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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.
| 351. |
In the `V-T` graph shown in figure: `{:(,"Column-I",,"Column-II"),((A),"Gas" A is ... and gas B is... ", E is ",(p),"monoatomic , diatomic"),((B),"P_(A)/P_(B) is ",(q),"diatomic , monoatomic"),((C),"n_(A)/n_(B) is " " ",(r),"gt1"),(,,(s),"lt1"),(,,(t),"cannot say any thing"):}` |
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Answer» From given `V-T` graph we cannot tell the nature of gas slope of `V-T "graph" = (nR)/(P)` From graph `((nR)/(P))_(A) gt ((nR)/(P))_(B) implies ((n)/(P))_(A) gt ((n)/(P))_(B)` `implies` Cannot say anything about `(n_(A))/(n_(B))` & `(P_(A))/(P_(B))` |
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| 352. |
A copper rod (initially at room temperature `20^(@)C`) of non-uniform cross section is placed between a steam chamber at `100^(@)C` and ice-water chamber at `0^(@)C`. `A` and `B` are cross sections are as shown in figure. Then match the statements in column -I with results in columns-II using comparing only between cross section `A` and `B`. (The mathematical expressions in column-I have usual meaning in heat transfer). `{:(,"Column-I",,"Column-II"),((A),"initially rate of heat flow" ((dQ)/(dt))"will be ",(p),"Maximum at section" A),((B),"At steady state rate of heat flow" ((dQ)/(dt)) "will be ",(q),"Maximum at section" B),((C),"At steady state temperature gradient" |((dT)/(dx))| "will be " ,(r),"Minimum at section B"), ((D), "At steady state rate of change of temperature"((dT)/(dt)) " at a certain point will be",(s),"Same for all section"):}` |
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Answer» Initially rate of heat flow will be maximum at `A` and minimum at `B` as there is no temperature difference across section `B`. In steady state `(dQ)/(dt)` will be same. In steady state `(dQ)/(dt) = -KA(dT)/(dx)`= same & `(dT)/(dt) = 0 implies ((dT)/(dx)) prop (1)/(A) implies ((dT)/(dx))` will be maximum at `B` & minimum at `A` |
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| 353. |
For one mole of a monoatomic gas :- `{:(,"Column-I",,"Column-II"),((A),"Isothermal bulk modulus, ",(p),-(RT)/(V^(2))),((B),"Adiabatic bulk modulus ",(q),-(5P)/(3V)),((C),"Slope of P-V graph in isothermal process " ,(r),"T/V"), ((D), "Slope of P-V graph in adiabatic process",(s),"4T/3V"),(,,(t),"None"):}` |
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Answer» Isothermal bulk modulus = `P = (RT)/(V)` Adiabatic bulk modulus =`gammaP = (5RT)/(3V)` Slope of `PV` graph in isothermal process =-`(P)/(V) = -(RT)/(V^(2))` Slope of `P-V` graph in adiabatic process = `-(gammaP)/(V) = -(5P)/(3V)` |
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| 354. |
A short linear object of length b lies along the axis of a concave mirror of focal length f at a distanee u from the pole of the mirror. The size of the image is approximately equal toA. `b((u-f)/(f))^(1//2)`B. `b((b)/(u-f))^(1//2)`C. `b((u-f)/(f))`D. `b((f)/(u-f))^(2)` |
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Answer» Correct Answer - D is the correct option. |
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| 355. |
For a prism of refractive index `1.732`, the angle of minimum deviation is equal to the angle of the prism. The angle of the prism isA. `80^(@)`B. `70^(@)`C. `60^(@)`D. `50^(@)` |
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Answer» Correct Answer - C `mu=(sin .(A+delta_(m))/(2))/(sin .(A)/(2))=(sin.(A+A)/(2))/(sin.(A)/(2))=(sinA)/(sin.(A)/(2))` `=(2sin.(A)/(2)cos .(A)/(2))/(sin.(A)/(2))=2 cos.(A)/(2)` So,`sqrt(3)=2 cos .(A)/(2)=(sqrt(3))/(2)=cos .(A)/(2)impliesA=60^(@)` |
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| 356. |
When a glass prism of refracting angle `60^(@)` is immersed in a liquid its angle of minimum deviation is `30^(@)`. The critical angle of glass with respect to the liquid medium isA. `42^(@)`B. `45^(@)`C. `50^(@)`D. `52^(@)` |
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Answer» Correct Answer - B `A=60^(@), delta_(m)=30` so`mu=(sin((A+delta_(m))/(2)))/(sin((A)/(2)))` `mu=(sin((600^(@)+30^(@))/(2)))/(sin((60^(@))/(2)))=(sin 45^(@))/(sin 30^(@))=sqrt(2)` Also `mu=(1)/(sinC)impliesC=sin^(-1)((1)/(mu))impliesC=45^(@)` |
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| 357. |
Consider a an equilateral prism ABC of glass `(mu= (3)/(2))` placed in water `(mu=(4)/(3))` `{:("Column"I, "Column"II), ((A) FG "is parralel to" BC , (P) "Maximum deviation"),((B) i_(1) = 90^(@) , (Q) "Minimum deviation"), ((C) i_(1)= i_(2) = sin^(-1)((9)/(16)) , (R) TIR " will take place at surface" AC), ((D) EF "is perpendicular to" AB , (S) "No" TIR "will take place at surface" BC):}` |
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Answer» Correct Answer - A::B::C::D At minimum deviation `i_(1) = i_(2)` , `EF||BC` , At maximum deviation `i_(1) = 90^(@) "or" i_(2) = 90^(@)` For `i_(1) = 0, TIR` will not take place at `AC` |
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| 358. |
An equilateral prism ABC is placed in air with its base side BC lying horizontally along x-axis as shown in the figure. A ray given by `sqrt(3)z +x=10` is incident at a point P on the face AB of the prism (a) Find the value `mu` from which the ray grazes the face AB (b) Find the direction of the initially refracted ray if `mu = (3)/(2)` (c) Find the equation of ray coming out of the prism if bottom BC is silvered ? |
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Answer» Correct Answer - (a) `(2)/(sqrt(3))`; (b) Along -z axis; (c) `sqrt(3)x + x = 10` |
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| 359. |
A spherical mirrorr forms an image of magnification 3. The object distance, if focal length of mirrorr is `24cm,` may beA. `32cm, 24 cm`B. `32cm, 16 cm`C. `32 cm ` onlyD. `16 cm` only |
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Answer» Correct Answer - B `m=(f)/(f-u)` `3=(-24)/(-24-u)` or `-24-u=-8` or `u+24=8` or `u=(8-24)cm=-16cm` If `m=-3`, then, `-3=(-24)/(-24-u) u+24=-8` or `u=-32 cm`. Note that the magnification is greater than one `S_(0)`, mirrorr cannot be convex. |
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| 360. |
Water is not considered suitable for use in thermometers. Give the reason for the same. |
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Answer» It is unsuitable due to following reasons: (i) Anomalous behaviours of its expansion (not uniform) (ii) Large specific heat and low thermal conductivity (iii) Temperature range (freezing & boiling ) ` 0^(@)C & 100^(@) ` are small . (iv) Transparent (so visible) |
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| 361. |
Statement-`1`: Water is considered unsuitable for use in thermometers. Statement-`2`: Thermal Expansion of water is non uniform.A. Statement -`1` is True , Statement -`2` is True , Statement -`2` is correct explanation for Statement - `1`B. Statement -`1` is True , Statement -`2` is True , Statement -`2` is not correct explanation for Statement - `1`C. Statement -`1` is True , Statement -`2` is False.D. Statement -`1` is False , Statement -`2` is True. |
| Answer» Correct Answer - A | |
| 362. |
In the figure shown, how many images of the star will an observer at O see? A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - D As angle between the two mirror less than `90^(@)` so number of images formed is greater than 3. |
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| 363. |
The distance of an object from a spherical mirror is equal to the focal length of the mirror. Then the image:A. must be at infinityB. may be at infinityC. may be at the focusD. none |
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Answer» Correct Answer - B Image formation by concave mirror when object is real & placed at focus f. forms image at infinity but not in other cases. |
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| 364. |
When two plane mirrors subtend an angle `theta` between them, then a ray of light incident parallel to one of them retraces its path afer n reflections such that `ntheta=c` (where c is constant) What is the angle of c?A. `pi`B. `(pi)/(2)`C. `(3pi)/(2)`D. None of these |
| Answer» Correct Answer - B | |
| 365. |
5 mole of oxygen are heated at constant volume from `10^(@)C "to" 20^(@)C`. What will be the change in internal energy of the gas? Gram molar specific heat of gas at constant pressure `=8cal. "Mole"^(-1).^(@)C^(-1)` and `R=8.36J "mole"^(-1).^(@)C^(-1)`. |
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Answer» ` because" "` `C_(V) = C_(P) -R= 8-2 = 6"cal"//"mole" . ^(@)C` `therefore` `" "` Heat absorbed by `5` moles of oxygen at constant volume ` Q=nC_(V) DeltaT = 5xx6(20-10) = 30xx10 = 300cal` At constant volume `Delta V= 0`. `" "` `therefore` ` " "` `DeltaW=0` `therefore` `" "` From first law of thermodynamic `Q=DeltaV + W implies 300 = DeltaU + 0 implies DeltaU = 300` `"cal"` |
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| 366. |
At normal pressure and `0^(@)C` temperature the volume of `1` kg of ice is reduced by `91 cm^(3)` on melting . Laters heat of melting of ice is `3.4xx 10^(5)J//kg`. Calculate the changes in the internal energy when `2`kg of ice melts normal pressure and `0^(@)C. (P=1.01 xx 10^(5) Nm^(-2))` |
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Answer» Heat energy absorbed by `2` kg of ice melting `Q= "m"L = 2xx3.4xx10^(5) = 6.8 xx 10^(5)J` Change in volume of `2` kg of ice = `2xx91 = 182cm^(3) = 182xx 10^(-6)m^(3)` `therefore` `" "` `W= PDelta V = 1.01 xx 10^(5) xx (-182xx10^(-6)) = -18.4J` Since, work is done on ice so work `W` is taken -ve . Now from first law of thermodynamics ` Q = DeltaU + W implies Delta U = Q-W = 6.8 xx 10^(5)- (-18.4) = (6.8xx10^(5) + 18.4)J = 680018.4J` |
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| 367. |
A sample of gas follow process represented by `PV^(2)` = constant . Bulk modulus for this process is `B`, then which of the following graph is correct ?A. B. C. D. |
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Answer» Correct Answer - A::B::C `PV^(2)` = constant `implies P prop V^(-2) implies (DeltaP)/(P) = -2(DeltaV)/(V)` `implies ` Bulk modulus ` K = (DeltaP)/(-(DeltaV)/(V)) = 2P` As `PV = nRT` So `" "` `K prop (1)/(V^(2))` and `K prop T^(2)` |
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| 368. |
n moles of an ideal triatomic linear gas undergoes a process in which the temperature changes with volume as `T=K_(1)V^(2)` where `K_(1)` is a constant. Choose incorrect alternative.A. At normal temperature `C_(v)= (5)/(2)R`B. At any temperature `C_(p)-C_(v) = R`C. At normal temperature molar heat capacity ` C=3R`D. At any temperature molar heat capacity `C=3R` |
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Answer» Correct Answer - D At normal temperature `C_(v) = (f)/(2)R = (5)/(2)R` At any temperature `C_(p) - C_(v) = ((f)/(2)+1) - (f)/(2)R = R` from process `" " T = k_(1)V^(2)` and Ideal gas equation `PV= nRT` we have `PV^(-1)` = constant `implies x = -1` `implies C =C_(v) + (R)/(1-x) = C_(v) + (R)/(1+1) = C_(v) + (R)/(2)` At normal temperature ` C = (5)/(2)R + (R)/(2) = 3R` |
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| 369. |
Statement-`1`: Absolute zero temperature is not the temperature of zero energy. Statement-`2` : Only the intenal kinetic energy of the molecules is represented by temperature.A. Statement -`1` is True , Statement -`2` is True , Statement -`2` is correct explanation for Statement - `1`B. Statement -`1` is True , Statement -`2` is True , Statement -`2` is not correct explanation for Statement - `1`C. Statement -`1` is True , Statement -`2` is False.D. Statement -`1` is False , Statement -`2` is True. |
| Answer» Correct Answer - D | |
| 370. |
Statement-1 : Change in internal energy in the melting process is due to change in internal potential energy. Statement-2 : This is because in melting, distance between molecules increases but temperature remains constant.A. Statement -`1` is True , Statement -`2` is True , Statement -`2` is correct explanation for Statement - `1`B. Statement -`1` is True , Statement -`2` is True , Statement -`2` is not correct explanation for Statement - `1`C. Statement -`1` is True , Statement -`2` is False.D. Statement -`1` is False , Statement -`2` is True. |
| Answer» Correct Answer - A | |
| 371. |
Statement-`1`: Potential energy of water at `0^(@)C` is more than ice at `0^(@)C`. Statement-`2`: Heat given to melt ice at `0^(@)C` is used up in increasing the potential energy of water molecules formed at `0^(@)C`.A. Statement -`1` is True , Statement -`2` is True , Statement -`2` is correct explanation for Statement - `1`B. Statement -`1` is True , Statement -`2` is True , Statement -`2` is not correct explanation for Statement - `1`C. Statement -`1` is True , Statement -`2` is False.D. Statement -`1` is False , Statement -`2` is True. |
| Answer» Correct Answer - A | |
| 372. |
Statement-`1`: When an electric fan is switched on in a closed room, the air of the room is cooled. Statement-`2`: When fan is switched on, the speed of the air molecules will increase.A. Statement -`1` is True , Statement -`2` is True , Statement -`2` is correct explanation for Statement - `1`B. Statement -`1` is True , Statement -`2` is True , Statement -`2` is not correct explanation for Statement - `1`C. Statement -`1` is True , Statement -`2` is False.D. Statement -`1` is False , Statement -`2` is True. |
| Answer» Correct Answer - D | |
| 373. |
In a compound microscope, the focal length of the objective and the eye lens are `2.5 cm` and `5 cm` respectively. An object is placed at `3.75cm` before the objective and image is formed at the least distance of distinct vision, then the distance between two lenses will be `( i.e.` length of the microscope tube )A. `11.67cm`B. `12.67cm`C. `13.00 cm`D. `12.00 cm` |
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Answer» Correct Answer - A `L_(D)=V_(o)=u_(e)` and for objective lens `(1)/(f_(o))=(1)/(v_(o))-(1)/(u_(o))` Putting the values with proper sign convention. `(1)/(+2.5)=(1)/(v_(o))-(1)/((-3.75))impliesv_(o)=7.5cm` For eye lens `(1)/(f_(e))=(1)/(v_(e))-(1)/(u_(e))` `rArr (1)/(+5)=(1)/((-25))-(1)/(u_(e))impliesu_(e)=-4.16cm` `=|u_(e)|=4.16cm` Hence `L_(D)=7.5+4.16=1.67 cm` |
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| 374. |
The internal energy of a gas is given by `U= 5 + 2PV`. It expands from `V_(0)` to `2V_(0)` against a constant pressure `P_(0)`. The heat absorbed by the gas in the process is :-A. `-3P_(0)V_(0)`B. `3P_(0)V_(0)`C. `2P_(0)V_(0)`D. `P_(0)V_(0)` |
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Answer» Correct Answer - B `DeltaU = 2P_(0) DeltaV` & `W = P_(0)DeltaV` So `Q = W + DeltaU = 3P_(0)DeltaV = 3P_(0)V_(0)` |
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| 375. |
Choose the correct mirror image of A. B. C. D. |
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Answer» Correct Answer - C `(C)` |
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| 376. |
A particle is dropped along the axis from a height `(f)/(2)` on a concave mirror of focal length f as shown in figure. Find the maximum speed of image . |
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Answer» `v_(M1) = -m^(2)v_(OM) = -m^(2)(gt)` where. `m = (f)/(f-u) = (-f)/(-f + ((f)/(2) - (ft^(2))/(2))) = (2f)/(f +gt^(2)) implies v_(1) = -((2f)/(f+ gt^(2)))^(2)(gt) = -(4f^(2)gt)/((f + gt^(2))^(2))` For maximum speed `(dv_(I))/(dt) = 0 implies t= sqrt((f)/(3g)) implies v_(Imax) = (3)/(4)sqrt(3fg)` |
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| 377. |
A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P) by the metal. The sensor has scale that displays `log_2,(P//P_0)`, where `P_0` is constant. When the metal surface is at a temperature of `487^@C`, the sensor shows a value 1. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to `2767^@C`? |
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Answer» Correct Answer - 9 `P = eAsigmaT^(4)` where `T` is in kelvin `"log"_(2)(eAsigma(487 + 273)^(4))/(P_(0)) = 1 …..(i)` `"log"_(2)(eAsigma(2767 + 273)^(4))/(P_(0)) = x ….. (ii)` (ii) - (i) `"log"_(2) ((3040)/(760))^(4) = x-1` `therefore x = 9` |
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| 378. |
Calculate the kinetic energy of a gram molecule of argon at `127^(@)C`. |
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Answer» Temperature , `T = 127^(@)C= 273+127 = 400K, R=8.31J//mol K` `K.E.` per gram molecule of argon = ` 3/2 R T = 3/2 xx 8.31 xx 400 = 4986J` |
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| 379. |
The figure shows the face and interface temperature of a composite slab containing of four layers of two materials having identical thickness. Under steady state condition, find the value of temperature `theta`. |
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Answer» Correct Answer - `5^(@)C` `(DeltaQ)/(Deltat) = "same So" (KA(20-10))/(l) = (2KA(10-0))/(l) implies 2 theta = 10 implies theta = 5 ^(@)C` |
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| 380. |
Which of the following graphs correctly represents the variation of `beta= -((dV)/(dP))//V` with `P` for an ideal gas at constant temperatureA. B. C. D. |
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Answer» Correct Answer - A At constant temperature `PV` = constant So `PdV + VdP = 0` `implies (dV)/(dP) = -(V)/(P)` `implies beta = -(dV//dP)/(V) = (1)/(P)` |
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| 381. |
In Figure, a fish watcher watches a fish through a 3.0 cm thick glass wall of a fish tank. The watcher is in level with the fish, the index of refraction of the glass is `8//5` and that of the water `4//3` . a. To the fish, how far away does the watcher appear to be ? b. To the watcher, how far away does the fish appear to be ? |
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Answer» a. Distance of the watcher as seen from fish `8xx(4)/(3)+3xx(4//3)/(8//5)+10=(139)/(6)cm` b. Distance of the fish as seen from watcher is `10/(4//3)+(3)/(8//5)+8=(139)/(8)cm` |
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| 382. |
A ray of light AO is incident on a diverging lens as shown in figure. The focal plane of the lens is also given. The possible direction of the refracted ray will be B. C. D. none of the above |
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Answer» Correct Answer - a. For diverging lens, all refracted rays corresponding to an incident beam consisting of parallel rays meet at a point in focal plane when produced backward. |
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| 383. |
Statement-I : Sun glasses have zero power even through their surfaces are curved. and Statement-II: Both surface of the sun glasses are curved in the same direction with same radii.A. Statement-I is true , Statement -II is true, Statement-II is correct explanation for Statement-I.B. Statement - I is true, Statement -II is true: Statement -II is NOT a correct explanation for statement -I .C. Statement-I is true, Statement-II is false.D. statement-I is false, statement- II is true. |
| Answer» Correct Answer - A | |
| 384. |
STATEMENT-`1` Hollow prism forms no spectra as a solid equilateral prism of glass. `STATEMENT 2` Neglecting the thickness of hollow glass surface.The media is same .So dispersion is not to take place.A. Statement-I is true , Statement -II is true, Statement-II is correct explanation for Statement-I.B. Statement - I is true, Statement -II is true: Statement -II is NOT a correct explanation for statement -I .C. Statement-I is true, Statement-II is false.D. statement-I is false, statement- II is true. |
| Answer» Correct Answer - A | |
| 385. |
`STATEMENT-1` A convex lens may be diverging `STATEMENT -2` The nature of a lens depends upon the refractive incides fo the material of lens and surrounding medium besides geometry.A. Statement-I is true , Statement -II is true, Statement-II is correct explanation for Statement-I.B. Statement - I is true, Statement -II is true: Statement -II is NOT a correct explanation for statement -I .C. Statement-I is true, Statement-II is false.D. statement-I is false, statement- II is true. |
| Answer» Correct Answer - D | |
| 386. |
Column I (optical system ) Column II (focal length ) `(P) 80 "cm" " " (Q) 40 "cm" " " (R) 30"cm" " " (S) 20"cm"` |
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Answer» For `(A) : (1)/(f) = (mu-1)((1)/(p_(1)) + (1)/(p_(2))) = (1.5-1) ((1)/(20) + (1)/(20)) = (1)/(20) implies f= 20"cm"` For `(B) : (1)/(f) = ((1.5)/(4//3) -1)((1)/(20) + (1)/(20)) = (1)/(80) implies f = 80"cm"` For `(C) : (1.5)/(v_(1)) - (4//3)/(oo) = (1.5-(4)/(3)) ((1)/(20))` & ` (1)/(f) - (1.5)/(v_(1)) = (1- 1.5)((1)/(-20)) implies f = 30 "cm" For `(D) (1.5)/(v_(1)) - (1)/(oo) = (1.5-1) ((1)/(20))` & ` (4//3)/(f) - (1.5)/(v_(1)) = ((4)/(3) - 1.5)(-(1)/(20)) implies f = 40 "cm" |
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| 387. |
STATEMENT-`1` For observing traffic at our back, we prefers to use a convex mirror `STATEMENT 2` A convex mirror has a more larger field of view than a plane mirror or concave mirror.A. Statement-I is true , Statement -II is true, Statement-II is correct explanation for Statement-I.B. Statement - I is true, Statement -II is true: Statement -II is NOT a correct explanation for statement -I .C. Statement-I is true, Statement-II is false.D. statement-I is false, statement- II is true. |
| Answer» Correct Answer - A | |
| 388. |
Column -I shows different situations in which an object O is shown in front of the optical element. Column-II gives the corresponding velocity vector of the image of the object. Object is moving with velocity v and diverging mirror is moving with velocity 2vA. `{:("P Q R S"),("1 2 3 4"):}`B. `{:("P Q R S"),("4 3 2 1"):}`C. `{:("P Q R S"),("4 1 3 2"):}`D. `{:("P Q R S"),("3 1 4 2"):}` |
| Answer» Correct Answer - (B) | |
| 389. |
A sample of oxygen with volume of `500` cc at a pressure of ` 2` atm is compressed to a volume of `400` cc. What pressure is needed to do this if the temperature is kept constant? |
| Answer» Temperature is constant ,so `P_(1) V_(1) = P_(2)V_(2) therefore P_(2) = P_(1)(V_(1))/(V_(2)) = 2[(500)/(400)] = 2.5`atm | |
| 390. |
An air bubble doubles in radius on rising from bottom of a lake to its surface. If the atmosphere pressure is equal to that due ot a column of `10` m of water, then what will be the depth of the lake. (Assuming that surface tension is negligible)? |
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Answer» Given that constant temperature , we use `P_(1)V_(1) = P_(2) V_(2)` `P_(2) = (10)` dg (for water column) ` P_(1) = (10+h)` dg (where h=depth of lake) ` V_(1) = (4pi)/(3) r^(3) , V_(2)= (4pi)/(3) (2r)^(3) = 8((4pi)/(3) r^(3)) = 8V_(1)` Thus for `P_(2)V_(2) = P_(1)V_(1)`, We have `10` dg `(8V_(1))= (10+h)` dg `V_(1) implies 80= 10+h implies h=70m` |
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| 391. |
Consider the situation shown in figure. Light from a point source S is made paralel by aconvex lens L. The beam travels horizontlly andfals on an `88^@-88^@-4^@` prism as shown in figure. It passes through the prism symmetrically. The transmited light falls on a vertial mirror. Through what angle should the mirror be rotated so that an image of S is formed as S itself? |
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Answer» The parallel beam after going through the prism will be deviated by an angle `delta`. If the mirror is also rotated by this angle `delta`. The rays will fal normally on it. The rays will be reflected back along the same path and form the image of S on itself. as the prismis thin, the angle `delta` is given by `delta=(mu-1)A` `=(1.5-10xx4^@=2^@` thus , the mirror should be rotated by `2^@` |
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| 392. |
When light of wavelength `lambda` on an equilateral prism, kept on its minimum deviation position, it is found that the angle of deviation equals the angle the angle of the prism itself. The refractive index of the material of the prism for the wavelength `lambda` isA. `sqrt(3)`B. `(sqrt(3))/(2)`C. 2D. `sqrt(2)` |
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Answer» Correct Answer - A `mu=(sin((A+delta_(m))/(2)))/(sin ((A)/(2)))=(sin((60^(@)+60^(@))/(2)))/sin ((60^(@))/(2))=sqrt(3)` |
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| 393. |
The length of the compound microscope is `14 cm.` The magnifying power for relaxed eye is `25`. If the focal length of eye lens is `5 cm`, then the object distance for objective lens will beA. `1.8cm`B. `1.5 cm`C. `2.1 cm`D. `2.4 cm` |
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Answer» Correct Answer - A `L_(oo)=v_(o)+f_(e)implies14=v_(o)+5impliesv_(o)=9cm` Magnifying power of microscope for relaxed eye `m=(v_(o))/(u_(o)).(D)/(f_(e))` or `25=(9)/(u_(o)).(25)/(5)`or `u_(o)=(9)/(5)=1.8cm` |
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| 394. |
The frequency of a light wave in a material is `2xx10^(14)Hz` and wavelength is `5000 Å`. The refractive index of material will beA. `1.40`B. `1.50`C. `3.00`D. `1.33S` |
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Answer» Correct Answer - C Velocity of light waves in material is `v=v lambda` ….`(i)` Refractive index of material is `mu=(c)/(v)` ....`(ii)` where `c` is speed of light in vacuum or air. or `mu=(c)/(vlambda)` ...`(iii)` Given, `v=2xx10^(14)Hz,` `lambda=5000 Å=5000 xx10^(-10)m,` `c=3xx10^(8)m//s` Hence, from Eq. `(iii)`, we get `mu=(3xx10^(8))/(2xx10^(14)xx5000xx10^(-10))=3.00` |
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| 395. |
Figure shows a process `ABCA` performed on an ideal gas. Find the net heat given to the system during the process. ` |
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Answer» Since the process is cyclic , hence the change in internal energy is zero. The heat given to the system is then equal to the work done by it. The work done in part `AB` is `W_(1) = 0` (the volume remains constant ) . The part `BC` represents an isothermal process so that the work done by the gas during this part is `W_(2) = nRT_(2) "ln"(V_(2))/(V_(1))` During the part `CA: Vprop T " "` So,`V//T` is constant and hence , `P=(nRT)/(V)` is constant The work done by the gas during the part `CA "is" W_(3) = P(V_(1)-V_(2)) = nRT_(1) -nRT_(2) = -nR(T_(2)-T_(1))`. The net work done by the gas in the process `ABCA` is `W= W_(1) + W_(2) + W_(3) = "n"R[T_(2)"ln"(V_(2))/(V_(1)) - (T_(2)-T_(1))]` The same amount of heat is given to the gas. |
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| 396. |
Calculate the work done when `1` mole of a perfect gas is compressed adiabatically . The initial pressure and volume of the gas are `10^(5)N//m^(2)` and ` 6` litre respectively. The final volume of the gas is `2` litres. Molar specific heat of the gas at constant volume is `(3R)/(2) [(3)^(5//3) = 6.19]` |
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Answer» For an adiabatic change `PV^(gamma)` = constant `P_(1)V_(1)^(gamma) = P_(2)V_(2)^(gamma)`. As molar specific heat of gas at constant volume `C_(v) = (3)/(2)R` `C_(P) = C_(V) + R = (3)/(2)R + R = (5)/(2)R implies gamma = (C_(P))/(C_(V)) = (5/2R)/(3/2R) = (5)/(3)` `therefore P_(2) = [(V_(1))/(V_(2))]^(gamma) P_(1) = [(6)/(2)]^(5//3) xx 10^(5) = (3)^(5//3) xx 10^(5) = 6.19 xx 10^(5) N//m^(2)` Work done ` W= (1)/(1-gamma) [P_(2)V_(2)- P_(1)V_(1)] = (1)/(1-(5//3)) [6.19 xx 10^(5) xx 2 xx 10^(-3) - 10^(-5) xx 6 xx 10^(-3)]` `= -[(2xx10^(2)xx3)/(2) (6.19-3)] = -3xx10^(2) xx 3.19 = -957` joules -ive sign shows external of work done on the gas |
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| 397. |
Two diverging lenses are kept as shown in figure. The final image formed will be A. virtual for any value of `d_(1)` and `d_(2)`B. real for any value of `d_(1)` and `d_(2)`C. virtual or real depends on `d_(1)` and `d_(2)` onlyD. virtual or real depends on `d_(1)` and `d_(2)` also on the focal lengths of the lens |
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Answer» Correct Answer - A for real object concave lens made the virtual image |
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| 398. |
In the figure (i) a thin lens of focal length 10cm is shown. The lens is cut into two equal parts, and the parts are arranged as shown in the figure (ii). An object AB of height 1cm is placed at distance of 7.5cm from the arrangement. The height of the final image will be: A. 0.5 cmB. 2 cmC. 1 cmD. 4 cm |
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Answer» Correct Answer - B The effective focal length is 5 cm The height of final image is `=v/uxxO=f/(u+f) xxO =5/(-7.5+5)xx1=2 cm` |
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| 399. |
In the figure shown an object O is placed in front of a thin lens of small aperture. The dotted curves represent the extensions of the respective curves in the figure. How many images will be formed finally. A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B this lens is behaves like as two different lense one converging and other is diverging. |
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| 400. |
One mole of an ideal gas at an initial temperature true of `TK` does `6R` joule of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is `5//3`, the final temperature of the gas will beA. `(T+2.4)K`B. `(T-2.4)K`C. `(T+4)K`D. `(T-4)K` |
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Answer» Correct Answer - D `W_("adiabatic") = (nR(T_(1)-T_(2)))/(gamma-1) = 6R` `implies (1xxR(T_(1)-T_(2)))/(5//3-1) = 6R implies T_(2) = (T-4)K` |
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