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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.
| 7401. |
A disc of mass m is moving with constant speed v_(0) on a smooth horizontal table. Another disc of mass M is placed on the table at rest as shown in the figure. If the collision is elastic, find the velocity of the disc after collision. Both disc lie on the same horizontal plane of the table. |
Answer» Solution :`sin THETA = (d)/(r+R)` For mass m : Momentum before impact  Momentum after impact  For mass M : Momentum before impact = 0 Momentum after impact  Net impulse on the system = 0 Momentum of the system is conserved. `mv_(0) COS theta + 0=mv_(1)+Mv_(3)` ....(i) `mv_(0)sin theta + 0 = mv_(2)+Mv_(4)` ....(ii) INDIVIDUALLY for mass m and M, the impulse along the tangent = 0 `therefore` For mass `m, mv_(0) sin theta = mv_(2)`....(iii) For mass M,`0 = Mv_(4)` .... (iv) As the collision is elastic `E=-(v_(3)-v_(1))/(0-v_(0)cos theta)=1`. `v_(3)-v_(1)=v_(0)cos theta` ....(v) SOLVING equations (i) through (v) `v_(2)=v_(0)sin theta,"" v_(4)=0` `v_(3)=(2mv_(0)cos theta)/(m+M), v_(1)=((m-M)/(m+M))v_(0)cos theta` It can be verified that, `(1)/(2)mv_(0)^(2)=(1)/(2)m(v_(1)^(2)+v_(2)^(2))+(1)/(2)M(v_(3)^(2)+v_(4)^(2))` `rArr`The kinetic energy is conserved in an elastic impact. Special case : If m =M `v_(1)=0, v_(0) sin theta` `v_(3)=v_(0)cos theta, v_(4)=0`. Note : The impulse along the common tangent during an impact may not be always negligible. For example, when normal reaction is impulsive and the effect of friction is to tbe considered. |
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| 7402. |
A particle starts moving from rest along y axis and a variable force F acts on it such that F infty y^-frac{1}{5} (y>0). The variation of power P of this force with y is given as |
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Answer» <P>`P infty y^frac{1}{5} ` |
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| 7403. |
Linear mass density of a rod PQ of length l and mass m is varying with the distance x (from P),as lambda = (m)/(2l)(1+ax)(i) Determine the value of a(ii)Also determine the distance of c.m. from the end P. |
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Answer» Solution :`dm = lambda DX` `m = int_(0)^(l)(m)/(2l)(1+ax)dx "" RARR a = 2//l` From P `x_(CM)=(INT xdm)/(m)=(int_(0)^(1)XX(m)/(2l)(1+(2x)/(l))dx)/(m)=(7)/(12)l` 
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| 7404. |
Two lighium .^(6)Li nuclei in a lithium vapour at room temperature do not combine to form a carbon .^(12)C nucleus because |
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Answer» a LITHIUM NUCLEUS is more tightly bound than a carbon nucleus |
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| 7405. |
The angle between incident ray and reflection ray and reflecting surface is |
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Answer» ANGLE of incidence |
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| 7406. |
अधिकतर पेड़-पौधे होते है- |
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Answer» स्वपोषी |
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| 7407. |
Derive an expression for the torque acting or a current carrying loop which subtends angle with uniform magnetic field. |
Answer» Solution :1. As shown in figure plane ABCD is not along the magnetic field but makes an angle with it.  2. We take the angle between the field and the normal to the coil to be angle `theta`. 3. The forces on the arms BC and DA are equal, opposite and act along the axis of the coil, which connects the centres of mass of BC and DA. Being collinear along the axis they cancel each other, resulting in no net force or torque. 4. The forces on arms AB and CD are `vecF_(1)andvecF_(2)` respectively. They too are equal and opposite with magnitude `F_(1)=F_(2)=IbB`.  5. In figure is a VIEW of the arrangement from the AD and it illustrates these two forces constituting a couple. The magnitude of the torque on the loop is, `TAU=tau_(1)+tau_(2)` `tau=F_(1)(a/2sintheta)+F_(2)(a/2sintheta)` `[becausetau=("magnitude of force")xx("perpendicular distance from reference points")]` `tau=(IbB)(a/2sintheta)+(IbB)(a/2sintheta)` `tau=I(ab)Bsintheta` `tau=IABsintheta""...(1)` `vectau=IvecAxxvecB""...(2)` 6. As `thetato0`, the perpendicular distance between the forces of the couple also approaches zero. This makes the forces collinear and the net force and torque zero. 7. But, magnetic moment of the current loop as, `vecm=IvecA""...(3)` 8. Where the direction of the area vector `vecA` is GIVEN by the right hand thumb rule and is directed into the plane of the paper. 9. Angle between `vecmandvecB" is "theta`, so from equation (2), (3) we can write, `vectau=vecmxxvecB""...(4)` 10. For N NUMBER of turns in coil, `vectau=NIvecAxxvecB` = NIAB, where `vecm=NIvecA` (This equation is analogous to the electrostatic case, `vectau=vecp_(e)xxvecE`) |
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| 7408. |
Two identical metal spheres having charges +q and +q respectively. When they are separated by distance r, exerts force of repulsion F on each other. The spheres are allowed to touch and then moved back to same separation. The new force of repulsion will be |
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Answer» `F/2` |
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| 7409. |
What stress would cause a wire to increases in length by 1/10th of 1% if Young's modulus for the wire is 12xx10^(5)N//m^(2) ? (Take g=10 N/kg) |
| Answer» SOLUTION :`12XX10^(7) N//m^(2)` | |
| 7410. |
In photoelectric effect experiment, the number of photoelectrons emitted by a photosensitive material is proportional to the ____ of incident radiation. |
| Answer» SOLUTION :INTENSITY. | |
| 7411. |
F(x)=sinx का परिसर है |
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Answer» (-1,1) |
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| 7412. |
The image formed by a convex mirror of a real object is larger than the object |
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Answer» when `mu LT 2F` |
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| 7413. |
The atomic bonding is same for which of the following pairs :– |
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Answer» AG and Si |
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| 7414. |
A long wire hangs vertically with its upper end clamped, when a torque of 2 Nm is applied to the free end, it is twisted through an angle of 30^(@). Then the potentialenergy of the twisted wire is |
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Answer» `PI` joules |
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| 7416. |
Two charged particles of charge -2q and +q have masses m and 2m respectively. They are kept in uniform electric field and allowed to move for the same time, find the ratio of their kinetic energies. |
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| 7417. |
The change in the value of g at a height h above the surface of earth is the same as at a depth X below its surface then |
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Answer» `X=H^(2)` `G(1-(X)/(R )) = g(1-(2h)/(R ))` `RARR X=2h`. Thus correct CHOICE is (c ). |
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| 7418. |
Statement I: The charge passing through a coil in time Deltat is (Deltaphi)/(R ), where R is the resistance of the coil and Deltaphi is the change in flux linked with the coil in time Deltat. Statement II: The induced emf in a conductor eprop(dphi)/(dt), where (dphi)/(dt) is the time rate of change of flux linked with the conductor. |
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| 7419. |
A plano convex lens has a thickness of 4 cm. When placed on a horizontal table with the curved surface in contact with it, the apparent depth of the bottom most point of the lens is found to be 3 cm. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the center of plane face is found to be 25/8 cm. Find the focal length of lens. Assume the thickness to be negligible while finding its focal length. |
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Answer» 75cm |
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| 7420. |
The following questions consists of two statements.Assertion: Zener diode works as a voltage regulator Reason: Zener voltage is independent of the Zener current variations and change of load resistance.Write the correct response from the following. |
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Answer» Both ASSERTION and reason are TRUE and the reason is not a CORRECT EXPLANATION of the assertion. |
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| 7421. |
A body is dropped from a height 39.2m. After it crosses half distance , the acceleration due to gravity ceases to act. The body will hit the ground with velocity. |
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Answer» `19.6m/s` |
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| 7422. |
Explain the working of a n-p-n transistor in CE mode as an amplifier. |
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Answer» SOLUTION :(1) COLLECTOR end is reverse biased and base - emitter ends forward biased. (2) A small variation in the base current produces a large variation in the collector current with a phase DIFFERENCE of `180^@` between them, without ALTERING the input frequency. This is the basis of an amplifier circuit.  Power gain of the amplifier in CE mode is given by, `P_"gain"=P_"output"/P_"input"=beta^2 R_0/R_i` Where , current gain `beta=I_0/I_1 = I_C/I_B`  (4)The input and output currents are as shown in the figure above. |
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| 7423. |
A TV Tower has a height of 100 m.How much population is covered by TV broadcast, if the average pupulation density around the tower is 1000 km^-2 :(Radius of earth = 6.4xx10^6m) |
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Answer» `10^3` |
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| 7424. |
Like poles……………………each other and…………..poles…………………..each other. |
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Answer» |
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| 7425. |
We call die separation of highest energy valence bond and lowest level of conduction bond as ? |
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| 7426. |
if a negatively charged rod is brought nearer to a neutral conducting sphere, then ...... |
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Answer» it will BECOME positively charged. |
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| 7427. |
A solid hemisphere and a solid cone have a common base and are made of same material. The centre of mass of the common structure coincides with the centre of the common base. If R is the radius of hemisphere and h is height of the cone, then |
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Answer» `(H)/(R )=SQRT(3)` `v=sqrt((k)/(M))l` MAXIMUM momentum `=Mv=Msqrt((k)/(M))l=sqrt(Mk)l` |
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| 7428. |
What is mean by Electric field lines ? |
| Answer» Solution :Electric FIELD vectors are VISUALIZED by the CONCEPT of electric field lines . THEYFORM a set of continuous lines which are the VISUAL representation of the electric in some region of space. | |
| 7429. |
An electron and a proton are in a uniform electric field the ratio of their acceleration will be .......... |
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Answer» zero `therefore m_(e)a_(e) = Ee`……(1) and force exerts on proton `F_(p) = E_(e)` `therefore m_(p)a_(p) = Ee`……(2) From EQUATION (1) and (2). `therefore a_(e)/a_(p) = m_(p)/m_(e)` |
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| 7430. |
Let barA=(1)/(sqrt(2))costhetahati+(1)/(sqrt(2)) sin thetahatjbe any vector. What will be the unit vector hatnin the direction of barA ? |
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Answer» `COS theta hati+sin theta hatj` `=costhetahati+sin thetahatj` |
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| 7431. |
Electron in a hydrogen atom makes a transition from n_1" to "n_2, where n_1 and n_2 are principal quantum numbers of two states. Assuming Bohr's model to be valid, time period of clectron in the n_1 state is found to be eight times the time period in the final state. Then possible values of principal quantum numbers n_1 and n_2 are: |
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Answer» Solution :`((T_(1))/T_(2))^(2) =r_(1)^(3)/r_(2)^(3)" Now "R alpha n^(2)` `((T_(1))/T_(2))^(2)=((n_(1))/(n_(2))^(6)" But "(T_(1)/T_(2))^(2)=(8/1)^(2)=(4/2)^(6)` `(n_(1)/n_(2))^(6)=(4/2)^(6)""n_(1)/n_(2)=4/2` |
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| 7432. |
A P-type semiconductor can be formed by doping Si or Ge with |
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Answer» Boron |
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| 7434. |
A meterbridge is in balance condition. Now if galvanometer and cell are interchanged, the galvanometer shows no deflection. Give reason. |
| Answer» SOLUTION :Galvanometer will SHOW no DEFLECTION. Proportionality of the arms are RETAINED as the galvanometer and CELL are interchanged. | |
| 7435. |
In the beginning the space between the plates of a parallel plate capacitor contains air and thereafter it is filled up with a medium of dielectric constant K. Then ...... |
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Answer» The electric field and the CAPACITANCE become K TIMES. Where `E_(0)` = Electric field for AIR medium. `C_(0)` = Capacitance of capacitor of air medium. |
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| 7436. |
भौतिकी के अध्ययन में शामिल है ? |
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Answer» पादप |
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| 7437. |
A bodyreleased from the top of a tower of height h takes T seconds to reacch the ground.The position of the body at T/4 seconds is |
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Answer» at `(h)/(16)` from the ground |
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| 7438. |
Whether em waves of different frequencies exist. |
| Answer» SOLUTION :Yes, All such waves can be PROPAGATED through space with SPEED of LIGHT. | |
| 7439. |
Suppose that the electric field amplitude of an electromagnetic wave is E_(0)=120N/C and that its frequency is v=50.0 MHz. Determine. B_(0), omega, k and lambda |
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Answer» Solution :`B_(0)=(E_(0))/(C)=(120)/(3XX10^(8))=4XX10^(-7)T` `omega=2piv=2xx3.14xx(50xx10^(6))=3.14xx10^(8)` rad/s K`=(omega)/(C)=(3.14)/(C)=(3.14xx10^(8))/(3xx10^(8))=1.05`rad/m `LAMBDA=(C)/(V)=(3xx10^(8))/(50xx10^(6))` =6.00 m |
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| 7440. |
Which of the following is not an electromagnetic wave? |
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Answer» X - RAY |
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| 7441. |
Energy released per fission of a nucleus is of the order of 200 MeV whereas that per fusion is of the order of 10 MeV. But a fusion bomb (Hydrogen bomb) is said to be more powerful than a fission bomb. Explain why? |
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Answer» Solution :(i) Energy produced per fusion is LESS but the number of nuclei per unit mass is larger on ACCOUNT of smaller mass number. (II) So, energy released per unit mass is GREATER. On the other hand, the fissionable materials have HIGH atomic weight. (iii) The number of fission nuclei per unit mass is less. So, energy released is less. |
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| 7442. |
When a bacteria takes up foreign DNA that leads to gene recombination is know as |
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Answer» Transformation |
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| 7443. |
Pressure at the Earth's surface is p_(a)=10^(5)Pa and the density of air at Earth's surface is rho_(0)=1.4kg//m^(3). At height h from the surface of Earth the density of air is reduced to (rho_(0))/(2) the value of h is (Assume that the temperature is constant through out the earth's atmosphere and let In (2)=0.7) |
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Answer» 10,500 m `p_(h)=p_(0)e^(-mgh//kT)` where, k= BOLTZMANN's constant g= gravitational acceleration and `p_(0)`= prsssure at the earth surface |
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| 7444. |
Consider the following statements A and B and identify the correct answer : A: A Zener diode is always connected in reverse bias. B: The potential barrier of a P-n junction lies in between 0.1 to 0.7 volts, approximately. |
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Answer» A is TRUE, B is false |
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| 7445. |
In a semiconducting material the mobilities of electrons and holes are, mu_eand mu_hrespectively. Which of the following is true? |
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Answer» `mu_e gtmu_h` |
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| 7446. |
A charged particle (change q ) is moving in a circle of radius R with uniform speed v. The associated magnetic moment mu is given by |
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Answer» `qvR^(2)` Since `T = (2piR)/v`.ALSO, `I=q/T=(qv)/(2piR)` `therefore mu=((qv)/(2piR))(piR^(2))=(qvR)/2` |
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| 7447. |
The displacement y of a particle executing periodic motion is given by y = 4 cos^(2) (1/2 t) sin(1000 t). This expression may be considered to be a result of the superposition of............independent harmonic motions |
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Answer» two |
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| 7448. |
The electric field in a plane electromagnetic wave propagating along + Y axis is given by E_(y)=30 sin[2xx10^(11)t+300 pi x]hat(j)Vm^(-1).(a)What is the wavelength and frequency of the wave ?(b) Write an expression for the magnetic field. |
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Answer» Solution :(a) `lambda=0.67m, v ~~ 3.18xx10^(10)Hz` (b) `vec(B)_(z)=10^(-7)sin[2XX10^(11)t+300 pi x]HAT(K)T` |
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| 7449. |
A particle of mass 0.3 kg is subjected to a forceF =- kx with k = 15 N//m. What will be its initial acceleration if it is released from a point 20 cm away from the origin? |
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Answer» `3 m//s^(2)` `F = 15 XX 0.2 = 3N` So, acceleration `= (f)/(m) =(3)/(0.3)=10ms//s^(2)` Hence CHOICE is (d) |
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| 7450. |
Two identical magnetic dipole of magnetic moments 1.0 Am^(2) each placed at a separation of 2m withthe resultant magnetic field at pointmidway between thedipole is |
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Answer» `5xx10^(-7) T` |
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