Saved Bookmarks
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.
| 39151. |
Preferential filtering of light consisting of different wavelengths can be done using total internal reflection. The concept of filtering lies in the fact that different wavelengths have different refractive indices in a medium The beam of light with different wavelengths are made incident on an isosceles right angled prism as shown in the figure. The wavelength or wavelengths which satisfy the condition of totai internal reflection, theta sin ge(1/mu), get reflected Others get refracted, hence gets separated. G is the angle of incidence and u. is the refractive index .of the particular wavelength in that particular medium. Thus, by changing the medium and by changing the angles of the prism we can satisfy the condition df total internal reflection for a particular wavelength, which gets separated out. Q Now instead of changing the material of prism, we change one of the angle of prism to 30^(0) Then, which of the colour gets separated? |
|
Answer» red |
|
| 39152. |
The earliest method of measuring refracting angle of a prism is to direct a parallel beams from the 2 faces of the prism as shown. Show that this angular separation is twice the angle of the prism (beta = 2alpha ). |
|
Answer» |
|
| 39153. |
Preferential filtering of light consisting of different wavelengths can be done using total internal reflection. The concept of filtering lies in the fact that different wavelengths have different refractive indices in a medium The beam of light with different wavelengths are made incident on an isosceles right angled prism as shown in the figure. The wavelength or wavelengths which satisfy the condition of totai internal reflection, theta sin ge(1/mu), get reflected Others get refracted, hence gets separated. G is the angle of incidence and u. is the refractive index .of the particular wavelength in that particular medium. Thus, by changing the medium and by changing the angles of the prism we can satisfy the condition df total internal reflection for a particular wavelength, which gets separated out. Q As the density of the material-of the prism keeps on increasing, then which of the following statement holds good |
|
Answer» First blue GETS SEPARATED from red and GREEN, then green from blue and red. and FINALLY red from green and.blue. |
|
| 39154. |
Preferential filtering of light consisting of different wavelengths can be done using total internal reflection. The concept of filtering lies in the fact that different wavelengths have different refractive indices in a medium The beam of light with different wavelengths are made incident on an isosceles right angled prism as shown in the figure. The wavelength or wavelengths which satisfy the condition of totai internal reflection, theta sin ge(1/mu), get reflected Others get refracted, hence gets separated. G is the angle of incidence and u. is the refractive index .of the particular wavelength in that particular medium. Thus, by changing the medium and by changing the angles of the prism we can satisfy the condition df total internal reflection for a particular wavelength, which gets separated out. Q An experiment was performed based on the above theory with red, green and blue light. Their refractive indices in the material were 1.39, 1.44 and 1.47, respectively. Which colour would get separated? |
| Answer» Answer :A | |
| 39155. |
Preferential filtering of light consisting of different wavelengths can be done using total internal reflection. The concept of filtering lies in the fact that different wavelengths have different refractive indices in a medium The beam of light with different wavelengths are made incident on an isosceles right angled prism as shown in the figure. The wavelength or wavelengths which satisfy the condition of totai internal reflection, theta sin ge(1/mu), get reflected Others get refracted, hence gets separated. G is the angle of incidence and u. is the refractive index .of the particular wavelength in that particular medium. Thus, by changing the medium and by changing the angles of the prism we can satisfy the condition df total internal reflection for a particular wavelength, which gets separated out. Q If hypothetically the density of the material is increased gradually, then which of the colour gets separated next? |
|
Answer» red |
|
| 39156. |
A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment? |
| Answer» SOLUTION : YES, because magnetic FORCE can change the direction of V, not its MAGNITUDE. | |
| 39157. |
An aluminium wire of cross-sectionbal area 1 xx 10^(-6)m^(2) is joined to a steel wire of same cross-sectional ara. This compound wire is stretched on a sonometer, pulled by a weight of 10 kg. The total length of the compound wire between the bridges is 1.5 m, of which the aluminium is 0.6 m and the rest is steel wire. Transverse vibrationas are set up in the wire by using an ecternal force of variable frequency. Find the lowest frequency of excitation for which standing waves are formed such that the joint in the wire is a node. What is the total number of nodes observed at this frequency, excluding the two at the ends of the wire? The density of aluminium is 2.6 xx 10^(3) kg//m^(3) and that of steel is 1.04 xx 10^(4) kg//m^(3). |
|
Answer» `sigma_(1) = 2.6 xx 10^(3)kg//m^(2) , sigma_(2) = 1.04 xx 10^(4) kg//m^(3)` `v_(1) = sqrt(100/(1xx10^(6)xx1.04xx10^(3)))=(10^(3))/(sqrt(26))m//sec` `V_(2)=sqrt(100/(1xx10^(-6)xx1.04xx10^(4)))=(10^(3))/(2sqrt(26))` `f_(1)=f_(2)` `(n_(1)xx10^(3))/(2 xx 0.6sqrt(26)) = (n_(2) xx 10^(3))/(2 xx 2 xx 0.9sqrt(26)) [(n_(1))/(n_(2)) = (1)/(3)]` Lowest FREQUENCY `= (n_(1)v_(1))/(2L_(1)) = (1 xx 10^(3))/(sqrt(26) xx 2 xx 0.6)` `= (10^(4))/(12sqrt(26)) = 162 "vibration"//"sec"`. `(n_(1))/(n_(2)) = ((1)/(3))` One side `1` loop and other side `3` loop  So EXCLUDING the two nodes at ends total nodes is `3` 
|
|
| 39158. |
Derive an expression for electrostatic potential due to a point charge . |
|
Answer» Solution :Electric POTENTIAL due to point charge : Comnsider a positive charge q kept fixed at the origin . Let P be a point at distance r from the charge q . The electric potential at the point P is `V=int_(oo)^(r )(-vecE).dvecr=-int_(oo)^(r )vecEdvecr "" cdots (1)`  Electric field due to positive point charge q is `vecE=(1)/(4piepsilon_(0))(q)/(r^(2))hatr "" V=(-1)/(4piepsilon_(0))int_(oo)^(r)(q)/(r^(2)) hatr.dvecr` The infinitesimal displacement vector `dvecr=drvecr` and using `hatr.hatr=1`, we have `V=-(1)/(4piepsilon_(0))int_(oo)^(r)(q)/(r^(2))hatr.drhatr=-(1)/(4piepsilon_(0))int_(oo)^(r)(q)/(r^(2))dr` After the integration `V=-(1)/(4piepsilon_(0))q{-(1)/(r)}_(oo)^(r) =(1)/(4piepsilon_(0))(q)/(r)` Hence the electric potential due to a point charge q at a distance r is `V=(1)/(4piepsilon_(0))(q)/(r)` Important points ( If asked in exam ) (i) If the source charge q is positive `V gt 0. ` If q is negative then V is negative and equal to `V = (1)/(4piepsilon_(0))(q)/(r)` (ii) The description of objects using the concept of potential or potential energy is simpler than that using the concept of field. (iii) From expression (2) it is CLEAR that the potential due to positive charge decreases as the distance increases but for a negative charge the potential increases as the distance is increased . At infinity ( `r=oo` ) electrostatic potential is zero( V=0). (IV) The electric potential at a point P due to a collection of charges `q_(1), q_(2), q_(3) cdots q_(n)` is equal to SUM of the electric potentials due to individual charges . `V_("tot")=(kq_(1))/(r_(1)) +(kq_(2))/(r_(2))+(kq_(3))/(r_(3))+cdots+(kq_(n))/(r_(n))=(1)/(4piepsilon_(0))sum_(i=0)^(n) (q_(i))/(r_(i)) ` where `r_(1), r_(2), r_(3), cdots, r_(n)` are the distances of `q_(1), q_(2), q_(3)cdots, q_(n)` respectively from P 
|
|
| 39159. |
In a single state transistor amplifier, when by 10muA and collector current by 1mA. If collector load R_(C )=2kOmega and R_(L)=10kOmega, Calculate : (i) Current Gain (ii) Input impedance, (iii) Effective ac load, (iv) Voltage gain and (v) Power gain. |
|
Answer» <P> Solution :(i) Current Gain `beta=(Deltai_(c))/(Deltai_(b))beta=(Deltai_(c))/(Deltai_(b))=(1mA)/(10muA)=100`(ii) INPUT impedance `R_(i)=(DeltaV_("BE"))/(Deltai_(b))=(0.02)/(10muA)=2000Omega=2kOmega` (iii) Effective (a.c) LOAD `R_(AC)==R_(C)||R_(L))` `therefore R_(AC)=(2xx10)/(2+10)=1.66kOmega` (iv) Voltage gain, `A_(v)=betaxx(R_(AC))/(R_("in"))=(100xx1.66)/(2)=83` (v) `"Power gain, "A_(p)="Current gain"xx"Voltage gain"` `=100xx83=8300` |
|
| 39160. |
The sum of the magnitude of two vectors is 18. the magnitude of their resultant is 12. If the resultant is perpendicular to one of the vectors, then the magnitudes of the two vectors are : |
|
Answer» 5 and 13 Also `A^(2)+B^(2)+2ABcostheta=(12)^(2)=144` Also `(Bsintheta)/(A+Bcostheta)=tan90^@=OO` or `A+Bcostheta=0` and `Bcostheta=-A ` Putting above `A^(2)+B^(2)+2A(-A)=144` or `B^(2)-A^(2)=144`or `(B-A)(B+A)=144` or `18(B-A)=144 or B-A=8` Now B + A = 18 and B - A = 8. Solving we get B = 13 and A = 5. |
|
| 39161. |
Two diodes are connected in the following fashion. Provision is made to connect either +5 V or ground (0 V) to the points A to B. The output Qwill act as |
|
Answer» OR gate  When both the terminal of A and B at 5V (high), then both diodes becomes in reverse bias , hence no voltage appears across due to input voltage ,therefore the voltage acorss R appears as 5 V high due to` V_2` This SITUATION is obtained only is AND gate, henceoutput Q will be ACT as AND gate . |
|
| 39162. |
Atomic clock generally used in the national standards, is based on the periodic vibrationsproduced in a |
| Answer» Answer :A | |
| 39163. |
(a) Distinguish between unpolarised light and linearly polarised light. How does one get linearly polarised light with the hel of a polaroid ? (b) A narrow beam of unpolarised light of intensity I_(0) is incident on a polaroid P_(1). The light transmitted by it is then incident on a second polaroid P_(2) with its pass axis making angle of 60^(@) relative to the pass axis of P_(1). find the intensity of the light transmitted by P_(2). |
|
Answer» Solution :(b) When a narrow beam of unpolarised light of INTENSITY `I_(0)` is incident on a polarised `P_(1)`, the transmitted light is plane polarised light whose intensity is `I_(1)=(I_(0))/(2)`. When the plane polarised light is incident on a SECOND polaroid `P_(2)` with its PASS axis making an ANGLE of `theta=60^(@)` relative to the pass axis of `P_(1)`, then intensity `I_(2)` of emergent light from `P_(2)` is `I_(2)=I_(1)cos^(2)theta=((I_(0))/(2))xx(cos60^(@))^(2)=(I_(0))/(2)xx((1)/(2))^(2)=(I_(0))/(8)`. |
|
| 39164. |
How does young's modules change with rise of temperature? |
| Answer» SOLUTION :The YOUNG's MODULUS DECREASE with rise of temperature. | |
| 39165. |
Speed of light in glass is 2x10^8 m//s. Refractive index of glass is ________ |
| Answer» SOLUTION :`n=c/v=(3xx10^8)/(2xx10^8)=1.5` | |
| 39166. |
A plane wave of monochromatic light is incident normally on a uniform thin film of oil that covers a glass plate. The wavelength of the source can be varied continuously. Fully destructive interference of the reflected light is observed for wavelengths of 500 nm and 700 nm and for no wavelengths in between. If the index of refraction of the oil is 1.26 and that of the glass is 1.50, find the thickness of the oil film. |
| Answer» SOLUTION :694 nm | |
| 39167. |
The dimensional formula for angular acceleration will be |
|
Answer» `M^0L^0T^-1` |
|
| 39168. |
An electron of charges e experiences a force equal to its weight 'mg' when placed in an electric field E. Value of E is |
|
Answer» `MGE` |
|
| 39169. |
If scattering angle theta and impact parameter b are related to each other by the relation b=(Z e^(2) cot ""theta/2)/(4pi epsi_(0) (1/2 mv^(2))) then angle of scattering for zero impact parameter will be |
|
Answer» `0^(@)` If b=0, then `ze^(2), cot theta//2=0` so cot `theta//2=0, theta/2=90^(@) therefore theta=180^(@)` |
|
| 39170. |
Show that the electron revolving around the nucleus in a radius 'r with orbital speed 'r' has magnetic moment evr/2. Hence, using Bohr's postulate of the quantization of angular momentum, obtain the expression for the magnetic moment of hydrogen atom in its ground state. |
|
Answer» Solution :Magnetic momentum, `mu=-((e)/(2m))L` where, (-) INDICATES that direction of u is opposite to L. As per Bohr.s ATOMIC model `L=mvr` `:. Mu=-((e)/(2m))xx mvr` `mu=(evr)/(2)` Energy levels of hydrogen atom `E_(N)=(2pi^(2)mK^(2)e^(4))/(n^(2)h^(2))` For LOWEST energy level, n = 1 `E_(n)=(2pi^(2)mK^(2)e^(4))/(n^(2)h^(2))` `E_(2)=-(2pi^(2)mK^(2)e^(4))/(n^(2)h^(2))` `B_(2)=-(2pi^(2)mK^(2)e^(4))/(h^(2))` `E_(1)=-((4pi^(2)mKe^(2))/(h^(2)))(ke^(2))/(2)` While, `r=-(h^(2))/(4pi^(2)Kme^(2))` `E_(n)=-(Ke^(2))/(2r)=(-3.6)/(n^(2))eV` `:.mu=(evr)/(2)` `E_(1)=-(Kev.er)/(2vr^(2))=-13.6` `(Kmue)/(vr^(2))=-13.6` `mu=13.6((vr^(2))/(Ke))` |
|
| 39171. |
When the velocity of an electron increase, its de Broglie wavelength |
|
Answer» increases |
|
| 39172. |
An extended object is placed at the principal focus of a lens. Where will the final image be formed? Comment on the nature of the image. |
|
Answer» Solution :`(1)/(f)= (1)/(u)+(1)/(V)` `(1)/(f)= (1)/(f)+(1)/(v) ` `therefore (1)/(v)=0 rArr v=oo` The image is FORMED at INFINITY and the image will be real and inverted. |
|
| 39173. |
Figure shows two identical parallel plate capacitors connected to a battery through a switch S. Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant3. The ratio of the initial total energy stored in the capacitors to the final total energy stored is |
|
Answer» ` 3:5` |
|
| 39174. |
A car C of mass m is initially at rest on the boat A of mass M tied to the identical boat B of some, mass M through a massless inextensible string as shown in the figure. The car accelerates from rest to velocity v_(0) with respect to boat A in time t_(0)sec. At time t=t_(0) the car applies brake and comes to rest relative to boat in negligible time. Neglect friction between boat and water find the velocity of boat A just after applying brake. |
|
Answer» Solution :`A` the car `C` ACCELERATES to a velocity `v_(0)` relative to the double-boat SYSTEM, the two BOATS accelerate to the left. `V_(C )("to right")+v_(A)("to left")=v_(0)` `Mv_(C )=2Mv_(A)` Solving, we find `v_(A)=(mv_(0))/(m+2M)`, `v_(C )=(2Mv_(0))/(m+2M)` After the car brakes to a stop, the tension in the string connecting `A`, `B` becomes ZERO. Applying We find the velocity of `A`(to right) : `v'_(A)=(mMv_(0))/((m+M)(2+2M))` |
|
| 39175. |
Advantage of HF transmission is |
|
Answer» That the length of ANTENNA is small |
|
| 39176. |
In a common emitter transistor circuit, the base current is 40 muA , then V_(BE) is (##AAK_P7_NEET_PHY_SP7_C29_E03_025_Q01.png" width="80%"> |
| Answer» Answer :B | |
| 39177. |
Angle of contact is obtuse for |
|
Answer» GLASS and water |
|
| 39178. |
Which of the following plots correctly represents the variation of the magnitude of acceleration |a_R|with time t for a particle projected at t=0 with speed v_0 at an angle thetaabove the horizontal? |
|
Answer»
The value of `theta` LIES between : `theta=0`, at highest point and `LT 90^@` at the point of PROJECTION , and so `|a_R|_(max)=g cos 0^@ =g`. 
|
|
| 39179. |
A polarised examinedtwo adjacent plane polarised A and B whose planes of polarisation are mutuallyperpendicular In the firstpositionof the analyser,beamB shows zero intensity .Fromthispositiona rotationof 30^(0) showsthat the twobeams have same intnsitythe ratio of intensities of thetwo beams I_(A) andI_(B) will be |
|
Answer» `1:3` |
|
| 39180. |
The nucleus of an atom consists of: |
|
Answer» ELECTRONS and PROTONS |
|
| 39181. |
Calculate the de-Broglie wavelength associated with the electron revolving in the first excited state of hydrogen atom. The ground state energy of the hydrogen atom is - 13.6 eV. |
|
Answer» Solution :As PER question ground state energy of hydrogen atom `E_(1) =-13.6 eV` `:.`Energy in 1st excited state (i.E., for n = 2 state) will be `E_(2)= (E_(1))/(2)^(2)= -(13.6)/(4) = -3.4 eV` `:.` K.E. of ELECTRON on 1st excited state `K_(2) = -E_(2) =+3.4 eV` `:.` de Broglie WAVELENGTH in 1st excited state `lambda_(2)` = `(1.227)/(sqrt(K("ineV")))nm= (1.227)/(sqrt(3.4))=0.647`nm |
|
| 39182. |
Which of the following can not be polarised? |
|
Answer» ULTRASONIC WAVES |
|
| 39183. |
A: Remote sensing satellites can send continuous pictures of the earth even when clouds are present or it is dark night. R: These satellites use infrared as well as microwaves apart from visible light photography. |
|
Answer» |
|
| 39184. |
If angular frequency of an A.C. source used in L-C-R circuit is increased, then inductive reactance ….... and capacitive reactance..... |
|
Answer» DECREASE, increase |
|
| 39185. |
Show, by giving a simple example, how EM waves carry energy and momentum. |
| Answer» Solution :Consider a plane perpendicular to the direction of propagation of the WAVE. An electric CHARGE, on the plane, will be SET in motion by the electric and magnetic fields of EM wave, incident on this plane. This illustrates that EM waves carry energy and momentum. | |
| 39186. |
Three constant forcesvecF_(1)=2hati-3hatj+2hatk,vecF_(2)=hati+hatj-hatk and j - 2k in newton displace a particle from (1,-1,2) to (-1,-1, 3) to (2,2,0) (displacement being in metres). The total work done by the forces is, if displacement is along straight path: |
|
Answer» 2 J Net DISPLACEMENT =`(2-1)hati+(2+1)hatj+(0-2)hatk=hati+3hatj-2hatk` Now work DONE W =`(6hati-hatj-hatk).(hati+3hatj-2hatk)=(6-3+2)=5J` |
|
| 39187. |
Normal electric induction. |
| Answer» SOLUTION :Normal electric induction is DEFINED as the number of tubes of induction per UNIT AREA passing normally througha surface around a point . | |
| 39188. |
The fundamental frequency of a string stretched with a weight of 4kg is 256 Hz. The weight required to produce its octave is |
|
Answer» 4 KG wt |
|
| 39189. |
For the beta^(+) (positron) emission from a nucleus, there is another competing process known as electron capture (electron from an inner orbit, say, the K - shell, is captured by the nucleus and a neutrino is emitted). e^(+) ._(Z)^(A)X to ._(Z-1)^(A)Y+v Show that if beta^(+) emission is energetically allowed, electron capture is necessarily allowed but not vice-versa. |
|
Answer» Solution :The `beta^(+)` emission from a nucleus `._(Z)X^(A)` may be represented as `._(Z)X^(A)= ._(Z-1)Y^(A)+ ._(1)e^(0)+ V + Q_(1)` ______ (i) The other competing process of electron CAPTURE may be represented as `._(-1)e^(0)+ ._(Z)X^(A)= ._(Z-1)Y^(A)+ v + Q_(2)` _______ (ii) The energy released `Q_(1)` in 1. is given by `Q_(1)=[m_(N)(._(Z)X^(A))-m_(N)(._(Z-1)Y^(A))-m_(e)]c^(2)` `=[m_(N)(._(Z)X^(A))+Zm_(e)-m_(N)(._(Z-1)Y^(A))-(Z-1)m_(e)-2m_(e)]c^(2)` `Q_(1)=[m(._(Z)X^(A))-m(._(Z-1)Y^(A))-2m_(e)]c^(2)` _______ (iii) Note that `m_(N)` here denotes mass ofnucleus and m denotes the mass of atom similarly from (ii) `Q_(2)=[m_(N)(._(Z)X^(A))+m_(e)-m_(N)(._(Z-1)Y^(A))-m_(e)]c^(2)` `= [m_(N)(._(Z)X^(A))+Zm_(e)-m_(N)(._(Z-1)Y^(A))-(Z-1)m_(e)-m_(e)]c^(2)` `Q_(2)=[m(._(X)Z^(A))+m-(._(Z-1)Y^(A))]c^(2)` `= [m_(N)(._(Z)X^(A))+Zm_(e)+m_(e)-m_(N) (._(Z-1)Y^(A))-(Z-1)m_(e)-m_(e)]c^(2)` `Q_(2)=[m(._(Z)X^(A))-m-(._(Z-1)Y^(A))]c^(2)` Ir `Q_(1)gt 0` then `Q_(2) gt 0`. i.e., If positron emission is energetically allowed electron capture is necessarily allowed. But `Q_(2)gt 0` does not necessarily mean `Q_(1)gt 0`. Hence the REVERSE is not TRUE. |
|
| 39190. |
Derive the expression for emf induced in a straight conductor moving perpendicular to a uniform magnetic field. |
Answer» Solution :Let .PQ. represent a conductor moving with a speed `nu` at right angles to the magnetic flux density .B. pointing PERPENDICULAR to the sheet of paper and pointing towards the paper. Let RQ= x be the DISTANCE coverered by the conductor. Magnetic flux `Phi_B =Blx`.  INDUCED emf `=-(d(Phi_E))/(dt)=-d/(dt)(Blx)` i.e., `e=-Bl(dx)/(dt)` But,`(-dx)/(dt)=nu` `thereforee=Blnu""` ...........(1) Note : The Lorentz force acts on a change in a conductor and acts along PQ. done in moving a charge .q. from P to Q is W= (Lorentz force ) (displacement of the charge .q.) i.e.,W =`(Bqnu)L` by DEFINITION electric p.d`=W/q=(Bqnul)/q=Bnul`...............(2) Equation (1) and (2) are identical. |
|
| 39191. |
If both the mass and radius of the earth decreases by 1%, the value of |
|
Answer» acceleration due to gravity would decrease by nearly 1% The new value of g would be `g.=(G(0.99" M"))/((0.99 R)^(2))=1.01(GM)/(R^(2))` Thus, g would increase by about `1%`. The new escape velocity would be `v_(e )=sqrt((2xx0.99" M"xxG)/(0.99 R))=sqrt((2MG)/(R ))=v_(e )` Thus, the escape velocity will remain unchanged. The potential energy of a body of mass m on the earth.s surface would be `-(Gm(0.99" M"))/((0.99 R))=-(GmM)/(R )`. Thus, the potential energy will also remain unchanged. Hence, the CORRECT CHOICE are (b) and (d). |
|
| 39192. |
The potential difference across 7Omega resistor is equal to_________and the current flowing through the battery is equal to_________(b) The equivalent resistance across A and Bis equal to__________ |
|
Answer» SOLUTION :`14V,8V (B)R//4)` |
|
| 39193. |
Four charges +q,+2q,+3q and +4q are placed at the four corners ABCD of a square of side a. Calculate the field and potential at the centre of the square. |
|
Answer» |
|
| 39194. |
Gas at a pressure P_(0) is contained in a vessel. If the masses of all the molecules are halved and their speeds are doubled, the resulting pressure P will be equal to |
|
Answer» `4P_(0)` `P_(0)=(1)/(3)(mNv_(rms)^(2))/(V)` when `m=(m)/(2),v_(rms)=2v_(rms)" then"` `P.=(1)/(3)((m)/(2))xxNxx(2v_(rms))^(2)=2P_(0)` |
|
| 39195. |
M.I. of the solid sphere about its diameter is 25 kgm^2.Its M.I. about the tangent is (M.I. of sphere about diameter=2//5(MR^2)? |
|
Answer» 87.5 `kgm^2` |
|
| 39196. |
What is the nature of motion of charge entered inclined to a magnetic field? |
| Answer» SOLUTION :HELICAL in PATH | |
| 39197. |
A 2.00 kg particle moves along an x axis in onedimensional motion while a conservative force along that axis acts on it. The potential energy U(x) associated with the force is plotted in Fig. 8-36a. That is, if the particle were placed at any position between x=0 and x=7.00m, it would have the plotted value of U. At x=6.5 m, the particle has velocity vec(v)_(0) = (-4.00 m//s)hat(i). (b) Where is the particle's turning point located? |
|
Answer» Solution :KEY IDEA The turning point is where the force momentarity STOPS and then reverses the particle.s motion. That is , it iswhere the particle momentarily has `v=0` and thus `K = 0`. CALCULATIONS: Because K is the difference between `E_("mec")` and U, we want the point in Fig. 8-36a where the plot of U rises to MEET the horizontal line of `E_("mec")`, as shown in Fig. 8-36b, we can draw nested right triangles as shown and then write the proportionality of distances `(16-7.0)/(d) = (20-7.0)/(4.0 -1.0)` Which gives us `d=2.08m`. Thus, the turning point is at `x=4.0 m -d=1.9m`. |
|
| 39198. |
In an A.C circuit the current... |
|
Answer» is in the PHASE with voltage |
|
| 39199. |
In Young's double slit experiemnt the wavelength of light was changed from 7000 Å to 3500 Å.While doubling the separation between slits, which of the following is not rue? |
|
Answer» width of the fringe changes `therefore beta = (lambda)/(2). (D)/(2d) = (beta)/(4)` The separation between dark and bright fringes decreases. |
|
| 39200. |
A 2.00 kg particle moves along an x axis in onedimensional motion while a conservative force along that axis acts on it. The potential energy U(x) associated with the force is plotted in Fig. 8-36a. That is, if the particle were placed at any position between x=0 and x=7.00m, it would have the plotted value of U. At x=6.5 m, the particle has velocity vec(v)_(0) = (-4.00 m//s)hat(i). (a) Form Fig. 8-36a, determine the particle's speed at x_(1)=4.5 m. |
|
Answer» Solution :KEY IDEA (1) The particle.s kinetic energy is given by Eq. 8-1 `(K=1//2mv^(2))`. (2) Because only a CONSERVATIVE force acts on the particle, the mechanical energy `E_("mec")(=K+U)` is conserved as the particle moves. (3) Therefore, on a plot of U(x) such as Fig. 8-36a, the kinetic energy is equal to the difference between `E_("mec")` and U. Calculations: At `x=6.5m`, the particle has kinetic energy `K_(0)=1/2 mv_(0)^(2)=1/2(2.00kg)(4.00 m//s)^(2)` `=16.0 J`.  Figure 8-36 (a) A plot of potential energy U versus position x. (b) A section of the plot used to find where the particle turns around. Because the potential energy there is `U=0`, the mechanical energy is `E_("mec")=K_(0)+U_(0)+16.0 J + 0 = 16.0 J`. This VALUE for `E_("mec")` is plotted as a HORIZONTAL line in Fig. 8-36 a. From that figure we SEE that at `x=4.5m`, the potential energy is `U_(1)=7.0J`. The kinetic energy `K_(1)` is the difference between `E_("mec")` and `U_(1)`: `K_(1) = E_("mec")-U_(1)=16.0J-7.0J = 9.0 J`. Because `K_(1) = 1//2 mv_(1)^(2)`, we `v_(1)=3.0 m//s.` |
|
