NSEP EXAMINATION

Transcription

1 NSEP EXAMINATION INDIAN ASSOCIATION OF PHYSICS TEACHERS NATIONAL STANDARD EXAMINATION IN PHYSICS Tota tie : 0 inutes (A-, A- & ) PART - A (Tota Marks : 80) SU-PART A- Q. The Schrodinger equation for a free eectron of ass and energy E written in ters of the d ψ 8π E wavefunction ψ is + = 0. The dx h diensions of the coefficient of ψ in the second ter ust be - [M L ] () [L ] [L ] [M L T ] [C] y diensiona anaysis the diensions of each ter in an equation ust be the sae. In the first ter the second derivative with respect to distance x indicates the diensions of the coefficient of φ to be [L ] and hence the answer. Q. In an experient to verify poiseuie's aw πλpa 4 η = where the sybos have their 8V usua eanings, the axiu error that enters in cacuating the coefficient of η is due to easureent of - pressure difference p () ength of the tube voue rate of fow V radius of the tube a. [D] Since a is raised to the power 4, it has the argest contribution to the tota error. Q. A unifor rope of ength L and ass M party ies on a horizonta tabe and party hangs fro the edge of the tabe. If µ is the coefficient of friction between the rope and the surface of the tabe (negecting the friction at the edge), the axiu fraction of the ength of the rope that overhangs fro the edge of the tabe without siding down is : µ µ () µ µ + µ µ + If is the ength of the hanging part of the rope, M its weight is g and hence the weight of L M the rope on the tabe is (L )g which is L aso equa to the nora reaction N. Then, the M force of friction is µn = µ (L ) g. L Equating this to the weight of the hanging part of the rope, we get the answer. Q.4 A thin unifor rod XY of ength is hinged at one end X to the foor and stands vertica. When aowed to fa, the anguar speed with which the rod strikes the foor is - g 4 g () g The oent of inertia of the rod about a horizonta axis passing through point X is I x = () + = 4 where is the ass of the rod. Now, since centre of gravity fas through a height, the change in gravitationa potentia energy (g) of the rod can be equated to the rotationa energy I x ϖ where ϖ is the anguar speed. Q.5 A 60 c eta rod M is joined to another 00 c eta rod M to for an L shaped singe piece. This piece is hung on a peg at the joint. The two rods are observed to be equay incined to the vertica. If the two rods are equay thick, the ratio of density of M to that of M is 5/ () /5 5/9 9/5 g, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: /

2 NSEP EXAMINATION [C] The ange ade by each rod with the vertica is 45. Appying the aw of oents, we get the ratio of weights and hence that of asses to be 5. Then, using ass = voue density, we get the ratio of densities as 5 : 9. Q.6 The string of a sipe penduu is repaced by a thin unifor rod of ass M and ength. The ass of the bob is. If it is aowed to osciate with sa apitude, the period of osciation is (M + )L π () π (M + )g (M + )L π π (M + )g (M + )L (M + )g (M + )L (M + )g [A] The restoring torque can be written as L Mg sin + glsin which reduces to L (M + )g if is sa. This is equated to Iα where the oent of inertia I = ML + L = L (M + ). After (M + )g substitution, we get α =. (M + )L Coparing with a = (ϖ ), expression for anguar frequency and hence period can be obtained. Q.7 Two stars each of ass and radius R approach each other to coide head-on. Initiay the stars are at a distance r (>> R). Assuing their speeds to be negigibe at this distance of separation, the speed with which the stars coide is G () G R r R r G + G + R r R r The change in potentia energy can be equated to the gain of tota kinetic energy : G r G = v R Q.8 A nonviscous iquid of density ρ is fied in a tube with A as the area of cross section, as shown in the figure. If the iquid is sighty depressed in one of the ars, the iquid coun osciates with a frequency () π π ρgasin + ρga(sin sin ) π ρga(sin sin ) ρgasin π [C] The force that is responsibe for restoring the iquid eves in the two ars of the tube is pa = (h + h ) ρga where p is the pressure difference and A is the area of cross section of the tube, h and h being the rise and fa of iquid eves in the two ars in vertica direction respectivey. Note that the change in ength of the iquid thread aong the tube wi be the sae, say x. Using this the force can further be written as { (x sin + x sin )ρga}. Writing force as ass ties acceeration, we get expression for period and then for frequency. Q.9 A wax cande foats verticay in a iquid of density twice that of wax. The cande burns at the rate of 4 c/hr. Then, with respect to the surface of the iquid the upper end of the cande wi fa at the rate of 4 c/hr () fa at the rate of c/hr rise at the rate of c/hr reain at the sae height, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: /

3 NSEP EXAMINATION Let the decrease in the height of the cande outside the iquid be x. Now, if d denotes the diaeter and L the ength of the cande respectivey, after one hour appying theaw of foatation gives d d L π ( L 4) ρwax = π xρiq. Note that ρ wax = ρiq akes ony haf of the cande of stand outside the iquid. Soving the equation one gets the answer. Q.0 A ass hangs at the end of a assess spring and osciates up and down at its natura frequency f. If the spring is cut at the idpoint and the ass reattached at the end, the frequency of osciation is : f () f f/ f/ [A] If k is the spring constant for the origina spring and k that for the haf-cut spring, we have k = k. Use the expression for frequency f = π k to get the answer. Q. The density of a soid at nora pressure is ρ. When the soid is subjected to an excess pressure ρ. The density changed to ρ. If the ρ buk oduus of the soid is k, then the ratio ρ is p + k p p + k [A] () + p k k p + k We use expression for density ρ = V so as to get d ρ = ρ dv V. Substituting this in the expression for p buk oduus, we get k =. Since dv V increase of pressure increases the density, ρ ρ p dρ = ρ ρ. Using this we get = and ρ k hence the answer, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: Q. A eta ba (density. g / cc) is dropped in water, whie another eta ba (density 6.0 g / cc) is dropped in a iquid of density.6 g / cc. If both the bas have the sae diaeter and attain the sae terina veocity, the ratio of viscosity of water to that of the iquid is :.0 () indeterinate due to insufficient data Use the expression for terina veocity of a body r ( ρ σ) g faing in a viscous iquid v = 9 η where sybos have their usua eanings. Q. A thin copper rod rotates about an axis passing through its end and perpendicuar to its ength with an anguar speed ω 0. The teperature of the copper rod is increased by 00 C. If the coefficient of inear expansion of copper is 0 5 / C, the percentage change in the anguar speed of the rod is % () 4% 0. % 0.4 % [D] The oent of inertia of the rod is. Let I 0, ϖ 0 denote the initia oent of inertia and initia anguar veocity respectivey and I, ϖ be the corresponding quantities after heating the rod. Appying conservation of anguar oentu, we I 0 ϖ0 I ϖ get 0 0 ϖ = = I ( + α) 0 = ϖ 0 ( α) = ϖ 0. Thus the change is 0.4% and is negative. Q.4 The interna energy of one gra of heiu at 00 K and one atospheric pressure is 00 J () 00 J 00 J 500 J [C] The heiu oecue is onatoic and hence interna energy per oecue is kt. The energy per oe is therefore is RT. One gra of heiu is one fourth oe and hence its energy is R 00 = 00 J, taking the 4 vaue of R to be approxiatey 8 J/oe-K. /

4 NSEP EXAMINATION Q.5 Voue of a onatoic gas varies with its teperature as shown. The ratio of work done by the gas to the heat absorbed by it when it undergoes a process fro A to is V O A / () /5 /7 / Since the curve passes through the origin, V T, the pressure being constant. Then, heat absorbed at constant pressure is dq = n C p dt whereas the change in interna energy is du = n C v dt where the sybos have their usua eanings. The work done dw = dq du = n (C p C v )dt. This gives the ratio of work done to the heat absorbed as and hence the resut. γ Q.6 A sa gass bead of ass initiay at rest starts fro a point at height h above the horizonta and ros down the incined pane A as shown. Then it rises aong the incined pane C. Assuing no oss of energy, the tie period of osciation of the gass bead is C A () h 8h g 4h 5g 8h g 8h 5g (sin + sin ) sin sin sin + sin + sin + sin T The speed with which a sphere roing down an incined pane reaches the botto is / / gh 0 v = = gh I + 7 r where sybos have their usua eanings. The 5 acceeration is gsin. With this, the tie 7 taken by the gass bead to ro down turns out to be 4h 5g sin other incined pane is. Siiary the tie to go up the 4h 5g sin. Twice the su of these two ties is the required tie of osciation. Q.7 Water at 0 C (coefficient of viscosity = 0.0 poise) fowing in a tube of diaeter c with an average veocity of 0 c/s has the Reynod nuber 500 () indeterinate due to insufficient data ρvd Reynod nuber is given by N = where ρ η is the density, v is the veocity and D is the diaeter of the tube. Use this to get the answer. Q.8 A biiard ba is hit by a cue at a point distance h above the centre. It acquires a inear veocity v 0. Let be the ass and r be the radius of the ba. The anguar veocity acquired by the ba is 5v0h v0h v0h v0h () r r 5r r [A] Equating the anguar oentu (v 0 h) about the center of ass to ω one gets the answer. Q.9 Two pipes are each 50 c in ength. One of the is cosed at one end whie the other is open at both ends. The speed of sound in air is 40 /s. The frequency at which both the pipes can resonate is 680 Hz () 50 Hz 85 Hz None of these, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: /

5 NSEP EXAMINATION [D] A pipe open ony at one end and another one of equa ength but open at both the ends have their fundaenta frequencies n and n respectivey. Again ony odd haronics are possibe for a pipe open at one end whereas a haronics are possibe for a pipe open at both ends. Thus, it is ipossibe to have a coon frequency at which they can resonate. Q.0 The work invoved in breaking a bigger size spherica oi drop into n saer size identica dropets is proportiona to n () n n n [D] Note that the work invoved in the process of breaking a bigger drop into saer drops is the change in surface area ties the surface tension. If r is the radius of saer drop and R that of the 4 4 bigger one, then πr = n πr where n is the nuber of saer drops. This gives R r =. If T is the surface tension, then the work n done wi be W = (n4πr 4πR ) T. Substituting for r gives the expected proportionaity. Q. A train oving towards a hi at a speed of 7 k/hr sounds a whiste of frequency 500 Hz. A wind is bowing fro the hi at a speed of 6 k/hr. If the speed of sound in air is 40 /s, the frequency heard by a an on the hi is 5.5 Hz () Hz 56.5 Hz None of the above [A] If n represents the apparent frequency and n the actua one, then use the reation (v ± w) n n (v ± w) v s where v is the veocity of sound, w is the veocity of wind and v s that of the source. Note that in this case the observer at rest. Q. A convex ens fors a rea iage with agnification on a screen. Now, the screen is oved by a distance x and the object is aso oved so as to obtain a rea iage with agnification on the screen. The, the foca ength of the ens is x () x x x ( ) ( ) [D] Use the usua ens forua. In the first case if v is the iage distance, (v + x) is the iage distance after the oveent. The agnifications and v in the two cases turn out to be and f v + x respectivey. This can be sipified f to get the expression for f. Q. Two eta wires of identica diensions are connected in series. If σ and σ are the conductivities of the etas respectivey, the effective conductivity of the cobination is σ + σ σ + σ () σ σ σσ σ + σ [D] Use the expression for resistance in ters of conductivity R = and note that the tota σa resistance is (R + R ) and tota ength of the wires is. Q.4 An aternating suppy of 0 vot is appied across a circuit with resistance oh and ipedance of 44 oh. The power dissipated in the circuit is 00 watt. () 550 watt. 00 watt. (00/) watt The current in the circuit is Z V = 5 A and hence the power consued is (I R) equa to 550 watt., CP Tower, Road No., IPIA, Kota (Raj.) Ph.: /

6 NSEP EXAMINATION Q.5 If T denotes the teperature of the gas, the voue thera coefficient of expansion of an idea gas at constant pressure is T () T T T The voue coefficient of expansion is given by dv. For an idea gas at constant pressure, we vdt write pdv = nrdt and substitute for p fro the usua reation pv = nrt to get the required expression for the coefficient. Q.6 A coi having N turns is wound in the for of a spira with inner radius a and outer radius b respectivey. When a current I passes through the coi, the agnetic fied at the centre is : µ NI 0 µ () 0 NI (a + b) ab µ NI b µ n 0 0 NI b n (b a) a (b a) a [D] Consider dn to be the nuber of turns in between N radii r and (r + dr) so that we get dn = dr. (b a) The agnetic induction d due to these any µ I(dN) turns at the centre is 0. After substituting r for dn and integrating between a and b, we get the resut. Q.7 A circuit is arranged as shown. Then, the current fro A to is : A + 0 V 0 Ω 0 Ω 5 Ω + 5 V A () + 50 A 50 A 500 A Use superposition theore. We get the potentia difference between A and to be +.75 vot when source of 5 vot is shorted, whereas.5 vot when source of 0 vot is shorted. Therefore, when both the sources are working the net potentia difference is +.5 vot so that current is 50 A fro A to. Q.8 Two identica thin rings, each of radius a are paced coaxiay at a distance a apart. Let charges Q and Q be paced unifory on the two rings. The work done in oving a charge q fro the centre of one ring to that of the other is zero q () (Q Q ) 4πε 0 a q( ) (Q Q ) 4πε 0 a q( ) (Q Q ) 4πε 0 a [C] The eectrostatic potentia at the centre of the first ring with charge Q is due to charge Q itsef as we as due to charge Q on the other ring. This Q Q turns out to be +. Siiary 4πε0 a 4πε0 a the eectrostatic potentia at the centre of the centre of the other ring is Q Q +. The difference between 4πε0 a 4πε0 a these potentias tie the charge q is the required word done. Q.9 An equiatera trianguar oop of wire of side carries a current i. The agnetic fied produced at the circucentre of the oop is µ 0 4π µ 0 8i 4π i () µ 0 9i 4π µ 0 6i 4π [C] If is the side of the triange, the distance of the circucentre fro each of the side of the triange µ i carrying a current i is 0 (sin 60 + sin 60 ) = 4π r µ 0 6i. Since the direction of agnetic fied in 4π each case in the sae, three ties this woud be the tota agnetic induction., CP Tower, Road No., IPIA, Kota (Raj.) Ph.: /

7 NSEP EXAMINATION Q.0 Consider a doube sit interference experient. Let E 0 be the apitude of the eectric fied of the waves starting fro the sits. If φ is the phase difference between the two waves reaching the screen, the apitude of resutant eectric fied at a point on the screen is E 0 cos φ () E 0 cos (φ/) E 0 cos (φ/) E 0 cos φ [C] Consider the agnitudes of the eectric fieds reaching the screen to be E 0 sin ϖt and E 0 sin (ϖt + φ). Then, the resutant eectric fied at the screen woud be su of the two, that is, φ φ E 0 cos sin ϖt +. Note that the apitude of the resutant eectric fied is the coefficient of the sine function. Q. In a doube sit experient, the coherent sources are spaced d apart and the screen is paced a distance D fro the sits. If n th bright fringe is fored on the screen exacty opposite to a sit, the vaue of n ust be d d d d () λd λ D λd 4λD λd The fringe width in this case is and the n th d bright fringe is fored at a distance d away fro D the centre. Therefore, d = n λ giving the vaue d of n. Q. When two sound sources of the sae apitude but of sighty different frequencies n and n are sounded siutaneousy, the sound one hears has a frequency equa to n + n n n () n n [n + n ] The resuting sound wave has a frequency equa to haf the su of the individua frequencies. Note that the resuting intensity varies at the beat frequency equa to difference of the individua frequencies., CP Tower, Road No., IPIA, Kota (Raj.) Ph.: Q. Four identica irrors are ade to stand verticay to for a square arrangeents as shown in a top view. A ray starts fro the idpoint M of irror AD and after two refections reaches corner D. Then ange ust be C A M D tan (0.75) () cot (0.75) sin (0.75) cos (0.75) Note that the ray starting fro point M at an ange reaches the corner D at the right aong a parae path. Refer to the figure. Let a be the ength of the side, so that tan x a x a = = =. Soving these equations (a / ) y a y a one gets x = and hence cot =. 4 a x x A y a y C M Q.4 The refecting surfaces of two irrors M and M are at an ange (ange between 0 and 90 ) as shown in the figure. A ray of ight is incident on M. The eerging ray intersects the incident ray at an ange φ. Then, M φ = () φ = 80 φ = 90 φ = 80 [D] If x is the ange of incidence when the ray strikes the irror M and y be that for irror M, then, using sipe properties of triange one gets φ = 80 (x + y) and (Students are expected to draw the ray diagra and check.) φ D 7 /

8 NSEP EXAMINATION Q.5 An unpoarized ight bea is incident on a surface at an ange of incidence equa to rewster's ange. Then, the refected and the refracted beas are both partiay poarized () the refected bea is partiay poarized and the refracted bea is copetey poarized and are at right anges to each other the refected bea is copetey poarized and the refracted bea is partiay poarized and are at right anges to each other both the refected and the refracted beas are copetey poarized and are at right anges to each other [C] Refer to the artice on poarization by refection when the ray is incident at rewster's ange fro any standard book. Q.6 Switch S is cosed at t =0. After sufficienty ong tie an iron rod is inserted into the inductor L. Then, the ight bub L R [D] When the switch is cosed, 5 oh resistance gets shorted. Thus a current of A fows fro b to a. Q.8 Two radioactive aterias A and have decay constants 5λ and λ respectivey. Initiay both A and have the sae nuber of nucei. The ratio of the nuber of nucei of A to that of wi be e after a tie 5λ 5 4λ () 4λ 4 5λ Using the aw of radioactive decay, one can write NA (t) N0 exp( 5λt) = =. Soving this one N(t) N0 exp( λt) e gets the resut. Q.9 The radius of the hydrogen ato in its ground state is a 0. The radius of a 'uonic hydrogen' ato in which the eectron is repaced by an identicay charged uon with ass 07 ties that of an eectron. is a µ equa to S gows ore brighty () gets dier gows with the sae brightness gets oentariy dier and then gows ore brighty As the rod is inserted, inductance increases and hence the votage across inductor increases. This caused a drop in the votage across the bub and hence it gets dier. Q.7 In the circuit shown beow, the current that fows fro a to b when the switch S is cosed is : a 0 Ω 5 Ω b.5 A +.0 A 0 V + () +.5 A.0 A 0 Ω, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: a 0 07 a 0 () 07 a 0 07 a 0 07 Use the expression for the first ohr radius for hydrogen ato This expression indicates that the radius is inversey proportiona to the ass and hence the resut. Q.40 The instantaneous agnitudes of the eectric fied (E) and the agnetic fied () vectors in an eectroagnetic wave propagating in vacuu are reated as E = () E = c C E = E = c c At every instant the ratio of the agnitude of the eectric fied to that of the agnetic fied in an eectroagnetic wave equas the speed of ight. 8 /

9 NSEP EXAMINATION SU-PART A- Q.4 A onkey hods a ight rope that passes over a sooth puey. A bunch of bananas of equa ass as that of the onkey is attached to the other end of the rope. The onkey starts cibing the rope to get to the bananas. Then, the bananas aso ove up () the bananas ove downwards the distance between the onkey and the bananas decreases the distance between the onkey and the bananas reains constant. [A, D] Note that the asses of the onkey and the bunch of bananas are equa and the puey is sooth. Q.4 Consider the curve representing the Maxwe- otzann speed distribution of gas oecues at soe teperature. Let v rs, v avg and v p be the rs, the average and the ost probabe speeds respectivey. Then, the curve has a axiu at v p () the area under the curve gives the tota nuber of oecues of the gaseous syste. v rs > v avg > v p v vg < v p < v rs [A,, C] kt kt Note that v rs =.7, v avg =.60, kt v p =.4 where the sybos have their usua eanings. The ost probabe speed v p is the speed at which the curve reaches the peak. The area under the curve is obviousy the tota nuber of oecues. Q.4 A partice of ass oves aong a straight ine under the action of a force f varying with t T tie as f = f 0 where f 0 and T are T positive constants. Then the speed of the partice after a tie T is 4f 0 T () after tie interva of T, the partice starts oving backwards between tie instants 0 and T, the acceeration first increases and then decreases. the partice stops at t = T [A,,CD] Use the given expression for force to get an expression for acceeration. Integrate this to get an expression for veocity. Uness otherwise stated about the initia conditions, the veocity f0 turns out to be v = [ t T t ]. Use this to T get the required resuts. Q.44 A sound wave of anguar frequency ϖ traves with a speed v in a ediu of density ρ and buk oduus. Let k be the propagation constant. If p and A are the pressure apitude and dispaceent apitude respectivey, then the intensity of sound wave is ϖ ka () vp p p ρv ρ [A,, C, D] Intensity, by definition, is the energy fowing per unit area per unit tie. The dispaceent P ϖ apitude is given by A = where k = k v is the propagation constant. The speed v =. ρ Use these reations to get the required expressions. Q.45 Two conducting pates A and are paced parae to each other at a sa distance between the. Pate A is given a charge q and pate is given a charge q. Then, the outer surfaces of A and (not facing each other) get no charge () the inner surfaces of A and (facing each other) get a the charge the inner surfaces of A and (facing each other) get equa and opposite charge of q q agnitude the outer surfaces of A and (not facing each other) get charge of the sae poarity q + q and of agnitude, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: /

10 NSEP EXAMINATION [C, D] Let a charge q be present on the inner surface of pate A so that on its outer surface the charge is (q q ). Obviousy a charge q wi get induced on the inner surface of the pate and a charge (q + q ) wi ove to its outer surface. With these charges, write the net eectric fied at a point inside the pate and equate it to zero. This reation can be sipified to get the vaue of q and hence the concusions. Q.46 A an with nora vision uses a agnifying ens foca ength 0 c. Then, agnification of any vaue is possibe () axiu agnification possibe is.5 iniu agnification possibe is.5 agnification depends upon the distance of the ens fro the eye [, C, D] In case of a icroscope the agnification is D + when the iage is fored at the f distance of distinct vision D. However if the iage is fored at infinity, the agnification is sipy f D. Q.47 An eectroagnetic wave is traveing through a ediu of refractive index n and is incident at the boundary of a ediu of refractive index n. If the wave refects at the boundary, the wave undergoes a phase change of 80, if n < n () the wave undergoes a phase change of 80, if n > n the wave undergoes no phase change, if n < n the wave undergoes no phase change, if n > n [A, D] Note that an eectroagnetic wave undergoes a phase change of 80 upon refection fro a ediu that has a higher index of refraction than the one in which it is traveing. However, there is no phase reversa if the case is opposite. Q.48 In cycotron (partice acceerator) an ion is ade to trave successivey aong seicirces of increasing radii under the action of a agnetic fied. The anguar veocity of the ion is independent of speed of the ion () radius of the circe ass of the ion charge of the ion [A, ] Use the expression qv = rϖ where v = rϖ. The sybos carry their usua eanings. This indicated that anguar veocity is independent of the radius of the circuar path and the speed of the ion. Q.49 Physica quantities A and have the sae diensions. Then. A ± ust be a eaningfu physica quantity. () A ± ay not be a eaningfu physica quantity A/ ust be a diensioness quantity both ust be either scaar or vector quantities. [, C] The ratio of two quantities having the sae diensions ust necessariy be a diensioness quantity. However, two quantities having the sae diensions ay not add to necessariy give a eaningfu quantity; for exape, work and torque have the sae diensions but their addition is eaningess. Q.50 A step votage V 0 is appied to a series cobination of R and C as shown. Then, V 0 V R 0 V R KΩ C µf t t = 0 after sufficienty ong tie V R = 0 () as tie passes V R decrease as (/t) V C after s, V C = 6. vot (approxiatey) initiay current through R is 0 A, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: /

11 NSEP EXAMINATION [A, C, D] After sufficienty ong tie since the charging current drops to zero, the drop across the resistance is zero. The tie constant is kω µ F = s. hence according to the definition of tie constant, the votage across the capacitor woud be about 6% of the axiu, that is, 6. vot after s. Initia current is obviousy (0 vot/kω) = 0 A. PART Marks : 60 * A questions are copusory. * A questions carry equa arks Q.5 A cube of side 0 c is rigidy joined to a thin rod of ength 40 c. The rod is pivoted at the other end so that the rod aong with the cube is abe to rotate freey about the pivot in a vertica pane. A buet of ass 50 g, oving horizontay hits a point of the cube 5 c fro the ower end and gets ebedded into it. Deterine the speed of the buet so that the syste just rises to a horizonta position. (ass of the rod = 00 g, ass of the cube = 750 g) Note that the coision between the buet and the cube is ineastic. Hence the kinetic energy is not conserved but the oentu is conserved. Equating the anguar oentu of the buet to that of the rod together with the cube about the pivot gives a reation anguar frequency ϖ = v, where v is the speed of the buet. Note 7.49 that, in genera, the oent of inertia of the rod of ength and ass about the pivot is and that of the cube of ass about an axis through its centre of ass is 6 a where a sit he side of the cube. The tota oent of inertia of the buet, the rod and the cube about the pivot coes out to be 0.68 kg-. After the coision the syste rises through a height ( ). The gain in gravitationa potentia energy can then be equated to the oss of rotationa kinetic energy. This gives ϖ = s. Fro this the speed of the buet can be cacuated to be /s. Q.5 Consider a unifor square pate of side ade of wood. A seicircuar portion is cut and attached to the right as shown. Deterine the centre of ass of the redesigned pate. L For the seicircuar pate of radius, the center 4 of ass ies at a distance of fro the centre. π Taking σ to be the ass per unit area, the position of centre of ass of the reaining piece of the square woud be at a distance of (π 4) fro the centre of the origina square (8 π) pate. Now, taking the centre of the origina square to the origin, the centre of ass of the new structure can be deterined. This turns out to be at a distance of to the right of the origin. Q.5 Consider two ong parae and oppositey charged thick wires of radius d with their centra axes separated by a distance D apart. Obtain an expression for the capacitance per unit ength of this pair of wires. Note that when the two wires for a capacitor, the charges reside ony on the inner side; positive on one and negative on the other. Let us consider a point P distance r fro the axis of one wire. Using Gauss fux theore, the eectric fied E at P due to the positive charge (of surface charge density σ) on this wire of unit ength is E (πr) = σ( πd) σd E =. The point P is at a ε0 ε 0 (r) distance (D r) fro the axis of the other wire carrying negative charge. Again using Gauss theore, the eectric fied E can be written as E σd =. Obviousy the two fieds are in ε 0 (D r) the sae direction so that net eectric fied is E = σd + ε. Integrate this between the 0 r D r iits d and (D d) to get the potentia σ D d difference d n. Then, the ε0 d capacitance per unit ength turns out to be πε0. D d n d, CP Tower, Road No., IPIA, Kota (Raj.) Ph.: /

12 NSEP EXAMINATION Q.54 Ferat's principe states that 'when ight ray traves between two points, the path is the one that required the east tie'. Use this principe to derive aw of refection regarding ange of incidence and ange of refection. You ay refer to the foowing figure. h d Consider the tota distance d to be ade up of x to the eft of the point of incidence and (d x) to its right. If n is the refractive index of the ediu and c the speed of ight in vacuu, then the speed c in the ediu under consideration is. The n tota tie of trave can be written as x + h (d x) + h t = +. According to c c n n Ferat principe, for the east tie, cacuate dt and equate it to zero. Using sipe dx x geoetry, we write = sin and x + h (d x) = sin, we get the aw of (d x) + h refection that the ange of incidence is equa to the ange of refection. Q.55 Two sape X and Y of a gas have equa voues and pressure. The gas in X is aowed to expand isotheray to.5 ties its initia voue, whie that in Y is aowed to expand adiabaticay to an equa voue. if the work done in the first expansion is.5 ties that in the second, show that the ratio of specific heats γ satisfies a reation γ ( γ )n = Let n and n be the nuber of oes of sapes X and Y of the gas, the initia teperatures of T and T respectivey and p 0 and V 0 be their initia pressure and voue. Then, p 0 V 0 = n RT = n RT. Now, the work done in isothera expansion is W = n RT n whereas that in n RT n RT adiabatic expansion is W =. γ Note that during adiabatic expansion the teperature fas to T. Aso in case of adiabatic process, TV γ = constant. Appying this, we get γ T = T. Using the fact that W = W, we get the required expression., CP Tower, Road No., IPIA, Kota (Raj.) Ph.: /