Physics-
General
Easy

Question

a) Name the experiment for which the adjacent graph, showing the variation of intensity of scattered electrons with the angle of scatter ing (q) was obtained.
b) Also name the important hypothesis that was confirmed by this experiment

  1. a) Davission and Germer experiment b) de Broglie hypothesis    
  2. a) Photo electric effect b) de Broglie hypothesis    
  3. a) Thermionic emission b) de Broglie hypothesis    
  4. a) Photocell    

The correct answer is: a) Photo electric effect b) de Broglie hypothesis

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Related Questions to study

General
physics-

Three identical particles are joined together by a thread as shown in figure. All the three particles are moving in a horizontal plane. If the velocity of the outermost particle is v subscript 0 end subscript, then the ratio of tensions in the three sections of the string is

Let omega is the angular speed of revolution

T subscript 3 end subscript equals m omega to the power of 2 end exponent 3 l
T subscript 2 end subscript minus T subscript 3 end subscript equals m omega to the power of 2 end exponent 2 l rightwards double arrow T subscript 2 end subscript equals m omega to the power of 2 end exponent 5 l
T subscript 1 end subscript minus T subscript 2 end subscript equals m omega to the power of 2 end exponent l rightwards double arrow T subscript 1 end subscript equals m omega to the power of 2 end exponent 6 l
T subscript 3 end subscript colon T subscript 2 end subscript colon T subscript 1 end subscript equals 3 blank colon 5 blank colon 6

Three identical particles are joined together by a thread as shown in figure. All the three particles are moving in a horizontal plane. If the velocity of the outermost particle is v subscript 0 end subscript, then the ratio of tensions in the three sections of the string is

physics-General
Let omega is the angular speed of revolution

T subscript 3 end subscript equals m omega to the power of 2 end exponent 3 l
T subscript 2 end subscript minus T subscript 3 end subscript equals m omega to the power of 2 end exponent 2 l rightwards double arrow T subscript 2 end subscript equals m omega to the power of 2 end exponent 5 l
T subscript 1 end subscript minus T subscript 2 end subscript equals m omega to the power of 2 end exponent l rightwards double arrow T subscript 1 end subscript equals m omega to the power of 2 end exponent 6 l
T subscript 3 end subscript colon T subscript 2 end subscript colon T subscript 1 end subscript equals 3 blank colon 5 blank colon 6
General
physics-

A ball of massopen parentheses m close parentheses 0.5 kg is attached to the end of a string having length left parenthesis L right parenthesis 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in rad/s) is

T cos invisible function application theta component will cancel m g.

T sin invisible function application theta Component will provide necessary centripetal force the ball towards center C.
therefore T sin invisible function application theta equals m r omega to the power of 2 end exponent equals m left parenthesis l sin invisible function application theta right parenthesis omega to the power of 2 end exponent
o r blank T equals m l omega to the power of 2 end exponent ⟹ omega equals square root of fraction numerator T over denominator m l end fraction end root r a d divided by s
o r blank omega subscript m a x end subscript equals square root of fraction numerator T subscript m a x end subscript over denominator m l end fraction equals square root of fraction numerator 324 over denominator 0.5 cross times 0.5 end fraction equals 36 blank r a d divided by s end root end root

A ball of massopen parentheses m close parentheses 0.5 kg is attached to the end of a string having length left parenthesis L right parenthesis 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in rad/s) is

physics-General
T cos invisible function application theta component will cancel m g.

T sin invisible function application theta Component will provide necessary centripetal force the ball towards center C.
therefore T sin invisible function application theta equals m r omega to the power of 2 end exponent equals m left parenthesis l sin invisible function application theta right parenthesis omega to the power of 2 end exponent
o r blank T equals m l omega to the power of 2 end exponent ⟹ omega equals square root of fraction numerator T over denominator m l end fraction end root r a d divided by s
o r blank omega subscript m a x end subscript equals square root of fraction numerator T subscript m a x end subscript over denominator m l end fraction equals square root of fraction numerator 324 over denominator 0.5 cross times 0.5 end fraction equals 36 blank r a d divided by s end root end root
General
maths-

The ellipse x squared over z squared plus y squared over h squared equals 1 and the straight line y = mx + c intersect in real points only if

fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator m to the power of 2 end exponent x to the power of 2 end exponent plus 2 m c x plus c to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction = 1
Þ (b2 + a2 m2) x2 + 2 mca2 x + a2 (c2 - b2) = 0
D > 0 Þ m2c2a4 - a2 (c2 - b2)(b2 + a2m2) > 0 Þ b2 + a2m2 > c2

The ellipse x squared over z squared plus y squared over h squared equals 1 and the straight line y = mx + c intersect in real points only if

maths-General
fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction plus fraction numerator m to the power of 2 end exponent x to the power of 2 end exponent plus 2 m c x plus c to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction = 1
Þ (b2 + a2 m2) x2 + 2 mca2 x + a2 (c2 - b2) = 0
D > 0 Þ m2c2a4 - a2 (c2 - b2)(b2 + a2m2) > 0 Þ b2 + a2m2 > c2
General
physics-

Four capacitors are connected as shown in figure. The equivalent capacitance between A and B is

fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 1 end fraction equals 3
C subscript s end subscript equals fraction numerator 1 over denominator 3 end fraction
Capacitance between A and B
C subscript p end subscript equals fraction numerator 1 over denominator 3 end fraction plus 1
fraction numerator 4 over denominator 3 end fraction mu F equals 1.33 mu F

Four capacitors are connected as shown in figure. The equivalent capacitance between A and B is

physics-General
fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 1 end fraction equals 3
C subscript s end subscript equals fraction numerator 1 over denominator 3 end fraction
Capacitance between A and B
C subscript p end subscript equals fraction numerator 1 over denominator 3 end fraction plus 1
fraction numerator 4 over denominator 3 end fraction mu F equals 1.33 mu F
General
physics-

A frictionless track A B C D E ends in a circular loop of radius R comma figure. A body slides down the track from point A which is at a height h equals 5 blankcm. Maximum value of R for the body to successfully complete the loop is

For successfully completing the loop,
h equals fraction numerator 5 over denominator 4 end fraction R rightwards double arrow R equals fraction numerator 2 h over denominator 5 end fraction equals fraction numerator 2 cross times 5 over denominator 5 end fraction equals 2 c m

A frictionless track A B C D E ends in a circular loop of radius R comma figure. A body slides down the track from point A which is at a height h equals 5 blankcm. Maximum value of R for the body to successfully complete the loop is

physics-General
For successfully completing the loop,
h equals fraction numerator 5 over denominator 4 end fraction R rightwards double arrow R equals fraction numerator 2 h over denominator 5 end fraction equals fraction numerator 2 cross times 5 over denominator 5 end fraction equals 2 c m
General
physics-

In a circuit shown in figure, the potential difference across the capacitor of 2 F is

therefore fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 1 end fraction equals fraction numerator 3 over denominator 2 end fraction
C subscript s end subscript equals fraction numerator 2 over denominator 3 end fraction F
Q equals C subscript s end subscript V equals fraction numerator 2 over denominator 3 end fraction cross times 12 equals 8 C
V subscript 1 end subscript equals fraction numerator Q over denominator C subscript 1 end subscript end fraction equals fraction numerator 8 over denominator 2 end fraction equals 4 V

In a circuit shown in figure, the potential difference across the capacitor of 2 F is

physics-General
therefore fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction plus fraction numerator 1 over denominator 1 end fraction equals fraction numerator 3 over denominator 2 end fraction
C subscript s end subscript equals fraction numerator 2 over denominator 3 end fraction F
Q equals C subscript s end subscript V equals fraction numerator 2 over denominator 3 end fraction cross times 12 equals 8 C
V subscript 1 end subscript equals fraction numerator Q over denominator C subscript 1 end subscript end fraction equals fraction numerator 8 over denominator 2 end fraction equals 4 V
General
physics-

The effective capacitance between points X and Y shown in figure. Assuming C subscript 2 end subscript equals 10 blank muF and that outer capacitors are all 4 blank muF is

The arrangement shows a Wheatstone bridge.
As fraction numerator C subscript 1 end subscript over denominator C subscript 3 end subscript end fraction equals fraction numerator C subscript 4 end subscript over denominator C subscript 5 end subscript end fraction equals 1 comma blanktherefore the bridge is balanced.
fraction numerator 1 over denominator C subscript s subscript 1 end subscript end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 4 end fraction equals fraction numerator 2 over denominator 4 end fraction equals fraction numerator 1 over denominator 2 end fraction comma C subscript s subscript 1 end subscript end subscript equals 2 mu blank F
Similarly, C subscript s end subscript subscript 2 end subscript equals 2 mu blank F
therefore effective capacitance
equals C subscript p end subscript equals C subscript s end subscript subscript 1 end subscript plus C subscript s end subscript subscript 2 end subscript equals 2 plus 2 plus equals 4 mu blank F

The effective capacitance between points X and Y shown in figure. Assuming C subscript 2 end subscript equals 10 blank muF and that outer capacitors are all 4 blank muF is

physics-General
The arrangement shows a Wheatstone bridge.
As fraction numerator C subscript 1 end subscript over denominator C subscript 3 end subscript end fraction equals fraction numerator C subscript 4 end subscript over denominator C subscript 5 end subscript end fraction equals 1 comma blanktherefore the bridge is balanced.
fraction numerator 1 over denominator C subscript s subscript 1 end subscript end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 4 end fraction equals fraction numerator 2 over denominator 4 end fraction equals fraction numerator 1 over denominator 2 end fraction comma C subscript s subscript 1 end subscript end subscript equals 2 mu blank F
Similarly, C subscript s end subscript subscript 2 end subscript equals 2 mu blank F
therefore effective capacitance
equals C subscript p end subscript equals C subscript s end subscript subscript 1 end subscript plus C subscript s end subscript subscript 2 end subscript equals 2 plus 2 plus equals 4 mu blank F
General
physics-

The four capacitors, each of 25 muF are connected as shown in figure. The DC voltmeter reads 200 V. the change on each plate of capacitor is

Charge on each plate of each capacitor
Q equals plus-or-minus C V equals plus-or-minus 25 cross times 10 to the power of negative 6 end exponent cross times 200
equals plus-or-minus 5 cross times 10 to the power of negative 3 end exponent C

The four capacitors, each of 25 muF are connected as shown in figure. The DC voltmeter reads 200 V. the change on each plate of capacitor is

physics-General
Charge on each plate of each capacitor
Q equals plus-or-minus C V equals plus-or-minus 25 cross times 10 to the power of negative 6 end exponent cross times 200
equals plus-or-minus 5 cross times 10 to the power of negative 3 end exponent C
General
physics-

For the circuit shown in figure the charge on 4muF capacitor is

Combined capacity of 1 blank mu F and 5 mu F blank= 1 + 5=6 blank mu F
Now, 4 mu F blankand 6 mu F are in series.
therefore blank fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 6 end fraction plus fraction numerator 3 plus 2 over denominator 12 end fraction equals fraction numerator 5 over denominator 12 end fraction
C subscript s end subscript equals fraction numerator 12 over denominator 5 end fraction mu F
Charge in the arm containing 4 mu F blankcapacitor is
q equals C subscript s end subscript cross times V equals fraction numerator 12 over denominator 5 end fraction cross times 10 equals 24 blank mu C

For the circuit shown in figure the charge on 4muF capacitor is

physics-General
Combined capacity of 1 blank mu F and 5 mu F blank= 1 + 5=6 blank mu F
Now, 4 mu F blankand 6 mu F are in series.
therefore blank fraction numerator 1 over denominator C subscript s end subscript end fraction equals fraction numerator 1 over denominator 4 end fraction plus fraction numerator 1 over denominator 6 end fraction plus fraction numerator 3 plus 2 over denominator 12 end fraction equals fraction numerator 5 over denominator 12 end fraction
C subscript s end subscript equals fraction numerator 12 over denominator 5 end fraction mu F
Charge in the arm containing 4 mu F blankcapacitor is
q equals C subscript s end subscript cross times V equals fraction numerator 12 over denominator 5 end fraction cross times 10 equals 24 blank mu C
General
Maths-

Polar of origin (0, 0) w.r.t. the circle x squared plus y squared plus 2 lambda x plus 2 mu y plus c equals 0 touches the circle x squared plus y squared equals r squared, if

Polar of origin (0, 0) w.r.t. the circle x squared plus y squared plus 2 lambda x plus 2 mu y plus c equals 0 touches the circle x squared plus y squared equals r squared, if

Maths-General
General
physics-

Electrons ejected from the surface of a metal, when light of certain frequency is incident on it, are stopped fully by a retarding potential of 3 volts Photo electric effect in this metallic surface begins at a frequency 6 x 1014s -1 The frequency of the incident light in s-1 is [h=6 x 10-34J-sec;charge on the electron=1.6x10-19C]

Electrons ejected from the surface of a metal, when light of certain frequency is incident on it, are stopped fully by a retarding potential of 3 volts Photo electric effect in this metallic surface begins at a frequency 6 x 1014s -1 The frequency of the incident light in s-1 is [h=6 x 10-34J-sec;charge on the electron=1.6x10-19C]

physics-General
General
physics-

Trajectories of two projectiles are shown in figure. Let T subscript 1 end subscriptand T subscript 2 end subscript be the time periods and u subscript 1 end subscript and u subscript 2 blank end subscripttheir speeds of projection. Then

Maximum height and time of flight depend on the vertical component of initial velocity
H subscript 1 end subscript equals H subscript 2 end subscript rightwards double arrow u subscript y end subscript subscript 1 end subscript equals u subscript y end subscript subscript 2 end subscript
Here T subscript 1 end subscript equals T subscript 2 end subscript
Range R equals fraction numerator u to the power of 2 end exponent sin invisible function application 2 blank theta over denominator g end fraction equals fraction numerator 2 left parenthesis u sin invisible function application theta right parenthesis left parenthesis u cos invisible function application theta right parenthesis over denominator g end fraction
equals fraction numerator 2 u subscript x end subscript u subscript y end subscript over denominator g end fraction
R subscript 2 end subscript greater than R subscript 1 end subscript therefore u subscript x subscript 2 end subscript end subscript greater than u subscript x subscript 1 end subscript end subscript blank o r blank u subscript 2 end subscript greater than u subscript 1 end subscript

Trajectories of two projectiles are shown in figure. Let T subscript 1 end subscriptand T subscript 2 end subscript be the time periods and u subscript 1 end subscript and u subscript 2 blank end subscripttheir speeds of projection. Then

physics-General
Maximum height and time of flight depend on the vertical component of initial velocity
H subscript 1 end subscript equals H subscript 2 end subscript rightwards double arrow u subscript y end subscript subscript 1 end subscript equals u subscript y end subscript subscript 2 end subscript
Here T subscript 1 end subscript equals T subscript 2 end subscript
Range R equals fraction numerator u to the power of 2 end exponent sin invisible function application 2 blank theta over denominator g end fraction equals fraction numerator 2 left parenthesis u sin invisible function application theta right parenthesis left parenthesis u cos invisible function application theta right parenthesis over denominator g end fraction
equals fraction numerator 2 u subscript x end subscript u subscript y end subscript over denominator g end fraction
R subscript 2 end subscript greater than R subscript 1 end subscript therefore u subscript x subscript 2 end subscript end subscript greater than u subscript x subscript 1 end subscript end subscript blank o r blank u subscript 2 end subscript greater than u subscript 1 end subscript
General
physics-

A body is projected up a smooth inclined plane with a velocity v subscript 0 end subscript from the point A as shown in figure. The angle of inclination is 45 degree and top B of the plane is connected to a well of diameter 40 m. If the body just manages to cross the well, what is the value of v subscript 0 end subscript? Length of the inclined plane is 20 square root of 2 m, and g equals 10 m s to the power of negative 2 end exponent

Let v be the velocity acquired by the body at B which will be moving making an angle 45 degree with the horizontal direction. As the body just crosses the well so fraction numerator v to the power of 2 end exponent over denominator g end fraction equals 40
or v to the power of 2 end exponent equals 40 g equals 40 cross times 10 equals 400
or v equals 20 blank m s to the power of negative 1 end exponent
Taking motion of the body from A to B along the inclined plane we have
u equals v subscript 0 end subscript comma a equals negative g sin invisible function application 45 degree equals negative fraction numerator 10 over denominator square root of 2 end fraction m s to the power of negative 2 end exponent
s equals 20 m comma v equals 20 m s to the power of negative 1 end exponent
As v to the power of 2 end exponent equals u to the power of 2 end exponent plus 2 a s
therefore blank 400 equals v subscript 0 end subscript superscript 2 end superscript plus 2 open parentheses negative fraction numerator 10 over denominator square root of 2 end fraction close parentheses cross times 20 square root of 2
or v subscript 0 end subscript superscript 2 end superscript equals 400 plus 400 equals 800 or v equals 20 square root of 2 m s to the power of negative 1 end exponent

A body is projected up a smooth inclined plane with a velocity v subscript 0 end subscript from the point A as shown in figure. The angle of inclination is 45 degree and top B of the plane is connected to a well of diameter 40 m. If the body just manages to cross the well, what is the value of v subscript 0 end subscript? Length of the inclined plane is 20 square root of 2 m, and g equals 10 m s to the power of negative 2 end exponent

physics-General
Let v be the velocity acquired by the body at B which will be moving making an angle 45 degree with the horizontal direction. As the body just crosses the well so fraction numerator v to the power of 2 end exponent over denominator g end fraction equals 40
or v to the power of 2 end exponent equals 40 g equals 40 cross times 10 equals 400
or v equals 20 blank m s to the power of negative 1 end exponent
Taking motion of the body from A to B along the inclined plane we have
u equals v subscript 0 end subscript comma a equals negative g sin invisible function application 45 degree equals negative fraction numerator 10 over denominator square root of 2 end fraction m s to the power of negative 2 end exponent
s equals 20 m comma v equals 20 m s to the power of negative 1 end exponent
As v to the power of 2 end exponent equals u to the power of 2 end exponent plus 2 a s
therefore blank 400 equals v subscript 0 end subscript superscript 2 end superscript plus 2 open parentheses negative fraction numerator 10 over denominator square root of 2 end fraction close parentheses cross times 20 square root of 2
or v subscript 0 end subscript superscript 2 end superscript equals 400 plus 400 equals 800 or v equals 20 square root of 2 m s to the power of negative 1 end exponent
General
Maths-

The equation of the circumcircle of the triangle formed by the lines y plus square root of 3 x equals 6 comma y minus square root of 3 x equals 6 and y=0 is

The equation of the circumcircle of the triangle formed by the lines y plus square root of 3 x equals 6 comma y minus square root of 3 x equals 6 and y=0 is

Maths-General
General
physics-

Which of the substance A comma blank B or C has the highest specific heat? The temperature v s time graph is shown

Substances having more specific heat take longer time to get heated to a higher temperature and longer time to get cooled.

If we draw a line parallel to the time axis then it cuts the given graphs at three different points. Corresponding points on the times axis shows that
t subscript C end subscript greater than t subscript B end subscript greater than t subscript A end subscript rightwards double arrow C subscript C end subscript greater than C subscript B end subscript greater than C subscript A end subscript

Which of the substance A comma blank B or C has the highest specific heat? The temperature v s time graph is shown

physics-General
Substances having more specific heat take longer time to get heated to a higher temperature and longer time to get cooled.

If we draw a line parallel to the time axis then it cuts the given graphs at three different points. Corresponding points on the times axis shows that
t subscript C end subscript greater than t subscript B end subscript greater than t subscript A end subscript rightwards double arrow C subscript C end subscript greater than C subscript B end subscript greater than C subscript A end subscript