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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.
| 37801. |
A convex lens of focal length 15 cm is split into two halves and the two halves are placed at a separation of 120 cm. Between, the two halves of convex lens a plane mirror is placed horizontal and at a distance of 4 mm below the principal axis of the lens halves. An object of length 2 mm is placed at a distance of 20 cm from one half lens as shown in figure. Find the position and size of thefinal image. Trace the path of rays forming the image. |
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Answer» `1/v-1/(-20)=1/15` `:. V=60 cm` Magnification, `m=v/u=60/(-20)=-3` The image formed by first half lens is shown if figure . `AB=2mm, A_1B_1=6mm, AO_1=20 cm,` `O_1F=15 cm` and `O_1A_1=60 cm`  Point `B_1` is `6 mm` below the principal axis of the lenses. Plane mirror is 4mm below it.  Hence, `4mm` lenght of `A_1B_1(i.e.A_1C_1)` acts as real OBJECT for mirror. Mirror forms its virtual image `A_2C_2. 2mm` length of `A_1B_1 (i.e.C_1B_1)` acts as virtual object for mirror. Real image `C_2B_2` is formed of this PART. Image formed by plane mirror is shown in figure (b) . For the second half lens, `1/v-1/(-60)=1/15` `:. v=+20` `m=v/u=20/(-60)=-1/3` So, length of final image `A_3B_3=1/3A_2B_2=2mm` Point `B_2` is 2mm below the optic axis of second half lens. Hence, its image `B_3` is formed `2//3 mm` above the principal axis. Somilarly, point `A_2` is 8mm below the principal axis. Hence, its image is `8//3 mm` above it. Therefore, image is at a distance of 20 cm behind the second half and at a distance of `2//3 mm` above the principal axis. The size of image is 2 mm and is INVERTED as compared to the given object. Image formed by second half lens is shown in figure  Ray diagram for final image is shown in figure 
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| 37802. |
Coherent sources are produced by division of wavefront in___ slit experiment. |
| Answer» SOLUTION :Young.s DOUBLE | |
| 37803. |
A geosynhronous satellite is |
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Answer» located at a height of 34860 KM to ensure global coverage |
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| 37804. |
(A) In Rutherford experiment only few alpha-particles get rebounded. (R) Most of the atomic space is empty. |
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Answer» Both 'A' and 'R' are true and 'R' is the correct explanation of 'A'. |
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| 37805. |
A particle travels in a circular path of radius 0.2 m with a constant kinetic energy of 4J . What is the net force on the particle? |
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Answer» 4N `F=(mv^(2))/(R)=(2((1)/(2)mv^(2)))/(r)=(2K)/(r)=(2(4J))/(0.2m)=40N`. |
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| 37806. |
For the wave y = 5 sin 30 pi [t -(x//240)], where x and y are in cm and t is in seconds, find the Displacement when t = 0 and x = 2cm |
| Answer» SOLUTION :`-3.535 CM` | |
| 37807. |
A streched string is vibratingin the secondovertone, then the numberof nodes and antinodes between the ends of the string are respectively |
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Answer» 2 and 3  Nodes+ 4 ANTI nodes : 3 |
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| 37808. |
A body is allowed to slide from the top along a smooth inclined plane of length 5m at an angle of inclination 30^(@). If g= 10ms^(-2), time taken by the body to reach the bottom of the plane is |
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Answer» `(SQRT3)/(2) s` |
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| 37809. |
Our old culture is ? |
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Answer» Burnt |
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| 37810. |
(a) A small compass needle of magnetic moment 'm' is free to turn about an axis perpendicular to the direction of uniform magnetic field 'B'. The moment of inertia of the needle about the axis is 'I'. The needle is slightly disturbed from its stable position and then released. Prove that it executes simple harmonic motion. Hence, deduce the expression for its time period . (b) A compass needle , free to turn in a vertical orients itself with its axis vertical at a certain place on the earth . Find out the values of (i) horizontal component of earth's magnetic field and (ii) angle of dip at the place. |
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Answer» Solution :(B) (i) When a compass NEEDLE, free to turn in a vertical plane, orients itself with its AXIS vertical at a CERTAIN place on the earth, it means that magnetic FIELD is along vertical direction at the place. Consequently, the horizontal component of earth.s magnetic field is zero at that place. (ii) As tan `delta = (B_V)/(B_H)and B_H` is zero, `tan delta = oo and ` so `delta = 90^@` . Thus , of dip `delta` at given place is `90^@`. |
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| 37811. |
A simple harmonic wave has the equationy_(1) = 0.3 sin(314t - 1.57 x) and another wave has equation y_(2) = 0.1 sin(314 t - 1.57 x + 1.57)where x,y_1 and y_2 are in metre and t is in second. |
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Answer» `v_(1)=v_(2) = 50 Hz` |
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| 37812. |
What is diffraction of light ? |
| Answer» Solution :The phenomenon of BENDING of LIGHT around the edges of a sharp OBSTACLE WHOSE size is comparable to WAVELENGTH of light is called diffraction. | |
| 37813. |
Figure (a)shows two wires, each carrying a current. Wire 1 consists of a circular are of radius R and two radial lengthsit carries current i_(t)=20A in the direction indicated. Wire 2 is long and straigth itcarries a current i_(2) that can be varried and it is at distance R//2 from the centre of the are. The net magnetic field vec(B) due to the current is measuredat the center of curvature of the areFigure (b) is a plot of the component of vec(B) in the direction prependicular to the figure as a function of current i_(2) What is the angle subtendedby the arc ? |
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Answer» |
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| 37814. |
Select wrong statement from the following - Electromagnetic waves |
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Answer» are transverse |
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| 37815. |
Define the magnifying power of a compound microscope when thefinal image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope has short focal lengths? Explain. |
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Answer» SOLUTION :Define of magnifying power Reason for short focal lenghts of OBJECTIVE and eyepeice. Magnifying power is DEFINED as theangle subtended at the eye by the image to the angle subtended (at unaided eye) by the object. (Alternatively: Also ACCEPT this definition int eh formj of FORMULA ) `m=m_(0)xxm_(e)=L/(f_(0))xxD/(f_(e))` To increase the magnifying power both the objective and eyepiece must have short focal lenghts (as `m=L/(f_(0))xxD/(f_(e))`) |
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| 37816. |
In the corn cob the tassels which wave in the wind to traP the pollen grains represents |
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Answer» STIGMA and style |
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| 37817. |
The velocity of a projectile when at its 2 greatest height is sqrt(2/5)of its velocity when at half of its greatest height find the angle of projection |
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Answer» Solution :Step 1: we know that, velocity of a projectile at HALF of maximum height: `=usqrt((1+cos^(2)theta)/2)` Step 2: given that `u cos theta =SQRT(2/5) xx u sqrt((1+cos^(2)theta)/2)` Squaring on both sides `u^(2) cos^(2)theta =2/5u^(2)(1+cos^(2)theta)/2` `10cos^(2)theta =2 + 2 cos^(2)theta RARR theta=60^(@)` |
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| 37818. |
Different hydrogen spectral series have different wave- lengths. In the given table, Column I shows the values of energy levels between which electronic transition takes place, Column II shows the formula of maximum wave- length of different spectral series and Column III shows the formula of minimum wavelength of different spectral series Determine the wavelengths, lambda_("max") and lambda_("min") for Balmer series. |
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Answer» `(II)(i)(L)` |
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| 37819. |
Different hydrogen spectral series have different wave- lengths. In the given table, Column I shows the values of energy levels between which electronic transition takes place, Column II shows the formula of maximum wave- length of different spectral series and Column III shows the formula of minimum wavelength of different spectral series Determine the wavelengths lambda_("max") and lambda_("min") for Pfund series. |
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Answer» `(I)(II)(M)` |
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| 37820. |
Different hydrogen spectral series have different wave- lengths. In the given table, Column I shows the values of energy levels between which electronic transition takes place, Column II shows the formula of maximum wave- length of different spectral series and Column III shows the formula of minimum wavelength of different spectral series Determine the wavelength, lambda_("max") and lambda_("min") for Paschen series. |
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Answer» `(III)(i)(L)` |
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| 37821. |
A spring of constant 100 N/m is stretched by applying equal forces each of magnitude Fat the two ends. The energy stored in the spring is 200 J. Now spring is cut into two equal parts and one of the part is stretched by applying equal forces each of magnitude F at the two ends. The energy stored is |
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Answer» 200 J `therefore` Elastic potential energy stored in the spring is, `U=(1)/(2) kx_(0)^(2)` or `200= (F^(2) )/( 2K)` If spring is CUT into two parts, spring constant of each part becomes twice. `therefore F=2k x_(0) rArr x_(0) = (F)/( 2k)` `therefore` The energy stored is, `U.= (1)/(2) (2k) x_(0)^(2) = k(F^2)/(4k^(2))` `= (F^(2))/(4k)= (200)/(2) = 100` J. |
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| 37822. |
Two identical heaters rated 220 V, 1000 W are placed in series with each other across 220 volt line, then the equivalent combined power in the circuit is : |
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Answer» 1000 W |
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| 37824. |
Two boys simultaenously aim theirgunsat a bridsittingon a tower. The firstboy releases hisshot with a speed of 100m/s at an angleof projection of 30^(@). The secondboy is ahead of the first by a distance of 50mand releases his shot with a speed of 80m/s. How musthe aim his gun so thatboththe shots hitthe bird simultaenously ? Whatis the distanceof thefoot of the towerfrom the first boy and the height of the tower ? |
| Answer» SOLUTION :`THETA = SIN^(-1)(5//8) . 179.26 m 82.5m` | |
| 37825. |
A block of mass 1 kg kept on a rough horizontal surface (mu=0.4) is attached to a light spring (force constant =200N//m) whose other end is attached to a verticle wall. The block is pushed to compress the spring by a distance d and released. Find the value(s) of 'd' for which (spring + block) system loses its entire mechanical energy in form of heat. |
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Answer» 4 cm  (From work energy theorem, `DeltaK=0`) Rightarrow `d-d_(1)=(2mumg)/k Rightarrow d_(1)=d-(2mumg)/k` Now, similarly if `d_(2),d_(3)` e.t.c. are the successive distances of the block from the mean position where it come to rest for the second, third times etc. then `d_(2)-d_(3)=d_(1)` `-d_(2)=d-d_(1)=(2mumg)/k Rightarrow d_(2)=d_(1)-(2mumg)/k=d-(4mumg)/k` `d_(3)=d_(2)-(2mumg)/k=d-(6mumg)/k` `d_(n)=d-(2mumg)/k` we WANT `d_(n)=0` Rightarrow `d=n((2mumg)/k)` i.e.d should be an integral MULTIPLE of`(2mumg)/k` |
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| 37826. |
Electromagnetic waves are transverse in nature as in evident by : |
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Answer» POLARISATION |
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| 37827. |
The potential difference across the terminals of a cell varies with the current drown from the cell accroding to the graph. |
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Answer»
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| 37828. |
The torque necessary to produce an angular acceleration of 25 rad//s^2 in flywheel of mass 50 kg and radius of gyration 50 cm about its axis is |
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Answer» 312.5 DYNE cm |
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| 37829. |
Which of the following is a correct statement? A: A zener diode is mainly operated in reverse biased condition B: In forward biased condition, the zener diodem acts like an ordinary P-n junction diode |
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Answer» Both A and B |
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| 37830. |
A solid sphere is rotating about its diameter. Due to increase in room temperature, its volume increases by 0.5 %. If no external torgue acts, the angular speed of the sphere will : |
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Answer» Increase by nearly `(1)/(2)%` `therefore (DeltaV)/(V)=(3Deltar)/(r)` `(Deltar)/(r)=(1)/(3)(DeltaV)/(V)=(1)/(3)xx0.5=(1)/(6)%` Since there is no external torque IMPLIES `Iomega` = constant `(2)/(5)mr^(2)omega` = constant implies `r^(2)omega` = CONST. Taking log on both sides, `2logr+logomega=logc` Differentiating, we have `2.(1)/(r)dr+(1)/(omega)domega=0` `(domega)/(omega)=-2(dr)/(r)=-2XX(1)/(6)=-(1)/(3)%` `-ve` sign show that `omega` DECREASES |
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| 37831. |
Statement I: When two coils are wound on each other, the mutual induction between the coils is maximum. Statement II: Mutual induction does not depend on the orientation of the coils. |
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Answer» |
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| 37832. |
Calculate the internal resistance of the given cell using the following data. Balancing length for in open circuit = l _(1) = 0.60m |
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Answer» Solution :`r = R [ (l _(1))/(l _(2)) -1 ]` Trial `(i) r = R [ (l _(1))/( l _(2)) -1 ] =3 (0.6)/(0.4) -1] =1.5 Omega` Trial (ii) `r = R [ ((l _(1))/( l _(2))-1] =4 [ (0.6)/(0.44)-1] =1.45 Omega` Trial (III) `r = R [ (l _(1))/( l _(2)) -1 ] =5 [ (0.6)/(0.46) -1] = 1.52 Omega` Result : Internal resistance of the given cell LIES between `1.45 and 1.52 Omega` |
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| 37833. |
A cyclotron’s oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating protons, if the radius of its dees is 60 cm ? What is the kinetic energy of the proton beam produced by the accelerator? Given e = 1.6 xx 10^(–19) C, m = 1.67 xx 10^(–27) kg. Express your answer in units of MeV [1MeV = 1.6 xx 10^(–13) J]. |
| Answer» SOLUTION :`B = 0.656T, E_("MAX") = 7.421` MEV | |
| 37834. |
The magnetic field inside a currentcarrying toroidal solenoid is 0.1 mT. What is the magnetic field inside the toroid if the current through it is doubled and it's radius is made half? |
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Answer» SOLUTION :B=`mu_0nI` It does not depend upon t. It is DOUBLED if I is doubled `THEREFORE B=0.2mT. |
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| 37835. |
Of the following graphs, the one that correctly represents the I-V characteristics of a Ohmic device is |
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Answer»
i.e., `V alpha I` V = IR `(V)/(I)=R` (constant) HENCE, the curve between V and I is straight line. 
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| 37836. |
n identical cells each of emfE and internal resistance r are connected in series. An external resistance R is connected in series with this combination.The current through R is : |
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Answer» `NE/(R+nr)` |
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| 37837. |
Two identical conducting balls having positive charges q_1 and q_2 are separated by a center to center distance r. If they are made to touch each other and then separated to the same distance, the force between them will be. |
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Answer» LESS than before |
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| 37838. |
A pulley of radius 2 m is rotated about its axis by a force F=(20t-5t^(2)) Newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotationis 10kgm^(2), the number of rotations made by the pulley before its direction of motion is reversed, is : |
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Answer» more than 3 but less than 6 `tau=Ialpha=Fror10xxalpha=(20t-5t^(2))xx2` or `alpha=4t-t^(2) and omega=(d theta)/(dt)` Also `(d theta)/(dt)=alpha.t` or `d theta=alpha.TDT` `d theta(4t-t^(2)).tdt=(4t^(2)-t^(3))dt` Integrating both SIDES `theta=(4t^(3))/(3)-(t^(4))/(4)` If n rotations are completed in 4s, then putting `t=4` `therefore theta=2pin=(4xx64)/(3)-64=(64)/(3)~=3.4` But `3lt3.4lt6` |
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| 37839. |
Which of the following device is full duplex ? |
| Answer» Answer :A | |
| 37840. |
Draw a plot ofthe variation of amplitude versusomega for an amplitudemodulated wave.Define modulationindex. State its importance for effective amplitude modulation. |
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Answer» Solution :A PLOT of amplitude versus `omega` for an amplitude MODULATED wave : Modulation INDEX : Modulation index is the ratio of amplitude of modulating voltage `(A_(m))` andamplitude of carrier voltage `(A_(c))`. Modulation index `MU` is `mu=(A_(m))/(A_(c))`. Importance : The amplitude modulation index `(mu)` determines the quality of the transmitted index. When modulation index is small, variation in carrier amplitude will be small. 
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| 37841. |
The instantaneous values of alternating current and voltage in a circuit are i=(1)/(sqrt(2))sin(100pit)" A and "v=(1)/(sqrt(2))sin(100pit+(pi)/(3))V. The average power in watts consumed in the circuit is |
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Answer» `(1)/(4)` |
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| 37842. |
State any rule which relates the direction of electric current and the direction of the accompanying mangnetic field due to circular coil carrying current. |
Answer» Solution :Right HAND rule. For STATEMENT, REFER to ART. 
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| 37843. |
What are step-up and step-down transformers ? |
| Answer» SOLUTION :If the transformer CONVERTS an ALTERNATING CURRENT with low voltage into an alternating current with high voltage, it is called step-up transformer. On the CONTRARY, if the transformer converts alternating current with high voltage into an alternating current with low voltage, then it is called step-down transformer. | |
| 37844. |
A and B are two points separated by a distance 5 cm. Two charges 10 muC and 20 muC are placed at A and B. the resultant electric intensity at a point P outside the charges at a distance 5 cm from 10 mu C is |
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Answer» `54 xx 10^(6)` N/C away from F `10 mu C` |
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| 37845. |
(a) A particle of mass 1 mg and having charges 1 muC is moving in a magnetic field vec(B) = 2hat(i) + 3hat(j) + kT, with velocity vec(v) = 2 hat(i) + hat(j) - hat(k) m//sec. Find magnetic force in vector form and magnitude of acceleration. (b) A magnetic field of 8 hat(k)T exerts a force 8 hat(i) + 6hat(j) N on a particle having a charge 2C and going in x-y plane. Find the velocity of the particle. ( c) A particle is moving in a magnetic field 3hat(i) + 4hat(j) T and acceleration of particle is lambda hat(i) + 3hat(j) m//sec^(2). Find the value of lambda. (d) When a proton has a velocity bec(v) = (2 hat(i) + 3hat(j)) xx 10^(6) m//sec it experiences a force vec(F) = (- 1.28 xx 10^(-13) vec(k)) N. When its velocity is along the z-axis, it experiences a force along the x-axis. What is the magnetic field ? (e) A circular loop of radius 20 cm carries a current of 10 A. An electron cross the plane of the loop with a speed of 2.0 xx 10^(6) m//sec. The direction of motion makes an angle centre. Find the magnitude of the magnetic force on the electron at the instant it crosses the plane. |
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Answer» Solution :(a) `m = 1 mg = 10^(-6)kg, q = 1 muC = 10^(-6C` `vec(B) = 2hat(i) + 3hat(j) + hat(k)T, vec(v) = 2hat(i) + hat(j) - hat(k) m//sec` `vec(F) = q vec(v) xx vec(B) = 10^(-6)|(hat(i),hat(j),hat(k)),(2,1,-1),(2,3,1)|` `vec(a) = (vec(F))/(m) = (10^(-6))/(10^(-6))(4hat(i) - 4hat(j) + 4hat(k))` `= 4hat(i) - 4hat(j) + 4hat(k) m//sec^(2)` `|vec(a)| = a = 4sqrt(3) m//sec^(2)` (b) The particle is moving in `x-y` plane, hence its velocity `vec(v) = v_(x) hat(i) + v_(y) hat(j) m//sec` `q = 2C, vec(B) = 8hat(k)T, vec(F) = 8 hat(i) + 6 hat(j) N` `vec(F) = q vec(v) xx vec(B) = 2 |(hat(i),hat(j), hat(k)),(v_(x), v_(y), 0),(0, 0, 8)|` `= 2 [ 8v_(y)hat(i) - 16v_(x) hat(j)]` `8 hat(i) + 6hat(j) = 16v_(y)hat(i) - 16v_(x)hat(j)` Comparing coefficient of `hat(i)` and `hat(j)` `16v_(y) = 8 rArrv_(y) = (1)/(2)` `- 16v_(x) = 6rArrv_(x) = -(3)/(8)` `vec(v) = v_(x)hat(i) + v_(y) hat(j) = -(3)/(8)hat(i) + (1)/(2) hat(j) m//sec` ( c) `vec(B) = 3hat(i) + 4 hat(j)T, vec(a) = lambdahat(i) + 3 hat(j) m//sec^(2)` SINCE `vec(F) = qvec(v) xx vec(B), vec(F)` is `_|_^(ar)` to `vec(v)` as wll as `vec(B)` `vec(F) = mvec(a)`, hence `vec(a)` is `_|_^(ar)` to `vec(B)` The dot product of perpendicular vectors is zero. `vec(B).vec(a) = 0` `(3hat(i) + 4hat(j)).(lambda hat(i) + 3 hat(j)) = 0` `lambda = -4` (d) `vec(v) = (2hat(i) + 3hat(j)) xx 10^(6) m//sec, vec(F) = -(1.28 xx 10^(-13) hat(k))` Since `vec(F)`is `_|_^(ar)` to `vec(B), vec(B)` should be in `x-y` plane Let `vec(B) = B_(x)hat(i) + B_(y)hat(j)` `vec(F) = qvec(v) xx vec(B) = (1.6 xx 10^(-9))|(hat(i), hat(j),hat(k)),(2 xx 10^(6), 3xx10^(6), 0),(B_(x), B_(y), 0)|` `-(1.28 xx 10^(-13)) hat(k) = (1.6 xx 10^(-19) xx 10^(6))[2B_(y) - 3B_(x)] hat(k)` `2B_(y) - 3 B_(x) = 0.8`...(i) Let`vec(v) = v_(0)hat(k), vec(F) = F_(0)hat(i)` `vec(F) = q vec(v) xx vec(B)` `F_(0)hat(i) = q(v_(0)hat(k)) xx (B_(x) hat(i)+B_(y)hat(j))` `= qv_(0) [B_(x)hat(j) - B_(y)hat(i)]` `= q v_(0)B_(x) hat(i) - q v_(0) B_(y) hat(i)` Comparing coefficient of `j` : `qv_(0) B_(x) = 0 rArr B_(x) = 0` `2B_(y) - 3B_(x) = 0.8rArr B_(y) = 0.4` `vec(B) = 0.4 hat(j) T` (e)  Magnetic field centre of loop `B = (mu_(0) i)/(2R) = ((4pi xx 10^(-7)) (10))/(2 xx 0.2) = pi xx 10^(-5) T`, along x-axis `vec(B) = pi xx 10^(-5) hat(i) T` Velocity of electron, `|vec(v)| = v = 2 xx 10^(6) m//sec` `vec(v) = v cos 30^(@) hat(i) + v SIN 30^(@) hat(j) = (sqrt(3)hat(i) + hat(j)) xx 10^(6) m//sec` `vec(F) = q vec(v) xx vec(B) = (-1.6 xx 10^(-19))|(hat(i), hat(j), hat(k)),(sqrt(3) xx 10^(6), 10^(6), 0),(10^(-5)pi, 0, 0)|` `= (1.6 xx 10^(-19)) (10^(6))(10^(-5)pi) hat(k)` `= 1.6 pi xx 10^(-18) hat(k) N` `|vec(F)| = 1.6pi xx 10^(-18)N` |
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| 37846. |
What is the probability that an electron in the ground state of the hydrogen atom will be found between two spherical shells whose radii are r and r + Deltar, (a) if r = 0.500a and Deltar = 0.010a and (b) if r = 1.00a and Deltar = 0.01a, where a is the Bohr radius? (Hint: Deltar is small enough to permit the radial probability density to be taken to be constant between r and r+ Deltar.) |
| Answer» SOLUTION :(a) `3.2xx10^(-3),` (B) `5.4xx10^(-3)` | |
| 37847. |
How is the resolving power of a compound microscope affected if (a) wavelength of light used is decreased, and (b) the diameter of its objective lens is increased? Justify your answers. |
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Answer» Solution :The resolving power of a compound microscope is GIVEN as : `R.P.=(2nsinalpha)/(1.22lamda)` where n = refractive index of medium present between the object and the objective lens. `2alpha` is the angle subtended by the microscope objective at the object and `lamda` = WAVELENGTH of LIGHT used to illuminate the object. (i) If wavelength `lamda` is decreased then as per above relation resolving power of microscope increases. (ii) If diameter of objective lens is increased, value of `.alpha.` and consequently sin `alpha` increases and so the resolving power increases. |
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| 37848. |
The pressure and density of gas (gamma =(7)/(5)) changes adiabatically from (P, rho)" to "(P',rho'). If (rho')/(rho)=32, then find the ratio of (P')/(P) : |
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Answer» 128 `therefore (P_(2))/(P_(1))=((V_(1))/(V_(2)))^(gamma) RARR (P.)/(P)=((m//rho)/(m/rho.))^(gamma)` `(P.)/(P) =((rho.)/(rho))^(7//5) =(32)^(7//5)=(2^(5))^(7//5)=2^(7)` `(P.)/(P) =132` THUS, CORRECT choice is (a) |
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| 37849. |
What is the significance of the negative total energyof an electronorbiting round the nucleus ? |
| Answer» SOLUTION :The ELECTRONIS revolvingaroundthe NUCLEUS . | |
| 37850. |
A radioactive isotope has a half-life of T years. How long will it take the activity to reduce to (a) 3.125%, (b) 1% of its original value ? |
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Answer» Solution :(a) Here `T_(1/2)` =T YEARS and `R/R(0)`=3.125% = `3.125/100=1/32` Let after n half-lives the activity is reduced to R. Then, we have `(1/2)^(n)=R/R_(0)=1/32 implies n=5` HENCE, after a time 5T years, the activity is reduced to 3.125% of its original activity. (b) Let after n half-lives the activity becomes 1% of its original value, `THEREFORE (1/2)^(n)=R/R_(0)=1/100 implies 100=(2)^(n) or LOG2 implies n=(log100)/(log2)=2.000/0.3010=6.65` `therefore time " "t=6.65T_(1/2)=6.65 T` years. |
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