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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.
| 1. |
A symmetric doule convex lens is cut in two equal parts by a plane perpendiculr to the pricipal axis. If the power of the original lens was 4D, the power of a cut lens will beA. 2DB. 3DC. 4DD. 5D |
| Answer» Correct Answer - A | |
| 2. |
Three perfect gases at absolute temperature `T_(1), T_(2)` and `T_(3)` are mixed. The masses f molecules are `m_(1), m_(2)` and `m_(3)` and the number of molecules are `n_(1), n_(2)` and `n_(3)` respectively. Assuming no loss of energy, the final temperature of the mixture isA. `(n_(1)T_(1)^(2) + n_(2)T_(2)^(2) + n_(3)T_(3)^(2))/(n_(1)T_(1) + n_(2)T_(2) + n_(3)T_(3))`B. `(n_(1)^(2)T_(1)^(2) + n_(2)^(2)T_(2)^(2) + n_(3)^(2)T_(3)^(2))/(n_(1)T_(1) + n_(2)T_(2) + n_(3)T_(3))`C. `(T_(1) + T_(2) + T_(3))/(3)`D. `(n_(1)T_(1) + n_(2)T_(2) + n_(3)T_(3))/(n_(1) + n_(2) + n_(3))` |
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Answer» Correct Answer - D `(n_(1)C_(V_(1))T_(1) = n_(2)C_(V_(2))C_(V_(2))T_(2) + n_(3)C_(V_(3))T_(3))` =`(n_(1)+n_(2) +n_(3))C_(V_("mtx"))T` As `C_(V_(1)) = C_(V_(2))"so" T = (n_(1)T_(1)+n_(2)T_(2)+n_(3)T_(3))/(n_(1)+n_(2) +n_(3))` |
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| 3. |
The critical angle for water is `48.2^@`. Find is refractive index. |
| Answer» `mu=1/(sintheta_c)=1/(sin48.2^@=1.34` | |
| 4. |
Find the size of the image formed in the situation shown in figure. |
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Answer» Here u=-40cm, R=-20cm, `mu_1=1, mu_2=1.33. We have` `mu_2/v-mu_1/u=(mu_2-mu_1)/R` `or, 1.33/v-1/(-40cm)=1.33/(-20cm)` or, `1.33/v=-1/(40cm)-0.3./(20cm)` `or v=-32cm` The magnification is `m=h_2/h_1=(mu_1v)/(mu_2u)` `or h_2/(1.0cm)=(-32cm)/(1.33x(-40cm)` or `h_2=+0.6cm` The image is erect. |
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| 5. |
The angle of minimum, deviation from a prism is `37^@`. If the angle of prism is `53^@`, find the refrence index of the material of the prism. |
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Answer» `mu=(sin(A+delta_m)/2)/(sinA/2)=(sin(53^@37^@)/2)/sin(53^@/2)=(sin45^@)/(sin26.5^@)` `=1.58` |
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| 6. |
A prism of refractive index `1.53` is placed in water of refractive index `1.33`. If the angle of prism is `60^@`, calculate the angle of minimum deviation in water. |
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Answer» Here `.^(a)mu_(g) = 1.33, .^(a)mu_(w) = 1.53, A=60^(@), delta_(m) = ? .^(w)mu_(g)= (.^(a)mu_(g))/(.^(a)mu_(w)) = (1.53)/(1.33) = 1.15 because .^(w)mu_(g) = (sin(A + delta_(m))/(2))/(sin(A)/(2))` `therefore (sin(A+delta_(m)))/(2) = .^(w)mu_(g)xx "sin"(A)/(2) = 1.15"sin"(60)/(2) = 0.575 implies (A+delta_(m))/(2) = sin^(-1)(0.575) = 35.1^(@)` `therefore delta_(m) = 35.1 xx 2-60 = 10.2 ^(@)` |
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| 7. |
Light passes symmetrically through a `60^(@)` prism of refractive index 1.54. After emergence out from the prism the light ray is incident on a plane mirror fixed to the base of the prism extending beyond it. Find the total deviation of the light ray after reflection from the mirror. |
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Answer» Correct Answer - `0^(@)` |
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| 8. |
A vertical beam of light of cross sectional radius `(R)/(2)` is incident symmetrically on the curved surface of a glass hemisphere of refractive index `mu = (3)/(2)`. Radius of the hemisphere is R and its base is on a horizontal table. Find the radius of luminous spot formed on the table. `sin 20^(@) = (1)/(3)` and `sin 80^(@) = 0.98` |
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Answer» Correct Answer - `0.34 R` |
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| 9. |
An obsever looks at a distant tree of height 10m with a telescope of magnifying power of 20. to the observer the tree appears:A. 20 times nearerB. 10 times tallerC. 10 times nearerD. 20 times taller |
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Answer» Correct Answer - D Angular magnification is 20 |
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| 10. |
A large transparent cube (refractive index = 1.5) has a small air bubble inside it. When a coin (diameter 2 cm) is placed symmetrically above the bubble on the top surface of the cube, the bubble cannot be seen by looking down into the cube at any angle. However, when a smaller coin (diameter 1.5 cm) is placed directly over it, the bubble can be seen by looking down into the cube. What is the range of the possible depths d of the air bubble beneath the top surface ? |
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Answer» Correct Answer - `(3sqrt(5))/(8) lt d lt (sqrt(5))/(2)cm` |
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| 11. |
In Fig., determine the apparent shift in the position of the coin. Also, find the effective refractive index of the combinatino of the glass and water slab. |
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Answer» Total apparent shift is `s=t_(1)(1-(1)/(mu_(1)))+t_(2)(1-(1)/(u_(2)))` `=8(1-(1)/(4))+4.5(1-((1)/(3))/2)` `=2+1.5=3.5cm` The apparent depth of coin from the top is `t=(8+4.5)-3.5=9cm` and, the real depth of the coin is `t_(1)+t_(2)=8+4.5=12.5` Therefore, the effective refractive index is `mu_(eff)=(" Real depth ")/(" Apparent depth ")` `=(t_(1)+t_(2))/(t)=(12.5)/9` `=1.39` |
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| 12. |
Two transparent plastic sheets of red and blue colour overlap as shown in the Fig. An observer looks at a clear sky through the sheets. What can you say about the colour and brightness of light coming through sections 1, 2 and 3 (see Fig.) |
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Answer» Correct Answer - Section 1: Red colour, less bright than usual Section 2: Blue colour, less bright than usual Section 3: Dark, almost no light. |
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| 13. |
A glass cube is cut symmetrically into two halves and the two parts are kept at a small separation between them. Calculate the angular deviation suffered by a light ray incident normally on one of the faces of the cube. |
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Answer» Correct Answer - Zero |
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| 14. |
A layer of oil 3 cm thick is floating on a layer of coloured water 5 cm thick. Refractive index of coloured water is `5//3` and the apparent depth of the two liquids appears to be `36//7`cm. Find the refractive index of oil. |
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Answer» Apparent depth (AI)=`(t_(1))/(mu_(1))+(t_(2))/(mu_(2))` `:. (36)/(7)=(5)/(5//3)+(3)/(mu_(2))rArr (3)/(mu_(2))=(36)/(7)-3=(15)/(7)` = `mu_(2)=7/5=1.4` |
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| 15. |
A small piece of wood is floating on the surface of a 2.5 m deep lake. Where does the shadow form on the bottom when the sun is just setting? Refractive index of water = 4/3 |
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Answer» Correct Answer - B::C::D Height of the lake =2.5 m When the sun is just setting q is approximately =900 `:. (sini)/(sinr)=mu_2/mu_1` `rarr 1/(sinr)=(4/3)/1` `rarr sinr=3/4 rarr r=49^@` Thus (x/2.5)=tanr=1.15` `rarr x=2.5xx1.15=2.8m` |
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| 16. |
`n_(1)` and `n_(2)` moles of two ideal gases (mo1 wt `m_(1)` and `m_(2))` respectively at temperature `T_(1)K` and `T_(2)K` are mixed Assuming that no loss of energy the temperature of mixture becomes . |
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Answer» Total energy of molecules of first gas =`3/2 n_(1)kT_(1)`, Total energy of molecules of second gas =`3/2 n_(2)kT_(2)` Let temperature of mixture be `T` then total energy of molecules of mixture =`3/2 k(n_(1) + n_(2))T` `therefore (3)/(2) (n_(1)+n_(2))kT =3/2 k(n_(1)T_(1) +n_(2)T_(2)) implies T=(n_(1)T_(1) + n_(2)T_(2))/((n_(1)+n_(2)))` |
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| 17. |
A wooden stick of length 100 cm is floating in water while remaining vertical. The relative density of the wood is 0.7. Calculate the apparent length of the stick when viewed from top (close to the vertical line along the stick) Refractive index of wate `=(4)/(3)`. |
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Answer» Correct Answer - `82.5 cm` |
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| 18. |
A double convex lens has two surfaces of equal radii R and refractive index m=1.5, we haveA. `f=R/2`B. `f=R`C. `f=-R`D. `f=2R` |
| Answer» Correct Answer - B | |
| 19. |
The first excited state of hydrogen atom is `10.2 eV` above its ground state. The temperature is needed to excite hydrogen atoms to first excited level is |
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Answer» `KE` of the hydrogen atom `3/2 kT` = `10.2 eV` = `10.2xx(1.6xx10^(-19))J` ` implies T=(2)/(3) xx (10.2xx1.6xx10^(-19))/(1.38xx10^(-23)) = 7.88xx10^(4)K` |
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| 20. |
At what temperature , will the root mean square velocity on hydrogen be double of its value at `S.T.P`., pressure remaining constant ? |
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Answer» Let `v_(1)` be the `r.m.s` velocity at `S.T.P.(=273K`) and `v_(2)` be the `r.m.s` velocity at unknown temperature `T_(2)`. `therefore (v_(1)^(2))/(v_(2)^(2)) = (T_(1))/(T_(2))` or `T_(2) =T_(1)[v_(2)/v_(1)]^(2) = 273xx(2)^(2) = 273xx4` `=1092 K = (1092-273) = 819^(@)C` |
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| 21. |
The velocities of ten particles in `ms^(-1)` are `0,2,3,4,4,4,5,5,6,9`. Calculate (i)average speed and (ii)rms speed (iii) most probable speed. |
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Answer» (i) average speed ,` V_(av) = (0+2+3+4+4+4+5+5+6+9)/10 = 42/10 = 4.2ms^(-1)` (ii) rms speed , `V_(rms) =[((0)^(2)+(2)^(2)+(3)^(2)+(4)^(2)+(4)^(2)+(4)^(2)+(5)^(2)+(5)^(2)+(6)^(2)+(9)^(2))/10]^(1//2)` =`[228/10]^(1//2) = 4.77ms^(-1)` (iii) most probable speed `V_(mp) = 4m//s` |
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| 22. |
The diameter of the objective lens of a telescope is `5.0m` and wavelength of light is `6000 Å`. The limit of resolution of this telescope will beA. `0.03 sec`B. `3.03 sec`C. `0.06 sec`D. `0.15 sec` |
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Answer» Correct Answer - A Limit of resolution `=(1.22lambda)/(a)xx(180)/(pi)`(in degree) `=((1.22xx(6000xx10^(-10)))/(5)xx(180)/(pi))^(@)=0.03 sec` |
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| 23. |
The diameter of objective of a telescope is `1m`. Its resolving limit for the light of wave length `4538 Å`, will beA. `5.54xx10^(-7)rad`B. `2.54xx10^(-4)rad`C. `6.54xx10^(-7)rad`D. None of these |
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Answer» Correct Answer - A Resolvinig limit `d theta=(1.22lambda)/(a)=(1.22xx4538xx10^(-10))/(1)=5.54xx10^(-7)rad.` |
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| 24. |
The velocity of light in a medium is half its velocity in air. If ray of light emerges from such a medium into air, the angle of incidence, at which it will be totally internally reflected, isA. `15^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)` |
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Answer» Correct Answer - B `mu=(c)/(v)impliesmu=(c)/(c//2)=2` also for total internal reflection `igtcimpliescsinigesincimpliessinige(1)/(mu)` Hence `igesin^(-1)((1)/(mu))` or `ige30^(@)` |
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| 25. |
Assertion `:` In a movie, ordinarily 24 frames are projected per second from one end to the other of the complete film. Reason `:` The image formed on retina of eye is sustained up to `1//10s` after the removal of stimulus.A. If both the assertion and reason are true and reason explains the assertion.B. If both the assertion and reason are true but reason does not explain the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - C When the movie is screened, the frames are displayed on after the other at the rate of `24` frames per second. The image of an object is formed on the retina. From where it is sent to the brain. If the object moves, the brain continues to show the old image for about one sixteenth of a second. |
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| 26. |
Assertion `:` The air bubble shines in water. Reason `:` Air bubble in water shines due to refraction of light.A. If both the assertion and reason are true and reason explains the assertion.B. If both the assertion and reason are true but reason does not explain the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A Shining of bubble is due to total internal reflection of light. |
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| 27. |
Assertion `:` Blue colour of sky appears due to scattering of blue colour. Reason `:` Blue colour has shortest wave length in visible spectrum.A. If both the assertion and reason are true and reason explains the assertion.B. If both the assertion and reason are true but reason does not explain the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A From Rayleigh criteria of scattering, we have scattering `prop (1)/(("wavelength")^(4))` Since wavelength of blue colour is least compared to other colours, it gets scattered by atomospheric particles such as dust, water vapour ets. Because of which sky appears blue. |
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| 28. |
To find the focal length of a convex mirror , a student records the following data: `{:("Object Pin","Convex Lens","Convex mirror","Image Pin"),(22.2 cm,32.2 cm,45.8 cm,71.2 cm):}` The focal length of the convex lens is `f_(1)` and that of mirror is `f_(2)` . Then taking index correction to be negligibly small , `f_(1)` and `f_(2)` are close to :A. `f_(1) = 7.8 cm " " f_(2) = 12.7 cm`B. `f_(1) = 15.6 cm " " f_(2) = 25.6 cm`C. `f_(1) = 12.7 cm " " f_(2) = 7.8 cm`D. `f_(1) = 7.8cm " " f_(2) = 25.4 cm` |
| Answer» Correct Answer - A | |
| 29. |
A point object is placed at distance of 30 cm from a convex mirror of local length 30 cm. The image will form atA. infinityB. poleC. focusD. 15 cm behind the mirror |
| Answer» Correct Answer - D | |
| 30. |
Statement I: Keeping a point object fixed, if a plane mirror is moved, the image will also move. Statement II: In case of a plane mirror, distance of object and its image is equal from any point on the mirrorA. Statement is True, Statement II is True, Statement II is correct explanation for Statement IB. 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. |
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Answer» Correct Answer - d. If the mirror is shifted parallele to itself such that the velocity of the mirror is parallel to its surface, the image shall not shift. Hence, Statement I is false. |
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| 31. |
State the following statements as TRUE or FALSE. a. A convex mirror cannot from a real image for a real object. b. The image formed by a convex mirror is always diminished and erect. c. Virtual image formed by a concave mirror is always enlarged. d. Only in the case of a concave mirror, it may happen that the object and its image move in same direction. e. In the case of a concave mirror, the image always move faster than the object. f. If an object is placed in front of a diverging mirror at a distance equal to its focal length, then the height of image formed is half of the height of object. g. For two positions of an object , a concave mirror can form englarged image. h. Concave mirror is used as a rear view mirror in motor vehicles. i. If some portion of the mirror ois covered, then complete image will be formed but of reduced brightness. j. A plane mirror always forms an erect iamge of same size as that of the object. k. The image formed by a plane mirror has left-right reversal. l. A virtual object means a converging beam. |
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Answer» a. True. According to the magnification formula, `m=(f_(0))/(f_(0)+x)` where x is distance of a real object. For all positive values of x, m is always positive. b. False. For a real object , the image formed by a convex mirror is always erect and diminished. `m=(f_(0))/(f_(0)+x)` For a virtual object lying between pole and focus, the image formed is real, enlarged, and erect. `m=(f_(0))/(f_(0)+x)` c. True. For a concave mirror, we know that `m=(f_(0))/(f_(0)+x)` where x is the object distance. If `x lt f_(0)`, then `m gt 1` , which implies a virtual englarged image. d. False. In case of mirrors (convex, concave, and plane), the object and its image always move in opposite directions. That is, `(V)/(U)=-((f)/(u-f))^(2)` where V=velocity of image and U=velocity of object. e. False. When the object is beyon the center of curvature, its image moves slower. f. True. For a convex mirror, `m=f_(0)//f_(0)+x` i.e., if `x=f_(0)`, then `m=1//2` g. True. A concave mirror forms an enlarged image in two situations: i. Object is placed between C and F. ii. Object is placed between F and P. h. False. Convex mirror is used as a rear view mirror. i. True. Complete image will be formed, but of less brightness because now the incident rays will be reflected from reduced area to form the image. j. True, `m =-v//u` . For a plane mirror , `v=-u` `rArr m=+1`. Hence, image is erect and of same size. k. True . We see our left hand as right hand and right as left hand in a plane mirror. l. True. Virtual object is that point where incident rays seem to be converging. |
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| 32. |
In Figure. , a plane mirror is moving with a uniform speed of `5ms^(-1)` along negative x-direction and observer O is moving with a velocity of `10ms^(-1)`. What is the velocity of image of a particle P, moving with a velocity as shown in the figure, as observed by observer O? Also find its direction. |
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Answer» Velocity of point P, `v_(p)=10(hati+hatj)` Velocity of mirror `v_(m)=-5hati` Velocity of partical relative to mirror is `vecv_(p m) =-25hati+10hatj` Velocity of image relative to observer `vecv_(IO)=vecv_(p m)-vecv_(Om)=(-15hat(i)+10hat(j))-(vecv_(O)-vecv_(m))` `=( -30hat(i) +10hat(j)) ms^(-1)` Angle with X-axis, `alpha= tan^(-1)(-1//3)` |
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| 33. |
Find the velocity of the image when the object and mirror both are moving towards each other with velocities 2 and `3ms^(-1)`. How are they moving? |
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Answer» Here, `v_(OM)=-v_(IM)` `v_(0)-v_(M)=-0(v_(1)-v_(M)` `(+2ms^(-1))-(-3ms^(-1))=-v_(1)+(-3)` `rArr v_(1)=-8ms^(-1)` |
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| 34. |
White light is incident on a glass prism as shown. Four easily identifiable colors – red, green, yellow and blue get separated as A, B, C and D. Which of the rays (A, B, C and D) correspond to which color? |
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Answer» Correct Answer - `A rarr` red, `B rarr` yellow, `C rarr` Blue, `D rarr` green. |
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| 35. |
A converging lens and a diverging mirror are placed at a separation of 15 cm. The focal length of the lens is 25 cm and that of the mirror is 40 cm. Where should a point source be placed between the lens and the mirror so that the light, after getting reflected by the mirror and then getting transmitted by the lens, comes out parallel to the principal axis ? |
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Answer» Correct Answer - A::C If the image in the mirror will from at the focus of the converging lens, then after transmission through the lens the rays of light wil go paralel. Let the object is at a distance x cm from the mirror. `u=-xcm` `v=25-15=+10cm` `(………focal lenghtof lens =25cm)` f=+40 cm `rarr 1/v-1/u=1/f` `rarr 1/x=1/10-1/40` `rarr x=40/3` The object is at distance `(15-10/3)=5/3` `=1.67cm from the lens |
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| 36. |
An object is placed `12 cm` to the left of a diverging lens of focal length `-6.0 cm.` A converging lens with a focal length of `12.0 cm` is placed at a distance d to the right of the diverging lens. Find the distance d that corresponds to a final image at infinity. |
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Answer» Correct Answer - 8 cm |
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| 37. |
One mole of gas is taken from state `A` to state `B` as shown in figure. Work done by the gas is `alpha xx 10^(beta)J`. Find the value of `alpha + beta`. (Given: ` T_(1) = 320 K , R = (25)/(3)`) |
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Answer» Correct Answer - 7 Work done = `(1)/(2) [P_(0) + 2P_(0)][2V_(0)-V_(0)] = (3)/(2)P_(0)V_(0)` and `P_(0)V_(0) = R xx 320` So work done `(3)/(2)xx R xx 320 = (3)/(2) xx (25)/(3) xx320 = 4000J = 4 xx 10^(3)J implies alpha + beta = 4+3 = 7` |
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| 38. |
The relation between internal energy `U`, pressure `P` and volume `V` of a gas in an adiabatic processes is : `U = a+ bPV` where a = b = `3` . Calculate the greatest integer of the ratio of specific heats `[gamma]` . |
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Answer» Correct Answer - 1 For an adiabatic process , `dQ = dU + PdV = 0` `implies d[a+ bPV] + PdV = 0 implies bPdV + bV dP + PdV = 0 implies (b+1)PdV + bV dP = 0 implies (b+1) (dV)/(V) + b(dP)/(P) = 0` `implies (b+1) "log"V + b"log"P ` = constant, `V^(b+1)P^(b)` = constant `implies PV^(b+1)/(b) ` = constant `therefore gamma = (b+1)/(b) = (4)/(3) = 1.33` |
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| 39. |
The variation of lengths of two moles rods `A` and `B` with change in temperature is shown in figure . The ratio of `(alpha_(a))/(alpha_(B))` is :- A. `(3)/(2)`B. `(2)/(3)`C. `(4)/(3)`D. `(3)/(4)` |
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Answer» Correct Answer - B ` l= l_(0) (1+alphaDeltaT) implies l-l_(0) = l_(0)alphaDeltaT` `implies (l_(0)alpha_(A)DeltaT)/(l_(0)alpha_(B)DeltaT) = (104-100)/(106-100) = (4)/(6) implies (alpha_(A))/(alpha_(B)) = (2)/(3)` |
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| 40. |
A thin prism of angle `A=6^(@)` produces a deviation `d=3^(@)`. Find the refractive index of the material of prism. |
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Answer» We know that `d=A(mu-1)` or `mu=1+(delta)/(A)` Here, `A= 6^(@), d_(min)=30^(@)` `mu=sin((60+30)/(2))/sin((60)/(2))=(sin45^(@))/(sin30^(@))=sqrt(2)=1.41` |
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| 41. |
Calculate the dispersive power for crown glass from the given data `mu_(v)=1.523` and `mu_(r)=1.5145`. |
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Answer» Here, `mu_(v)=1.523` and `mu_(r)=1.5145` Mean refractive index, `mu=(1.523+1.5145)/(2)=1.51875` Dispersive power is given by, `omega=(mu_(v)-mu_(r))/((mu-1))=(1.523-1.5145)/((1.51875-1))=0.1639` |
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| 42. |
A fish in an aquarium approaches the left wall at a rate of `3ms^(-1)` observes a fly approaching it at `8ms^(-1)`. If the refractive index of water is `(4//3)`, find the actual velocity of the fly. |
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Answer» For the fish, the apparent distance of the fly from the wall of the aquarium is `mux,` if x is the actual distance. Then apparent velocity will be `d(mux)//dt rArr (v_("app"))_("fly")= muv_("fly")` Now the fish observes the velocity of the fly to be `8ms^(-1)` `rArr` Apparent relative belocity will be `= 8ms^(-1)` `rArr v_("fish")+(v_("app"))_(" fly ")=8` `3+muv_("fly")=8 rArr v_("fly" )=5xx3/4=3.75ms^(-1)` |
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| 43. |
Most substances contract on freezing . However, water does not belong to this category. We know that water expands on freezing. Further , coefficient of volume expansion of water in the temperature range from `0^(@)C` to `4^(@)C` is negative and above `4^(@)C` it is positive . This behaviour of water shapes the freezing of lakes as the atmospheric temperature goes down and it is still above `4^(@)C`. As the atmospheric temperature goes below `4^(@)C`A. Cooled water at the surface flows downward because of its greater densityB. Cooled water at the surface does not flow downward and remains at the surface because its smaller densityC. Cooled water at the surface , through it remains on the surface because of its smaller density, will conduct heat from the interior to the atmosphereD. Cooled water at the surface flows to the bottom because of its smaller density |
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Answer» Correct Answer - A At temperature above `4^(@)C`, temperature of water above is less as compared to below as water is heated by radiation of longer wavelength. |
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| 44. |
Three identical adiabatic containers `A, B and C` Contain helium, neon and oxygen respectively at equal pressure. The gases are pushed to half their original volumes.A. The final temperature of the gas in each container is sameB. The final pressure of the gas in each container is sameC. The final temperature of both helium and neon is sameD. The final pressure of both helium and neon is same |
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Answer» Correct Answer - C::D He and Ne are monatomic gas. |
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| 45. |
Two closed Identical conducting containers are found in the laboratory of an old scientist. For the verification of the gas some experiments are performed on the two boxes and the results are noted. Experiment `1` When the two containers are weighed `W_(A) = 225g , W_(B) = 160g` and mass of evacuated container `W_(C) = 100g`. Experiment 2. when the two containers are given same amount of heat same temperature rise is recorded . The pressure change found are `DeltaP_(A) = 2.5"atm"`. `" "` `DeltaP_(B) = 1.5"atm"` Identify the type of gas filled in container `A` and `B` respectively A. Mono , MonoB. Dia ,DiaC. Mono , DiaD. Dia , Mono |
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Answer» Correct Answer - C As `Q = n_(A) C_(VA)DeltaT = n_(8)C_(VB) DeltaT implies n_(A)C_(VA)= n_(B) C_(B)` But volume is constant So `DeltaP_(A)V_(A) = n_(A)RDeltaT` & `DeltaP_(B)V_(B) = n_(B)RDeltaT` `implies (n_(A))/(n_(B)) = (DeltaP_(A))/(DeltaP_(B)) = (2.5)/(1.5) = (5)/(3) implies (C_(VB))/(C_(VA)) = (5)/(3) = (5//2R)/(3//2R)` `implies "Gas B is diatomic & gas A is monoatomic"` |
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| 46. |
Two closed Identical conducting containers are found in the laboratory of an old scientist. For the verification of the gas some experiments are performed on the two boxes and the results are noted. Experiment `1` When the two containers are weighed `W_(A) = 225g , W_(B) = 160g` and mass of evacuated container `W_(C) = 100g`. Experiment 2. when the two containers are given same amount of heat same temperature rise is recorded . The pressure change found are `DeltaP_(A) = 2.5"atm"`. `" "` `DeltaP_(B) = 1.5"atm"` Identify the gas filled in the container `A` and `B`A. `N_(2) , Ne`B. `He, H_(2)`C. `O_(2), Ar`D. `Ar , O_(2)` |
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Answer» Correct Answer - D AS `n_(A) = (5)/(3)n_(B) "so" (125)/(M_(A)) = (5)/(3)((60)/(M_(B)))` `implies 5 M_(B) = 4M_(A) implies "Gas A " = Ar , "gas B" = O_(2)` |
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| 47. |
An object is present on the principal axis of a concave mirror at a distance 30cm from it. Focal length of mirror is 20cm. Q. Image formed by mirror isA. At a distance of 60cm in front of mirrorB. At a distance 60cm behind the mirrorC. At a distance 12 cm in front of mirrorD. At a distance 12cm behind the mirror |
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Answer» Correct Answer - a. `(1)/(v)+(1)/(u)=(1)/(f)` `rArr(1)/(v)+(1)/(-30)=(1)/(-20)rArrv=-60cm` |
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| 48. |
An object is present on the principal axis of a concave mirror at a distance 30cm from it. Focal length of mirror is 20cm. Q. If object starts moving with `2cms^(-1)` perpendicular to principal axis above the principal axis then,A. image moves with velocity `4cms^(-1)` below the principal axisB. image moves with velocity `4cms^(-1)` above the principal axisC. image moves with velocity `8cms^(-1)` below the principal axisD. image moves with velocity `8cms^(-1)` above the principal axis |
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Answer» Correct Answer - a. `m=-(v)/(u)=-(-60)/(-3)=-2,` so image will move twice distanc in same time as compared to object. Hence, velocity of image will be double. |
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| 49. |
An object is present on the principal axis of a concave mirror at a distance 30cm from it. Focal length of mirror is 20cm. Q. If object starts moving with `2cms^(-1)` along principal axis towards the mirror then,A. image starts moving with `4cms^(-1)` away from the mirrorB. image starts moving with `4cms^(-1)` towards the mirrorC. image starts moving with `8ms^(-1)` towards the mirrorD. image starts moving with `8cms^(-1)` away from the mirror |
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Answer» Correct Answer - d. Let `u=-x, v=-y` `(dx)/(dt)=-2cms^(-1),(1)/(-y)+(1)/(-x)=(1)/(f)` `rArr (1)/(y^(2))(dy)/(dt)+(1)/(x^(2))(dx)/(dt)=0` `rArr (dy)/(dt)=-((y)/(x))^(2)(dx)/(dt)rArrv_(I)=-((60)/(30))^(2)(-2)=8cms^(-1)` |
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| 50. |
A man uses a concave mirror for shaving. He keeps his face at a distance of 25 cm from the mirror and gets an image which is 1.4 times enlarged. Find the focal lengt.1 of the mirror |
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Answer» Correct Answer - C `u=-25cm` `=m=(A_1B_1)/(AB)=u/v` `rarr 1.4 =-(v)/(-25)` `rarr 14/10=v/25` `rarr v=(25xx14)/10=35 cm `Now, 1/f=1/v+1/u` `rarr 1/f=1/35-1/25 `=(5-7)/175=-2/175` `rarr f=-87.5` so, focal length of the concave mirror is 87.5 cm. |
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