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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.
| 38601. |
The capacity of a parallel plate condenser with air as dielectric is 2muF . The space between the plates is filled with dielectric slab with K = 5. It is charged to a potential of 200V and disconnected from cell. Work done in removing the slab from the condenser completely |
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Answer» 0.8 J |
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| 38602. |
A 30kg block is to be moved up an inclined plane at an anglue 30^(@) to the horizontal with a velocity of 5ms^(-1). If the frictional force retarding the motion is 150 N find the horizontal force required to move the block up the plane.(g=10 ms^(-2)). |
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Answer» Solution :The force REQUIRED to a body up an INCLINED plane is `F=mg sin theta+` frictional force `= 30(10)sin 30^(@)+150=300N`. If P is the horizontal force, `F=P COS theta` `P=(F)/(cos theta)=(300)/(cos theta)=(300xx2)/(sqrt(3))` `= 200sqrt(3)=346N`. |
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| 38603. |
Dimension of electrical resistance is |
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Answer» `ML^2T^-3A^-2` |
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| 38604. |
UsingBohr'spostulates obtain the expression for theradiusof the nthe orbitin hydrogen atom. |
| Answer» SOLUTION :SEE Short Answer QuestionNu,ber 10. | |
| 38605. |
electrical conductivity, has the SI unit "ohm-metre”. Identify the physical quantity. |
| Answer» SOLUTION :ELECTRICAL RESISTIVITY. | |
| 38606. |
A beam of light converges towards a point O, behind a convex mirror of focal lnegth 20cm. Q. Find the magnification and nature of the image when point O is 10cm behind the mirror. |
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Answer» 2(virtual,inverted) Here, `u=+10cm ` and `f=+20cm` So, `(1)/(v)+(1)/(+10)=(1)/(+20)rArri.e.,v=-20cm` i.e., the IMAGE will be at a distance of 20cm in front of the mirror and will be rea, erect and enlarged with `m=-(20//10)=+2`  b. For this situation also, object will be virtual as shown in the figure. Here, `u=+30cm ` and `f=+20cm` `(1)/(v)+(1)/(+30)=(1)/(+20) i.e., v=+60cm` i.e., the image will be at a distance of 60cm BEHIND the mirror and will be virtual, inverted, and enlarged with `m=-(+60//30)=-2` 
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| 38607. |
यदिबहुपद x^2 -2x+5 के शून्यक alpha ,betaहो तो alphabetaका मान होगा - |
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Answer» 2 |
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| 38608. |
Energy supplied to convert unit mass of substnace from solid to liquid state at its melting point is called |
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Answer» LATENT HEAT of fusion |
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| 38610. |
In the figure a hemispherical bowl of bowl of radius R is shown Electric field of intensity E is present perpendicular to the circular cross section of the hemisohere. |
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Answer» The MAGNITUDE of electric FLUX through the hemispherical surface in SITUATION I will be `EPIR^(2)` |
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| 38611. |
When a certain quantity of liquid bismuth at its melting point of 271^(@)C is transferred to a calorimeter contaiing oil, the temperature of oil from 13.4^(@)C to 28.5^(@)C. When the experiment under identical conditions except that bismuth is in solid form, the temperature of oil rises to 19^(@)C if the specific heat of bismuth is 0.134Jg^(-1).^(@)C^(-1), fine the metal heat of fusion of bismuth |
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Answer» |
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| 38612. |
How many times the proton mass is heavier than an electron ? |
| Answer» SOLUTION :`1.672xx10^(-19)C`, 1836 TIMES | |
| 38613. |
The following table gives the length of three copper wires, their diameters, and the applied potential difference across their ends. Arrange the wires in increasing order according to the following: The magnitude of the electric field within them. |
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Answer» `E2 LT E3 lt E1` |
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| 38614. |
Using the Rydberg formula, calculate the wavelengths of the beta-line in the Lyman series of the hydrogen spectrum. |
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Answer» `1218Å` For ANSWER see Physics DARPAN, Part-2, Section-B, Illustration of TEXTBOOK example no. 12.6 |
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| 38615. |
A lens behaves as a converging lens in air and diverging lens in water. The refractive index of the material of the lens is |
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Answer» EQUAL to that of WATER |
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| 38616. |
Three liquids of equal masses are taken in three identical cubical vessels A, B and C. Their densities are P_(A), P_(B) and P_(C) are respectively. But P_(A) < P_(B) < P_(C)· The force exerted by the liquid on the base of the cubical vessel is : |
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Answer» the same in all the VESSELS So CORRECT choice is (a). |
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| 38617. |
The following table gives the length of three copper wires, their diameters, and the applied potential difference across their ends. Arrange the wires in increasing order according to the following: The drift speed of electrons through them, and |
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Answer» `V_(d2) lt V_(d3) lt V_(d1)` |
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| 38618. |
Consider the arrangement shown in figure. By some mechanism, the separation between the slits S_3 and S_4 can be changed. The intensity is measured at the point P which is at the common perpendicular bisector of S_1S_2 and S_3 S_4. When z=(Dlambda)/(2d), the intensity measured at P is I. Find the intensity when z is equal to (a)(Dlambda)/d(b)(3Dlambda)/(2d)(c) (2Dlambda)/d. |
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Answer» When, `z=(Dlambda)/(2D), z/2=y =(Dlambda)/(4d) ` `:.Deltax = (yd)/D =lambda/4` and we have seen in the above example that, at `Deltax = lambda/4`, intensity is `2I_0`. `:.I_S_3 = I_S_4 = 2I_0` Now, P is at the perpendicular bisector of `S_3 S_4`. Therefore, intensity at P will be four times of `2I_0` or `8I_0`. `8I_0 =I` Hence, `I_0 =I/8` (a)When `z=(Dlambda)/d` `y=z/2 = (Dlambda)/(2d)` ` :.Deltax = (yd)/D = lambda/2` or` I_S_(3) = I_S_(4)=0` Hence, ` I_p=0` (b) When ` z = (3Dlambda)/(2d)` ` y=z/2 =(3Dlambda/4d)` `Delta x = (yd)/D = (3lambda)/4` `:. Deltaphi or phi = 2pi/lambda(Deltax) = 3pi/2` Using`I=4I_0cos^2phi/2` We have, `I_(S_3) =I_(S_4)=2I_0 ` ` :. I_p = 4(2I_0) = 8I_0 = I` (c) When `z=(2Dlambda)/d` `y = z/2 = (Dlambda)/d` `:. Delta x = yd/D = lambda` `:. I_(S_3) = I_(S_4) = 4I_0 ` ` I_p = 4(4I_0) = 16I_0 = 2I` . |
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| 38619. |
If the distance between nuclei is 2xx10^(-10) cm. The density of nuclear material is |
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Answer» `3.2xx10^(-12)"kg.m"^(3)` |
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| 38620. |
Consider two identical masses m attached to 3 identical spring as shown in the figure. They can be set in motion in two different ways as follows : (I) x_(1)(t)=x_(0)cos omega t and x_(2)(t)=x_(0)cos omega t (II) x_(1)=(t)=x_(0)cos omega t and x_(2)(t)=-x_(0)cos omega t Here x_(1)(t) and x_(2)(t) denote the displacements from the unstreched positions. Then the potential energy stored in the system is |
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Answer» LARGER for I than II at t = 0 |
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| 38621. |
The potential of outer shell is |
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Answer» `(Q)/(32piepsilon_(0)a)` |
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| 38622. |
The following table gives the length of three copper wires, their diameters, and the applied potential difference across their ends. Arrange the wires in increasing order according to the following: The current density within them. |
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Answer» `J=ne V_(d)` `J2 lt J3 lt J1`. |
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| 38623. |
Obtain the resonant frequency and Q factor of a series LCR circuit with L= 3.0 H, C = 27 muF and R = 7.4 omega. It is desired to improve the sharpness of resonance of the ciecuit by reducing it's full width at half maximum by a factor 2. Suggest a suitable resistance. |
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Answer» SOLUTION :`omega _0 = 1/(sqrtLC) = 1/sqrt(3.0 xx 27 xx 10^(-6))rads^-1` `= 1000/9 = 111.1 rad//s`. `Q = (omega_0L)/R = (111.1 xx 3.0)/7.4 = 45.04` |
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| 38624. |
Fig. E.I show s magnet coil experiment of electromagnetic induction. What happens when a. i. The number of turns of the coil is inceased. ii. The streangth of the magnet is increased. iii. The speed of motion of the magnet is increased. b. An induced e.m.f. has no direction of its own. Why? |
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Answer» Solution :a. i. Induced e.m.f increases ii. Induced e.m.f increases iii. Induced e.m.f increases b. The direction of induced e.m.f or current is such as to OPPOSE the cause produced it. In this experiment, if the direction of motion of magnet reverses, the e.m.f.s direction ALSO REVERSED. HENCE an induced e.m.f. has no direction of its own. |
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| 38625. |
A system consists of a thin charged wire ring of radius R and a very long uniformly charged thread oriented along the axis of the ring. With one its ends coinciding with the centre of the ring. The total charge of the ring is equal to q. The charge of the thread (per unit lenght) is equal to lambda. Find the interaction force between the ring and the thread. |
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Answer» Zero |
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| 38626. |
How electromagnetic waves propagate ? |
| Answer» Solution :They do not require any MATERIAL medium for their propagation.They can tracel through VACCUM as well as through solids,LIQUIDS and gases. | |
| 38627. |
A 4 muF capacitor is charged to 400 V and then its plates are joined through a resistance of 1 K Omega. The heat produced in the resistance is |
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Answer» 0.16 J ` U = H = (1)/(2) CV^(2) = (1)/(2) xx 4 xx 10^(6) xx (400)^(2) = 0.32 J ` |
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| 38628. |
Which of the following four graphs may best represent the current deflection relation in a tangent galvanometer? (##HCV_VOL2_C36_E01_010_Q01##) |
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Answer»
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| 38629. |
A particle is moving 1.5 times as fast as an electron.The ratio of the de-Broglie wavelength of the particle to that of the electron is 1.813xx10^(-4) Calculate the particle's mass and identify the particle |
| Answer» SOLUTION :m`=3.35xx10^(-27)` KG PARTICLE can be DEUTRON | |
| 38630. |
Displacement current exists only when |
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Answer» ELECTRIC field is changing. |
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| 38631. |
A ray of light passes from denser medium to rarer medium a. What happens to the refracted ray as the angle of incidence increases from 0^(@) ? b.What is the importance of the angle C (marked in the figure above )? c. How is 'C' related to the refractive index 'n' of the medium ? Explain. |
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Answer» Solution :a. The refracted ray propagates with an INCREASED ANLE of REFRACTION. b. At the angle of incidence .C., the refracted ray just GRACES the surface separting the two media. Angle .C. is CALLED critical angle . c.Sin C = `(1)/(n) ` |
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| 38632. |
It is required to pass only one-tenth of the main current through a galvanometer having a resistanec of 27Omega. Calculate the length of the wire of specific resistance 48xx10^(-6)Omegacm and area of cross-section 0.2mm^(2) required to make a shunt for this purpose. |
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Answer» |
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| 38633. |
A magnet is hung horizontally in the magnetic meridian by a wire without any twist. If the supporting wire is given a twist of 180° at the top, the magnet rotates by 30°. Now if another magnet is used, then a twist of 270° at the supporting end of wire also produces a rotation of the magnet by 30°. Compare the magnetic dipole moments of the two magnets. |
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Answer» Solution :If resultant TWIST in the wire `=delta`, `delta_(1) = 180^@ - 30^@ = 150^@ = 150 xx (pi)/( 180) "rad" and` `delta_(2) = 270^(@) - 30^(@) = 240^(@) xx (pi)/( 180) "rad"` If the twist - CONSTANT for the wire is k then Rotating torque `tau_(1) = k delta_(1) and tau_(2) = kdelta_(2)` `therefore (tau_(1) )/( tau_(2) ) = (delta_(1) )/( delta_(2))` Here `alpha` is the angle made by the magnetic dipole moment with the magnetic meridian `tau_(1) .= m_(1) B_(H) sin alpha` SINCE the second magnet is also rotated by the same angle `tau_(2). = m_(2) B_(H) sin alpha` `therefore (tau_(1) .) /( tau_(2) .) = (m_1)/( m_2)` At equilibrium, `tau_(1) = tau_(1). and tau_(2) = tau_(2).` `(tau_(1).) /( tau_(2) .) = (tau_(1) )/( tau_(2) )` `(m_1)/( m_2) = (delta_1)/(delta_2) = (150)/(240) = (5)/(8)` |
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| 38634. |
An X-ray tube operates at a voltage of 40 kV. Find the continuous spectrum limit of the X-ray spectrum. |
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Answer» `lamdalehc//evarphi` |
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| 38635. |
A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to : |
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Answer» 0.5 required % =`(5/3mv^2)/(2mv^2+mv^2)xx100=56%` 
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| 38636. |
A hollow sphere of radius 2R is charged to V volts and another smaller sphere of radius R is charged to V/2 volts. Now the smaller sphere is placed inside the bigger sphere without changing the net charge on each sphere. Find the potential difference between the two spheres. |
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Answer» |
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| 38637. |
The potential differences across the resistance, capacitance and inductance are 80V, 40V and 100V respectively in an L-C-R circuit. The power factor of this circuit is |
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Answer» 0.4 |
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| 38638. |
A biconvex lens of focal length 40 cm is placed in front of an object at a distance of 20 cm . Now a slab of refractive index4/3 is placed somewhere in between the lens and the object. The shift in the image formed after the introduction of slab equals (thickness of the slab is 2 mm) |
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Answer» 1 mm |
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| 38639. |
The same spectral line undergoing anomalous Zeeman splitting is observed in direction1 and, after reflection from the mirror M (figure), in direction 2. How many Zeeman components are observed in both direction if the spectral lines is caused by transition (a).^(2)P_(3//2) rarr .^(2)S_(1//2), (b) .^(3)P_(2) rarr .^(3)S_(1) ? |
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Answer» Solution :The difference arise because of different selection rules in the TWO cases. In (1) the line is emitted perpendicular to the field. The selection rules are then `Delta M_(Ƶ)=0 +- 1` In (2) the light is emotted along the direction of the field. Then the selection rules are `DeltaM_(Ƶ)=+- 1` `DeltaM_(Ƶ)=0` is forbidden (a)In the transition `.^(2)P_(3//2) rarr .^(2)S_(1//2)` This has been considered above. In (1) we get all the SIX lines shown in hte problem above In (2) the line corresponding to `(1)/(2) rarr(1)/(2)` and `-(1)/(2) rarr -(1)/(2)` is forbidden. Then we get four lines (b) `.^(3)P_(2)` level `g=1+(2xx3+1xx2-1xx2)/(2xx2xx3)=(3)/(2)` so the energies of the sublevels are `E'(M_(Ƶ))=E'_(0)-(3)/(2)mu_(B) BM_(Ƶ)` where `M_(Ƶ)= +-2, +-1,0` For the `.^(3)S_(1)` line `g=2` and the enrgies of the sublevels are `E(M_(Ƶ))= E_(0)-2mu_(B)BM_(Ƶ)` where `M_(Ƶ)= +-1, 0`. The lines are `DeltaM_(Ƶ)=M_(Ƶ)-M_(Ƶ)= +1 : -2 rarr -1,-1 rarr 0` and `0 rarr 1` `DeltaM_(Ƶ)= 0-1 rarr -1,0 rarr0,1 rarr 1` `DeltaM_(Ƶ)= 1, rarr 1,1 rarr0, 0 rarr -1` All energy difference are unequal beacuse the two `g` values are unequal. There are then nine lines if viewed along (1) and six lines if viewed along (2) |
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| 38640. |
A block of mass 4 kg is kept over a rough 6 horizontal surface. The coefficient of friction between the block and the surface is 0.1 At t = 0, (3 hati) (m)/(s)velocity is imparted to the block and simultaneously (-2 hati) N force starts acting on it. Its displacement in first 5s is |
| Answer» Answer :C | |
| 38641. |
A plane light wave with wavelength lambda = 0.60 mu m falls normally on a sufficiently large glass plate having a round recess on the oppiste side (Fig.) For the observation point P that recess correcponds to the first one and and a half Fresnel zones. Find the depth h of the recess at which the intensity of light at the point P is (a) maximum, (b) minimum, (c ) equal to the intensity of incident light. |
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Answer» Solution :We could require the comtribution to the amplitude of a wave at a point from half a Fresnel zone. For this we proceed DIRECTLY from the Fresnel Huyghens principle. The complex amplitude is written as `E = int K (varphi) (a_(0))/(r ) e^(-ikr) dS` Here `K(varphi)` is a factor which depends on the angle `varphi` between a normal `overset rarr(n)` to the area `dS` and the direction from `dS` to the point `P` and `r` is the distance from the element `dS` to `P`. We see that for the first Fresnal zone  (using `r = B + (rho^(2))/(2b)` (for `sqrt(rho^(2)b^(2))))` `E = (a_(0))/(b) int_(0)^(sqrt(b lambda)) e^(-ikb - ik rho^(2)//2b) 2pi rho d rho (K(varphi) ~~1)` For the first Fresnel zone `r = b+ lambda//2` so `r^(2) = b^(2) + b lambda rho` and `rho^(2) = b lambda` Thus `E = (a_(0))/(b)e^(-ikb) 2pi int_(0)^((b lambda)/(2)) e^(-(kx)/(b)) dx` `= (a_(0))/(b) 2pi e^(ikb) (e^(iklambda//2) - 1)/(-ik//b)` `= (a_(0))/(k) 2piie^(-ikb) (-2) = - (4PI)/(k)ia_(0)e^(-ikb) = A_(1)` For the next half `E = (a_(0))/(b)e^(-ikb) 2pi int_((b lambda)/(2))^((3b lambda)/(4)) e^(-ikx//b) d xx` `= (a_(0))/(k) 2pi ie^(-ikb) (e^(-i)(3klambda)/(4) - e^(-iklambda//2))` `= (a_(0))/(k) 2pi ie^(-ikb) (+1+i) =- (A_(1)(1+i))/(2)` If we calaclate the contribution of the full `2^(nd)` Fresnel zone we will get `-A_(1)`. If we take account of the factors `K(varphi)` and `(1)/(r )` which decrease monotonically we expect the contribution to change to `-A_(2)`. Thus we write for the combtribution of the half zones in the `2^(nd)` Fresnel zone as `-(A_(2)(1 + i))/(2)` and `-(A_(2) (1 -i))/(2)` The part lying in the RECESS has an extra phase difference equal to `-del =- (2pi)/(lambda) (n - 1)h`. Thus the full amplitude is (note that the correct from is `e^(-ikr)`) `(A_(1)-(A_(2))/(2) (1+i))e^(+idel) - (A_(2))/(2) (1-i) + A_(3) - A_(4) +........` `~~((A_(1))/(2)(1-i))e^(+idel) - (A_(2))/(2) (1-i) + (A_(3))/(2)` `~~ ((A_(1))/(2)(1-i))e^(+i del) + i(A_(1))/(2)` (as `A_(2)~~A_(3)~~A_(1))` and `A_(3) - A_(4) + A_(5)......= (A_(3))/(2)`. The corresponding intensity is `I = (A_(1)^(2))/(4) [(1-i)e^(+i del) + (i)/(e)][(1 + i)e^(-i del) -i]` `I_(0)[3 -2 cos del + 2sin del] = I_(0) [3 xx 2sqrt(2) sin (del-(pi)/(4))]` `(a)` For maximum intensity `sin (del - (pi)/(4)) = +1` or `del - (pi)/(2) = 2kpi + (pi)/(2), k = 0, 1, 2,....` `del = 2kpi + (3pi)/(4) = (2pi)/(lambda) (n - 1) h` so `h = (lambda)/(n -1) (k + (3)/(8))` `(b)` For minimum intensity `sin(delta - (pi)/(4)) = 1` `del -(pi)/(4) = 2kpi+(3pi)/(2)` or `del = 2k pi + (7pi)/(4)` so `h = (lambda)/(n - 1) (k + (7pi)/(8))` (c ) `{:(For I=I_(0)cos delta=0),(sin delta=-1):}}` or `{:{(sin delta=0),(cos delta=+1):}` Thus `delta = 2k pi + (3pi)/(2) h = (k lambda)/(n - 1)` or `delta = 2kpi + (3pi)/(2), h (lambda)/(n - 1) (k + (3pi)/(4))` |
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| 38642. |
The first and second resonating lengths of a resonance tube are 17 cm and 55 cm. The end-correction is given by |
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Answer» 1.7 CM `x = (55-3 xx 17)/2 = (55-51)/2 = 2 cm` |
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| 38643. |
An electronis released from rest in a region of uniform electric and magnetic fields acting parallel to each other. The electron will |
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Answer» move in a STRAIGHT LINE. |
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| 38644. |
If alpha particle, proton and electron move with the same momentum, then their respective de Broglie wavelengths lambda_alpha,lambda_p,lambda_e are related as |
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Answer» `lambda_alpha=lambda_p-lambda_e` |
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| 38645. |
A glass rod has ends as shown in figure. The refractive index of glass is mu. The object O is at a distance 2R from the surface of larger radius of curvature. The distance between apexes of ends is 3R. Find the distance of image formed of the point object from right hand vertex. What is the condition to be satisfied if the image is to be real? |
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Answer» |
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| 38646. |
Two blocks A and B of same material and of masses m and 2 m moving with same kinetic energy on a surface enter in a region of identical rough surface. Choose the correct statement from the following options. |
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Answer» A will MOVE more distance before STOPPING |
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| 38647. |
The angle field of given length of wire for single turn coil at its centre is B then its value for two turns coil for the same wire is |
| Answer» ANSWER :C | |
| 38648. |
An alternating current of frequency f is flowing in a circuit containing only a inductor L. If V_0 and I_0 represent peak values of voltage and current respectively , the average power given by source to inductor is equal to …… |
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Answer» <P>0 `therefore delta=pi/2` `therefore` Power `P=(V_0I_0)/2 cos delta=0` |
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| 38649. |
If the density of earth is doubled keeping its radius constant then acceleration due to gravity will be (g=9.8" m"//s^(2)) |
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Answer» `19.6" m"//s^(2)` `rho_(2)=2rho_(1)` (given) `:. (g_(1))/(g_(2))=(4piG)/(3). Rho_(1)R_(1)xx(3)/(4pi" G "rho_(2)R_(2))=(rho_(1))/(rho_(2)).(R_(1))/(R_(2))=(Rho_(1))/(rho_(2))=(rho_(1))/(2rho_(1)).(R )/(R )=(1)/(2)` `RARR g_(2)=2g_(1)=2xx9.8=19.6" m"//s^(2)` |
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| 38650. |
Obtain the expressionfor the force betweentwo infinitely long straight parallelconductors carryingcurrent . Hencedefine"ampere" the SI unit of electric current. |
Answer» Solution :X and Y are two long straightparallelconductors carryingcurrents`I_1` and `I_2`respectively, and PLACED close to each other .d is the separationbetweenthe two conductorsand L is the lengthof the conductors.  The magnetic fieldat any point on theconductorY dueto current`I_1` in the conductor X , is given by `B_1=(mu_0)/(4pi) (2I_1)/d`. `vecB_1` acts in adirectionperpendicularto the planecontainingthe two conductors . The conductorY whichforce due to `B_1` acting on it and this forceis given by `F_1 = B_1 I_2 L sin q`. `therefore F_1=(mu_0/(4pi) (2I_1)/d)I_2L sin theta` `F_1=mu_0/(4pi) (2I_1 I_2)/d L`...(1) Accordingto Fleming.s left hand rule , thedirectionof the force `F_1` on X is perpendicularto `B_2` and is towardsthe conductorY. The magneticfieldat any pointon theconductorX due tothe current`I_2` in y ,is given by `B_2=mu_0/(4pi)(2I_2)/d. vecB_2` acts on X and oppositeto `B_1` . The mechanicalforce actson X due to`B_2` is `I_1 L sin q.` `therefore F_2=(mu_0/(4pi)(2I_2)/d)I_2 L sin 90^@` `F_1=mu_0/(4pi) (2I_1I_2)d L` ....(2) Accordingto Flemings left hand side rule ,the directionof the force `F_2` on X isperpendicularto X and it is towardsthe conductory if the current`I_1` is inwards (oraway fromthe conductorY if the current`I_1` isoutwards). The force `F_1` actingon a certainlength of theconductorY due to the CURRENTIN the conductorX is equal in magnitudeto the force`F_2` actingon the same length of X due to the currentinconductorY. If the two conductors carry the currents in the same direction (parallelcurrents ) then the forcesattract each other . If the two conductorscarry the currentsin the oppositedirections (anti-parallel currents ), thenthey are foundto repel each otherbecausethe two forcesact awayfrom each other. The force per unit lengthon each conductoris `F_L=F_1 /L=F_2/L=(mu_0/(4pi)(2I_1I_2)/d L)/L` i.e., `F_L=mu_0/(4pi)(2I_1I_2)/d` One ampere of currentcan be defined as thatconstantcurrentwhich when maintainedthrougheach of the two infinitelylong staightparallelconductorsof negligiblearea of crosssection in the same directionplaced 1 metreapart in VACUUM,causesan attractiveforce of2. `10^(-7) N m^(-1)`lengthon eachconductor . |
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