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InterviewSolution
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.
| 601. |
A capacity is made of a flat plate of area A and a second plate having a stair-like structure as shown in figure.The width of each stair is .a. and the height is .b.. Find the capacitance of the assembly . |
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Answer» `(epsilon_(0)A(3D^2+6bd+2b^2))/(3d(d+b)(d+2b))` |
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| 602. |
Magnitude of acceleration of an electron having speed 2.5xx10^(6)ms^(-1) in a magnetic field of value 2.0 G, is (given : specific charge of the electron =1.76xx10^(11)Ckg) |
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Answer» `6.6xx10^(23)CMS^(-2)` |
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| 603. |
Assertion: Though radiation of a single frequency are incident on a metal surface, the energies of emitted photoelectrons are different. Reason: The energy of electrons emitted from inside the metal surface is lost in collision with other atom is the metal. |
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Answer» If both ASSERTION and REASON are ture and the reason is the correct EXPLANATION of the assertion. |
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| 604. |
In the given figure, the electron enters into the magnetic field. It deflects in ...... .. direction . |
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Answer» `+` ve X DIRECTION |
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| 605. |
Pure Si at 300 K has equal electron (n_(e)) and hole (n_(h)) concentration of 1.5xx10^(16)m6(-3). Doping by indium increases into 4.5xx10^(22)m^(-3). Calculate n_(e) in the doped silicon . |
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Answer» Solution :Here `n_(i)=1.5xx10^(16)m^(-3),n_(H)=4.5xx10^(22)m^(-3)` But `n_(e)n_(h)=n_(i)^(2)` `therefore n_(e)=(n_(i)^(2))/(n_(h))=((1.5xx10^(16))^(2))/(4.5xx10^(22))=5xx10^(9)m^(-3)` |
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| 606. |
A 1 mu F capacitor C is connected to a battery of 10 V through a resistance 1 M Omega. The voltage across C after 1 sec is approximately |
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Answer» Solution :`C=1 mu F = 10^(-6)F, R = 1M Omega = 10^(6)Omega` `V_(0)=10 V` Time constant of circuit, `tau_(c )=RC=10^(6)xx10^(-6)=1s` Voltage across C in time `t, V=V_(0)(1-e^(-t//tau_(c )))` Fort = 1 s, and `tau_(c )=1s` `therefore V=10(1-(1)/(e ))=6.3 V` |
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| 607. |
Curie is the unit of : |
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Answer» ENERGY of `gamma-rays` |
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| 608. |
Two lenses of power - 15D and + 5D are in contact with each other. The focal length of the combination is ..... |
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Answer» `+10 cm ` `P=P_1+P_2` `therefore 1/F=-15+5` `therefore 1/f=-10` `therefore f=-(1)/(10)` m `therefore f=-(100)/(10) cm` `therefore f=-10` cm |
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| 609. |
In the grid circuit of a triode a signal E = 2 sqrt(2) cos omega t is applied. If mu = 14 and r_(p) = 10 k Omega then root mean square current flowing through R_(L)=12k Omega will be |
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Answer» `1.27 mA` |
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| 610. |
If a 10 kg body falls to the ground from a height of 15 m and if all its mechanical energy is convetred into heat the heat produced will be : (5 = 4.2 J/cal) |
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Answer» 6 cal Q = 350 CALORIES. |
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| 611. |
What will be the image of an object placed before a convex mirror-erect or inverted? |
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Answer» |
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| 612. |
What is the principle of production of electromagnetic wave ? |
| Answer» Solution :An ACCELERATED charge PRODUCES ELECTRIC and MAGNETIC field, which VARY both in space and time | |
| 613. |
A semiconductor is known to have an electron concentration of 8 xx 10^(13) cm^(-3) and a hole concentration of 5 xx 10^(12) cm^(-3) . a . Is the semiconductor n- type of p-type ? b. What is the resistitivity of the simple , if the electron mobility is 23,000 cm^(2) V^(-1) s^(-1) and hole mobility is 100 cm^(2) V^(-1) s^(-1) ? Take charge on electron , e = 1.6 xx 10^(-19) C . |
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Answer» Solution :Data SUPPLIED , `n_(e) = 8 xx 10^(13) cm^(-3) = 8 xx 10^(19) m^(-3)` `n_h = 5 xx 10^2 cm^(-3) = 5 xx 10^(18) m^(-3)` `mu_(e) = 23,000 cm^(2) V^(-1) s^(-1) = 2.3 m^(2) V^(-1) s^(-1)` `mu_(H) = 100 cm^(2) V^(-1) s^(-1) = 0.01 m^(2) V^(-1) s^(-1)` a. Here semiconductor has GREATER electron concentration . So it is n- type semiconductor . b. `(1)/(rho) = e(n_(e) mu_(e)+ n_(h) mu_(h)) = 1.6 xx 10^(-19) (8 xx 10^(19) xx 2.3 +5 xx 10^(18) xx 0.1) = 30 m HO m^(-1)` `rho = (1)/(30) = 3.3 xx 10^(-2) Omega m` |
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| 614. |
An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is closed to: |
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Answer» 80 H Now, `I=V_(rms)/"|z|"` (For A.C. current ) `10=220 /sqrt(R^2+X_L^2)` `100=(220xx220)/((8)^2+X_L^2)` `64+X_L^2=484` `therefore X_L^2`= 420 `therefore X_L = sqrt420` `therefore 2pif L=sqrt420` `therefore 2xx3.14xx50xxL = sqrt420` 314 L = `sqrt420` `therefore L=20.49/314` `therefore` L=0.6526 `APPROX` 0.065 H |
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| 615. |
The ratiobetweentheradiiof nucleiwithmassnumbers27and125is |
| Answer» Answer :B | |
| 616. |
Clarify your elementary notions about current in a metallic conductor by answering the following queries : (a) The electron drift speed is estimated to be only a few mm s^(-1) for currents established almost the instant a circuit is closed? (b ) The electron drift arises due to the force experienced by electons in the electric field inside the conductor. But force should cause acceleation. Why then do the electrons acquire a steady drift speed? ( c )If the electron drift speed is so small, and electron's charge is small, how can we still obtain large amounts of current in a conductor? (d ) When electrons drift in a metal from lower to higher potential, does it mean that all the ''free'' electrons of the metal are moving in the same direction ? (e ) Are the paths of electons straight lines between successive collisions (with the positive ions of the metal ) in the (i) absence of electic filed, (ii) presence of electic field. |
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Answer» Solution :(a) When we close the circuit, the electric field is set up in the entire closed circuit instantly with the speed of eleectromagnetic wave which causes electron drift at every portion of the circuit. Due to it, the current so set up in the entire circuit instantly. The current so set up does not wait for the electrons to flow from one end of the conductor to other end. However, the current does take a little time to reach its steady value. ( b) When a potential difference is applied across the ends of a conductor, the electric field is set up inside the conductor. Due to it, each FREE electron in the conductor experiences a force and is accelerated towards the positive end of the conductor, resulting in the increase in its speed until it collides with a positive ion of the metal. After the collision, the free electorn loses its speed but again starts to accelerate by virute of force on it due to electric field. Its speed inceraess again only to suffer a collision again and so on. Therfore, on the average, the free electrons acquire only a drift speed in the conductor, under the effect of electic field. ( c) In a conductor, we obtain large amount of current EVEN if the electorn drift speed is small and electron's charge is small, because the electron number density in the conductor is very large, (of the order of `10^(29) m^(-3)`). (d) When a potential difference is applied across the ends of a conductor, the drifting of free electrons takes PLACE from lower to higher potential. These drifiting electrons also have thermal velocity. At an INSTANT, each free electron moves in a direction of resultant velocity of its thermal velocity and drift velocity, which is different for various free electrons in a conductor. Hence, all the free electrons in the metal conductor are not moving in the samedirection. (e) (i) In the absence of electric field, the paths of electrons are straight lines between successive collisions with positive ions of the metal, since the electrons are moving with their thermal velocity like molecules in a gas. (ii) In the presence of electric field, the paths of the electrons between two successive collisisons with positive ions are curved as each electron is having two velocities (a) thermal velocity and (b) velocity by virtue of force due to electic field. |
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| 617. |
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter ? Hence explain why aluminium wires are preferred for overhead power cables.(rho_(Al) = 2.63 xx 10^(-8) Omega m , rho_(Cu) = 1.72 xx 10^(-8) Omega m, Relative density of Al = 2.7 , of Cu = 8.9 ) |
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Answer» Solution :Light weight CABLES are chosen for overhead power cables. If VOLUME density of MASS for GIVEN wire is d then, `d = (m)/(V) = (m)/(Al) = (m)/(((rho L)/(R))l)"" (because R = (rho l)/(A) rArr A = (rho l)/(R) )` `therefore d = (mR)/(rho l^(2)) ` ` therefore m = (rho l^(2) d)/(R)` Here for given wires I and R are same and so, m `proprho`d `therefore (m_(1) )/(m_(2)) = (rho_(1) d_(1))/(rho_(2) d_(2))` Taking aluminium and copper wires as first and second wire respectively, `(m_(1))/(m_(2)) =((2.63 xx 10^(-8))( 2.7 xx 10^(3)))/((1.72 xx 10^(-8) ) (8.9 xx 10^(3)))` ` therefore (m_(1))/(m_(2))` = 0.4638 `therefore m_(1) lt m_(2)` Aluminium wire is lighter than copper wire. Hence, aluminium wires are chosen as overhead power cables. |
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| 618. |
A small portion of Indium is incorporated is germanium . The crystal will be : |
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Answer» <P>N - TYPE |
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| 619. |
The embryo sac of a typical dicotyledonous plant at the time of fertilization is |
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Answer» 7 cell |
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| 620. |
Doping in silicon with iodium leads to which type of semiconductor ? |
| Answer» SOLUTION :It LEADS to p-type SEMICONDUCTOR. | |
| 621. |
A planeconvex lens is made of refraction index of 1.6 The radius of curvature of the curved surface is 60 cm. The focal leght of the lens is : |
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Answer» Solution :(c ) Here, `(1)/(f) = (mu - 1) [(1)/(R_(1)) - (1)/(R_(2))]` `(1)/(f) = (1.6 - 1) [(1)/(60) - (1)/(oo)] = (1)/(100)` |
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| 622. |
Waves of same amplitude and same frequency from two coherent sources overlap at a point. The intensity when they arrive in phase to that when they arrive with 90^(@) phase difference is |
| Answer» ANSWER :C | |
| 623. |
Considering the circuit and data given in the diagram, calculate the currents flowing in the diodes D_(1) and D_(2). Forward resis- tance of D_(1) and D_(2) is 20Omega |
Answer» Solution :Since the positive terminal of battery is connected to P-type of both DIODES `D_(1) and D_(2)` they are FORWARD biased. These diodes are replaced by with their forward RESISTANCES as shown in FIG.  The resistance of `20Omega` in PARALLEL `(1)/(R)=(1)/(20)+(1)/(20)=(2)/(20)` Therefore, total current I in the circuit `I=(1)/(100+10)=(1)/(110)" amp and "` `I_(1)=I_(2)=(1)/(2)xx(1)/(110)=(1)/(220)" amp"` |
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| 624. |
The force experienced by a charge of 2 muC in an electric field is 3xx10^(-3)N. The intensity of the electric field |
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Answer» `1.5 XX 10^3 N//C` |
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| 625. |
Two identical capacitors are connected in the circuit shown. C_(2) is charged to 100 V and switch S is shifted from position 1 to position 2. The negative terminal of C_(2) is connected with the positive terminal of C_(1). Find the work done by the battery. If the polarity of C_(2) were reversed, what work would have been done by the battery ? |
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Answer» Solution :Case I :- `KVLrArr100-(2(1000+q))/(C )=0` `Q=500muC` WORK done by the BATTERY, `W=0.5xx10^(-3)xx100=0.05J` Case II :- `KVL` `rArr 100-(((q-1000))/(10))-(((q+1000))/(10))=0` `Q=500muC` `:. ` Work done by the battery, `W=DeltaqE=0.05J` 
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| 626. |
The real image formed by a concave mirror is larger than the object when the object is : |
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Answer» at a DISTANCE equal to RADIUS of curvature |
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| 627. |
In a Young's double slit experiment set-up source S of wavelength 5000A^(@) illuminates two slits S_(1) and S_(2). Which act as two coherent sources. The source S oscillates about its shown position according to the equation y = 0.5 sin pi t. where y is in millimeters and t in second. Find positions of central maxima as a function of time with respect to 'O' |
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Answer» `y = (cos 2PI t)MM` |
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| 628. |
The distance between diminshed and magnified images of slits for twopositions are d_(1) and d_(2) respectively of the slite is |
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Answer» `d=d_(1)d_(2)` |
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| 629. |
A moving coil galvanometer has a circular coil of area 16 cm^(2) and 30 turns. It is mounted on a torsional spring of spring constant 100 Nm/rad. The coil turns by angle 21^(@) when a current 40 mA is passed through it. The magnitude of the uniform magnetic field produced by the permanent magnet inside the galvanometer is _________________T. [use cos 21^(@)=15/16, 21^(@)~~11/30 rad] |
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Answer» `vec(tau)=vecmxxvecB` Here, `vecm`is the magnetic moment of the coil and `vecB`is the magnetic field. The construction of the galvanometer is such that initially the magnetic moment vector is perpendicular to the magnetic field, and the torsional spring is in a RELAXED state. As the torque acts on the coil, the rotation of the coil leads to a deformation of the spring. An equilibrium is achieved where the torque due to the magnetic field is balanced by the restoring torque due to the spring. Let the angular displacement of the coil from the original position be . Then, the restoring torque due to the spring is given by:`tau_(S)=C theta` , Here C is the torsional constant of the spring. In this condition, the torque due to the magnetic field is, `|vec(tau)_(B)|=|vecmxxvecB|=m B sin (90^(@)-theta)=m B cos theta` SINCE in equilibrium `tau_(B)=tau_(S)` `m B cos theta =C thetaimpliesB=(C theta)/(m cos theta)=((5.4xx10^(-3))(11/30))/((0.04)(16xx10^(-4))(30)(15/16))=1.1T` |
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| 630. |
Find the force between two small circular coils of radii r_(1) = 1 cm, r_(2) =2 cm and currents I_(1) = 1 A and I_(2) = 2A placed at d = 20cm apart with their planes parallel to each other. |
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Answer» |
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| 631. |
At equatorial line horizontal factor of the magnetic field of the earth is ........ . |
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Answer» zero `therefore B_(h) = B cos phi` `therefore` At on EQUATORIAL LINE `phi = 0^(@)` `therefore = B_(h)= B`becomes maximum. |
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| 632. |
calculate the maximum possible force that the negative charge may experience if it is to be placed somewhere on X-axis, What will be the distance of negative charge from origin where it experiences the maximum electrostatic force? |
| Answer» SOLUTION :`(QQ)/(piepsilon_0)1/(3sqr3d^2),d/sqrt2` | |
| 633. |
The peak voltage of an ac supply is 300 V. What is the rms voltage ? |
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Answer» SOLUTION :(a) `(300)/(SQRT(2))=212.1V` (B) `10sqrt(2)=14.1A` |
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| 634. |
A ray of light passes through an equilateral glass prism such that the angle of incidence is equal to the angle of emergence. The angle of emergence is 3/4 times the angle of prism. Calculate the refractive index of the glass prism. |
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Answer» Solution :Since the angle of incidence is equal to angle of emergence therefore the ray of LIGHT passes symmetrically through the prism. So, the prism is in minimum DEVIATION position. Now, `A + delta_m = i + e ` ` delta_m = e + e - A = 2 (3/4 A) -A` ` = 3/2 A - A = A/2 = (60^@)/(2)= 30^@` ` n_(21) = (sin ((A+ delta_m)/(2)))/(sin (A/2))= (sin ((60^@ + 30^@)/(2)))/(sin (60^@)/(2))` ` = (sin 45^@)/(sin 30^@) =(1)/(sqrt2) xx 2/1 = SQRT 2`= 1.414 |
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| 635. |
An electric field line emerges from a positive point charge +q at angle alpha to the straight line connecting it to a negative point charge -2q as shown in figure. At what angle beta with the field line enter the charge-2q ? |
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Answer» `alpha` `L(B_(0)IL)=3(lambdaL)g((L)/(2)+(L)/(6))=3lambdaLg((2L)/(3))` `B_(0)=(2lambdag)/(I)` No external FORCE, so COM cannot displace initial coordinate of `COM=(3(lambdaL)("zero")+2lambdaL((L)/2)+lambdaL(L))/(6lambdag)=(L)/(3)` FINAL coordinate of `COM=(L)/(3)("same")` But COM displaces with respect to QT by `(2L)/(3)`. So displacement of `QPUT=(2L)/(3)`. Initial magnetic dipole moment M makes an angle of `(pi)/(4)` ACW from +ive x-axis and finally it makes angle `(3pi)/(4)` ACW from `+"ive x-axis" (M=sqrt(2)IL^(2))` So change in `PE=-MB(cos theta_(2)-costheta_(1))=MB_(0)[cos135^(@)-cos45^(@)]=2B_(0)IL^(2)="Gain in KE"=(1)/(2)I omega ^(2)` `I("about QT")=2[(2lambdaL^(3))/(3)+lambdaL^(3)]=(10lambdaL^(3))/(3),2B_(0)IL^(2)=(1)/(2)(Iomega^(2))=(1)/(2)((10lambdaL^(3))/(3))omega^(2),omega^(2)=(6B_(0)I)/(5lambdaL)` |
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| 636. |
The question has a paragraph followed by two statements, Statement - 1 and Statement - 2. Of the given four alternatives after the statements, choose the one that describes the statements. (A.I.E.E.E. 2011) A thin air film is formed by putting the convex surface of a plane - convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film. Statement - 1 : When light reflects from the air - glass plate interface, the reflected wave suffers a phase change of pi. Statements - 2 : The centre of the interfrence pattern is dark. |
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Answer» Statement - 1 is TRUE, Statement - 2 is true, Statement - 2 is the correct explanation of Statement - 1. |
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| 637. |
Why did the boy come early? |
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Answer» To have fun |
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| 638. |
Earth's magnetic field always has horizontal component except at |
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Answer» equator |
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| 639. |
At what angle should an object be projected so that the maximum height reached is equal to the horizontal range. |
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Answer» Solution :VELOCITY at the HIGHEST POINT `v = cos 45^@ =v/ /SQRT2` ` (k.E)_h =1/2 m ( v/sqrt2) =1/2xx 1/2 mv^2= E//2`. |
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| 640. |
What is photoelectric work function and how it is related to threshold frequency ? |
| Answer» Solution :The MINIMUM amount of RADIANT energy needed to PULL an electron from a METALLIC SURFACE is called work function of the metal. m | |
| 641. |
An atom in the state .^(2)P_(3//2) is located on the axis of a loop of radius r=5 cm carrying a current, I=10 A. The distance between the aotm and the centre of the loop is equal to the radius of the latter. How great may be the maximum force that the magnetic field of that current exerts on the atom? |
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Answer» Solution :The force on an atom with magnetic moment `vec(mu)` in a magnetic field of induction `vec(B)` is given by `vec(F)=(vec(mu).vec(GRAD))vec(B)` In the PRESENT case, the maximum force arise when `vec(mu)` is along the axis of close to it. Then`F_(Ƶ)=(mu_(Ƶ))_(max)(del B)/(delƵ)` Here `(mu_(Ƶ))_(max)=gmu_(B)J`. The Lande FACTOR `g` is for `^(2)P_(1//2)` `g=1+((1)/(2)xx(3)/(2)+(1)/(2)xx(3)/(2)-1xx2)/(2(1)/(2)xx(3)/(2))=1-(1//2)/(3//2)=(2)/(3)` and `J=(1)/(2) so (mu_(Z))_(max)=(1)/(3)mu_(B)`. The magnetic field is given by `B_(Ƶ)=(mu_(0))/(4 pi).(2Ipi R^(2))/((r^(2)+Ƶ^(2))^(3//2))` or `(del B_(Ƶ))/(delƵ)=-(mu_(0))/(4pi)6I pir^(2)(Ƶ)/((r^(2)+Ƶ^(2))^(5//2))` Thus `((del B_(Ƶ))/(delƵ=r))=(mu_(0))/(4 pi)(3I pi)/(sqrt(8)r^(2))` Thus maximum force is `F=(1)/(3)mu_(B)(mu_(0))/(4 pi)(3 pi)/(sqrt(8))(I)/(r^(2))` Substitution gives (using data in `MKS` units) `F=4.1xx10^(-27)N` |
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| 642. |
What were the conclusions drawn from the alpha-particle experiments. |
| Answer» Solution :(i) The positive charge of the ATOM is concentrated in a small core called nucleus. (II) The SPACE round the nucleus is EMPTY. | |
| 643. |
S_1 and S_2are two hollow concentric thin spherical shells enclosing charges Q and 2Q respectively as shown in What is the ratio of the electric flus through S_1 and S_2? (ii) How will the eletric flux through the shell S_1change, If a medium of dielectric constant 5 is introduced in the space inside S_1in place of air ? (##U_LIK_SP_PHY_XII_C01_E09_024_Q01.png" width="80%"> |
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Answer» SOLUTION :(i) According to Gauss.s law electric flux through spherical shell` S_1 ` `phi_1=(1)/(in_0) Q ` and flux through outer shell ` S_2 phi_2=(1)/(in_0)(Q+ 2Q)=(1)/(in_0).3Q ` ` rArr "" (phi_1)/(phi_2) =(1)/(3) ` (ii)When a medium of DIELECTRIC constant K=5 is introduced in the space inside the shell `S_1 ` in place or AIR , flux through `S_1 ` will be modified to ` phi_1 =(1)/(in ) .Q =(1)/(K in _0)Q =(phi_1)/(5) i.e. `flux will be reduced to `(1)/(5)`th of its previous VALUE. |
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| 644. |
If C_(p) and C_(v) denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then |
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Answer» `C_(p)-C_(v)=(R )/(14)` `C_(p)-C_(v)=r," where "r=(R )/(M)` is gas constant. For 1 g of nitrogen Molecules WT. of nitrogen M=28 g `therefore C_(p)-C_(v)=(R )/(28)` So, CORRECT CHOICE is (d). |
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| 645. |
An arbitary surface enclose an electric dipole of dipole moment 20 xx 10^(-6)C-m, What is the electric flux through this surface? |
| Answer» Solution :ZERO, because NET charge WITHIN the closed surface zero. | |
| 646. |
An angular magnification (magnifying power) of 30 X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope ? |
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Answer» Solution :In normal ADJUSTMENT of microscope, the final image is formed at the least distance of distinct vision i.e., D = 25 cm. In that case, angular magnification of eyepiece: `m_(e) =(1+ D/f_(0)) = 1+ 25/5 = 6``[therefore f_(e) = + 5 cm]` As magnification of microscope m = 30 and `m = m_(0) XX m_( e)` `rArr m_(0) =m/m_( e) = 30/6 -5, therefore m_(0) =v_(0)/u_(0) =5` or `v_(0) = 5u_(0)` and as per sign CONVERSION `u_(0)` is `-ve` but `v_(0)` is `+ve` and `f_(0) = + 1.25 cm` Hence, we have `1/v_(0) - 1/u_(0) = 1/f_(0)` or `1/(5u_(0)) -1/(-u_(0)) = 1/1.25` or `6/(5 u_(0)) = 1/1.25` `rArr u_(0) = 1.5 cm` and hence, `|v_(0| = |5u_(0)| = 7.5 cm` Moreover `m_( e) =|v_(e)/u_( e)| =D/|u_(e)|`, hence, `|u_(e)| = D/m_(e) = 25/6 = 4.17 cm` Separation between the objective and eye-lens = 7.5 + 4.17 = 11.67 cm and the object should be placed 1.5 cm from the objective lens to obtain the DESIRED magnification. |
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| 647. |
Uncertainty of momentum of a particle is 10^(-30) kg ms^(-1),minimum uncertainty in its position will be…….. |
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Answer» `10^(-8)m` `THEREFORE Deltax=(h)/(2piDeltap)` `=(6.62xx10^(-34))/(6.28xx10^(-30))=1.054xx10^(-4)m~~10^(=-4)m` |
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| 648. |
A body of mass 2 kg is thrown vertically with K.E. 490 J. Taking g=10 m//s^(2). The height at which K.E. is reduced to half will be : |
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Answer» 10 m `h=(245)/(2xx9.8)=12.25 m` |
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| 649. |
Find out values of i_1, i_2 and i_3 |
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Answer» ` 0,1//5A ,1//5A ` |
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| 650. |
If the capacitors are uncharged prior to the switch being closed, the amount of heat dissipated in resistors R_(1) and R_(2) after the switch is closed would be in the ratio of |
| Answer» Answer :B | |