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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.
| 37601. |
Explain reflection of a plane wave by a plane surface. |
Answer» Solution : Let 'v' be the SPEED of wave in the medium , `tau` be the TIME taken by the wavefront to move from the point B to C. `BC=t tau` Let AE be the radius of an arc drawn at E and EC be the tangent drawn at E. `AE=BC=v tau` The triangles AEC and CBA are congruent and therefore `OVERSET(^)(i)=overset(^)(r )` Hence the LAW of reflection is proved. |
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| 37602. |
Consider four equal charges q_(1),q_(2),q_(3) and q_(4)=q=+1muC located at four different points on a circle of radius 1m , as shown in the figure. Calculate the total force acting on the charge q_(1) due to all the other charges . |
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Answer» Solution :According to the superposition principle the total electrostatic force on charge `q_(1)` is the vector sum of the forces due to the other charges, `vecF_(1)^("tot")= vecF_(12)+vecF_(13)+vecF_(14)` The following diagram shows the direction of each force on the charge `q_(1)`  The charges `q_(2)` and `q_(4)` are equi-distant from `q_(1)` . As a result the strengths ( magnitude ) of the forces `vecF_(12)` and `vecF_(14)` are the same even though their directions are different . Therefore the vectors representing these two forece are drawn with equal length . But the charge `q_(3)` is located farther compared to `q_(2)` and `q_(4)` . Since the strength of the electrostatic force decreases as distance increases the strength of the force `vecF_(13)` is drawn with smaller length compared to that for forces `vecF_(12)` and `vecF_(14)` . From the figure `r_(21)= sqrt(2)m= r_(41) "and " r_(31)= 2m` The m agnitudes of the forces are given by `F_(13)=(kq^(2))/(r_(31)^(3))= (9xx10^(9)xx10^(-12))/(4)=2.25xx10^(-3)N` `F_(12)=(kq^(2))/(r_(21)^(2))=F_(14)=(9xx10^(9)xx10^(-12))/(2)=4.5xx10^(-3)N` From the figure the ANGLE `theta = 45^(@)` . In terms of the components we have `vecF_(12)=F_(12)cos thetahati-F_(12) SIN theta hatj= 4.5xx10^(-3)xx(1)/(sqrt(2))hati-4.5xx10^(-3)xx(1)/(sqrt(2))hatj` `vecF_(13)=F_(13)hati=2.25xx10^(-3)Nhati` `vecF_(14)=F_(14)costhetahati+F_(14) sin theta hatj=4.5xx10^(3)xx(1)/(sqrt(2))hati+4.5xx10^(-3)xx(1)/(sqrt(2))hatj` Then the total force on `q_(1)` is `vecF_(1)^("tot")=(F_(12)costheta hati-F_(12)sin thetahatj)+ F_(13)hati+(F_(14)cos thetahati+F_(14)sintheta hatj)` `vecF_(1)^("tot")=(F_(12) cos theta+F_(13)+F_(14)costheta) hati+(-F_(12)sintheta+F_(14)sin theta)hatj` Since `F_(12)=F_(14)` the `j^(th)` component is zero . Hence we have `vecF_(1)^("tot")=(F_(12)costheta+F_(13)+F_(14)costheta)hati` Substituting the values in teh above equation `=(4.5)/(sqrt(2))+2.25+(4.5)/(sqrt(2))hati=(4.5sqrt(2)+2.25)hati` `vecF_(1)^("tot")=8.61xx10^(-3)Nhati` The resultant force is ALONG the positive x axis . |
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| 37603. |
In given region electric and magnetic field are prpendicular to each other.Electron moving in this region do not get deflected,charge on cathode will be….. [V=potential between anode and cathode] |
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Answer» `(B^(2))/(2VE^(2))` `impliesv(E )/(B)` ALSO `(1)/(2)mv^(2)=eV` `therefore (e )/(m)=(v^(2))/(2V)=(E^(2))/(2VB^(2))` |
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| 37604. |
A pole 5m high is situated on a horizontal surface. Sun rays are incident at an angle 30^(@) with the vertical. The size of shadow on horizontal surface is |
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Answer» `5M` `TAN30^(2)=(BC)/5` 
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| 37605. |
Why are coherent sources necessary to produce sustained interference pattern ? |
| Answer» Solution :Coherent SOURCES have a constant phase difference. This ensures that positions of maxima and minima do not changes with TIME i.e., a SUSTAINED INTERFERENCE pattern is obtained . | |
| 37606. |
Statement I: For distant communication of audio signal, microphone and loudspeaker are used at transmitting end and receiving end respectively. Statement II: In communication of audio signal, the sound wave is to be converted to a similar electromagnetic wave. |
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Answer» STATEMENT I is true, statement II is true, statement II is a CORRECT explanation for statement I. |
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| 37607. |
In a biprism experiment with sodium light (lambda = 5893 Å), the micrometer reading is 2.32mm when the eyepiece is placed at a distance of 100 cm from the source. If the distance between two virtual sources is 2 cm, find the new reading of micrometer when the eyepiece is moved such that 20 fringes cross the field of veiw. |
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Answer» |
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| 37608. |
The energy of an electron in an excited hydrogen atom is -3.4 eV The corresponding excited state (n) is |
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Answer» 2 For `H_(2) z=1` `E_(n)=(-2pi nk^(2)e^(4))/(n^(2)h^(2))` `=-(2pi mk^(2)e^(4))/(ch^(3)) (ch)/(n^(2))=-(R ch)/(n^(2))` `n^(2)=-(Rch)/(E)=4.012 "R=Rydberg.s CONST."` `rArr n=2 =1.09737 XX 10^(7)` |
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| 37609. |
ZnS shows .............. defect |
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Answer» FRENKEL Defect |
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| 37610. |
निम्नलिखित फलनों के द्वितीय अवकल गुणांक ज्ञात कीजिए ?log_ex |
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Answer» `-1/x^2` |
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| 37611. |
For the circuit shown in the figure, potential difference between points A and B is 16 V. Find the current passing through 2Omega. |
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Answer» SOLUTION :`V_(A)-V_(B)=16V` `therefore 4i_(1)+2(i_(1)+i_(2))-3+4i_(1)=16V"…(1)"` USING Kirchhoff.s second LAW in the closed loop, we have `9-i_(2)-2(i_(1)+i_(2))=0"….(3)"` SOLVING eqs (1) and (2), we GET `i_(1)=1.5A and i_(2)=2A` `therefore " Current through "2Omega" resistor "=2+1.5=3:5A` |
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| 37612. |
How will you define Q-factor ? |
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Answer» SOLUTION :If is defined as the RATIO of voltage ACROSS L or C to the applied voltage. `"Q-factor "=("Voltage across L or C")/("Applied voltage")` |
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| 37613. |
Figure shows the results of a series of observations of a migrating Brownian particle. The observations were made at intervals of 30 s, the temperature of water was 25^@C, the radius of a Brownian particle is 4.4 xx 10^(-7)m. Measuring the "steps" of the particle in the scale specified, find the square of the r.m.s. displacement for a given time, and calculate the Boltzmann constant and the Avogadro number. The scale is as follows: 1 mm on the graph corresponds to a i displacement of 1.25 mu m. |
Answer» Solution :Consider 10 successive steps starting with the lower left. Measure their lengths in millimetres as accurately as possible and transform them to scale into actual dimensions (see Table).  Hence the square of the r.m.s displacement is `Delta^2 = 1.04 xx 10^(-9) m^2`. Substituting `Delta^2 and t = 300 s`. into Einstein.s formula we obtain `k (pi eta r Delta^2)/(Tt) = (pi xx 8.9 xx 10^(-4) xx 4.4 xx 10^(-7) xx 1.04 xx 10^(-9))/(300 xx 300) = 5.9 xx 10^(26) "k mole"^(-1)`. We WOULD ADVISE the reader to carry out SIMILAR calculations using other parts of the graph and to assess the inherent error in the method. |
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| 37614. |
The velocity of a particle performing S.H.M. at extreme position is, |
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Answer» Zero |
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| 37615. |
An electromagnetic wave propagating along north has its electric field vector upwards. Its magnetic field vector point towards. |
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Answer» North |
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| 37616. |
Figure 22-8c is identical to Fig. 22-8a except that particle 3 now lies on the x axis between particles 1 and 2. Particle3 has charge q_(2)= -3.20 xx 10^(-19)C and is at a distance (3//4)R from particle 1. What is the net electrostatic force vecF_("1.net") on particle 1 due to particles 2 and 3? |
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Answer» Solution :KEY IDEA The presence of particle 3 does not alter the electrostatic force on particle 1 from particle 2. THUS, force `vecF_(12)` still acts on particle 1. Similarly, the force `vecF_(13)` that acts on particle 1 DUE to particle 3 is not affected by the presence of particle 2. Because particles 1 and 3 have charge of opposite signs, particle 1 is attracted to particle 3. Thus, force `vecF_(13)` is directed toward particle 3, as indicated in the free-body diagram of Fig. 22-8d. Three particles : To find the magnitude of `vecF_(13)`, we can rewrite Eq. 22-4 as `F_(13)=(1)/(4pi epsilon_(0))(|q_(1)||q_(3)|)/(((3)/(4)R)^(2))` `=(8.99xx10^(9)N*m^(2)//C^(2))xx ((1.60xx10^(-19)C)(3.20xx10^(-19)C))/(((3)/(4))^(2)(0.0200 m)^(2))` `=2.05xx10^(-24)N`. We can also write `vecF_(13)` in unit-VECTOR notation: `vecF_(13)=(2.05xx10^(-24)N)hati`. The net force `vecF_("1,net")` on particle 1 is the vector sum of `vecF_(12)`, and `vecF_(13)`, that is, from Eq. 22-7, we can write the net force `vecF_("1,net")` on particle 1 in unit-vector notation as `vecF_("1,net")=vecF_(12)+vecF_(13)` `=-(1.15xx10^(-24)N)hati+(2.05xx10^(-24)N)hati` `=(9.00xx10^(-25)N)hati`.(ANSWER) Thus, `vecF_("1, net")` has the following magnitude and direction (relative to the positive direction of the x axis): `9.00xx10^(-25)N and 0^(@)`. (Answer) |
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| 37617. |
The most commonly used material for making transistor is: |
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Answer» copper |
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| 37618. |
Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better. Keeping other factors fixed, it is found experimentally that for small thickness t, the number of a-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide? |
| Answer» Solution :It suggests that the scattering is predominantly due to a single COLLISION, because the chance of a single collision increases linearly with the number of target ATOMS, and hence linearly with THICKNESS. | |
| 37619. |
Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better. In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of a-particles by a thin foil? |
| Answer» Solution :In Thomson’s MODEL, a single collision causes very little deflection. The observed AVERAGE SCATTERING angle can be explained only by considering multiple scattering. So it is wrong to ignore multiple scattering in Thomson’s model. In Rutherford’s model, most of the scattering COMES througha single collision and multiple scattering effects can be ignored as a first APPROXIMATION. | |
| 37620. |
A potentiometer wire of 5m length and having 20Omega resistance is connected in series with a battery and a resistance of 3980 Omega . A cell of emf 1.1V is balanced across the potential difference of the external resistance. If the emf of the thermocouple is 2.2mV, the corresponding balancing length is |
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Answer» 398cm |
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| 37621. |
Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better. Is the probability of backward scattering (i.e., scattering of a-particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model? |
| Answer» SOLUTION :MUCH LESS. | |
| 37622. |
A man of mass 50kg is running on a plank of mass 150 kg with a speed of 8m/s relative to the plank as shown [both are initially at rest and the velocity of the mas w.r.t ground any how remains constant]. Plank is placed on smooth horizontal surface. The man, while running whistles with frequency fQ. A detector (D) placed on plank detects frequency. The man jumps off with same velocity from point D and slides on the smooth horizontl surface [Assume fl =0 between plunk and horizontal surface]. The speed of sound in still medium is 330 m/s Choose the correct plot between the frequency detected by D verses position of the man relative to detector (x) |
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Answer»
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| 37623. |
Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better. (a) Is the average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model? |
| Answer» SOLUTION :About the same. | |
| 37624. |
A man of mass 50kg is running on a plank of mass 150 kg with a speed of 8m/s relative to the plank as shown [both are initially at rest and the velocity of the mas w.r.t ground any how remains constant]. Plank is placed on smooth horizontal surface. The man, while running whistles with frequency fQ. A detector (D) placed on plank detects frequency. The man jumps off with same velocity from point D and slides on the smooth horizontl surface [Assume fl =0 between plunk and horizontal surface]. The speed of sound in still medium is 330 m/s The frequency of sound detector D, before man jumps off the plank |
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Answer» `332/324 f_(0)` |
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| 37625. |
A man of mass 50kg is running on a plank of mass 150 kg with a speed of 8m/s relative to the plank as shown [both are initially at rest and the velocity of the mas w.r.t ground any how remains constant]. Plank is placed on smooth horizontal surface. The man, while running whistles with frequency fQ. A detector (D) placed on plank detects frequency. The man jumps off with same velocity from point D and slides on the smooth horizontl surface [Assume fl =0 between plunk and horizontal surface]. The speed of sound in still medium is 330 m/s The frequency of sound detected %v detector D, after man jumps of the plank |
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Answer» `332/324 f_(0)` |
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| 37626. |
What was the first indication of disaster? |
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Answer» around 6 PM when WINDS DROPPED and sky grew darker |
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| 37627. |
Calculate the mutual inductance between two coils when a current of 2A changes to 6A in 2 seconds and induces an emf of 20 mV in the secondary coil. |
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Answer» Solution :`|E| = M (dI)/(dt)` `20 xx 10^(-3) = M((6-2))/(2) " or " M = 10 mH` |
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| 37628. |
The work function of potassium is 2.0 eV. When it is illuminated by light of wavelength 3300A^(@) photo electrons are emitted. The stopping potential of photo electrons is [Plank’s constant h=6.6xx10^(-34)Js,1eV=1.6xx10^(-19)J,C=3xx10^(8)ms^(-1)] |
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Answer» 0.75V |
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| 37629. |
A man is riding an elevator which is rising up with a uniform acceleration of 2 ms^(-1) He tosses a coin vertically upwards with a speed of 20 ms. What time the coin would take to fall back into his hands? (Take g = 10 ms^(-2)) |
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Answer» `(20)/(2)sec` `=(10)/(3)`seconds Hence (C) is the CORRECT CHOICE |
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| 37630. |
Can electric potential at any point in space be zero while electric field is not zero ? |
| Answer» SOLUTION :At the CENTRE of a DIPOLE, there is field, but POTENTIAL will be zero. | |
| 37631. |
The equivalent capacitance between the points A and B of five identical capacitors each of capacity C_(1) will be |
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Answer» `(C )/(2)` |
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| 37632. |
the fernetion of starchinto ethanol involes threestages . Theenzymesthatcatalysethese steps are respectively - |
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Answer» AMYLASE , MALTASE , ZYMASE 
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| 37633. |
In a series RC circuit having battery of 12 V, capacitor is charged from O to 6 V in 0.1 s. Find value of resistance R. |
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Answer» Solution :`V=V_(0)[1-e^(t//1)]` `6=12[1-e^(-0.1//t)]` `e^(0.1//t)=2` `(0.1)/(tau)=LN(2)` `R_(C)=(0.1)/(C L n(2))` |
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| 37634. |
A rod of length 10 cm lies along the principal axis of a concave mirror of focal length10 cm in such a way that the end closer to the pole is 20 cm away from it. The length of the image is |
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Answer» 5cm |
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| 37635. |
Who had observed magnetic effect of electric current first ? |
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Answer» Oersted |
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| 37636. |
In the ground state of .......... electrons are in stable equilibrium, while in ......... electrons always experience a net force. (Thomson's model/ Rutherford's model.) |
| Answer» SOLUTION :Thomson.s MODEL, Rutherford.s model | |
| 37637. |
There are 27 drops of a conducting fluid. Each has a radius r and they are charged to a potential V_(0) These are combined to form a bigger drop. Its potential will be |
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Answer» `V_(0)` |
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| 37638. |
एक कण स्थिरावस्था से एक धनात्मक x-अक्ष की दिशा में मूलबिंदु 0 से नियत त्वरण से चलता है। वह सभी चित्र ज्ञात कीजिये जो इस कण की गति को गुणात्मक रूप में सही दर्शाते है (a = त्वरण, v = वेग,x = विस्थापन, tसमय) |
| Answer» Answer :C | |
| 37639. |
A particle moves with simple harmonic motion in a straight line. In first tau s, after starting from rest it travels a distance a, and in next tau s it travels 2a, in same direction, then : |
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Answer» time period of oscillations is `6tau` `cos omega tau=(1-(a)/(A)),cos 2 omega tau=(1-(3a)/(A))` `2(1-(a)/(A))^(2)-1=1-(3a)/(A)` Solving the EQUATION `(a)/(A)=(1)/(2)impliesA=2a` `cos omega tau=(1)/(2),T=6tau` So correct choice is (a). |
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| 37640. |
When a coil is connected to a cell, its resistance is R if it is connected to a source of alternating e.m.f. the resistance of the coil : |
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Answer» will not change |
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| 37641. |
The convex surface of a thin concaveo-convex lens of glass of refractive index 1.5 has a radius of curvature 20cm. Theconcave surface has a radius of curvature 60cm. The convex sidei is silvered and placed on a horizontal surface. Q. Where should a pin be placed on the optice axis such that its image is formed at the same place? |
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Answer» `X=5cm` `(1)/(F)=(1)/(f_(g))+(1)/(f_(m))+(1)/(f_(g))` where `f_(x)` is th efocal length of the lens without silvering and `f_(m)` is the focal length of the mirror. `(1)/(f_(g))=(n-1)((1)/(R_(1))-(1)/(R_(2)))` `=(1.5-1)((1)/(20)-(1)/(+60))=(1)/(60)` `f_(g)=60cm` `f_(g)=R_(1)//2=20//2=10cm` `(1)/(F)=(1)/(60)+(1)/(10)+(1)/(60)=(8)/(60)` `F=(60)/(8)=7.5cm` For the IMAGE to be formed at THEPLACE of the object, `X=R=2F=7.5xx2` `=15cm`  Method2: We use the relation `(n_(2))/(x_(2))-(n_(1))/(x_(1))=(n_(2)-n_(1))/(R)` For the object and the image to coincide, the rays fall normal on the reflecting surface. i.e., on the silvered face of the lens. Then, the rays retrace backward and meet at the object point again (optical reversibility). For the refraction at the upper surface of the lens, `n_(1)=1.0, n_(2)=1.5, x_(1)=20, R=+60` `(x_(2)=+20 ` ensures that the rays fall on the silvered face normally. ) `(1.5)/(20)-(1.0)/(x_(1))=(1.5-1.0)/(+60)` `(1.0)/(x_(1))=(1.5)/(20)-(0.5)/(60)=(3.0)/(60)` `x_(1)=15cm`  Method 3: We use lensmaker's FORMULA and the equation `(1)/(f)=(1)/(x_(2))-(1)/(x_(1))` The given optical arrangement can be visualised as a convex lens of focal length 60cm and a concave mirror of focal length 10cm kept in contact as shown in the figure. If the rays fall normally on the mirror after the refraction through the lens, they will retrace backward and meet at the point of the pin again. For the lens, `x_(1)=?` `x_(2)=+20` (for normal incidence on the mirror ) `f=-60` (using cartesian-coordinate sign convention) `(1)/(-60)=(1)/(+20)-(1)/(x_(1))` 
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| 37642. |
A heavy mass M resting on the ground is connected to a lighter mass m through a light inextensible string passing over an ideal pulley. The string connected to mass M is loose. Let lighter mass m be allowed to fall freely through a height h such that the string becomes taut. If t is the time from this instant onward when the heavier mass again makes contact with the ground and Delta E is change in kinetic energy then |
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Answer» `t=(2m)/(M+m)SQRT((2h)/(g))` |
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| 37643. |
Which is the corrective vision for shortsighted defect? |
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Answer»
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| 37644. |
Three identical capacitors and two identical batteries of negligible internal resistance are connected as shown in figure. Neglect the resistance of the wires and all the three switches are initially open. There is no charge on any capacitor initially All the switches are closed for long time. Total energy stored in all the capacitors is U_1 . Now S_1 as open and total energy stored in all the capacitors is U_2What is the value of U_1 -U_2 ? |
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Answer» ZERO |
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| 37645. |
Consider the situation shown in figure. Both the pulleys and the strings are light and all the surfaces are frictionless. Calculate (a) the acceleration of mass M, (b) tension in the string PQ and (c ) force exerted by the pulley P. |
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Answer» Solution :As pully Q is not fixed so if it moves a distance d athe length of string between P and Q will changes by 2nd ( d from above and d from below), i.e., M will move 2d. This in turn implies that if a is the acceleration of M the acceleration of Q and of 2M will be `a/2.` now if we CONSIDER the motion of mass M, it is accelerated DOWNWARD, so T=M(g-a)........(i) And for the motion of Q `2T-T'0xxa/2=0impliesT'=2T"".......(ii)` And for the motion of mass `2MT'=2M((a)/(2)),impliesT'=Ma""(III)`   (a) From equation (ii) and (iii) as `T=1/2Ma,` so equation (i) reduces to `T=1/2Ma(g-a)impliesa=2/3g` while tension in the string PQ from equation (1) will be `T=M"("g-2/3g")"=1/3Mg` (C ) Now from figure (b), it is clear that force on pulley by the CLAMP will be equal and opposite to the resultant of T and T at `90^(@)` to each other, i.e., `(N_(2))=sqrt(T^(2)+T^(2))=sqrt(2)T=(sqrt2)/(3)Mg` |
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| 37646. |
Three identical capacitors and two identical batteries of negligible internal resistance are connected as shown in figure. Neglect the resistance of the wires and all the three switches are initially open. There is no charge on any capacitor initially S_(2) and S_(3)are closed and after some time S_1is also closed. What is work done by battery 1 due to closing of switch S_1 ? |
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Answer» `CV^2` |
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| 37647. |
Three identical capacitors and two identical batteries of negligible internal resistance are connected as shown in figure. Neglect the resistance of the wires and all the three switches are initially open. There is no charge on any capacitor initially In the above situation, what is work done by battery 2 due to closing of switch S_1 ?. |
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Answer» ZERO |
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| 37648. |
The area of cross section of a current carrying conductor is A_(0) and (A_(0))/(4) at section (1) and (2) respectively. If v_(alpha_(1)),v_(alpha_(2)) and E_(1), E_(2) be the drift velocity and electric field at sections 1 and 2 respectively then : |
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Answer» `v_(alpha_(1)):v_(alpha_(2))=1:4` |
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| 37649. |
A thin and circular disc of mass m and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity omega . If another disc of same dimension but mass ml4 is placed gently on the first disc co-axially, then the new angular velocity of the system is |
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Answer» `5omega//4` `1/2mR^(2)OMEGA=[1/2mR^(2)+1/2(m/4)R^(2)]omega_(2)=1/2mR^(2)(1+1/4)omega_(2)` `1/2mR^(2)omega=1/2mR^(2)xx5/4omega_(2)rArr=5/4omega_(2)thereforeomega_(2)=4/5omega` |
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| 37650. |
A heat engine is a device |
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Answer» Which converts MECHANICAL energy into heat energy |
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