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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.
| 201. |
A spherical surface of radius R separates air from a medium of refractive index `mu`. Parallel beam of light is incident, from medium side, making a small angle `theta` with the principal axis of the spherical surface. Find the co-ordinates of the point where the rays will focus in air. |
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Answer» Correct Answer - `x = (R)/(mu-1); y = R ((mu)/(mu-1))` |
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| 202. |
An equilateral glass prism can produce a minimum deviation of `30°` to the path of an incident ray. A transpar- ent slab of refractive index 1.5 is placed in contact with one of the refracting faces of the prism. Thickness of the slab is 3 cm. Now calculate the minimum possible deviation which can be produced by this prism-slab combination. |
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Answer» Correct Answer - `30^(@)` |
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| 203. |
A prism of refractive index sqrt2 has refractive angle `60^@.` In the order that a ray suffers minimum deviation it should be incident at an angle ofA. `45^(@)`B. `60^(@)`C. `90^(@)`D. `180^(@)` |
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Answer» Correct Answer - A `mu=(sini)/(sin((A)/(2)))impliessqrt(2)=(sini)/(sin((60)/(2)))impliesi=45^(@)` |
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| 204. |
The angle of minimum deviation caused by a prism is equal to the angle of the prism. What are the possible values of refractive index of the material of the prism? |
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Answer» Correct Answer - `sqrt(2) lt mu lt 2` |
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| 205. |
When light rays are incident on a prism at an angle of `45^(@)`, the minimum deviation is obtained. If refractive index of the material of prism is `sqrt(2)`, then the angle of prism will beA. `30^(@)`B. `40^(@)`C. `50^(@)`D. `60^(@)` |
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Answer» Correct Answer - D `(sin.(A+deltam)/(2))/(sin .(A)/(2))=mu, `But` (A+delta_(m))/(2)=i=45^(@)` So,` (sin 45^(@))/(sin(A//2))=sqrt(2)implies(1)/(2)=sin.(A)/(2)impliesA=60^(@)` |
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| 206. |
The angle of minimum deviation measured with a prism is `30^(@)` and the angle of prism is `60^(@)`. The refractive index of prism material isA. `sqrt(2)`B. `2`C. `3//2`D. `4//3` |
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Answer» Correct Answer - A `mu=(sin((A+delta_(m))/(2)))/(sin(A//2))=(sin 45^(@))/(sin30^(@))=sqrt(2)` |
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| 207. |
The minimum deviation produced by a hollow prism filled with a certain liquid is found to be `30^(@)`. The light ray is also found to be refracted at angle of `30^(@)`. The refractive index of the liquid isA. `sqrt(2)`B. `sqrt(3)`C. `sqrt((3)/(2))`D. `(3)/(2)` |
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Answer» Correct Answer - A For minimum angle of deviation for a prism `A=2r, :. A=60^(@)` Now `mu=(sin((60+30)/(2)))/(sin((60)/(2)))=(sin45^(@))/(sin30^(@))=(1)/(sqrt(2))xx(2)/(1)=sqrt(2)` |
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| 208. |
If the refractive indices of a prism for red, yellow and violet colours be `1.61, 1.63` and `1.65` respectively, then the dispersive power of the prism will beA. `(1.65-1.62)/(1.61-1)`B. `(1.62-1.61)/(1.65-1)`C. `(1.65-1.61)/(1.63-1)`D. `(1.65-1.63)/(1.61-1)` |
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Answer» Correct Answer - C `omega=(mu_(V)-mu_(R))/(mu_(Y)-1)=(1.65-1.61)/(1.63-1)` |
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| 209. |
A thermodynamic system undergoes cyclic process `ABCDA` as shown in figure. The work done by the system is A. `P_(0)V_(0)`B. `2P_(0)V_(0)`C. `(P_(0)V_(0))/(2)`D. zero |
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Answer» Correct Answer - D `W_(OBC) = W_(ODA)` `therefore` Net work done `W_(OBC) - W_(ODC) = 0` |
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| 210. |
The focal length of field achromatic combination of a telescope is `90 cm` .The dispersive powers of lenses are `0.024` and `0.036` respectively .Their focal lengths will be-A. `30 cm and 60 cm`B. `30 cm and -45 cm`C. `45 cm and 90 cm`D. `15 cm and 45 cm` |
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Answer» Correct Answer - B For achromatic combination `(f_(1))/(f_(2))=-(omega_(2))/(omega_(1))=-(0.036)/(0.024)=-(3)/(2)` and `(1)/(f_(1))-(1)/(f_(2))=(1)/(90)` Solving above equations we get `f_(1)=30cm, f_(2)=-45cm` |
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| 211. |
The dispersive powers of crown and flint glasses are `0.02` and `0.04` respectively. In an achromatic combination of lenses the focal length of flint glass lens is `40 cm`. The focal length of crown glass lens will beA. `-20 cm`B. `+20 cm`C. `-10cm`D. `+10 cm` |
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Answer» Correct Answer - A By using `(omega_(1))/(f_(1))+(omega_(2))/(f_(2))=0implies(0.02)/(f_(1))+(0.04)/(40)=0` `f_(1)=-20cm` |
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| 212. |
An achromatic telescope objective of focal length 1.5m is consists of two thin lenses of dispersive power 0.050 and 0.075, respectively, placed in contact. Find the focal length of each lens. |
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Answer» Correct Answer - For convex lens: `f = 50 cm, omega = 0.050`, for concave lens: `f = 75 cm, omega = 0.075` |
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| 213. |
Two lenses in contact made of materials with dispersive powers in the ratio`2:1`behaves as ab achromatic lens of focal length `10cm`.The individual focal length of the lenses are : |
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Answer» Correct Answer - 10 cm for concave and 5 cm for convex |
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| 214. |
Two lenses in contact made of materials with dispersive powers in the ratio`2:1`behaves as ab achromatic lens of focal length `10cm`.The individual focal length of the lenses are :A. 5 cm, -10cmB. `-5cm , 10cm`C. 10 cm, -20 cmD. `-20 cm, 10 cm` |
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Answer» Correct Answer - B `(omega_(1))/(f_(1)) + (omega_(2))/(f_(2)) = 0` (condition for achromatism) `omega_(1) : omega_(2) = 2 :1 ` (Given) `implies f_(1) : f_(2) = -2 :1` …..(i) ( for fulfills condition) `(1)/(f_(1)) + (1)/(f_(2)) = (1)/(-10)implies (1)/(f_(1)) (1 + (f_(1))/(f_(2))) = (1)/(-10)` ....(ii) by (i) & (ii) `f_(1) = 10 cm` & `f_(2) = -5 cm` |
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| 215. |
A light ray is incident perpendicularly to one face of a `90^circ` prism and is totally internally reflected at the glass-air interface. If the angle of reflection is `45^circ`, we conclude that the refractive index n A. Less than `1.41`B. Equals to `1.41`C. Greater than `1.41`D. None of the above |
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Answer» Correct Answer - C For total internal reflection `thetagtC` `impliessin thetagtsin Cimpliessin thetagt(1)/(mu)` or`mugt(1)/(sin theta)implies mugt(1)/(sin 45^(@))implies mugtsqrt(2)impliesmu gt 1.41` |
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| 216. |
In the above question, the magnification isA. `4//9`B. `-4//9`C. `9//4`D. `8//13` |
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Answer» Correct Answer - a. `m=-(v)/(u)=-((4)/(9))/(-1)=(4)/(9)` |
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| 217. |
A small piece of wire bent into an L shape, with upright and horizontal portions of equal lengths, is placed with the horizontal portion along the axis of the concave mirror whose radius of curvature is 10cm. I fthe bend is 20cm from the pole of the mirror, then the ration of the lengths of the images of the upright and horizontal portions of the wire isA. `1:2`B. `3:1`C. `1:3`D. `2:1` |
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Answer» Correct Answer - b. For upright portion, `m=(f)/(f-u)=((-10)/(2))/((-10)/(2)-(-20))=(-5)/(-5+20)=(-5)/(15)=-(1)/(3)` For horizontal portion, magnification is `(-(1)/(3))^(2)` i.e., `(1)/(9)` Required ration is `(-1//3)/(1//9)=-3:1` |
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| 218. |
A convex lens of focal length 40cm is held at a distance 12cm coaxially above a concave mirror of focale length 18cm. Q. If the convex lens is replaced by a glass plate of thickness 6cm, refractive index `mu=(3)/(2)` and gives rise to an image coincident with itself, then the value of d is equal to A. 20cmB. 30cmC. 40cmD. 60cm |
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Answer» Correct Answer - a. Shift produced by the glass plate `Deltax=(1-(1)/(mu))t=(1-(2)/(3))6=2cm` For the mirror, the object is placed at a distance `[12+6+d-Deltax]=(16+d)` The image is coincident with the object itself if `(16+d)=36cmrArrd=20cm` |
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| 219. |
A convex mirror and a concave mirror of radius 10cm each are placed 15cm apart facing each other. An object is placed midway between them. If the reflection first takes place in the concave mirror and then in convex mirror, the position of the final image isA. on the pole of the convex mirrorB. on the pole of the concave mirrorC. at a distance of 10cm from the convex mirrorD. at a distance of 5cm from the concave mirror |
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Answer» Correct Answer - a. For a concave mirro, `u=-(15)/(2)cm, v=?` `f=(10)/(2)cm=-5cm` `(1)/(v)=(1)/(f)-(1)/(u)-(1)/(-5)-(1)/(-15//2)` `=(1)/(5)+(2)/(15)=(-1)/(15)` or `v=-15cm` Clearly, the position of the final image is on the pole of the convex mirror. |
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| 220. |
Which one of the following spherical lenses does not exhibit dispersion? The radii of curvature of the surfaces of the lenses are as given in the diagrams. `A. B. C. D. |
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Answer» Correct Answer - C `(1)/(f) = (mu -1) ((1)/(R_(1)) - (1)/(R_(2)))` For no dispersion `d{(1)/(f)} = 0 implies "d"mu{(1)/(R_(1)) - (1)/(R_(2))} = 0 implies R_(1) = R_(2)` |
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| 221. |
Two spherical concave mirrors of equal focal length are put against each other with their reflecting surfaces facing each other. The upper mirror has an opening at its centre. A small object (O) is kept at the bottom of the cavity so formed (see Figure). The top mirror produces a virtual image of the object and the lower mirror then creates a real image of the virtual image. The second image is created just outside the cavity mouth. This creates an optical illusion as if the object is raised above its original position. (a) Prove that the height h shown in the Figure is related to focal length of the mirror as `(1)/(f) = (1)/(h) + (1)/(h+((1)/(h)-(1)/(f))^(-1))` (b) Rewrite the above equation in terms of `x = (h)/(f)` and solve if for x. |
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Answer» Correct Answer - (b) `x = 1` |
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| 222. |
A mirror of radius of curvature 20cm and an object which is placed at a distance of 15cm are both moving with velocities `1ms^(-1)` and `10ms^(-1)` as shown in Fig. Fing the velocity of image at this situation. |
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Answer» Using `1/v+1/u=2/R,` we get `1/v-1/(15)=-1/(10)rArr v=-30cm` Now, using `V_(im) = -(v^(2))/(u^(2))V_(OM)` `(V_(i)-v_(m))=-(v^(2))/(u^(2))(V_(o)-V_(m))` `rArr V_(i)-(1)=-((-30)^(2))/((-15)^(2))[(-10)-(1)]` `rArr V_(i)=45cm s^(-1)` So, the image will move with veloxity `45 cms^(-1)`. |
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| 223. |
Two thin sysmmetrical lenses of different nature and of different material have equal raii of curvature `R=15cm`. The lenses are put close together and immersed in water `(mu_(w)=4//3)` . The focal length of the system ini water is 30cm. The difference between refractive indices of tthe two lenses isA. `1//2`B. `1//4`C. `1//3`D. `3//4` |
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Answer» Correct Answer - c. Let `f_(1)` and `f_(2)` be the focal length in water. Then `(1)/(f_(1))=((mu_(1))/(mu_(w))-1)((1)/(R)+(1)/(R))=((mu_(1))/(mu_(w))-1)((2)/(R))` (i) `(1)/(f_(2))=((mu_(2))/(mu_(w))-1)(-(1)/(R)-(1)/(R))=((mu_(2))/(mu_(w))-1)((-2)/(R))` (ii) Adding (i) and (ii), we get `(1)/(f_(1))+(1)/(f_(2))=(2(mu_(1)-mu_(2)))/(mu_(w)R)` or `(1)/(30)=(2(mu_(1)-mu_(2)))/(mu_(W)R)` `:. (mu_(1)-mu_(2))=(mu_(w)R)/(60)` Substituting the values, `(mu_(1)-mu_(2))=(4xx15)/(3xx60)=(1)/(3)` |
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| 224. |
Each quarter of a vessel of depth `H` is filled with liquids of the refractive indices `n_(1),n_(2), n_(3)` and `n_(4)` from the bottom respectively. The apparent depth of the vessel when looked normally isA. `(H(n_(1)+n_(2)+n_(3)+n_(4)))/(4)`B. `(H((1)/(n_(1))+(1)/(n_(2))+(1)/(n_(3))+(1)/(n_(4))))/(4)`C. `((n_(1)+n_(2)+n_(3)+n_(4)))/(4H)`D. `(h((1)/(n_(1))+(1)/(n_(2))+(1)/(n_(3))+(1)/(n_(4))))/(2)` |
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Answer» Correct Answer - B Apparent depth of bottom `=(H//4)/(mu_(1))+(H//4)/(mu_(2))+(H//4)/(mu_(3))+(H//4)/(mu_(4))` `=(H)/(4)((1)/(u_(1))+(1)/(mu_(2))+(1)/(mu_(3))+(1)/(mu_(4)))` |
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| 225. |
Two concave lenses `L_1 and L_2` are kept in contact with each other. If the space between the two lenses is filled with a material of smaller refractive index, the magnitude of the focal length of the combinationA. becomes undefinedB. remains unchangedC. increasesD. decreases |
| Answer» Correct Answer - C | |
| 226. |
consider three congerging lenses `L_1, L_2 and L_3` having identical geometrical construction. The index of refraction of `L_1 and L_2 are mu_1 and mu_2` respectively. The upper half of the lens `L_3` has a refractive index `mu_1` and the lower half has `mu_2`. A point object O is imaged at `O_1` by the lens `L_1 and at O_2` by the lens `L_2` placed in same position . If `L_2` is placed at the same place. A. there will be n image at OB. there will be an image t `O_2`C. the only imge will form somewhere between `O_1 and O_2`.D. the only image will form away form `O_`. |
| Answer» Correct Answer - A::B | |
| 227. |
In an isosceles prism of angle `45^@`, it is found that when the angle of incidence is same as the prism angle, the emergen ray grazes the emergent surface. Find the refractive index of the material of the prism. For what angle of incidenc, the angle of deviation will be minimum? |
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Answer» As the ray of light grazes the second surface, `r_(2)` is the critical angle. `sinr_(2)=(1)/(mu)` `r_(2)=(45^@-r_(1))` `sinr_(1)=(sin45^@)/(mu)=(1)/(sqrt(2)mu)` `sinr_(2)=sin(45^@-r_(1))` `=(1)/(sqrt(2))[cosr_(1)-sinr_(1)]` `(1)/(mu)=(1)/(sqrt(2))[sqrt(1-(1)/(2mu^(2)))-(1)/(sqrt(2)mu)]` `(1)/(mu)=(1)/(sqrt(2))[sqrt(2mu^(2)-1)/(sqrt(2)mu)-(1)/(sqrt(2)mu)]` `2=sqrt(2mu^(2)-1)-1` `2mu^(2)-1=9` `2mu^(2)=10` `rArr mu^(2)=5` or `mu=sqrt(5)` At minimum deviation, `r_(1)=r_(2)(45^(@))/(2)=22.5^(@)` `mu=(sini_(1))/(sinr_(1))rArrsini_(1)=(sqrt(5))sin(22.5^(@))` `i_(1)=58.8^(@)` |
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| 228. |
The refracting angle of a prism is A and refractive index of the material of prism is `cot(A//2)` . The angle of minimum deviation will beA. `180^(@)-3A`B. `180^(@)+3A`C. `90^(@)-3A`D. `180^(@)-2A` |
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Answer» Correct Answer - d. From `sin((A+delta_(m))/(2))=mu "sin"(A)/(2)` `sin((A+delta_(m))/(2))=("cos"(A)/(2))/("sin"(A)/(2))xx"sin"(A)/(2)` `(A+delta_(m))/(2)=(pi)/(2)-(A)/(2)` `delta_(m)=pi-2A=180^(@)-2A` |
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| 229. |
The refracting angle of a prism is `A` and refractive index of the material of the prism is `cos(A//2)`. The angle of minimum deviation isA. `180^(@)-3A`B. `180^(@)-2A`C. `90^(@)-A`D. `180^(@)+2A` |
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Answer» Correct Answer - B The refractive index of the material of the prism `mu=(sin .(A+delta_(m))/(2))/(sin .(A)/(2))` `impliescot.(A)/(2)=((sin(A+delta_(m)))/(2))/(sin .(A)/(2))` `sin((A+delta_(m))/(2))=cot.(A)/(2)sin .(A)/(2)` `sin ((A+delta_(m))/(2))=(cos.(A)/(2))/(sin .(A)/(2))sin .(A)/(2)` `sin ((A+delta_(m))/(2))=sin ((pi)/(2)-(A)/(2))` `implies(A+delta_(m))/(2)=(pi)/(2)-(A)/(2)` `A+delta_(m)=pi -Aimpliesdelta_(m)=180-2A` |
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| 230. |
A prism is made of glass of refractive index 1.5. If the angle of minimum deviation is equal to the refracting angle of the prism, calculate the angle of the prism. |
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Answer» Given, `mu=1.5 ` and `A=delta_(m i n)` At minimum deviation, `r=A//2` and `delta_(min)=2i-A` Substituting for A, we get `i=delta_(min)` But `mu=(sin((delta_(m)+A)/(2)))/(sin((A)/(2)))=(sinA)/(sin[(A)/(2)])=2cos[(A)/(2)]` Therefor, `1.5=2cos[(A)/(2)]` or `A =cos^(-1)((3)/(4))` |
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| 231. |
The refractive indices of flint glass for red and violet lights are 1.613 and 1.632, respectively. Find the angular dispersion produced by a thin prism of flint glass having refracting angle `5^@`. |
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Answer» Correct Answer - `0.095^(@)` |
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| 232. |
Refracting angle o f a prism `A=60^@` and its refractive index is `n=3//2`. What is angle of incidence i to get minimum deviation. Also, find the minimum deviation. Assume the surrounding medium to be air `(n=1)`. |
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Answer» Correct Answer - `sin^(-1) ((3)/(4)), 2 sin^(-1) ((3)/(4)) -(pi)/(3)` |
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| 233. |
In an astronomincal telescope, the distance between the objective and the eyepiece is 36cm and the final image is formed at infinity. The focal length `f_(0)` of the objective and the focall length `f_(e)` of th eeyepiece areA. `f_(0)=45` cm and `f_(e)=-9cm`B. `f_(0)=50` cm and `f_(e)=10cm`C. `f_(0)=7.2` cm and `f_(e)=5cm`D. `f_(0)=30` cm and `f_(e)=6cm` |
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Answer» Correct Answer - a.,b. Usin `M=(f_(0))/(f_(e))` and `L=f_(0)+f_(e)` |
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| 234. |
A simple telescope, consisting of an objective of focal length `60 cm` and a single eye lens of focal length `5 cm` is focussed on a distant object is such a way that parallel rays comes out from the eye lens. If the object subtends an angle `2^(@)` at the objective, the angular width of the image.A. `10^(@)`B. `24^(@)`C. `50^(@)`D. `1//6^(@)` |
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Answer» Correct Answer - B Since `m=(f_(o))/(f_(e))` Also `m=("Angle subtended by the image" )/("Angle subtended by the object")` `:. (f_(o))/(f_(e))=(alpha)/(beta)impliesa=(f_(o)xxbeta)/(f_(e))=(60xx2)/(5)=24^(@)` |
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| 235. |
Assertion : A concave mirror and convex lens both have the same focal length in air. When they are submerged in water, they will have same focal length. Reason refractive index of water is smaller than be refractive index of air.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 - D If a mirrorr is placed in a medium other than air, its focal length does not change as `f=R//2`. But for the lens `(1)/(f_(a))=(._(a)n^(g)-1)((1)/(R_(1))-(1)/(R_(2)))` and `(1)/(f_(w))=(._(w)n^(g)-1)((1)/(R_(1))-(1)/(R_(2)))` As `._(w)n^(g)lt ._(a)n^(g)` , hence focal length of lens in water increases. The refractive index of water `4//3` and that of air is 1. Hence, `mu_(w)gtmu_(a)` |
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| 236. |
A simple telescope, consisting of an objective of focal length 60 cm and a single eye lens of focal length 5 cm is focussed on a distant object in such a way that parallel rays comes out from the eye lens. If the object subtends an angle `2^(@)` at the objective, the angular width of the image |
| Answer» `MP = (f_(0)/(f_(e)) = (beta)/alpha) implies beta = alpha(f_(0))/(f_(e)) = 2^(@) xx (60)/(5) = 24^(@)` | |
| 237. |
The diameter of the moon is `3.5 xx 10^(3) km` and its distance from the earth is `3.8 xx 10^(5)km`. It is seen through a telescope having focal lengths of objective and eye-piece as `4m` and `10 cm` respectively. Calculate (a) magnifying power of telescope (b) length of telescope tube and (c) anngular size of image of moon. |
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Answer» `MP = (f_(0))/(f_(e)) = -(400)/(10) = -40` . Angle subtended by the moon at the objective = `(3.5xx10^(3))/(3.8 xx 10^(5)) = 0.009 "radian"`. Thus angular diameter of the image = `MP xx "visual angle" = 40 xx 0.009 = 0.36 "radian" = (0.36 xx 180)/(3.14) = 21^(@)` |
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| 238. |
A diatomic gas obeys the law `pV^x=` constant. For what value of x, it has negative molar specific heat?A. ` x gt 1.67`B. `x lt 1.67`C. `1 lt x lt1.4`D. `1 lt x lt 1.67` |
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Answer» Correct Answer - D `C = C_(V) + (R)/(1-x) = (3)/(2)R = + (R)/(1-x) lt 0 ` `implies (5-3x)/(1-x) lt 0 implies 1 lt x lt 1.67` |
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| 239. |
One mole of a monatomic ideal gas is taken along two cyclic processes `E to F to G to E and E to F to H to E` as shown in the PV diagram. The processes involved are purely isochoric , isobaric , isothermal or adiabatic . Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists. A. `{:(P,Q,R,S),(4,3,2,1):}`B. `{:(P,Q,R,S),(4,3,1,2):}`C. `{:(P,Q,R,S),(3,1,2,4):}`D. `{:(P,Q,R,S),(3,1,2,4):}` |
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Answer» Correct Answer - A Process `FG` is isothermal so work done =`nRT "In"((P_(t))/(P_(f))) = 32 P_(0)V_(0) "In" ((32P_(0))/(P_(0)))= 160 P_(0)V_(0) "In"2`. Process `GE` is isobaric So work done = `P|DeltaV| = P_(0)|(V_(G)- V_(E))|` =`P_(0)|(32V_(0) - V_(0))|` =`31P_(0)V_(0)` Process `FH` is adiabatic so `(32P_(0))V_(0)^(5//3) = (P_(0))V_(H)^(5//3) implies V_(H) = 8V_(0)` Since process `FH` is adiabatic so `|((P_(H)V_(H) - P_(F)V_(F)))/((8-1))|= (|(P_(0)8V_(0))-32P_(0)V_(0)|)/(((5)/(3)-1))=36P_(0)V_(0)` Process `G rarr H` is isobaric so work done =`P_(0)|(32V_(0)- 8V_(0))| = 24 P_(0)V_(0)` |
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| 240. |
Molar heat capacity of an ideal gas in the process `PV^(x)` = constant , is given by : `C = (R)/(gamma-1) + (R)/(1-x)`. An ideal diatomic gas with `C_(V) = (5R)/(2)` occupies a volume `V_(1)` at a pressure `P_(1)`. The gas undergoes a process in which the pressure is proportional to the volume. At the end of the process the rms speed of the gas molecules has doubled from its initial value. Heat supplied to the gas in the given process is :A. `7P_(1)V_(1)`B. `8P_(1)V_(1)`C. `9P_(1)V_(1)`D. `10P_(1)V_(1)` |
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Answer» Correct Answer - C `Q = nCDeltaT = n(3R)DeltaT = n(3R)(3T) = 9nRT = 9 P_(1)V_(1)` |
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| 241. |
An ideal gas expands in such a way that `PV^2 = `constant throughout the process.A. This expansion is not possible without heatingB. This expansion is not possible without coolingC. Internal energy remains constant in this expansionD. Internal energy increases in this expansion |
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Answer» Correct Answer - B `PV= nRT` & `PV^(2)` = constant `implies V prop (1)/(T) implies ` gas can expand only if it cods As temperature decreases during expansion so internal energy will decrease. |
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| 242. |
An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume `V_1` and contains ideal gas at pressure `P_1` and temperature `T_1`.The other chamber has volume `V_2` and contains ideal gas at pressure `P_2` and temperature `T_2`. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will beA. `(T_(1)T_(2)(P_(1)V_(1) + P_(2)V_(2)))/(P_(1)V_(1)T_(2) + P_(2)V_(2)T_(1))`B. `(P_(1)V_(1)T_(1) + P_(2)V_(2)T_(2))/(P_(1)V_(1) + P_(2)V_(2))`C. `(P_(1)V_(1)T_(2) + P_(2)V_(2)T_(1))/(P_(1)V_(1) + P_(2)V_(2))`D. `(T_(1)T_(2)(P_(1)V_(1) + P_(2)V_(2)))/(P_(1)V_(1)T_(1) + P_(2)V_(2)T_(2))` |
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Answer» Correct Answer - A `U = U_(1) + U_(2) implies (n_(1) + n_(2)) C_(V)T = n_(1)C_(V)T_(1)+ n_(2)C_(V)T_(2)` `implies T = ((P_(1)V_(1) + P_(2)V_(2)) T_(1)T_(2))/((P_(1)V_(1)T_(2) + P_(2)V_(2)T_(2)))` |
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| 243. |
Two thin lenses of focal length `f_(1)` and `f_(2)` are in contact and coaxial. The power of the combination isA. `sqrt((f_(1))/(f_(2)))`B. `sqrt((f_(2))/(f_(1)))`C. `(f_(1)+f_(2))/(f_(1)f_(2))`D. `(f_(1)+f_(2))/(f_(1)f_(2))` |
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Answer» Correct Answer - D If two thin lenses of focal lengths` f_(1)` and `f_(2)` are placed in contact coaxially, then equivalent focal length of combination is `(1)/(F)=(1)/(f_(1)) + (1)/(f_(2))+(0)/(f_(1)f_(2))=(1)/(f_(1))+(1)/(f_(2))` Power for the combination is `P=(1)/(F)=(1)/(f_(1))+(1)/(f_(2))=(f_(1)+f_(2))/(f_(1)f_(2))` |
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| 244. |
For the given process :- `{:("Column-I","Column-II"),((A)Process " "JtoK,(P) Wgt0),((B)Process" "KtoL ,(q) Wlt0),((C)Process " "LtoM,(r)Qgt0),((D)Process" "MtoJ,(s)Qlt0):}` |
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Answer» Correct Answer - A::B::C::D Process `J rarr K` (isochoric) : `W= 0 , DeltaU lt 0 implies Q lt 0 ` Process `K rarr L` (isochoric) : `Wgt 0 , DeltaU gt 0 implies Q gt 0 ` Process `L rarr M` (isochoric) : `W= 0 , DeltaU gt 0 implies Q gt 0 ` Process `M rarr J` (isochoric) : `Wlt 0 , DeltaU lt 0 implies Q lt 0 ` |
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| 245. |
Column I gives some devices and Column II gives some processes on which the functioning of these devices depend. Match the devices in Column I with the processes in Column II and indicate your answer by darkening appropriate bubbles in the `4 xx 4 ` matrix given in the ORS. |
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Answer» Correct Answer - A::B::C::D (A) Bimetallic strip : Works on the thermal expansions of solids (different solids expands by different length for the same rise of temperature). The energy is converted to kinetic energy. (B) Steam engine Energy is converted (heat - mechanical) (C) Electric fuse Works on melting of fuse wire on heating. Heat `rarr P.E.` of molecules . |
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| 246. |
When a system is taken from state i to state f along the path iaf, it is found that `Q=50 cal and W=20 cal`. Along the path ibf `Q=36cal`. W along the path ibf is A. `6` calB. `16` calC. `66` calD. `14` cal |
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Answer» Correct Answer - A As a thermodynamic system is taken from state to state f, then the internal energy of the system remains the same irrespective of the path followed `(Q-W)_(faf) = (Q-W)_(lbf) implies 50 -20 = 36 W implies W = 6` cal |
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| 247. |
Two thin similar watch glass pieces are joined together,front to front ,with near convex portion silvered and the combination of glass pieces is placed at a distance `a=60cm` from a screen.A small object is placed normal to the optical axis of the combination such that its image is two time the object. Now water is filled in between the glass. water`mu=4//3`,calculated the distance through,which the object must be displayed so that a sharp image is again formed on the screen. |
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Answer» Correct Answer - 15 cm towards the combination |
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| 248. |
In the figure shown, find the relative speed of a approach/ separation of the two final images formed after the light rays pass through the lens, at the moment when u = 30 cm . The speed of object = 4 cm/s . The two lens halves are placed symmetrically w.r.t. the moving object |
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Answer» Correct Answer - `((8)/(5)) cm//s` `(I)/(0) = m = (v)/(u) = (f)/(f+u) , "rel" (2dI)/(dt) = [-(f_(0))/((f+u)^(2)) (du)/(dt)] xx 2` =`(-40 xx(1)/(2))/(10 xx 10) xx 4 xx 2 = (-8)/(5)"cm"//"sec"` |
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| 249. |
For a small angled prism, angle of prism `A` of minimum deviation`(delta)` varies with the refractive index of the prism as shown in the graph A. Point `P` corresponds to `m=1`B. Slope of the line `PQ=A//2`C. Slope `=2A`D. None of the above statements is true |
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Answer» Correct Answer - A At `P,delta=0=A(mu-1)impliesmu=1` Also `delta_(m)=(mu-1)A=Amu_(m)-A` Comparing it with `y=mx+c` Slope of the line `=m=A` It mean that medium 2 is denser medium. So total internal reflection cannot occur. |
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| 250. |
The graph between angle of deviation `(delta)` and angle of incidence (i) for a triangular prism is represented byA. B. C. D. |
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Answer» Correct Answer - A For a prism, as the angle of incidence increases, the angle of deviation first decreases, goes to a minimum value and then increases. |
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