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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 36951. |
Which of the following statements are true? (I) All radioactive element decay exponentially with time. (II) Half life time of a radioactive element is the time required for one half of the radioactive atoms to distintegrate. (III) Age of the earth can be determined by radioactive dating. (IV) Half life time of a radioactive element is fifty percent of its average life period. Select the correct answer using the codes given below. |
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Answer» I and II |
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| 36952. |
The wavelength of a monochromatic light in vacuum is 4000^@A. Its wave number in glass of refractive index 1.6 is: |
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Answer» `1xx10^6 m^(-1)` |
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| 36953. |
The electron drift arises due to the force experienced by electrons in the electric field Inside the conductor. But force should cause acceleration. Why then do the electrons acquire a steady average drift speed ? |
| Answer» Solution :In a current carrying conductor, each free electron is accelerated instantaneously by acceleration a = `(EE)/(m)`which increases the m drift speed of free electron until it collides with RANDOMLY oscillating positive ION in the metal. After collision, free electron comes to rest Again it gets accelerated, again its drift speed increases, again it undergoes anothercollision, again it comes to rest and so on. Here FREQUENCY of such collision is very large and so TIME lapsed between two successive collisions (known as relaxation time) is very small. Hence, on average we can consider all the free electrons drifting continuously. | |
| 36954. |
An object is placed at a distance of 25 cm on the axis of a concave mirror, having focal length 20 cm. Find the lateral magnification of an image. |
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Answer» -4 `therefore m =(-20)/(5)=-4` |
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| 36955. |
Define photoelectric effect and state the laws of photoelectric emission. |
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Answer» Solution :Laws of photoelectric emission are as FOLLOWS : (1) The velocity of electrons emitted is independent of the intensity of light and DEPENDS only UPON the frequency (or wavelength) of the incident light. (2) Photoelectric current (or the number of photoelectrons ejected per second) is proportional to intensity of incident light. (3) For a given metal, there exists a certain MINIMUM frequency below which the emission of electrons stops completely. This frequency is called the threshold frequency. (4) Photoelectric emission is an INSTANTANEOUS process. |
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| 36956. |
Obtainan expressionfor energy of an electronin Bohrorbit. Hence obtain the expression for its bindingenergy. A rectangular coilhaving 100 turnseach of length 1 cm and breadth0.5 cmis suspended in radialmagnetic induction0.002 T. The torsional constant ofsuspensionfiber is 2 xx 10^(-8) Nm/degree. Calculate the current sensitivity of a moving coil galvanometer. |
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Answer» SOLUTION :Energy of electronin Bohr orbit Accordingto Bohr'spostulate,electronrevolvesaround nucleusin circularorbit. As this motion has two energies. `:.` Total energypossessedby an electronis given by Totalenergy`(E ) = P.E. +K.E."...."(i)` Consider the electronrevolvingin the `n^(th)` orbit inhydrogen nucleus . Let `m` and -e be the MASS andcharge on the electron,r be the radius of orbit and v be LINEAR SPEED of electron. The electricpotential at a distance'r' from a charg`+e` is given by : `V = 1/(4 piepsi_(0)) .(e )/r` P.E. = Potential `xx` charge of electron P.E. `= ((1)/(4piepsi_(0)). (e)/(r ))(-e)` P.E. `= (-e^(2))/(4piepsi_(0)r)` By Bohr's `1^(st)`postulate, `(mv^(2))/(r) = (1)/(4piepsi_(0)) .(e^(2))/(r^(2))` `1/2 mv^(2) = (e^(2))/(8piepsi_(0)r)` `:.` K.E. `= (e^(2))/(8piepsi_(0)r)` `:.` Substitutingthe valuesK.E. and P.E. in equation (i), `E = (-e^(2))/(4piepsi_(0)r) + (e^(2))/(8piepsi_(0)r)` `E = (e^(2))/(8repsi_(0)r)` `E = (-me^(4))/(8epsi_(0)^(2)h^(2)n^(2))` Energy of electronin `n^(th)`orbit is - `E_(n) = -((me^(4))/(8epsi_(0)^(2)h^(2))).(1)/(n^(2))` The bindingenegy of electronis the minimum energyrequiredto make it free from the nucleus. i.e, `E_(n) = -((me^(4))/(8epsi_(0)^(2)h^(2))).1/(n^(2))` `:.` It must receive energy `= + ((me^(4))/(8epsi_(0)^(2)h^(2)). (1)/(n^(2))` to make `E_(oo) = 0` Numerical: Given : `n = 100, L = 1.0 cm = 0.01m` `b = 0.5 cm = 0.005 m, B= 0.002T` `S_(i) = ?` As we know, `S_(i) = (nBA)/(c)` `S_(i) = (100 xx 0.002 xx 0.01 xx 0.005)/(2 xx10^(-8))` `S_(i) = 0.5 xx 10^(3)` div/A `S_(i) = 500` div/A |
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| 36957. |
STATEMENT-1: When an uncharged parallel-plate capacitor is charged by conneting it to a cell, the heat produced in the circuit is equal to the energy stored in the capacitor. STATEMENT-1: The charge on a parallel-plate capacitor means the equal and opposite charges, on its inner surfaces. |
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Answer» Statement-1 is TRUE, Statement-2 is True, Statement-2 is a CORRECT EXPLANATION, for Statement-1. |
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| 36958. |
The ratio of the wavelength of H_alpha and H_beta lines of Balmer series is of the order of: |
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Answer» `10:1` |
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| 36959. |
A total charge of 5 muC is distributed uniformly on the surface of the thin walled semispherical cup. If the electric field strength at the centre of the hemisphere is 9xx10^(8) NC^(-1), then the radius of the cup is (1/(4pi epsi_(0))=9xx10^(9) N-m^(2) C^(-2)) |
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Answer» 5 mm |
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| 36960. |
A slit whose width is 5.0 cm is irradiated with microwaves of wavelength 2 cm . The angular spread of the central maximum ? (Assume that the incidence is along the normal ) |
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Answer» `2 SIN^(-1) (2//5)` |
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| 36961. |
(A): Out of the measurements A = 10.00 and B = 10.000 B is more accurate . (R): Percentage error in B is less than the percentage error in A. |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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| 36962. |
Which of the following four statements is false |
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Answer» A body can have zero VELOCITY and STILL be ACCELERATED. |
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| 36963. |
Three identical charges are placed on three vertices of a square. If the force acting between q_(1) and q_2 is F_(12) and between q_(1) and q_(3) is F_(13), then= .......... |
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Answer» `1/SQRT(2)`  `F_(12) = (k(q)(q))/a^(2) = (KQ^(2))/a^(2)` and `F_(13) = (k(q)(q))/(sqrt(2)a)^(2) = (kq^(2))/(2a^(2))` `therefore F_(13)/F_(12) = 1/2` |
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| 36964. |
The space around the test charge, in which it experiences a force is known as |
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Answer» |
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| 36965. |
For the hydrogen atom in its ground state, calculate (a) the probability density y (r) and (b) the radial probability density Psi^(2)(r) for r= a, where a is the Bohr radius. |
| Answer» SOLUTION :(a) `291nm^(-3)`, (B) `10.2nm^(-1)` | |
| 36966. |
The unit of magnetic flux is______in SI system and____in CGS system. |
| Answer» SOLUTION :WEBER or N. `mA^(-1)` and MAXWELL. | |
| 36967. |
State wheather the following two statement are true or false |
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Answer» TF |
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| 36968. |
The silver polishing the walls of a thermos bottle minimizes the transfer of heat by |
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Answer» conduction |
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| 36969. |
A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If a force F is applied at the free end of the rope, the net force exerted on the block will be: |
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Answer» `(PM)/(M+m)` `F=(PxxM)/(M+m)` HENCE choice is (a) |
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| 36970. |
A Red LED emits light 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is : |
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Answer» (A) 1.73 V/m `RHO c=(P)/(4pi r^(2))[because I =rho c]` `(1)/(2)in_(0)E_(0)^(2)c=(P)/(4pi r^(2))` `THEREFORE E_(0)^(2)=(2P)/(4pi r^(2)in_(0)c)` `therefore E_(0) sqrt((2P)/(4pi r^(2)in_(0)c))` `= sqrt((2xx0.1xx9xx10^(9))/((1)^(2)xx3xx10^(8)))` `= sqrt((18)/(3))=sqrt(6)=2.449 V//m ~~ 2.45 V//m` |
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| 36971. |
If velocity of sound at room temperature is 35078 cm^(-1), then velocity of sound at 0^(@) C |
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Answer» `33200 CMS^(-1)` |
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| 36972. |
Figure shows an experimental set up to find the value of an unknown resistance (R_x) using a meter bridge. AB is the uniform meter bridge wire of length L = 100 cm. When the sliding jockey is placed at J (AJ = x), the galvanometer shows zero deflection. AJ = x is known as balancing length and is measured using a scale having 1 mm as least count. (a) In one experiment known resistance R was taken to be 20 Omega and balancing length was measured as x = (20.0 pm 0.1) cm. find the value of R_(x). (b) Show that fractional error in calculated value of R_(x) is least when x =(L)/(2). What shall we do to ensure that x is close to L//2 ? |
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Answer» |
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| 36973. |
What are (a) the speed and (b) the period of a 400 kg satellite in an approximately circular orbit 900 km above the surface of Earth? Suppose the satellite loses mechanical energy at the average rate of 1.4 xx 10^5 J per orbital revolution. Adopting the reasonable approximation that the satellite's orbit becomes a "circle of slowly diminishing radius," determine the satellite's (c) altitude, (d) speed, and (e) period at the end of its 1500th revolution. (f) What is the magnitude of the average retarding force on the satellite? Is angular momentum around Earth's center conserved for (g) the satellite and (h) the satellite-Earth system (assuming that system is isolated)? |
| Answer» SOLUTION :(a) `7.4 km//s`, (b) `6.2 xx 10^3s`>>103 min, (c ) `7.6 xx 10^3m`, (d) `~~7.5km//s,` (e) `6.0xx10^3s,` (f) `3.1 xx 10^(-3)N,` (g) The ANGULAR momentum of the satellite is not conserved, (h) the angular momentum of the saltellite -Earth SYSTEM is verynearly conserved. | |
| 36974. |
An alternating voltage E = 200sqrt(2)sin(100t)V is connected to a 1muF capacitor through an a.c. ammeter, what will be the reading of the ammeter |
| Answer» Answer :A | |
| 36975. |
According to classical physics, potential energy of electron moving in circular orbit of radius r is……… |
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Answer» `-(1)/(4piin_(0)).(ZE^(2))/(r)` `U=(1)/(4piin_(0)).(q_(1)q_(2))/(r)impliesU=(1)/(4piin_(0)).((Ze)(-E))/(r)` `:.U=-(1)/(4piin_(0)).(Ze^(2))/(r)` |
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| 36976. |
A particle executes simple harmonic oscillation with an amplitude a.The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is : |
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Answer» `(T)/(8)` When `""Y=( r )/(2)` then `""( r )/(2)=r sin omega t` `(1)/(2)=sin omega t` `impliessin((pi)/(6))=sin omega t` `IMPLIES""(pi)/(6)=omega t` or `(pi)/(6)=(2PI)/(T).timpliest=(T)/(12)`. Correct choice is (B). |
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| 36977. |
Draw a ray diagram to show how the image is formed when the object is placed between/and 2/distances from a convex lens. Deduce the relation between the distances of the object and the image from the lens and the focal length of the lens under this condition. |
Answer» SOLUTION :The RAY diagram is shown in FIG. 9.65. The image A.B. is real, inverted and magnified one. As `triangleA.B.C`and `triangleABC`are similar, HENCE `(A.B.)/(AB) = (CB.)/(CB)`…….(i) Again as `triangleA.B.F` and `triangleALCF` are similar, hence `(CB.)/(CB) = (B.F)/(CF) = (CB. - CF)/(CF)` As per sign conventions used, `CB =-u, CB. = +v` and `CF = +f` `therefore v/(-u) = (v-f)/f` or `vf =-uv + uf` Dividing both sides by uvf, we get `1/u = -1/f + 1/v` or `1/v - 1/u = 1/f` |
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| 36978. |
A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil? |
| Answer» SOLUTION :`PI XX 10^(-4) T ~~ 3.1 xx 10^(-4) T` | |
| 36979. |
Two bar magnets A and B are placed one over the other and are allowed to vibrate in a vibration manetometer. They of A and B are on the same side, while opposite poles lie on the same side. If M_(A) and M_(B) are the magnetic moments of A and B and if M_(A)gtM_(B), the ratio of M_(A) and M_(B) is |
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Answer» `4:3` `n=(1)/(2pi)sqrt((MB_(H))/(I)` `(n_(1))/(n_(2))=sqrt((M_(A)+M_(B))/(M_(A-M_(B))` `n_(1)=20, n_(2)=15` (GIVEN) `20/15=sqrt((M_(A)+M_(B))/(M_(A-M_(B))` `implies(M_(A)+M_(B))/(M_(A)-M_(B))=(4/3)^(2)` `implies9[M_(A)+M_(B)]=16[M_(A)-M_(B)]` `implies7 M_(A)=25 M_(B)` `M_(A):M_(B)=25:7` |
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| 36980. |
In the first excited state of hydrogen atom, its radius is found to be 21.2xx10^(-11) m. Calculate its Bohr radius in the ground state. Also calculate the total energy of the atom in the second excited state. |
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Answer» SOLUTION :First excited STATE means, n=2 `r_(0)=n^(2)r_(1)` `r_(1)=(21.2xx10^(-11))/(4)` `=5.3xx10^(11)m` `E=-(13.6)/(n^(2))=-3.4eV`. |
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| 36981. |
Air is thrown on a sail attached to a boat from an electric fan placed on the boat. Will the boat start moving? |
| Answer» SOLUTION :No, From Newton's 2nd law of motion we know that a body comes to ACCELERATION when some external force is applied over the body. When the FAN pushes the sail by the air, then air ALSO pushes the fan in opposite direction. The fan is also part of the boat and hence boat cannot be propelled. | |
| 36982. |
An astronomical telescope has an angular magnification of magnitude 5 for distant objects. The separation between the objective and an eye piece is cm and the final image is formed at infinity. Calculate the focal length of the objective and the focal length of the eye piece? |
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Answer» SOLUTION :`m=-f_(0)//f_(E)` `f_(0)=5f_(e)` `L=f_(0)+FE` `f_(e)=36//6=6cm` `f_(0)=30cm` |
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| 36983. |
When an object placed between the pole and focus of a concave mirror the image formed is ------ |
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Answer» a. real, INVERTED and SMALL For an object between pole and the focus, f `lt u lt ` 0 i.e., `(1)/(f) - (1)/(u) gt 0 "" i.e., (1)/(v) lt 0 ` Hence v is positive and lies to the right of the pole and so the image is virtual . m = `(-v)/(u ) = (v)/(|u|) "since" (1)/(v) lt (1)/(|u|) , VGT |u|` . , Hence .m. is positive and`gt` 1. so the image is enlarged. |
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| 36984. |
A metal sphere is kept on an insulating stand. A negatively charged rod is brought near it , then the sphere is earthed as shown. On removing the earthing, and taking the negatively charges rod away , what will be the nature of charges on the sphere? Give reason for your answer. |
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Answer» Solution :The CHARGES on the sphere is positive charge. The +ve charge is DEVELOPED on account of electric induction in different STEPS shown in following figure. ` (##U_LIK_SP_PHY_XII_C01_E08_036_S01.png" width="80%"> |
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| 36985. |
A sheet of cellophane acts as a half-wave plate for light of wavelength 4000Å. If the index of refraction did not change with wavelength, how would the sheet behave with respect to light of wavelength 8000Å? |
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| 36986. |
A charged 8mF capacitor having charge 5mC is connected to a 5mH inductor. What is : iv) the magnetic energy in the inductor at the instant when charge on capacitor is 4mC? |
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Answer» Solution :(iv) Magnetic ENERGY in the inductor when charge on CAPACITOR is 4mC. `U_(L) = U - U_(C ) = (1)/(2) (q_(0)^(2))/(C ) - (1)/(2) (Q^2)/( C ) = (q_(0)^(2) - q^2)/(2 C)` Here `q_(0) =5 m C, q = 4 m C` |
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| 36987. |
For a transistor the current amplification factor is 0.8 . If the transistor is now in common emitter configuration , the change in collector current when the base current changes by 6 mA is : |
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Answer» Solution :As `ALPHA= 0.8 and BETA = alpha/(1-alpha) =(0.8)/(1-0.8)=4` `(DeltaI_c)/(DeltaI_b) = 4 " or "DeltaI_c=4xx6 mA = 24 mA` |
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| 36988. |
A radioactive ncleus A finaly transforms into a stable nucleus. B Then A and B can be |
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Answer» ISOBARS |
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| 36989. |
In a certain region of space the electrostatic field depends only on the coordinates x and y as follows. E=0"for"sqrt(x^(2)+y^(2)) lt r_(0) E=a(xhati+yhatj)(x^(2)+y^(2))"for"sqrt(x^(2)+y^(2)) gt r_(0) Where a is a positive constant and hati and hatj are the unit vectors along the X-and Y-axes. find the charge within a sphere of radius 2r_(0) with the centre at the origin. |
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Answer» |
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| 36990. |
A closed coil consists of 500 turns wound on a rectangular frame of area 4.0 cm^(2) and has a resistances of 50Omega. It is kept with its plane perpendicular to a unifom magnetic field of 0.2 weber/"meter"^(2). Calculate the amount of charge flowing through the coil if it is turned over (rotated through 180^(@)). Will this answerdepend on the speed with which the coil is rotated? |
| Answer» SOLUTION :`1.6 XX 10^(-3)C`, No | |
| 36991. |
You are given the circuits as shown in figure, which consist of NAND gates. Identify the logic operation carried out by the two circuits. |
Answer» Solution :(a) The output of the first NAND gate is GIVEN on the input of the NOT gate made by connecting both of the NAND gate input, soits TRUTHTABLE is given as follows:  From the TRUTH table it can be said that `Y=A+B` is obtained so the given circuit act as AND gate. (b) The output of two NOT GATES made from the NAND gate given in input of the NAND gate, so its truth table is given as follows.  From the truth table it can be said that given circuit act as OR gate. |
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| 36992. |
Which ofthe followingcompound has highest enol content ? |
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Answer»
 This COMPOUNDCAN frommoststableenolamongall the givencompounddue toaromticstabilityof BENZENE RING . |
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| 36993. |
A man in a car at location Q on a straight highway is moving with speed v. He decides to reach a point P in a field at a distance d from the highway (point M) as shown in the figure. Speed of the car in the field is half to that on the highway. What should be the distance RM, so that the time taken to reach P is minimum ? |
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Answer» d |
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| 36994. |
(a) In the following diagram 'S' is a semiconductor . Would you increase or decrease the value of R to keep the reading of the ammeter A constant when S is heated ? Given reason for you answer . (b) Draw the circuit diagram of photodiode and explain its working . Draw its I-V characteristics. |
| Answer» SOLUTION :On heating S, resistance of SEMICONDUCTORS S is decreased so to COMPENSATE the value of resistance in the circuit R is increased. | |
| 36995. |
A ray of light falls on a transparent sphere sphere with centre C as shown in the figure. The ray emerges from the sphere parallel to the line AB. Find the angle of refraction at A, if refractive index of the material of the sphere is sqrt(3). |
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Answer» Solution :`(SIN i)/( sin R ) = MU` ` ( sin 60^(@))/( sin r ) = sqrt( 3) RARR r = 30^(@)` |
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| 36996. |
A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, E = 6.3 hat j V/m. What is B at this point? |
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Answer» Solution :Using Eq. (8.10), the magnitude of B is `B =(E)/(C )` `=(6.3V//m)/(3xx10^(8) m//s)=2.1xx10^(-8) T` To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using VECTOR algebra, `E XX B` should be along x-direction. SINCE, `(+ hat J) xx (+ hat k ) = hat i` , B is along the z-direction. THUS, `B = 2.1 xx 10^(-8) hat k T` |
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| 36997. |
An electric dipole of dipole moment vecP is placed parallel to the uniform electric field of intensity vecE. On rotating it through 180^(@) the amount of work done is ..... . |
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Answer» Solution :` W = -" p E cos " theta_(2) - (- p E cos theta_(1))` `= p E( cos 0^(@) - cos 180^(@)]` p E [ 1-(-1)] = 2 pE |
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| 36998. |
A galvanometer of coil resistance 100Omegais connected to a shunt of resistance 10 Omega. The current through the galvanometer is i_1 , the current through the shunt is i_2and the total current into the combination is i_3 , then the ratioi_1 : i_2 : i_3is |
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Answer» `1:10:11` |
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| 36999. |
In the circuits show epsilon=15V, R_(1)= 1 Omega , R_(2) = 1 Omega, R_(3) = 2 Omega and L= 15 H . Findthe current i_(1),i_(2),i_(3) (i)immediately after the switch is closed (ii)immediatelyafter theopening from the closed position (iii) sufficientlylongafter , the switch is openedf rom thecolsed position, |
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Answer» |
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| 37000. |
The temp. at which centigrade and fahrenheit scales have same reading is: |
| Answer» Answer :A | |
