice that of electron in the second non-degenerative energy state of a side then E₂-E₁

Solution:

For cubical box , the eigen values of energy

E₁→ state (1 ,2 ,3 ), when side of cubical box is ‘a’ .

Now for the E₂ [ in non -degeneratic state (2,2,2)] mass is twice and side of
cubical box is

QUESTION: 24

Three polarisers P,Q and R are placed parallel to each with their planes perpendicular to the z-axis. Q is placed between P and R Initially, the polarising directions of P and Q are parallel but that of R is perpendicular to them . In this arrangement when unpolarised light of density l₀ is incident on P , the R intensity coming out of R is zero, the polariser Q is now rotated about the z-axis. The maximum intensity of light coming out of R is

A.
I₀
B.
C.
D.

Solution:

Consider three polarisers P,Q and R Initially, polarising direction of P and Q are parallel but that of R is perpendicular to them.
Intensity of light passing through P is

Intensity of light passing through Q , according to law of malus I_Q = I_pcos² φ) Intensity of light coming out of R is

sin2 φ = l( I_R will be maximum , when φ= π/4)

QUESTION: 25

A diffraction grating having N slits , each of width b and period d , is illuminated normally by a monochromatic plane wave of wavelength λ. What is the phase difference between waves from first and Nth slit in the heighest diffraction order?

A.
B.
C.
D.
None

Solution:

Highest number of orders with grating
dsinθ = nλ

QUESTION: 26

The speed of sound propagating in air as a function of temperature T is given by v = αT, where , α is a constant of appropriate dimensions. Calculate the time taken for a sound wave to travel a distance L between two points A and B , If the air temperature between the points varies linearly from T₁ to T₂.

Solution:

Given that speed of sound v = αT

taking double integrations on both sides

Putting V₁= αT₁ and V₂= αT₂

QUESTION: 27

The extension in a string, obeying Hooke’s law, is x. The speed of transverse wave in the stretched string is V. If the extension in the string is increased to 1.5x, the speed of transverse wave will be

A.
1.22V
B.
0.61V
C.
1.50V
D.
0.75V

Solution:

From Hooke’s law
Tension in a string (T) α extension (x) and speed of sound in string

Therefore
x is increased to 1.5 times speed will increase by times or 1.22 times.
Therefore , speed of sound in new position will be 1.22V.

QUESTION: 28

Two radio stations broadcast their programs at the same Amplitude A and at slightly different frequencies ω₁ and ω₂ respectively, where ω₁ - ω₂= 10³Hz. A detector receives the signals from the two stations simultaneously. It can only detect signals of intensity ≥2A². Find the time interval between successive maxima of the intensity of the signal received by the detector.

Solution:

If the detector is at x = 0, the two radio waves can be represented as
y₁ = A sin ω₁t {Given that A₁ = A₂}
y₂ = A sin ω₂t
By the principle of superposition

Now we can consider that,

So , intensity will be maximum when

Therefore , time interval any two successive maxima is

QUESTION: 29

A thin convex lens made from crown glass as focal length f. When it is measured in two different liquids having refractive indices It has the focal length f₁ and f₂, respectively. The correct relations between the focal length is ________ .

A.
B.
f₁> f and f₂ becomes negative
C.
f₂ > f and becomes negative
D.
f₁ and f₂ both becomes negative

Solution:

By the lens maker’s formula

According to lens maker’s formula, when lens in the air

In case of liquid, where refractive index is
we get
focal length in first liquid

⇒ f₁ = 4 f
focal length in second liquid

f₂ is negative.

QUESTION: 30

In a YDSE bi-chromatic light of wavelength 400 nm and 560 nm are used. The distance between the slits is 0.1 mm and the distances between the plane of the slits and the screen is 1m. The minimum distance between two successive regions of complete darkness is _______

A.
4 mm
B.
5.6 mm
C.
14 mm
D.
28 mm

Solution:

Let nth minima of 400 nm coincides with mth minima of 560 nm, then

4th minima of 400 nm coincides with 3rd minima of 560 nm.
Location of this minima is

Next 11th minima of 400nm will coincide with 8 minima of 560 nm
Location of this minima is

∴ Required distance = y₂ - y₁ =28

*Multiple options can be correct

QUESTION: 31

Which of the following statement’s are correct ?

A.
For a mechanical isolated system kept at constant temperature the Helmholtz free energy never increases.
B.
For a mechanical isolated system kept at constant temperature the state of equilibrium is the state of minimum Helmholtz free energy.
C.
For a system kept at constant temperature and pressure the Gibbs potential never increases.
D.
For a system kept at constant temperature and pressure the state of equilibrium is the state of minimum Gibb’s potential

Solution:

We know that the Helmholtz free energy A and the Gibb’s thermodynamics potential G are defined as
A = U - TS
G = A + PV
Let a system undergo A to state B. Then we have from the second law that

since T is constant

where Δq is the amount of heat absorbed during the transformation and ΔS =
S(B) - S(A)
W ≤ -Δu + TΔS
where W is the work done by the system, since
W ≤ -ΔA
∴ Fora mechanically isolated system kept at constant temperature the Helmholtz free energy never increases.
or the state of equilibrium is the state of minimum Helmholtz free energy.
Again if T is constant w ≤ - ΔA
PΔV + ΔA ≤ 0
ΔG ≤ 0 ∴ minimum Gibbs potential
dG = -SdT + VdP

H = U + PV ⇒ dH = TdS + VdP
which follows that

*Multiple options can be correct

QUESTION: 32

Let us defined the following quantities, which are experimentally measurable
(coefficient of thermal expansion )
(isothermal compressibility )
(adiabatic compressibility )
Then which of the following relations are correct ?

Solution:

As we know that

This shows that C_p - C_v> 0, if K_T > 0

*Multiple options can be correct

QUESTION: 33

An expression for two un-normalized wave functions that describe a particle in a box that has 1/3 change of being in the n = 4 static and 3/4 change of being in the n = 5 state.

Solution:

We write the un-normalized wave functions for n = 4 and n = 5 states as

We have .Hence , the wave functions can be written as

*Multiple options can be correct

QUESTION: 34

A bullet of mass 50g is fired from below into the bob of mass 450g of a long simple pendulum as shown in figure. The bullet inside the bob and the bob rises through a height of 1.8 m Then

A.
The linear momentum is conserved.
B.
The linear momentum is not conserved .
C.
The speed of bullet is 60 m/sec.
D.
The speed of bullet is 49 m/sec.

Solution:

etthe speed of the bullet be Let the common velocity of the bullet and the bob, after the bullet is embedded into the bob, is V. By the principle of conservation of linear momentum.

The string becomes loose and the bob will go up with a declaration of g = 10 m/ sec². As it comes to rest at a height of 1.8 m , using the equation
= u² + 2ax

*Multiple options can be correct

QUESTION: 35

If (a, b and g being constants) and x = a’ and when t = 0 then

Solution:

The given equation is

its auxiliary equation is

The complete solution will be

It is given that at t = 0, x = a’

*Multiple options can be correct

QUESTION: 36

The coordinates of a particle moving in a plane are given by x(t) = a cos p(t) & y(t) = b sin(pt) where a,b (<a) and ‘p’ are positive constants of appropriate dimensions. Then:-

A.
The path of the particle is an ellipse
B.
The velocity & acceleration of the particles are normal to each other at
C.
The acceleration of the particle is always directed towards a focus.
D.
The distance travelled by the particle in time interval t = 0 to t = is ‘a

Solution:

x = a cos (pt) & y = b sin (pt)

squaring & adding eqⁿ(1) & (2) we get:
— [.-.path of the particle is an ellipse]
from the given eqⁿ we can find.

⇒ a₁ & V₁ becomes zero (because cos π/y
0)
Only V_x & a_y are left
or we can say that velocity is along negative x-axis and acceleration along y - axis.
Hence at t = velocity & accleration of the particle are normal to each other.
So option B is correct
At t = t₁ position of the particle

& acceleration of the particle is

∴ acceleration of the particle is always directed towards the origin. Hence option C is correct
At t = 0, particle is at (a,0) & at t = particle is at (0,b)
distance covered in th of the elliptical path not ‘a’.
Hence Option A, B,C, are correct

*Multiple options can be correct

QUESTION: 37

Let V_rms & V_p respectively denote the mean speed , root mean square speed and most probable speed of the molecules in an ideal monatomic gas at absolute temperature (T). The mass of the molecule is .m.. Then:

A.
no molecule can have a speed greater than
B.
no molecule can have a speed less than
C.
D.
The average kinetic energy of a molecule is

Solution:

From these expression we can see that
Secondly,

*Multiple options can be correct

QUESTION: 38

A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without touching it
The correct statements is(are) -

A.
The emf induced in the loop is zero if the current is constant.
B.
The emf induced in the loop is finite if the current is constant.
C.
The emf induced in the loop is zero if the current decreases at a steady rate.
D.
The emf induced in the loop is finite if the current decreases at a steady rate

Solution:

Due to the current in the straight wire, net magnetic flux from the circular loop is z e ro . Because in half of the circle magnetic field is inwards and in other half , magnetic field is outwards. Therefore, change in current will notcause any charge in magnetic flux from the loop. Therefore, induced emf under all conditions through the circular loop is zero.

*Multiple options can be correct

QUESTION: 39

A series R-C circuit is connected to AC voltage source consider two cases (A) When C is without a dielectric medium and (B) when C is filled with dielectric of constant 4. The current IR through the resister and voltage V_c across the capacitor are compared in the two cases. Which of the following is/are true ?

Solution:

In case (b) capacitance C will be more. Therefore, impedance Z will be less.
Hence, current will be more .
option (b) is correct.
Further

In case (b), since current I is more.
Therefore, V_c will be less.

*Multiple options can be correct

QUESTION: 40

A steady current I flows along an infinitely long hallow cylindrical conductor of radius R. This cylinder is placed coaxially inside an infinite solenoid of radius 2R. The solenoid has n turns per unit length and carries a steady current i. Consider a point P at a distance R from the common axis. The correct statements is/are-

A.
In the region 0 < r < R, the magnetic field is non-zero.
B.
In the region R < r < 2 R, the magnetic field is along the common axis.
C.
In the region R < r < 2R, the magnetic field is tangential to the circle of radius r, centered on the axis.
D.
In the region r > 2R the magnetic field is non-zero.

Solution:

Conside that
P-Hollow cylindrical conductor
Q - Solenoid
case - 1 → In the region , 0 < r < R
Bp = 0
B_Q ≠ 0, along the axis
B_net ≠ 0
case-II⇒ln the region R < r < 2R
B_p ≠ 0, tangential to the circle of radius r, centered on the axis.
B_Q ≠ 0, along the axis
B_net ≠ 0 neither in the directions mentioned in option(b) or (c)
case - → In region r > 2R
B_p ≠ 0
b_Q ≠ 0
B_net ≠ 0

QUESTION: 41

What is the probability that a particle be found between 0.49 L and 0.51 L in 1D box of length L for n = 1 ?

Solution:

The wave function is given by

The probability is given by

for n = 1 andx = 0.5 L

QUESTION: 42

The moment of inertia of a uniform sphere of radius ‘r’ about an axis passing through its centre is given by .A rigid sphere of uniform mass density ‘p ’ & radius ‘R’ has two smaller spheres of Radii hollowed out of it as shown in the figure given below.

The moment of inertia of the resulting body about y-axis if R = 2.5 m & p = 4.6 kg/m³ is

Solution:

Let I is the moment of inertia of bigger sphere.
Then ;
for small sphere of R/2 about its centre,

Moment of inertia for this sphere about y-axis

For second sphere
total

I = 620

QUESTION: 43

In a cylindrical conductor of radius 2mm, the current density varies with the distance from the axis according to J = 10³ (e)^-400r A/m². What is the value of the total current (I) (in mA) ?

Solution:

Total current (I)

I = 7.51 mA

QUESTION: 44

A person weighs W_p at Earth's pole and We at the equator. Treating the Earth as a perfect sphere of radius 6400 km, the value 1000 x

Solution:

We know that, gravity acceleration due to rotation is

⇒

=

QUESTION: 45

A particle constrained to move along the x-axis is described by the wave function :-
ψ ( x ) = 2x ; 0 < x <1
ψ (x)= 0 ; else where
What is the probability of finding the particle within the interval (0,0.1)?

Solution:

If ψ (x) is the wave function then probability of finding the particle in width ‘dx’ is given as:-

We have ψ(x) = 2x , 0< x <1
ψ(x) = 0 .... else here

QUESTION: 46

The pulley shown in figure has a radius 10 cm and moment of inertia 0.5kg.m² about its axis. Assuming the inclined planes to be frictionless .The acceleration of the 4.0 kg block is _____ cm/sec².

Solution:

QUESTION: 47

Steam at 100°C is mixed with 1500 gms of water at 15°C so that the final temperature of the mixture is 80° C. What will be the value of mass of steam ?

Solution:

m₁ = ? ; m₂ = 1500 gm.
T₁ =373 k, T₂ = 273+15 = 288 k.
Find temperature of mixture be (T) = 273+80 = 353 k
∴ m₁ x 540 + m₁ (373-353) = 1500 (353-288)
(540+20)m₁ = 1500x65
= 175g = 0.175 kg

QUESTION: 48

The vapour pressure (P) (in mm of Mercury) of solid ammonia is given by :- while that of liquid ammonia is given by where T is in K. What is the value of the pressure at which solid, liquid and vapour phase of ammonia exists simultaneously (in mm of Hg) ?

Solution:

44.87
At triple point, the vapour pressure of the substance in each of the 3 states is identical.
Hence equating the vapour pressure of solid ammonia

with that of liquid ammonia log_e P =
we have ;

on solving we get :- T = 195.2 K

we get log_e P=
2.303 log₁₀ P = 3.8

QUESTION: 49

The heat absorbed by a system in going through the following cyclic process is ------ mJ.

Solution:

Heat absorbed = workdone
= Area under the graph in the given case.
Heat absorbed = area of the circle
= πr² = 3.14x10⁴ x10^-6 x 10³J
Heat absorbed = 3.14 x10 = 31.4 J

QUESTION: 50

The number of photons emitted per second by a 10 W sodium vapour lamp is -_____ x 10¹⁹. Assume that 60% of the consumed energy is converted into light. Wavelength of sodium light = 590nm.

Solution:

P = 10W ∴E in 1 sec = 10J
% used to convert into photon = 60%
Energy used = = 6J
Energy used to take out 1 photon =
No. of photons used

QUESTION: 51

In young’s double slit experiment using monochromatic light the fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 & thickness 1.964 microns is introduced in the path of one of the interfering waves. The mica sheet is then removed & the distance between the slits & the screen is doubled. It is found that the distance between the maxima (or minima) now is the same as the observed fringe shift upon the introduction of the miashat. Calculate the wavelength of the monochromatic light used in the experiment (in A°) ?

Solution:

Shifting of fringes due to introduction of slab in the path of one of the slits, comes out to be;

Now, the distance between the screen & slits is doubled.
Hence the new fringe width will become,

Given Δy = ω'

= 0.5892 x 10^-6 m
= 5892 A°

QUESTION: 52

For the common emitter configuration of transistor, as shown in the figure find the voltage between collector & emitter.

Solution:

The Given figure can be simplified as

Base current
Collector current ( I_B )

QUESTION: 53

The load resistance of a full wave rectifier is 10³ Ω. And the input AC voltage (rms) is 200-0-200. Calculate the ripple factor for the following arrangement (while considering that diode do not show any significant resistance itself)?

Solution:

The peak value of d.c. current

Average d.c current is given by
Average d.c voltage
ripple factor

QUESTION: 54

What will be the value of V_EC in figure,

Solution:

First check to see if R_in(base) can be neglected.

since 150 kΩ is more than ten times “ R₂” . The condition is net and R_in(base) can be neglected.

= 6.88 V.
Then V_E = V_B + V_BE = 6.88 V + 0.7 V = 7.58 V

QUESTION: 55

A binary star is observed to consist of a blue star B (Peak wavelength 800 nm) and a red star R (Peak wavelength 400 nm) orbiting each other. As observed from the earth, B and R appear bright. Assuming that the stars radiate as perfect black bodies, it follows that the ratio of volumes V_B/V_R of the two stars is_________.

Solution:

For blackbody, According to wien’s displacement law

and energy per unit area per unit time is E =

So, that the volume

QUESTION: 56

A stone is dropped vertically from the top of a town of height 40 m. At the same time a gun is aimed directly at the stone from the ground at a horizontal distance 30 m from the base of the tower and fired. If the bullet from the gun is to hit the stone before it reaches the ground the minimum velocity (m/sec) of the bullet must be approximately _____ .

Solution:

To solve this question we will think about two cases

time taken by the stone to cover the x distance
and time taken by the bullet to reach the highest point

for case - II
vcosθ.t = 30
v = 17.7 m/sec
So minimum speed is 17.7 m/sec

QUESTION: 57

A variational calculation is done with the normalized trial wave function
for the one-dimensional potential well
The ground state energy is estimated to be .The value of n is ______ .

Solution:

The normalized trial wave function is ψ(x) =
The potential will is,
The ground state energy E₀ is given by

QUESTION: 58

The dispersion relation of phonons in a solid is given by . The velocity of phonons at large wavelength is nω₀a .
The value of ‘n’ is _______ .

Solution:

The dispersion relation of phonons in a solid is given by

at large wavelength
at large wavelength
at large wavelength

The velocity of the phonons at large wavelength is

QUESTION: 59

A circular disc of radius ‘a ’ on the x-y plane has a surface change density . The electric dipole moment of this charge distribution is __________ ( in the unit of

Solution:

The electric dipole moment is,

QUESTION: 60

The spherical surface of a piano convex lens of radius of curvature R = 1 m is gently placed on aflat plate. The space between then is filled with a transparent liquid of refractive index 1.57. The refrective indices of the lens and the flat plate are 1.52 and 1.62 respectively. The radius of the sixteenth dark newton’s ring in the reflected light of wave length λ is found to be √5 mm. Determine the wavelength λ of the light_______ (in ).

Solution:

For the formation of dark fringe , we can write

t = thickness of the liquid
λ = wave length of the light
μ₁ = refractive index of the liquid
From the geometry of the figure , we have