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1151.

In Newtonian mechanics, mass, time, length and space were treated as "___________".

Answer»

relative
absolute
variables
infinity

Answer :B
Discussion
1152.

A small 0.500 kg object moves on a frictionless horizontal table in a circular path of radius 1.00 m. The angular speed is 6.28 "rad"//s. The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than 105 N, what is the radius of the smallest possible circle on which the object can move?

Answer»

0.376 m
0.434 m
0.573 m
0.659 m

Answer :C
Discussion
1153.

Thepolarising angle for a medium is 60^(@). Determine (i) the refraction index of the medium and (ii) the refracting angle.

Answer»


ANSWER :(i) 1.732 (II) `30^(@)`
Discussion
1154.

Two identical conducting rods are first connected independently to two vessels, one containing water at 100^(@)C, and the other containing ice at 0^(@)C. In the second case, the rods are joined end to end and connected to the same vessels. Let q_(1),q_(2) gram/sec be the rate of melting of ice in the two cases respectively. The ratio q_(1)//q_(2) is :

Answer»

`1//2`
`2//1`
`4//1`
`1//4`.

Solution :Rate of melting of ICE `qprop(1)/(R )`, where R is thermal resistance.
`:.(q_(1))/(q_(2))=(R_(2))/(R_(1))`
Since `R_(1)=(R)/(2)` being in SERIES & `R_(2)=R+R=2R`.
`:.(q_(1))/(q_(2))=(4)/(1)`
Correct choice is (C).
Discussion
1155.

A long straight wire carries a current of 50 A. What is the magnitude of intensity of magnetic induction at a point 15 cm from the wire.

Answer»

SOLUTION :`6.7 XX 10^(-5) T`
Discussion
1156.

Thin films, including soap bubbles and oil show patterns of alternative dark and bright regions resulting from interference among the reflected ligth waves. If two waves are in phase, their crests and troughs will coincide. The interference will be cosntructive and the amlitude of resultant wave will be greater then either of constituentwaves. If the two wave are not of phase by half a wavelength (180^(@)), the crests of one wave will coincide width the troughs of the other wave. The interference will be destructive and the ampliutde of the resultant wave will be less than that of either consituent wave. 1. When incident light I, reaches the surface at point a, some of the ligth is reflected as ray R_(a) and some is refracted following the path ab to the back of the film. 2. At point b, some of the light is refracted out of the film and part is reflected back through the film along path bc. At point c, some of the light is reflected back into the film and part is reflected out of the film as ray R_(c). R_(a) and R_(c) are parallel. However, R_(c) has travelled the extra distance within the film fo abc. If the angle of incidence is small, then abc is approxmately twice the film's thickness . If R_(a) and R_(c) are in phase, they will undergo constructive interference and the region ac will be bright. If R_(a) and R_(c) are out of phase, they will undergo destructive interference and the region ac will be dark. I. Refraction at an interface never changes the phase of the wave. II. For reflection at the interfere between two media 1 and 2, if n_(1) gt n_(2), the reflected wave will change phase. If n_(1) lt n_(2), the reflected wave will not undergo a phase change. For reference, n_(air) = 1.00. III. If the waves are in phase after reflection at all intensities, then the effects of path length in the film are: Constrictive interference occurs when 2 t = m lambda // n, m = 0, 1,2,3,... Destrcutive interference occurs when 2 t = (m + (1)/(2)) (lambda)/(n), m = 0, 1, 2, 3,... If the waves are 180^(@) out of the phase after reflection at all interference, then the effects of path length in the film ara: Constructive interference occurs when 2 t = (m + (1)/(2)) (lambda)/(n), m = 0, 1, 2, 3,... Destructive interference occurs when 2 t = (m lambda)/(n) , m = 0, 1, 2, 3,... The average human eye sees colors with wavelength between 430 nm to 680 nm. For what visible wavelength (s) will a 350 nm thick (n = 1.35) soap film produce maximum destructive interference?

Answer»

945 nm
473 nm
315 nm
None of these

Solution :Choices (a) and (c ) are eliminated immediately since are outside the visible range. For soap, `n_("air") gt n_("soap")`. The reflected wave, ray a, undergoes a PHASE change at the air-soap water interface. Ray `R_(b)` does not change phase at the soapy water-air interface. Based on interface REFLECTIONS, the two rays are out of phase. To maintain this, the optical path difference, 2t, must not produce a phase change in `R_(b)`. Therefore, the path must be an INTEGER multiple of `lambda`
`2 t = m lambda//N implies lambda = 2 t n//m = 2(35nm) (1.35)//m = 945// m`
For `m = 1, lambda = 945 nm`
For `m = 2, lambda = 473 nm` [this is choice (b)]
For `m = 3, lambda = 315 nm`
Choice (b), `lambda = 473 nm`, is the only choice in the visible range.
Discussion
1157.

In an experiment it was found that a sonometer in its fundamental mode of vibration and a tuningfork gave 5 beats when length of wire is 1.05 metre of 1 metre. The velocity of tranverse waves in sonometer wire is

Answer»

400 m/s
210 m/s
420 m/s
450 m/s

ANSWER :C
Discussion
1158.

Thin films, including soap bubbles and oil show patterns of alternative dark and bright regions resulting from interference among the reflected ligth waves. If two waves are in phase, their crests and troughs will coincide. The interference will be cosntructive and the amlitude of resultant wave will be greater then either of constituentwaves. If the two wave are not of phase by half a wavelength (180^(@)), the crests of one wave will coincide width the troughs of the other wave. The interference will be destructive and the ampliutde of the resultant wave will be less than that of either consituent wave. 1. When incident light I, reaches the surface at point a, some of the ligth is reflected as ray R_(a) and some is refracted following the path ab to the back of the film. 2. At point b, some of the light is refracted out of the film and part is reflected back through the film along path bc. At point c, some of the light is reflected back into the film and part is reflected out of the film as ray R_(c). R_(a) and R_(c) are parallel. However, R_(c) has travelled the extra distance within the film fo abc. If the angle of incidence is small, then abc is approxmately twice the film's thickness . If R_(a) and R_(c) are in phase, they will undergo constructive interference and the region ac will be bright. If R_(a) and R_(c) are out of phase, they will undergo destructive interference and the region ac will be dark. I. Refraction at an interface never changes the phase of the wave. II. For reflection at the interfere between two media 1 and 2, if n_(1) gt n_(2), the reflected wave will change phase. If n_(1) lt n_(2), the reflected wave will not undergo a phase change. For reference, n_(air) = 1.00. III. If the waves are in phase after reflection at all intensities, then the effects of path length in the film are: Constrictive interference occurs when 2 t = m lambda // n, m = 0, 1,2,3,... Destrcutive interference occurs when 2 t = (m + (1)/(2)) (lambda)/(n), m = 0, 1, 2, 3,... If the waves are 180^(@) out of the phase after reflection at all interference, then the effects of path length in the film ara: Constructive interference occurs when 2 t = (m + (1)/(2)) (lambda)/(n), m = 0, 1, 2, 3,... Destructive interference occurs when 2 t = (m lambda)/(n) , m = 0, 1, 2, 3,... A 600 nm light is perpendicularly incident on a soap film suspended air. The film is 1.00 mu m thick with n = .37. Which statement most accurately describes the interference of ligth reflected by the two surfaces of the film?

Answer»

The waves are close to destructive interference.
The waves are closed to constructive interference.
The waves how COMPLETE destructive interference.
The waves how complete constructive interference.

Solution :Again, ray `R_(a)` is reflected with PHASE changed and ray `R_(b)` is not. For the two reflected waves to interact,
`2 t = M lambda // n`, where M is EITHER m or m `+1//2`
`M = 2t n // lambda = 2 (1.00 XX 10^(-6) m) (1.35)//600 xx 10^(-9) nm = 4.5`
`M = m + 1//2` for `m = 4`. This describe the complete constuctive interaction.
Discussion
1159.

Thin films, including soap bubbles and oil show patterns of alternative dark and bright regions resulting from interference among the reflected ligth waves. If two waves are in phase, their crests and troughs will coincide. The interference will be cosntructive and the amlitude of resultant wave will be greater then either of constituentwaves. If the two wave are not of phase by half a wavelength (180^(@)), the crests of one wave will coincide width the troughs of the other wave. The interference will be destructive and the ampliutde of the resultant wave will be less than that of either consituent wave. 1. When incident light I, reaches the surface at point a, some of the ligth is reflected as ray R_(a) and some is refracted following the path ab to the back of the film. 2. At point b, some of the light is refracted out of the film and part is reflected back through the film along path bc. At point c, some of the light is reflected back into the film and part is reflected out of the film as ray R_(c). R_(a) and R_(c) are parallel. However, R_(c) has travelled the extra distance within the film fo abc. If the angle of incidence is small, then abc is approxmately twice the film's thickness . If R_(a) and R_(c) are in phase, they will undergo constructive interference and the region ac will be bright. If R_(a) and R_(c) are out of phase, they will undergo destructive interference and the region ac will be dark. I. Refraction at an interface never changes the phase of the wave. II. For reflection at the interfere between two media 1 and 2, if n_(1) gt n_(2), the reflected wave will change phase. If n_(1) lt n_(2), the reflected wave will not undergo a phase change. For reference, n_(air) = 1.00. III. If the waves are in phase after reflection at all intensities, then the effects of path length in the film are: Constrictive interference occurs when 2 t = m lambda // n, m = 0, 1,2,3,... Destrcutive interference occurs when 2 t = (m + (1)/(2)) (lambda)/(n), m = 0, 1, 2, 3,... If the waves are 180^(@) out of the phase after reflection at all interference, then the effects of path length in the film ara: Constructive interference occurs when 2 t = (m + (1)/(2)) (lambda)/(n), m = 0, 1, 2, 3,... Destructive interference occurs when 2 t = (m lambda)/(n) , m = 0, 1, 2, 3,... A thin of liquid polymer, n = 1.25, coats a slab of pyrex, n = 1.50. White light is incident perpendicularly on the film. In the reflection, full destructive interference occurs for lambda = 600 nm and full constructive interference occurs for lambda = 700 nm What is the thickness of the polymer film?

Answer»

120 nm
280 nm
460 nm
840 nm

Solution :Both rays `R_(a)` and `R_(b)` are reflected with a CHANGE of phase Therefore, the net change of phase with reflection is zero. For constructive interference , `2t = (m + 1//2) lambda//n`
`t = (m + 1//52) (600nm)//2 (1.25) = m xx 700 // 2(1.25)`
`600 m + 300 = 700 m implies 300 = 100 m implies m = 3`
SOLVING EITHER equation for t, we have `t = 3(700 nm)// 2 (1.25)`
`= 840 nm`
Discussion
1160.

Two charges 2muC and -2muC are placed at points A and B is 6 cm apart:-Identify an equipotential surface of the system.

Answer»

Solution :If two lines of force INTERSECT, at intersecting point there will be two DIFFERENT directions for electric field, which is not possible. Hence electric lines of force NEVER intersect ONE another
Discussion
1161.

A car moving at a speed v is stopped in a certain distance when the brakes produce a deceleration a. If the speed of car was nv , what must be the decceleration of the car to stop it in the same distance and in the same time?

Answer»

`n^(3)a`
NA
`n^(2)a`
`sqrt(na)`

Solution :When SPEED is made n times then DISTANCE require to STOP the body by same retarding force become `n^(2)` TIME.
Discussion
1162.

In a projectile motion, the maximum height reached equals to the horizontal range covered by the body, the angle of projection with the horizontal equals to

Answer»

`tan^-1 1`
`tan^-1` 1/4
`tan^-1` 2
`tan^-1`(4)

SOLUTION :MAXIMUM height =`(u^2 sin^2 theta)/(2G)`
HORIZONTAL Range = `(u^2 sin2 theta)/(g)`
GIVEN `(u^2 sin^2 theta)/(2g)` = `(u^2 sin2 theta)/(g)` or `sintheta = 2 sin 2 theta` or `tantheta` = 4 or `theta tan^-1(4)`.
Discussion
1163.

Which of the following can react with both HCl and NaOH ?

Answer»

ss
BeO
`Al_(2)O_(3)`
All of these

Solution :FACT
Discussion
1164.

Co-efficient of cubical expansion of water is - Ve between 0^@ and

Answer»

a)`4^@C`
B)`10^@C`
c)`15.5^@C`
d)`100^@C`

Answer :A
Discussion
1165.

An iron plug is to be placed in a ring made of brass. At room temperature 28^(@)C the diameter of the plug is 9.114 cm and that of the inside of the ring is 9.097 cm. To what common temperatue these both be brought in order to fit? Give that coefficient of linear expansion of iron is 1.1xx10^(-5).^(@)C^(-1) and that of brass sis 1.9xx10^(-5).^(@)C^(-1)

Answer»


ANSWER :`262.2^(@)C`
Discussion
1166.

A particle is projected vertically up and another is let fall to meet at the same instant. If they have velocities equal in magnitude when they meet, the distance travelled by them are in the ratio of

Answer»

`1:1`
`1:2`
`3:1`
`2:3`

ANSWER :C
Discussion
1167.

पराग नली का अध्यावरण द्वारा बीजाण्ड में प्रवेश कहलाता है :

Answer»

निभागी युग्मन
अण्डद्वारी प्रवेश
उपर्युक्त दोनों
इनमे से कोई नहीं

Answer :D
Discussion
1168.

Which of the following functions are not rational

Answer»

`F(X)=X^2-3X+2`
`f(x)=x/2`
`f(x)=(X^2-3X+2)/SQRTX`
None of these

Answer :C
Discussion
1169.

What is the name of the level formed due to the impurity atom in the forbidden gap near the valence band of a p-type semiconductor ?

Answer»

CONDUCTION LEVEL
FORBIDDEN Level
Donor Level
Acceptor Level

Answer :D
Discussion
1170.

A transversal cuts side AB and AD and diagonal AC of parallelogram ABCD in points P,Q and R respectively such that AP: PB =1 :2 and AQ: QD=2:3 . Find the ratio AR : RC

Answer»

`(7)/(2)`
`(3)/(7)`
`(2)/(7)`
`(2)/(9)`

ANSWER :D
Discussion
1171.

What is coordination number in hexagonal close packing

Answer»

8
6
12
4

Answer :C
Discussion
1172.

The K, L, and M energy levels of platinum lie roughly at 78, 12, and 3 KeV, respectively. The ratio of wavelength of K_(alpha) line to that of K_(beta) line in x-ray spectrum is

Answer»

`(22)/(3)`
`(3)/(22)`
`(22)/(25)`
`(25)/(22)`

ANSWER :D
Discussion
1173.

The waves used for the line - of - sight (LOS) communication is

Answer»

SKY waves
sound waves
space waves
ground waves

Answer :C
Discussion
1174.

Find the Q-value of the emitted alpha-particle in the alpha-decay of ._(88)^(226)Ra.m_(alpha)=4.00260u m(._(88)^(226)RA)=226.02540u m(._(86)^(222)Rn)=222.01750u

Answer»

8.93 MeV
4.93 MeV
12.3 MeV
None of these

Solution :`DELTAM=(226.02540)-(222.01750+4.00260)`
`Deltam=.0053am U`
`Q=Deltamxx931.25`
`Q=4.93MeV`
Discussion
1175.

The range of a projectile fired at angle of 15^@ is 50m. If it is fired with the same speed at an angle of 45^@, it's range will be

Answer»

25 m
50 m
100 m
77.6 m

Solution :`u^2/gsin2xx12^@`=50 `u^2/g = 50/sin30 = 50/(1/2)` = 100
H.R = `u^2/gxxsin2xx45^@=u^2/g =100
Discussion
1176.

An e.m.f. of 16V is induced in a coil of self-induc- tance 4H. The rate of change of current must be

Answer»

64 A/s
32 A/s
16 A/s
4 A/s

Answer :D
Discussion
1177.

Which of following is suitable for making permanent magnet ?

Answer»

Copper
Steel
Nickel
Soft iron

Answer :B
Discussion
1178.

At what angle to the horizon the sum be for is refelcted rays from the surfaceof pond to be completely polarised (mu of water =1.33)

Answer»

`16^(@)56`
`26^(@)56`
`36^(@)56`
`46^(@)56`

ANSWER :C
Discussion
1179.

What magnets are useful in an extremely fast train running magnetically levitate ?

Answer»

SOLUTION :SUPER CONDUCTOR MAGNETS
Discussion
1180.

mu_0 is permeability of vacuum, chi_(m) is susceptibility, then permeability of a material…

Answer»

`MU= mu_(0) (1+ chi_(m) )`
`mu= mu_(0) (chi_(m) -1)`
`mu= mu_(0) (1- chi_(m) )`
`mu= ( mu_0)/( 1+ chi_m) `

SOLUTION :Relative permeability `(mu_0)/( mu) = 1 + chi_(m)`
`RARR mu= mu_(0) (1+ chi_(m) )`
Discussion
1181.

To observe diffraction, the size of the obstacle :

Answer»

should be `(lambda)/(2)`, where `lambda` is the wavelength.
should be of the ORDER of wavelength
has no relation to wavelength.
should be much larger than the wavelength.

Solution :Condition of diffraction is that size of OBSTACLE is EQUAL to wavelength of light of the SOURCE.
Discussion
1182.

A sonometer wire fixed at one end has a solid mass M hanging from its other end to producetension in it. It is found that a 70cm length of wire produces a certain fundamental frequency when plucked. When the same mass M is hanging in water completely submerged in it, it is found that the length of the wire has to be changed by 5cm in order that it will produced the same fundamental frequency. The density of the material of mass M hanging from the wire is 7.26 xx 10^n . Find n .

Answer»


ANSWER :3
Discussion
1183.

The force between the two poles is reduced to 'x' newtons when their separation is increased n times. It is increased by 'y' newtons when their separation is made 1//n^(th) of their original value. What is the relation between x and y?

Answer»

SOLUTION :`y=(N^(4)-n^(2))X`
Discussion
1184.

For a BJT circuit shown , assume that the 'beta' of the transistor is very large and V_(BE)=0.7V. The mode of operation.

Answer»

Solution :`V_(BE)=0.7V`
Input junction is a forward BIASED .
SINCE ,
`V_(BE)=0.7V`
`V_(CE)=V_(BE)+V_(CB)`
`V_(CB)=V_(CE)-V_(BE)`
To determine `V_(CB)` we find `I_(C)`
`I_(C) -=I_(C)=(2-V_(BE))/(R_(2))=(2-0.7)/(1kOmega)`
`I_(C)=1.3mA`
`V_(CE)=V_("CC")-I_(C)(R_(1)+R_(2))`
`=10-1.3mA (10K+1K)`
`V_(CE)=-4.3V`
`V_(CE)=-4.3V-0.7`
`V_(CB)=-5V`
Discussion
1185.

Two identical balls each of mass m moving on straight track approaching towards each other with same speed. Find kinetic energy of the two ball system is equal to the total energy loss during collision E is the total kinetic energy of the balls before collision and E' is after collision and coefficient of restitution is e. Then choose the correct options.

Answer»

`(E)/(E') = 2`
`e=(1)/(2)`
`e=(1)/(sqrt(2))`
`(E)/(E')=sqrt(2)`

SOLUTION :
`E = m u^(2)`
`DELTA E=(1)/(2)((m)/(2))(2U)^(2)(1-e^(2))=m u^(2)(1-e^(2))`
`E'=E=DeltaE = m u^(2) (1-1+e^(2))=m u^(2)e^(2)`
since `E' = Delta E`
`m u^(2)e^(2)=m u^(2)(1-e^(2))`
`e^(2)=1-e^(2)rArr e^(2)=(1)/(2)rArr e=(1)/(sqrt(2))`
and `(E)/(E')=(1)/(e^(2))=2`
Discussion
1186.

When drift velocityis so small, how is it that an electric bulb lights up as soon as we turn the switch on?

Answer»

Solution :A particular electrons does not have to reach the electric bult to make current flow in the bulb. As soon as the switch is turned on, the potential difference gets applied at the ends of the TERMINALS of the bulb and an electric field is set up.
Changes occur at the initial state of SETTING of electric field which give rise to a changing magnetic field. the changing magnetic field gives rise to another changing electric field and so on. this ELECTROMAGNETIC phenomena takes place at the speed of electromagnetic waves which is the same as velocity of LIGHT `3xx10^(8)ms^(-1)`.
The set electric field make current flow through the bulb instantly.
Discussion
1187.

The uranium ore mined today contains only 0.72% of fissionable ""^(235)U, too little to make reactor fuel for thermal-neutron fission. For this reason, the mined ore must be enriched with ""^(235)U. Both ""^(235)U (T_(1//2) = 7.0 xx 10% y) and ""^(235)U (T_(1//2) = 4.5 xx 10^(9) y) are radioactive. How far back in time would natural uranium ore have been a practical reactor fuel, with a ""^(235)U//""^(235)U ratio of 3.0%?

Answer»


ANSWER :`1.7xx10^(9)y`
Discussion
1188.

An object is kept in front of a spherical mirror and it is found that the image of height double to that of the object is formed by the mirror. If radius of curvature of spherical mirror is 40 cm then what are the possible positions of object in front of the mirror and which type of mirror is this?

Answer»

Solution :Height of the image is more (double) than that of the object , hence mirror in use MUST be concave. Radius of curvature is 40 CM, hence focal length can be written as half of radius of curvature.
`f=-20cm` (Concave Mirror)
Here image obtained is enlarged. But we know that in case of concave mirror enlarged image can be real as welll as concave mirror enlarged image can real as well as VIRTUAL. Hence, there are two possible cases in given situation.
Real image :
`m=-2`
`"i.e."-(v)/(u)=-2 rArr v=2u`
`(1)/(v)+(1)/(u)=(1)/(f)rArr (1)/(2u)+(1)/(u)=(1)/(-20)`
`rArr""(3)/(2u)=(1)/(-20)rArru=-30cm`
Hence, when object is kept at , 30 cm in front of the mirror then its real and inverted image will be formed of double size.
Virtual Image :
`m=2`
`"i.e."-(v)/(u)=2 rArr v=-2u`
`(1)/(v)+(1)/(u)=(1)/(f)rArr (1)/(-2u)+(1)/(u)=(1)/(-20)`
`rArr""(1)/(2u)=(1)/(-20)rArru=-10cm`
Hence, when object is kept at 10 cm in front of the mirror then its virtual and erect image will be formed of double size.
Discussion
1189.

A composite string is made up by joining two strings of different masses per unit length rightarrow mu and 4mu the composite string is under the same tension A transverse wave pulse Y=(6mm) sin (5t+40x) where is in seconds and x in metres is sent along the lighter string towards the joint the joint is at x=0 the equation of the wave pulse reflected from the joint is

Answer»

`(2mm)sin (5t_40x)`
`(4mm)sin (40x-5t)`
`-(2mm)sin (5t-40x)`
`(2mm)sin(5r-10x)`

Solution :`V_(1)=SQRT(T/mu) , V_(2)=sqrt(T/(4mu))V_(2)leV_(1)`
`Rightarrow2^(nd) is DANSER Rightarrow phase change of pi`
WAVE REFLECTED from denser medium
`Rightarrow A_(r)=(V_(2)_V_(1))/V_(2)+V_(1)xx6=(v_(1)/2-V_(1))/(V_(2)/2+V_(1))xx6=-2mm`
`EQ^(n) Rightarrow -(2mm)sin (5t-40x)`
Discussion
1190.

A screen receives 3 watt of radiant flux of wavelength 6000 Å . One lumen is equivalent to watt 1.5 xx 10^(-3) "watt" of monochromatic light of wavelength 5550 Å . If relative luminosity for 6000 Å is 0.685 while that for 5550 Å is 1.00, then the luminous flux of thesource is

Answer»

`4 XX 10^(3) lm`
`3 xx 10^(3) lm`
`2 xx 10^(3) lm`
`1.37 xx 10^(3) lm`

Answer :d
Discussion
1191.

The velocity of a car travelling on a straight road is 36 km h^(- 1) at an instant of time. Now travelling with uniform acceleration for 10s, the velocity becomes exactly double. If the wheel radius of the car is 25 cm, then which of the following numbers is the closest to the number of revolutions that the wheel makes during this 10 s?

Answer»

84
95
126
135

Solution :Angle rotated before corning to rest.
`theta=(((V_(f)^(2))/(R^(2))-(v_(i)^(2))/(r^(2))))/(((2a)/r))=(v_(f)^(2)-v_(i)^(2))/((2ar))`
`therefore` NUMBER of rotation,
`N=(theta)/(2pi)=(v_(f)^(2)-v_(i)^(2))/((2ar)(2pi))=95`
Discussion
1192.

A compound microscope has a magnifying power of 100 when the image is formed at infinity. The objective has a focal length of0.5 cm and the tube length is 6.5 cm. What isthe focal length of the eyepiece.

Answer»

Solution :GIVEN data : Magnifying power of compound microscope = m = 100
The focal LENGTH of objective `f_0 = 0.5 cm` (or)
` 5 xx 10^(-3) m`
Tube length = 6.5 cm
To find :The focal length of eye PIECE `f_e` = ?
Since the image is formed at infinity, the real image produced by the objective lens should lie on the focus of the eye piece.
So ` v_0 + f_e = 6.5 = 6.5cm "" ....(1)`

Formula : Magnifying power = `(v_0)/(u_0) xx (D)/(F_e)`(for normal adjustment)
` rArrm = - [ 1 - (v_0)/(f_0) ] (D)/(F_e) [ therefore (v_0)/(u_0) =1 - (v_0)/( f_0)]`
``rArr 100= - [ 1 - (v_0)/(0.5) ] xx (25)/(f_e) `
(Taking D = 25 cm)
` rArr 100 f_e = - 1 ( 1-2 v_0) xx 25`
` rArr 2v_0 -4f_e = 1 "" ...(2)`
Solving (1) & (2)
` v_0 =4.5 cm `
` f_e = 2 cm `
The focal length of eye piece = 2 cm
Discussion
1193.

A biprism experiment is performed using yellow light of wavelength 5600overset@A. The yellows light was then replaced by red light of wavelength 6400overset@A. Find the value of n for which (n+1)th yellow bright band coincides with the nth red bright band for the same setting.

Answer»

4
5
6
7

Answer :D
Discussion
1194.

A man moving on track ABC, getting rain drops vertical for the path AB, and along same line of its motion for path BC then actual speed of rain drop is

Answer»

`10sqrt3ms^(-1)`
`(10)/(sqrt3)ms^(-1)`
`5sqrt3ms^(-1)`
Can't find

Solution :`bar(V)_(R ) = V_(X)i +V_(y)i`

For path AB `V_(x) = 5 ms^(-1)`
For path BC velocity of rain `_|_r` to BC should be zero hence
`V_(x)sin30^(@) = V_(y)COS30^(@) V_(y) = (5)/(sqrt3)ms^(-1)`
`V_(R ) = sqrt(V_(x)^(2)+V_(y)^(2))= (10)/(sqrt3)ms^(-1)`
Discussion
1195.

Assertion : For a given applied voltage, conduction current in n-type semiconductor is more than in p-type semiconductors. Reason : Mobility of electrons is greater than that of holes.

Answer»

If both assertion and reason are true and the reason is the correct explanation of the assertion
If both assertion and reason are true but the reason is not the correct explanation of the assertion
If assertio is true statement but reason is false.
If both assertion and reason are false statement

Solution :Electron EFFECTIVE mass is LESSER than hole effective mass. The mobility is proprtional to the carrier RELAXATION TIME and inversly proprtionnal to the carrier effective mass.
Discussion
1196.

Embryonal axis above the cotyledon is known as

Answer»

Hypocotyl
Funicle
Epicotyl
Raphe

Answer :C
Discussion
1197.

Derive expression for the de-Broglie wavelength of an electron moving under a potential difference of V volt. Name an experiment which verified the wave nature of electrons.

Answer»

Solution :Wavelength associated with a moving electron
Suppose an electron starting from rest falls through a potential difference V
Let charge on electron = e
Mass of electron = m
Velocity acquired by the electron in falling through a potential difference, V = v
WORK done on the electrons `=1/2mv^(2)`
Obviously, `1/2mv^(2) =EV`
or `v=sqrt((2eV)/m)`...........(1)
If `LAMBDA` is the wavelength associated with the electron, then according to de-Broglie WAVE equation.
`lambda =h/(mv)`
`=h/(msqrt(2eV)/m) = h/sqrt(2meV)`
Substituting the values of h,m and e. we get
`lambda =(12.27)/sqrt(V) Å, (10^(-10) m =1 Å)`
The wave NATURE of electrons was verified by Davisson and Gerner experiment.
Discussion
1198.

How does an increase in doping concentration affect the width of depletion layer of a p-n junction diode?

Answer»


Answer :On INCREASING doping CONCENTRATION the WIDTH of DEPLETION LAYER of a p-n junction diode decreases.
Discussion
1199.

In Fig., PQ represents a plane wavefront and AO and BP the corresponding extreme rays of monochromatic light of wavelength lambda. The value of angle theta for which the ray BP and the reflected ray OP interfere constructively is given by :

Answer»

`cos theta =(lambda)/(2D)`
`cos theta =(lambda)/(4d)`
`SEC theta =(lambda)/(3D)`
`sec theta =(2lambda)/(3d)`

Answer :B
Discussion
1200.

Which of the following reation is correct for a trasistor.

Answer»

`I_C=I_B+I_E`
`I_B=I_C+I_E`
`I_E=I_B-I_C`
`I_E=I_B+I_C`

SOLUTION :`I_E=I_B+I_C`
Discussion