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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Calculate the energy required to remove an electron from He_(e)^(+) ion. |
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Answer» Solution :Required ENERGY `E=(nZ^(2)e^(4))/(8 epsi_(0)^(2)n^(2)h^(2))` `E=-(me^(4))/(8 epsi_(0)^(2))*(Z^(2))/(n^(2))` Here `(me^(4))/(8 epsi_(0)^(2)h^(2))=13.6eV` `E=13.6(Z^(2))/(n^(2))eV` for `H_(e)^(+)` ION, Z=2 and n=1 `E=(13.6xx4)/(1)` `=54.4eV` |
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| 2. |
A freshly prepared sample of a certain radioactive isotope has an activity of 10 mCi. After 4.0 hr its activity is 8.00 mCi. (a) Find the decay constant and half life (b) How many atoms of the isotope were contained in the freshly prepared sample. (c) What is the sample's activity 30.0 hr after it is prepared. |
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| 3. |
The current gain of a transistor in common base and common-emitter configurations, called alphaandbetaare related as |
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Answer» `(1)/( ALPHA)- (1)/( BETA) =1` |
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| 4. |
A nucleus of amss 20 u emits a gamma photon of energy 6 MeV . If the emission assume to occur when nuclues is free and at rest then the nulceus will have kintetic energy nearest to (take 1u=1.6xx10^(-27)Kg) |
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Answer» 10 KeV Momentum of photon = momentum of NUCLEUS `(E)/(c )=sqrt(2m(KE))` `(6xx10^(6)xx1.6xx10^(-19))/(3xx10^(8))=sqrt(2xx20xx1.6xx10^(-27)xxKE)` `3.2xx10^(-21)=sqrt(64xx10^(-27)xxKE)` `((3.2)^(2)xx10^(-42))/(64xx10^(-27))=KE` `KE=1.6xx10^(-16)J` `KE=(1.6xx10^(-16))/(1.6xx10^(-19))eV=10^(3)eV=1keV` |
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| 5. |
A block of mass 4m is attached to a vertical spring of spring constant k. The block is made by gluing two blocks of mass 3m and m, respectively. Initially, the block is in equilibrium and at rest (Fig. 15.13a). At t=0, the part of the block having ass 3m falls down. Considering hanging point on ceiling as y=0 and downward direction as positive y, find y-coordinate of mass m as a function of time. |
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Answer» Solution :(1) As the original block of mass 4m is in equilibrium, the combined mass will be STRETCHING the spring by 4mg/k. Thus, the spring will be 4mg/k below the natural length. As the bigger block falls, the smaller block will experience a FORCE larger than its weight in upward direction. (2) Theblock remaining attached to the spring will be undergoing SHM. To find equation of SHM, we need to find angular frequency, amplitude, and initial phase  Calculations: When the EXTENSION in the spring is x, the net force on the block is `mg- kx = (md^(2)x)/(DT^(2))` Again as x= 0 is not position of equilibrium, we have to shift origin to the position of equilibrium. Defining `x= (mg//k) +s`, as at s=0, x = mg/k is the position of equilibrium. Substituting for x in the equation we get `mg- k ((mg)/(k) + s) = (md^(2)s)/(dt^(2))` On solving we get, `(d^(2)s)/(dt^(2)) = - ((k)/(m))s` `s= x_(m) sin (omega t+ phi)` (with `omega= sqrt((k)/(m))`) Now we have to find `x_(m)`. We have defined `x_(m)` as distance between v=0 and a=0.At t=0, the block was at rest with extension of spring being 4mg/k and at the new equilibrium position extension of spring is mg/k. Thus, `x_(m) = (3mg)/(k)` Now we have to find the initial phase. As the particle is at rest at t=0 on positive side, the initial phase will be `pi//2`. `phi = (pi)/(2)`  Thus, the y coordinate of the block `y= l_(0) + (mg)/(k) + s` Substituting, all the values from above, we get `y= l_(0) + (mg)/(k) + (3mg)/(k) sin [sqrt((k)/(m)) t + (pi)/(2)]` Note: At TIME t=0, the distance of the block from the ceiling is `l_(0) + 4 mg//k`, which can be verified by putting t=0 in the above equation. |
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| 6. |
A Tennis player wants to slam serve a that the bulluet just inside the opposite comer of the shuld be ratio (u^(y))/(u_(x)) of component of initial velocity 'u' in y and x direction? |
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Answer» `(OMEGA)/(L)` |
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| 7. |
Using the Hund rules, find the total angular monentum of the atom in the ground state whose partially filled subshell contains (a) three d electrons, (b) seven d electrons. |
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Answer» Solution :(a) The minimum spin angular momentum of three electrons can be `S=(3)/(2)`. This state is totally symmertic and hence the conjugate angular wavefunction must be antisymmetric By Pauli's exclusion priciple the totally antisymmetric state must have DIFFERENT MAGNETIC quantum numbers. It is easy to see that for `d` electrons the maximum value of the magnetic quantum number for orbital angular momentum`|M_(LZ)|=3` (from `2+1+0`). Higher value violate Pauli's priciple.THUS the state of highest orbital angular momentum consistent with Pauli's principle is `L=3`. The state of the ATOM is then `.^(4)F_(J)` where `J=L-S` by Hund's rule. Thus we get `.^(4)F_(3//2)` The magnitude of the angular momentum is ` ħ sqrt((3)/(2).(5)/(2))=( ħ)/(2)sqrt(15)` (b) Seven `d` electrons mean three holes. Then `S(3)/(2)` and `L=3` as before. But `J=L+S=(9)/(2)` by Hund's rule for more than half filled shell. Thus the state is `.^(4)F_(9//2)` Total angular momentum has the magnitude `ħsqrt((9)/(2).(11)/(2))=(3 ħ)/(2)sqrt(11)` |
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| 8. |
In a uniformly charged dielectric sphere, a very thin tunnel has been made along the diameter as shown in figure. A charge partivel -q having mass m is released from rest at one end of the tunnel. For the situation described, mark the correct statement(s), (negalect gravity). |
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Answer» Charge particle will perform SHM about center of the SPHERE as mean position. At a center, the FORCE experienced by the charge particle is zero, so it is a position of equilibrium. As we displace the charge so it is a position of equilibrium. As we displace the chare from the equilibrium position, electric force starts acting on it toward equilibrium position and hence equilibrium is a stable one. At a distance `x` from the center of sphere (equilibrium position), force experienced by the charge particle is `F=(qQx)/(4piepsilon_(0)R^(3))` [for `xltR`] As `F=x`, it performs `SHM` about the center. Time period can be calculated by using `F=mw^(2)x`. |
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| 9. |
The distance travelled by a body moving along a line in time t is proportional to t^3. The acceleration - time (a, t) graph for the motion of the body will be |
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Answer»
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| 10. |
A hose pipe lying on the ground shootsstream of water upward at an angle 60^(@) to the horizontal at a speed of 20ms^(-1). The water strikes a wall 20m away at a height of (g=10ms^(-2)) |
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Answer» 14.64 m |
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| 11. |
A fish rising vertically to the surface of water in a lake uniformly at the rate of 3 ms^(-1)observes a bird diving vertically towards the water at a rate of 9 ms^(-1)vertically above it. If the refractive index of water is 4/3, the actual velocity of the dive of the bird is |
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Answer» 6 |
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| 12. |
Assertion : If a proton and an alpha-particle enter a uniform magnetic field perpendicularly, with the same speed, then the time period of revolution of the -particleis double than that of proton. Reason : In a magnetic field, the time period of revolution of a charged particle is directly proportional to mass. |
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Answer» If both ASSERTION and REASON are TRUE and the reason is the CORRECTEXPLANATION of the assertion. |
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| 13. |
In the given circuit, assuming point A to be at zero potential, use Kirchhoff's rules to determine the potential at point B. |
Answer» SOLUTION : USING Kirchhoff.s first rule, we find that current FLOWING in branch DC is 1 A from D to C. Now, APPLYING Kirchhoff.s second rule to the circuit ACDB, we have `V_A + 1+1 xx 2 -2 = V_B` `V_B = V_A + 1` But `V_A = 0` volt , hence `V_B = 0 + 1 = 1V` |
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| 14. |
Find equivalent capacitance across AB (all capacitancesin muF) |
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Answer» `(20)/3muF` |
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| 15. |
Clones' are individuals that have exactly the same |
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Answer» lifespan |
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| 16. |
An inductor of 10mH is connected to a 18V battery through a resistor of 10kOmega and a switch After a long time, when the maximum current is set up in the circuit, the current is switched off. Calculate the cur- rent in the circuit after 2 mus. |
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Answer» SOLUTION :GIVEN that L=10 mH = `10^(2)H, R=10k OMEGA= 10^(4)Omega` and E=18V , We know that `I_(0)=(E)/(R)=(18)/(10^(4))A` Time constant, `tau_(L)=(L)/(R)=(10xx10^(-3))/(10xx10^(3))=10^(-6)` sec We know that, `I=I_(0)e^(-(R//L)t)` Here `t = 2 mu s = 2xx10^(-6) s, I_(0) = 18 XX 10^(-4)A` `I=18 xx 10^(-4)e^(-2)=2.48 xx 10^(-4)A` |
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| 17. |
Two parallel wires 1 m apart carry currents of 1 A and 3 A respectively in opposite directions. The force per unit length acting between these two wires is |
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Answer» `6 xx 10^(-5) NM^(-1)` attractive `F= (2 xx 10^(-7) xx 1 xx 3)/(1) = 6 xx 10^(-7) Nm^(-1)` |
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| 18. |
Two monoatomic ideal gases 1 and 2 of molecular masses m_(1) and m_(2) respectively are enclosed in separate containers ketp at the same temperature. The ratio of the speed of sound in gas 1 t that in gas 2 is : |
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Answer» `sqrt((m_(1))/(m_(2)))` |
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| 19. |
You are asked to compare the emf of two cell using a potentiometer . Draw the circuit diagram for this and explain how will you determine the ratio of emfs. |
| Answer» SOLUTION :will not GET a BALANCING POINT . | |
| 20. |
In gametes cell |
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Answer» MEIOSIS TAKES place |
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| 21. |
In a meter bridge, the value of resistance in the resistance box is 10Omega. The balancing length is l_(1)=55 cm. Find the value of dunknown resistance. |
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Answer» <P> Solution :`Q=10Omega``(P)/(Q)=(l_(1))/(100-l_(1))=(l_(1))/(l_(2))` `P=Q XX (l_(1))/(100-l_(1))` `P=(10xx55)/(100-55) RARR P=(550)/(45)=12.2Omega` |
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| 22. |
Two identical charged protrudes moving with the same speed enter a region of uniform magnetic field . If one of these enters normal to the fielddirection and the other enters along a direction at 30^(@) with the field .What would be the ratio of their angular frequencies. |
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Answer» Solution :`W = (qB)/m` INDEPENDENCE of ANGLE of entrance with the MAGNETIC field. ` w_(1) : w_(2) = 1: 1` |
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| 23. |
In Young.s double slit experiment,when light of wavelength 4000 A^(0) is used 90 fringe are seen on the screen , when lightof 3000 A is used , the number of fringes seen is |
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Answer» 70 |
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| 24. |
Twelve charges of charge q are situated at the corners of the 12 sided polygon of side a. What is the net force on the charge Q at the centre |
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Answer» ZERO |
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| 25. |
AB and CD are long straight conductor, distance d apart, carrying a current l. The magnetic field at the midpoint of BC is |
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Answer» `(-mu_(0)I)/(2pid)HATK` |
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| 26. |
Figure represents a crystal unit of cesium chloride, CsCl. The cesium atoms, represented by open circles are situated at the corners of a cube of side 0.40 mm, whereas a Cl atom is situated at the centre of the cube. The Cs atoms are deficient in one electron while the Cl atom carries an excess electron. ltBrgt (i). What is the net electric field on the Cl atom due to eight Cs atoms? (ii) Suppose that the Cs atom at the corner A is missing. what is the net force now on the Cl atom due to seven remaining Cs atom? |
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Answer» Solution :(I )From the given figure, we can analy se the chlorine atom is at the centre of the cube i.E., at equal DISTANCE from all the eight coerners of cube where cesium atoms are placed. Thus due to symmetry the FORCES due to all Cs tons, on Cl atom will cancel out. Hence, `E=(F)/(q)` where F=0 `thereforeE=0` (ii) Thus, net force on Cla atom at A would be: where r=distance between Cl ion and Cs ion. Applying Pythagorous theorem we get `r=sqrt((0.20)^(2)+(0.20)^(2)+(0.20)^(2))xx10^(-9)m` `=0.346xx10^(9)m` Now, `F=(q^(2))/(4piepsi_(0)r^(2))=(e^(2))/(4piepsi_(0)r_(2))` `=(9xx10^(9)(1.6xx10^(-19))^(2))/((0.346xx10^(-9))^(2))=1.92xx10^(-9)N` |
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| 27. |
The given graph [Figure] shows variation of charge 'q' versus potential difference V for two capacitors C_1 and C_2. Both the capacitors have same plate separation but plate area of C_2 is greater then that of C_1. Which line (Aor B) corresponds to C_1 and why ? |
| Answer» SOLUTION :As PLATE area of `C_2` is greater than that of `C_1`, HENCE `C_2 gtC_1`. As `C = q/V` and from given graphs we CONCLUDE that `C_AgtC_B`. THUS, line A represents `C_2` and line B represents `C_1`. | |
| 28. |
What is called series connection of cell ? Derive equation of equivalent emf of two cell with emfepsilon_(1) and epsilon_(2) connected in series. |
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Answer» <P> Solution :`rArr `When one terminal of two cell connected with each other and another terminal of each cell are kept free then such connection is called series connection of cell. `rArr` In figure, battery having emf `epsilon_(1)` and internal resistance `r_(1)` is connected between A arid B and battery with emf `epsilon_(1)` and internal resistance `r_(2)` is connected between B and C. `rArr ` Let equivalent emf between A and C be `E_(eq)` and equivalent resistance of `r_(1) and r_(2)` be `r_(eq)` . `rArr ` Let potential at point A, B and C be V(A), V(B) and V(C) respectively. P .d. between positive and negative terminal of frist cell `V_(AB) = V(A) - V(B)` . `rArr ` p.d. between positive and negative terminal of second cell = V(B) - V(C). `V(A) - V(B) = epsilon_(1) - Ir_(1) ""` ... (1) `V(B) - V(C)= epsilon_(2) - Ir_(2) "" `... (2) `rArr` p.d. between A and C , `V_(AC) = V_(AB) + V_(BC)` = [V(A) - V(B)) + [V(B) - V(C)] `= epsilon_(1)- Ir_(1) + epsilon_(2) - Ir_(2) ` = ` epsilon_(1) + epsilon_(2) - Ir_(1) - Ir_(2)` ` = epsilon_(1) + epsilon_(2) - I(r_(1) + r_(2)) "" ` ... (3) For given combination equivalent emf between A and C be `e_(eq)` and internal resistance `r_(eq)` , `V_(AC) = V(A) - V(C) = e_(eq) - Ir_(eq) "" `... (4) Comparing (3) and (4) , `e_(eq) = epsilon_(1) + epsilon_(2)` and `r_(eq) - r_(1) + r_(2)` If cells are connected in series in opposing CONDITION then, `V_(BC) = V(B) - V(C) ` = `epsilon_(2) - Ir_(2)` `therefoe e_(eq)= epsilon_(1) - epsilon_(2)"" (epsilon_(1) gt epsilon_(2))` `rArr` If .n. cells are connected in series helping condition then, `epsilon_(eq) = epsilon_(1) + epsilon_(2) + ... epsilon_(n)` equivalent internal resistarice of combination will be equal to summation of individual internal resistance of each cell `r_(eq) = r_(1) + r_(2) + ... + r_(n)` If given cell is connected in opposing condition then emf of given cell will be considered as negative. |
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| 29. |
A rod is moving with velocity 9 m/s perpendicular to external magnetic field of 1T and there is a conducting sphere having charge 30 C and radius R. If electric field inside rod is same as electric field on the surface of sphere and radius R = x xx 10^(5) m then find x. |
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Answer» electric field on surface of sphere `= (KQ)/(R ^(2))` `(KQ)/(R ^(2)) =-vB IMPLIESR^(2)=(kQ)/(vB) =(9xx10^(9)xx30)/(9xx1)` `R= sqrt3xx10^(5) =1.73 xx10^(5)` |
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| 30. |
A TV transmitting antenna is 80 m tall. If the receiving antenna is on the ground The service area is |
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Answer» `12 pi SQ km` |
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| 31. |
A person looking through a telescope focuses lens at a point on the edgeof the bottom of an empty cylindrical vessel. Next he fills the entire vessel with a liquid of refractive index mu, without disturbing the telescope. Now, he observes the mid point of the vessel. Determine the radius to depth ratio of the vessel. |
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Answer» `1/2 SQRT((1-mu^2)/(mu^2 + 1))` |
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| 32. |
Electromagnetic waves are produced by |
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Answer» ATOMS andmolecules in an electrical discharge |
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| 33. |
What speed should a galaxy move with respect to us so that the sodium line at 589.0 nm is observed at 589.6 nm? |
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Answer» Solution :Since `v LAMBDA=c, (Deltav)/(v)=-(Deltalambda)/(lambda)""("for small changes in vand "lambda)` . For `DELTA lambda=589.6-589.0=+0.6nm` We get `["USING Eq. "(10.9)]` `(Deltav)/(v)=-(Delta lambda)/(lambda)=-(v_("radial"))/(c)` or, `v_("radial")~=+c((0.6)/(589.0))=+3.06xx10^(5)ms^(-1)` `=306km//s` THEREFORE, the galaxy is moving away from US |
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| 34. |
A rod CD of length b carryinga current l, isplaced in a magnetic field due to a thin log wire AB carrying currrent I as shown in fig. Then find the net force experienced by the wire |
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Answer» SOLUTION :Magnetic induction due to a straight wire at a position of small ELEMENT in at a DISTANCE x from the conductor `dB=(mu_(0))/(2pi)(I_(1))/x` force on the current element is `vecdF=dBI_(2)dxsin90^(@)` `dF=(mu_(0))/(2pi)(I_(1)I_(2))/xdx` Net force on conductor is `F=int_(a)^(a+b)(mu_(0))/(2pi)(I_(1)I_(2))/xdx` `=(mu_(0))/(2pi)I_(1)I_(2)[logx]_(a)^(a+b)=(mu_(0))/(2pi)I_(1)I_(2)log(1+b/a)` 
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| 35. |
A stone of relative densityK is released fromrest on the stone sinksinwaterwithan accleration of - |
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Answer» `G(1-k)` `"a" =("mg"-F_(B))/(m)` `="g"-("Vgp")/("Vd")` `="g"[1-(1)/(K)]` |
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| 36. |
A solenoid length 0.4cm, radius 1 cm and 400 turns of wire. If a current of 5 A is passed through this solenoid, what is the magnetic field inside the solenoid ? |
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Answer» `6.28 xx 10^(-4)T` `= 4 xx 3.14 xx 10^(-7) xx (400)/(0.4 xx 10^(-2)) xx 5` `0.628 T` |
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| 37. |
The relation between half life (T) and decay constant (lambda) is |
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Answer» `LAMBDA T = 1` |
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| 38. |
Light of wavelength 5000 A^(0) is incident on a slit. The first minimum of the diffraction pattern is observed to lie at a distance of 5 mm from the central maximum on a screen placed at a distance of 3 m from the slit. Then the width of the slit is |
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Answer» `3 CM` |
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| 39. |
Velocity of light in diamond, glass and water decreases in the following order. |
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Answer» WATER `gt` glass `gt` diamond `v prop 1/n` If the refractive index of diamond, glass and water are `n_D, n_g` and `n_w` respectively as `n_D gt n_g gt n_w` the velocity of light in these medium will be `v_w gt v_g gt v_D`. |
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| 40. |
The capacity of an isolated sphere of radius 9 cm is C. When it is connected to an earthed concentric thinhollow sphere of radius R, the capacity becomes 10 C. Then the valueof R is |
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Answer» 9cm |
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| 41. |
The frequency of the third harmonic of a closed organ pipe is equal to which of overtones: |
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Answer» FIRST Hence the correct choice is (a) . |
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| 42. |
In the circuit C=6muF. The charge stored in the capacitor of capacity C is : |
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Answer» ZERO |
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| 43. |
Which of the following are not electromagnetic waves ? |
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Answer» COSMIC RAYS |
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| 44. |
A plane flies at 2.00 times the speed of sound. Its sonic boom reaches a man on the ground 35.4 s after the plane passes directly overhead. What is the altitude of the plane? Assume the speed of sound to be 330 m/s. |
| Answer» SOLUTION :`1.34 xx10^4 m` | |
| 45. |
I amount of current is passed through a coil with N turns. Magnetic flux linked with the coil is ...... |
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Answer» LI where `Phi` = total MAGNETIC flux linked with ENTIRE coil I=current PASSING through the coil . `therefore Phi=LI` |
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| 46. |
In a meterbridge, the null point is found at a distance of 33.7cm from A. If a resistance of 12W is connected in parallel with S, the null points occurs at 51.9cm Determine the values of R and S. |
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Answer» Solution :Before connecting 12 `Omega`RESISTANCE, NULL point is OBTAINED at 33.7 CM and so according to condition of balanced Wheatstone bridge `(R)/(S) = (l_(1))/(100 - l_(1)) = (33.7)/(100 - 33.7)` `therefore (R)/(S)= (33.7)/(66.3) "" ` .... (1) After connecting `12 Omega` resistance in parallel with S, equivalent resistance of this parallel combination would be, `S_(eq) = (12 xx S)/(12 + S)` In this CASE, null point is obtained at 51.9 cm and so for balancing of Wheatstone bridge, `(R)/(S_(eq)) = (l_(1))/(100 - l_(1)) = (51.9)/(100 - 51.9)` `therefore (R(12 + S))/(12 xx S) = (51.9)/(48.1) ` `therefore ((12 + S)/(12) )((R)/(S)) = (51.9)/(48.1)` `therefore ((12 + S)/(12) ) ((33.7)/(66.3)) =(51.9)/(48.1) ` [ From equation (1) ] `therefore 12 + S = (51.9)/(48.1) xx (66.3)/(33.7) xx 12` `therefore 12 + S = 25.47` `therefore S = 13.47 Omega approx13.5 Omega ""` .... (2) From equations (1) and (2), `(R)/(13.5) = (33.7)/(66.3) ` `therefore R = 13.5 xx (33.7)/(66.3)` `therefore R = 6.86 Omega` |
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| 47. |
Let A=R-{3},B=R-{1}let f: Ararr B be defined by f (x) = (X-2)/(X-3) AAx epsilonR,then f is- |
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Answer» One-one |
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| 48. |
Show that the de-Broglie's wavelength of electrons accelerated through a potential V volts can be expressed as lamda=(h)/(sqrt(2meV)). What is the de-Broglie wavelength of a 2 kg object moving with a speed of 1 ms^(-1) ? |
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Answer» Solution :Consider an electron of mass m and charge e and initially at REST. If it is accelerated through a potential V volts, then KINETIC energy gained by electron=work done on the electron `therefore (1)/(2)mv^(2)=EV implies v=sqrt((2eV)/(m))` `therefore`de-Broglie wavelength of electron ,br> `lamda=(H)/(p)=(h)/(mv)=(h)/(m*sqrt((2eV)/(m)))=(h)/(sqrt(2meV))` In the numerical problem given we have `m=2kg and v=1ms^(-1)` `therefore lamda=(h)/(mv)=(6.63xx10^(-34))/(2xx1)=3.31xx10^(-34)m`. |
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| 49. |
The cells of the mammary gland that secrete and store milk are called |
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Answer» Alveoli |
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| 50. |
A parallel plate capacitor , each with plate area A and separation d , is charged to a potential difference V . The battery used to charge it remains connected . A dielectric slab of thickness of and dielectric constant K is now placed between the plates . What change , if any , will take placein : capacitance of the capacitor ? Justify your answer . |
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Answer» Solution :Given that plate area of either plate of parallel plate CAPACITOR = A distance between the plates = d and potential difference between the plates = V Hence initially CAPACITANCE ` C = (in_(0) A)/(d)` and CHARGE on plates Q = CV As the BATTERY remains connected THROUGHOUT the potential difference between the plates remains unchanged (V. = V)On placing a dielectric slab of the thickness .d. and dielectric constant .K. between the plates Electric field intensity between the plates `E. = (V.)/(d) = (V)/(d) = E` |
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