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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.
| 51. |
A particle goes in a circle of radius 2.0 cm. A concave mirror of focal length 20 cm is placed with its principal axis passing through the centre of the circle and perpendicular to its plane. The distance between the pole of the mirror and the centre of the circle is 30 cm. Calculate the radius of the circle formed by the image. |
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Answer» Correct Answer - C::D u = -30 cm, f= -20 cm We know, `(1)/(upsilon) +(1)/(upsilon) = (1)/(f)` `rArr = (1)/(upsilon) + -(1)/(30) =-(1)/(20)` `rArr (1)/(upsilon) = (1)/(30) - (1)/(20) = (1)/(60)` `rArr upsilon = 60cm` Image of the circle is formed at a distance 60 cm in front of the mirror. `:. m = -(upsilon)/(u) = (R_(image))/(R_(object))` `rArr - ((-60))/((-30)upsilon) = (R_(image))/(2)` `rArr R_(image) = 4 cm` Radius of image of the circle is 4 cm. |
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| 52. |
Find the diameter of the image of the moon formed by a spherical concave mirror of focal length 7.6 m. The diameter of the moon is 3450 km and the distance of the earth and the moon is `3.8 xx 10^5 km.` |
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Answer» Correct Answer - C `u=-30cm` `f=-20cm` We know `1/v+1/u=1/f` `rarr 1/v+-1/30=-1/20` `rarr 1/v=1/30-1/2=1/60` `rarr v=60 cm ` image of the circle si formed at a distance 60 cm in front of the mirror. `=:. m=-v/u=R_(image)/R_(object)` `rarr =(-60)/((-30)v)=R_(image)/R_(object)` `rarr R_(image)=4 cm` Radius of image of the circle is 4 cm. |
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| 53. |
In order to increase the magnifying power of a telescopeA. The focal powers of the objective and the eye-piece should be largeB. Objective should have small focal length and the eye-piece largeC. Both should have large focal lengthD. The objective should have large focal length and the eye-piece should have small |
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Answer» Correct Answer - D `M=(F_(0))/(f_(theta))` |
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| 54. |
A region bounding water has air on two sides. Tell the nature (real or virtual) of the image for following cases- (The object is real and lies on the principal axis (see fig) in all cases.) (a) The object is to the left of surface 1 and the image to be considered is formed after the first refraction. (b) The object is to the left of surface 1 and the image to be considered is formed after two refractions. (c) The object is to the right of second surface and the image to be considered is formed after one refraction. |
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Answer» Correct Answer - (a) Virtual (b) May be real or virtual (c) May be real or virtual |
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| 55. |
Photograph of the ground are taken form an air-craft ,flying at an altitude of 2000 m by a camera with a lens of focal length `50 cm`. The size of the film in the camera is `18xx18cm`.What area of the ground can be photography by this camera at any one time. |
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Answer» As here `u =-2000"m" "f", =0.50 "m"` , so from lens formula `(1)/(v) - (1)/(u) = (1)/(f)`, we have `(1)/(v) - (1)/((-2000)) = (1)/(0.5) implies (1)/(v) = (1)/(0.5) - (1)/(2000)~=(1)/(0.5) ["as" (1)/(0.5)gt gt(1)/(2000)] implies v=0.5"m" = 50"cm" = f` Now as in case of a lens , `m = (v)/(u) = (0.5)/(-2000) = -(1)/(4)xx 10^(-3)` So `I_(1)` = (ma)(mb) = `m^(2)A` `[because A= "ab"]` `A = (I_(1))/(m^(2)) = (18"cm" xx 18"cm")/([(1//4) xx 10^(-3)]^(2)) = (720"m" xx 720"m")` |
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| 56. |
Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index `n_(1)` surrounded by a medium of lower refractive index `n_(2)` . The light guidance in the structure takes place due to successive total internal reflections at the interface of the media `n_(1)` and `n_(2)` as shown in the figure . all rays with the angle of incidence i less than a particular value of `i_(m)` are confined in the medium of refractive index `n_(1)` . The numerical aperture (NA) of the structure is defined as `"sin"i_(m)` ` For two structures namely `S_(1)` with `n_(1) = sqrt(45) //4 "and" n_(2) = 3//2 , "and" S_(2)` with `n_(1) = 8//5 "and" n_(2) = 7//5` and taking the refractive index of water to be 4/3 and that of air to be 1 , the correct option (s) is (are)A. NA of `S_(1)` immersed in water is the same as that of `S_(2)` immersed in liquid of refractive index `(4)/(sqrt(5))`B. NA of `S_(1)` immersed in liquid of refractive index `(6)/(sqrt(5))` is the same as that of `S_(2)` immersed in water.C. NA of `S_(1)` placed in air is the same as that of `S_(2)` immersed in liquid of refractive index `(4)/(sqrt(15))`D. NA of `S_(1)` placed in air is the same as that of `S_(2)` placed in water . |
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Answer» Correct Answer - A,C Let the whole structure is placed in a medium of refractive index n , then n sini = `n_(1) "sin"(90- theta)` ` " n sini" = n_(1) "cos" theta …..(i)` Here for `i_(m) , theta = C "and " "sin" C = (n_(2))/(n_(1))` from eq. (i) , `n"sin"i_(m) = n_(1) sqrt((1 - n_(0)^(2))/(n_(1)^(2))) = sqrt(n_(1)^(2) - n_(2)^(2))` `implies "sin"i_(m) = sqrt(n_(1)^(2) - n_(2)^(2))/(n)` Now , for (A) `(NA)_(s_(1)) = (3)/(4)sqrt((45)/(16) - (9)/(4)) = (3)/(4) xx (3)/(4) = (9)/(16)` `(NA)_(s_(2)) = (3sqrt(15))/(16) sqrt(64)/(25) - (49)/(25)) = (3sqrt(15))/(16) (1)/(5)sqrt(15) = (9)/(16)` For (B) `(NA)_(s_(1)) = sqrt(15)/(6) xx (3)/(4) = sqrt((15))/(8)` `(NA)_(s_(2)) = (3)/(4) = sqrt(15)/(5)` Not equal For (C) `" " (NA)_(s_(1)) = 1 xx (3)/(4) = (3)/(4)` `(NA)_(s_(2)) = sqrt(15)/(4) xx sqrt((15))/(5) = (15)/(4 xx 5) = (3)/(4)` For (D) `" " (NA)_(s_(1)) = (3)/(4)` `(NA)_(s_(2)) = (3)/(4) sqrt(15)/(5)` Not equal |
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| 57. |
Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index `n_(1)` surrounded by a medium of lower refractive index `n_(2)` . The light guidance in the structure takes place due to successive total internal reflections at the interface of the media `n_(1)` and `n_(2)` as shown in the figure . all rays with the angle of incidence i less than a particular value of `i_(m)` are confined in the medium of refractive index `n_(1)` . The numerical aperture (NA) of the structure is defined as `"sin"i_(m)` ` If two structures of same cross - sectional area, but different numerical apertures `NA_(1)` and `NA_(2)(NA_(2) lt NA_(1))` are joined longitudinally , the number aperture of the combined structure is |
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Answer» Correct Answer - D It is given that `implies i_(m2) lt i_(m1)` Hence if the combination can be placed both ways i.e., `1^(st)` structure & then `2^(nd)` structure and then reversed also , then the condition of TIR is satisfied for lower `i_(m)` then it can satisfied for all other less angler as well. Hence `NA_(2)` will be the numerical aperture of the combined structure . |
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| 58. |
A hemispherical glass body of radius 10 cm and refractive index 1.5 is silvered on its curved surface . A small air bubble is 6 cm below the flat surface inside it along the axis. The position of the image of the air bubble made by the mirror is seen :A. 16 cm below flat surfaceB. 20 cm below flat surfaceC. 30 cm below flat surfaceD. 14 cm below flat surface |
| Answer» Correct Answer - B | |
| 59. |
One mole of a monoatomic gas is enclosed in a cylinder and occupies a volume of `4` liter at a pressure `100N//m^(2)` . It is subjected to process `T = = alphaV^(2)`, where `alpha` is a positive constant , `V` is volume of the gas and `T` is kelvin temperature. Find the work done by gas (in joule) in increasing the volume of gas to six times initial volume . |
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Answer» Correct Answer - 7 `W= intPdV ` where `P = (nRT)/(V) = alphanRV implies W = n alphaR underset(V_(0))overset(6V_(0))int VdV= (nalphaR)/(2)[(6V_(0))^(2) - (V_(0))^(2)]` `implies W=(nalphaR)/(2) (35) V_(0)^(2) = (P_(0)V_(0) xx 35)/(2) (because alpha= (P_(0))/(nRV_(0)))` |
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| 60. |
Two taps `A` and `B` supply water at temperature `10^(@)C` and `50^(@)C` respectively . Tap `A` alone fills the tank in `1` hour and tap `B` alone fills the tank in `3` hour. If we open both the taps together in the empty tank , if the final temperature of the water in the completely filled tank is found to be `5alpha ("in" "^(@)C)` . Find the value of `alpha`. Neglect loss of heat to surrounding and heat capacity of the tank.A.B.C.D. |
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Answer» Correct Answer - 4 `m(T-10)s = (m)/(3)S(50-T) implies T = 20^(@)` |
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| 61. |
Two identical square rods of metal are welded end to end as shown in figure (i), 20 calories of heat flows through it in 4 minutes. If the rods are welded as shown in figure (ii), the same amount of heat will flow through the rods in A. `1` minuteB. `2` minuteC. `4`minuteD. `16` minute |
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Answer» Correct Answer - A Fig `A` : `20 = ((100 -0)/((L)/(kA) + (L)/(kA))) xx 4 …. (i)` Fig `A` : `20 = Q = (100-theta) [(kA)/(L) + (kA)/(L)]t …..(ii)` Equation (i) -:(ii) `1= (4//2)/(2//t) t = 1` min. |
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| 62. |
Two identical metal plates are welded end to end as shown in - (i) . `20` cal of heat flows through it in `4` minutes. If the plates are welded as shown in figure-(ii) , find the time (in minutes) taken by the same amount of heat to flow through the plates. |
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Answer» Correct Answer - 1 Rate of heat flow `(DeltaQ)/(Deltat) = (kA(T_(1)-T_(2)))/(l) implies Deltat prop (l)/(A)` |
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| 63. |
Air is filled at `60^(@)C` in a vessel of open mouth. The vessle is heated to a temperature `T` so that `1//4th` of air escapes. Assuming the volume of vessel remaining constant, the value of `T` isA. `80^(@)C`B. `444^(@)C`C. `333^(@)C`D. `171^(@)C` |
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Answer» Correct Answer - D `(3)/(4)`th volume of air at `0^(@)C` occupies entire volume at `theta` , As `(V_(1))/(T_(1)) = (V_(2))/(T_(2)) implies (3//4V)/(273 + 60) = (V)/(273 + theta) implies theta = 171^(@)C` |
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| 64. |
The coefficient of apparent expansion of a liquid when determined using two different vessle A and B are `gamma_(1)` and `gamma_(2,)` respectily. If the coefficient of linerar expansion of vesel A is `alpha.` Find the coefficient of linear expension of the vessel B.A. `(alpha_(1)gamma_(1)gamma_(2))/(gamma_(1)+ gamma_(2))`B. `(gamma_(1)-gamma_(2))/(2alpha_(1))`C. `(gamma_(1) + gamma_(2) + alpha)/(3)`D. `(gamma_(1) - gamma_(2) + 3alpha_(1))/(3)` |
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Answer» Correct Answer - D `gamma_(r) = gamma_(1) + 3 alpha_(1) = gamma_(2) + 3 alpha_(2) implies alpha_(2) = (gamma_(1) - gamma_(2) + 3alpha_(1))/(3)` |
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| 65. |
Ice starts forming in lake with water at `0^(@)C` and when the atmospheric temperature is `-10^(@)C`. If the time taken for `1 cm` of ice be `7` hours. Find the time taken for the thickness of ice to change from `1 cm` to `2 cm`A. `7` hoursB. `14` hoursC. less than `7` hoursD. more than `7` hours |
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Answer» Correct Answer - D `because t prop(x_(2)^(2) - x_(1)^(2))` For `x_(1) = 0 , X_(2) = 1`cm ` 7 prop (1^(2)-0^(2))` For `x_(1) = 1` cm , `x_(2) = 2` cm `t prop(2^(2) - 1^(2)) implies (7)/(t) = (1)/(3) implies t = 21` hrs |
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| 66. |
Three rods of the same dimensions have thermal conductivities `3`k , `2`k and k . They are arranged as shown, with their ends at `100^(@)C, 50^(@)C` and `0^(@)C`. The temperature of their junction is :- A. `75^(@)C`B. `(200)/(3)"^(@)C`C. `40^(@)C`D. `(100)/(3)"^(@)C` |
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Answer» Correct Answer - B Let `theta` = junction temperature Net heat current at junction is zero `3k(100-theta) + k(0-theta) +2k(50-theta) = 0 implies theta= (200)/(3) "^(@)C` |
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| 67. |
Two long ,thin,solid cylinders are identical in size, by they are made of different substances with two different thermal conductivities. The two cylinders are connected in series between a reservoir at temperature `T_("hot")` and a reservoir at temperature `T_("cold")`.The temperature at the boundary between the two cylinders is `T_("b")`. One can conclude that:-A. `T_(b)` is closer to `T_("hot")` than it is to `T_("cold")`.B. `T_(b)` is closer to `T_("cold")` than it is to `T_("hot")`.C. `T_(b)` is closer to the temp. of the reservoir that is in contact with the cylinder with the lower thermal conductivity.D. `T_(b)` is closer to the temp. of the reservoir that is in contact with the cylinder with the higher thermal conductivity. |
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Answer» Correct Answer - D For same rate of heat transfer the body having higher conductivity will have lower temperature difference. If cylinder with higher conductivity is connected with hot reservoir first then the function temperature `T_("b")`, will be closer to hot reservoir temperature. |
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| 68. |
A ligth of wavelength 6000 `Å` , in air, enters, a medium with refractive index 1.5. Inside the medium, its frequency is `"________________"` Hz and its wavelength is `"________________" Å` |
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Answer» Frequency remains the same, i.e., `f=(c)/(lambda)=(3xx10^(3))/(6000xx10^(-10))=5xx10^(14)Hz` `mu=(V_(1))/(V_(2))=(vlambda_(1))/(vlambda_(2))rArrlambda_(2)=(lambda_(1))/(mu)` The wavelength gets modified to `lambda_(2)=(lambda_(1))/(mu)=(6000Å)/(1.5)=4000Å` |
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| 69. |
Statement I: Two solid cylindrical rods of identical size and different thermal conductivity `K_1` and `K_2` are connected in series. Then the equivalent thermal conductivity of two rods system is less than that value of thermal conductivity of either rod. Statement II: For two cylindrical rods of identical size and different thermal conductivity `K_1` and `K_2` connected in series, the equivalent thermal conductivity K is given by `(2)/(K)=(1)/(K_1)+(1)/(K_2)`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 | |
| 70. |
Statement-`1`: High thermal conductivity of metals is due to presence of free electrons. Statement-`2`: Electrons at same temperature have very high average velocity than atoms.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 - C | |
| 71. |
Statement-`1`: A sphere, a cube and a thin circular plate made of same material and of same mass are initially heated to `200^(@)C`,the plate will cool at fastest rate. Statement-`2`: Rate of cooling =`(rhoAsigma)/(ms)(T^(4) - T_(0)^(4))alpha "surface area"`. Surface area is maximum for circularplate.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 | |
| 72. |
During an adiabatic process, the pressure of gas is found to be proportional to the cube of its absolute temperature. The ratio of `(C_(p,m)//C_(v,m))` for gas is :A. `4//3`B. `2`C. `5//3`D. `3//2` |
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Answer» Correct Answer - A Given ` P alpha T^(3)` `because PV = muRT` `therefore P alpha (PV)^(3) implies P^(3)V^(3)alphaP implies P^(2)V^(3)` = constant `implies PV^(s//2)` = constant `implies gamma = (C_(P))/(C_(V)) = (3)/(2)` |
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| 73. |
One mole of ideal monoatomic gas `(gamma=5//3)` is mixed with one mole of diatomic gas `(gamma=7//5)`. What is `gamma` for the mixture? `gamma` Denotes the ratio of specific heat at constant pressure, to that at constant volumeA. `3//2`B. `23//15`C. `35//23`D. `4//3` |
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Answer» Correct Answer - B Monoatomic `" "` `n_(1) = 1` Diatomic `n_(2) = 1 gamma_(2) = (7)/(5)` `(n_(1) + n_(2))/(gamma_("mix") - 1) = (n_(1))/(gamma_(1) - 1) + (n_(2))/(gamma_(2) - 1) implies gamma_("mix") = (3)/(2)` |
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| 74. |
Assertion : The ratio `C_(P)// C_(upsilon)` for a diatomic gas is more than that for a monoatomic gas. Reason : The moleculess of a monoatomic gas have more degrees of freedom than those of a diatomic gas.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 - C | |
| 75. |
The refractive indicates of flint glass for red and violent colours are `1.644 "and" 1.664`. Calculate its dispersive power. |
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Answer» Here , `mu_(r) = 1.644, mu_(v) = 1.664, omega = ?` Now `mu_(y) = (mu_(v) + mu_(r))/(2) = (1.664+1.644)/(2) = 1.654 " " because omega = (mu_(v) - mu_(r))/(mu_(y) - 1) = (1.664-1644)/(1654-1) = 0.0305` |
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| 76. |
The graph shown relationship between object distance and image distance for a equiconvex lens. Then focal length of the lens is ` A. `0.50+-0.05cm`B. `0.50+-0.10cm`C. `5.00+-0.05cm`D. `5.00+-0.10cm` |
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Answer» Correct Answer - c. From the formula `(1)/(f)=(1)/(v)-(1)/(u)`, we have `(1)/(f)=(1)/(10)-(1)/(-10)rArrf=+5` Further, `Deltau=0.1` and `Deltav=0.1` (from the graph) Now, differentionting the lens formula, we have `(Deltaf)/(f^(2))=(Deltav)/(v^(2))+(Deltau)/(u^(2))` `Deltaf+((Deltav)/(v^(2))+(Deltau)/(u^(2)))f^(2)` Substituting the values, we have `Deltaf=((0.1)/(10^(2))+(0.1)/(10^(2)))(5)^(2)=0.05` `:.f+-Deltaf=5+- =0.5` |
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| 77. |
There is a spherical glass ball of refractive index `mu_(1)` and another glass ball of refractive index `mu_(2)` inside as shown in figure . The radius of the outer ball is `R_(1)` and that of inner ball is `R_(2)`. A ray is incident on outer surface of the ball at an angle `i_(1)`, Find the value of `r_(1)`A. `sin^(-1)((sini_(1))/(mu_(1)))`B. `sin^(-1)(mu_(1) sini_(1))`C. `sin^(-1)((mu_(1))/("sin"i_(1)))`D. `sin^(-1)((1)/(mu_(1)"sin"i_(1)))` |
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Answer» Correct Answer - A `mu_(1) "sin"r_(1) = "sin"i_(1) implies r_(1) = sin^(-1)(("sin"i_(1))/(mu_(1)))` |
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| 78. |
Let `f_(v)` and `f_(r)` are the focal lengths of a convex lens for violet and red lights respectively. If `F_(v)` and `F_(r)` are the focal lengths of a concave lens for violet and red light respectively, thenA. `f_(v) lt f_(r)` and `F_(v) gt F_(r)`B. `f_(v) lt f_(r)` and `F_(v) lt F_(r)`C. `f_(c) gt f_(r)` and `F_(v) gt F_(r)`D. `f_(v) gt f_(r)` and `F_(v) lt F_(r)` |
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Answer» Correct Answer - B According to lens makers formula `(1)/(f)=(mu-1)((1)/(R_(1))+(1)/(R_(2)))implies(1)/(f)prop (mu-1)` Since `mu_(Red) lt mu_("violet") rArr f_(v) lt f_(r)` and `F_(v) lt F_(r)` Always keep in mind that whenever you asked to compare ( greater than or less than ) `u,v ` or f you must not apply sign conventions for comparison. |
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| 79. |
A ray of light falls on a transparent sphere with center at C as shown in Figure . The ray emerges fromk the sphere paralllel to line AB. Find the refractive index of the sphere. A. `sqrt(2)`B. `sqrt(3)`C. `3//2`D. `1//2` |
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Answer» Correct Answer - B Deviation by a sphere is `2(i-r)` Here deviation `delta=60^(@)=2(i-r)` or `i-r=30^(@)` `:.r=1-30^(@)=60^(@)-30^(@)=30^(@)` `:. mu=(sin i)/(sin r)=(sin 60^(@))/(sin 30^(@))=sqrt(3)` |
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| 80. |
A thin made of glass of refractive index `1.5` has a front surface `+11 D` power and back surface `-6D` . If this lens is submerged in a liquid of refractive index `1.6`, the resulting power of the lens isA. `-0.5 D`B. `+0.5 D`C. `-0.625 D`D. `+0.625D` |
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Answer» Correct Answer - C Total power`P=P_(1)+P_(2)=11-6=5D` Also `(f_(1))/(f_(a))=((._(a)mu_(g)-1))/((._(I)mu_(g)-1))implies(P_(a))/(P_(I))=((._(a)mu_(g)-1))/((._(I)mu_(g)-1))` `implies(5)/(P_(I))=((1.5-1))/((1.5//1.6-1))impliesP_(I)=-0.625D` |
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| 81. |
An isosceles right angled triangular glass `(mu = 1.6)` prism has a cavity inside it in the shape of a thin convex lens whose both surfaces have radius of curvature equal to 20 cm. The cavity has been filled with a transparent liquid of refractive index 2.4. S is a point source and `sigma` is an opaque sheet having a small hole such that the source and the hole both lie on the principal axis of the lens. The small hole in the opaque sheet is just to ensure that only paraxial rays are incident on the optical system. However, the size of hole is large enough to neglect diffraction effects. Will the observers at P and Q be able to see the image of source S? Where is the image located? |
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Answer» Correct Answer - Observer at Q sees the image at a distance 6.88 cm from face BC, inside the prism |
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| 82. |
A hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquids `L_(1)` or `L_(2)` having refractive indices `n_(1)` and `n_(2)`, respectively `(n_(2) gt n_(1) gt 1)`. The lens will diverge parallel beam of light if it is filles withA. air and placed in airB. air and immersed in `L_(1)`C. `L_(1)` and immersed in `L_(2)`D. `L_(2)` and immersed in `L_(1)` |
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Answer» Correct Answer - d. If the refractive index of the material of the lens is greater thatn the refractive index of the surrounding medium, then a concave lens would behave as a concave lens. |
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| 83. |
A hollow double concave lens is made of very thin transparent material. It can be filled with air or either of two liquid `L_(1)` and `L_(2)` having refractive indices `mu_(1)` and `mu_(2)` respectively `(mu_(2)gtmu_(1)gt1)`. The lens will diverge a parallel beam of light if it is filled withA. air and placed in airB. air and immersed in `L_(1)`C. `L_(1)` and immersed in `L_(2)`D. `L_(2)` and immersed in `L_(1)` |
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Answer» Correct Answer - d. We know that `(1)/(f)=((mu_(2))/(mu_(1))-1)((1)/(R_(1))-(1)/(R_(2)))` For divergence, `mu_(2)gtmu_(1)` Here, `((1)/(R_(1))-(1)/(R_(2)))` is negative. Therefore, (d) is the correct option. |
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| 84. |
The refractive index of the material of a prism of refracting angle `45^@` is `1.6` for a certain monochromatic ray. What will be the minimum angle of incidence of this ray on the prism so that no TIR takes place as the ray comes out of the prism. |
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Answer» Correct Answer - `10.1^(@)` |
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| 85. |
An isosceles glass prime ( refractive index `=(3)/(2)`) has its base just submerged in water ( refractive index `= (4)/(3)`) . The base of the prism is horizontal. A horizontal light ray AB is incident on the prism and takes a path shown in figure to emerge out of the prism. Find the maximum value of base angle `theta` of the prism for which total internal reflection can take place at the base. |
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Answer» Correct Answer - `theta_(max) = cos^(-1) (sqrt((17)/(21)))` |
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| 86. |
A glass sphere with `10 cm` radius has a `5 cm` radius spherical hole at its centre. A narraow beam of parallel light is directed into the sphere. Where, if anywhere, will the sphere produce an image? The index of refraction of the glass is `1.50.` |
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Answer» Correct Answer - 5 cm of the left of the surface of sphere |
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| 87. |
When an object is kept at a distance of 30cm from a concave mirror, the image is formed at a distance of 10 cm. If the object is moved with a speed of `9cm s^(-1)` the speed with which the image moves isA. `0.1ms^(-1)`B. `1ms^(-1)`C. `3ms^(-1)`D. `9ms^(-1)` |
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Answer» Correct Answer - b. `(1)/(v)+(1)/(u)=(1)/(f)` `-(du)/(u^(2))-(dv)/(v^(2))=0` or `-(dv)/(u^(2))-(du)/(v^(2))` or `dv=-(v^(2))/(u^(2))du` `=-(10xx10)/(30xx30)xx9ms^(-1)=-1ms^(-1)` |
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| 88. |
The image produced by a concave mirror is one-quarter the size of object. If the object is moved 5cm closer to the mirror, the image will only be half the size of the object. The focal length of mirror isA. `f=5.0cm`B. `f=2.5cm`C. `f=7.5cm`D. `f=10cm` |
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Answer» Correct Answer - b. In first case, if object distance is x, mage distance`=x//4` . While it is `2^(nd)` case, object distance becomes`(x-5cm)` and image distance `(x-5cm)//2`. Using mirror formula, we get In first case, `(1)/(f)=-(5)/(x)` In second case, `(1)/(f)=-(3)/((x-5))` Solving we get, `x=12.5cm` and `f=2.5cm` and hence `|f|=2.5cm` |
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| 89. |
An ideal gas is heated from termperature `T_(1)` to `T_(2)` under various conditions. The correct statements(s) is/are:-A. `DeltaU = nC_(V)(T_(2)-T_(1))` for isobaric , isochoric and adiabatic processB. Work is done at expense of internal energy in an adiabatic process and both have equal valuesC. `DeltaU = ` for an isothermal processD. `C=0` for an adiabatic process |
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Answer» Correct Answer - A::B::C::D For any process `DeltaU = nC_(V)DeltaT` In adiabatic process `Q = DeltaU + W = 0 implies DeltaU = -W` For any process `C=(DeltaU)/(nDeltaT) + (W)/(nDeltaT) = (Q)/(nDelta) = (DeltaU)/(nDeltaT) +(P)/(n)((DeltaV)/(DeltaT))` For `Q = 0 ,C=0` (adibatic process) |
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| 90. |
Total radiations incident on body = `400 J`, `20%` radiation reflected and `120` J absorbs. Then find out `%` of transmittive power |
| Answer» ` Q=Q_(t)+Q_(r) + Q_(a) implies 400=80+120+Q_(t)implies Q_(t)=200 " "` So `%` of transmittive power is `50%` | |
| 91. |
An object is immersed in a fluid. In order that the object becomes invisible, it shouldA. behave as a perfect reflectorB. have refractive index exactly matching with that of the surrounding fluidC. absorb all light falling on itD. have refractive index one |
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Answer» Correct Answer - B Objects are invisible in liquid of `R.I.` equal to that of object. |
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| 92. |
An object is immersed in a fluid. In order that the object becomes invisible, it shouldA. behave as a perfect reflectionB. absorb all light falling on itC. have refractive index oneD. have refractive index exactly matching with that of the surrounding fluid |
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Answer» Correct Answer - D An object is seen when reflected light from object enters have into eyes when refractive index exactly same then no reflection from the object. |
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| 93. |
A ray of light is incident upon an air/water interface ( it passes from air into water) at an angle of `45^(@)` . Which of the following quantities change as the light enters the water? (I) wavelength `" "` (II) frequency `" "` (III) speed of propagation (IV) direction of propagationA. I , III onlyB. III, IV onlyC. I , II ,IV onlyD. I , III , IV only |
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Answer» Correct Answer - D As Speed of light in vaccum/air `mu` = Speed of light if medium `mu = (c)/(v) implies ` velocity is different in different medium From v= n`lambda` But frequency is the fundamental property . It never change by changing the medium hence `lambda` is also changed as v is change. |
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| 94. |
Look at the ray diagram shown ,what will be th focal length of th e`1^(st)`and the `2^(nd)`lens,if the incident light ray passes without any deviation? |
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Answer» Correct Answer - C From figure `1^(st)` behave as diverging and `2^(nd)` behave converging In `1st` incident ray appear to pass at 5 cm distance point and after refraction it becomes parallel to principal axis i.e. the focal length = -5cm and for second f=5 cm |
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| 95. |
A man of height H is standing in front of an inclined plane mirror at an angle `theta` from horizontal. The vertical separation between man and inclined plane is x. Man can see its complete image in length `(H(H+x)"sin"theta)/(H +2x)` of mirror . (Given : x = H = 1m and `theta = 45^(@)`) If man starts moving with velocity `1 ms^(-1)` along vertical, find out length of mirror at t = 1s in which he can see his complete image.A. `(3)/(10)m`B. `(3sqrt(2))/(5)m`C. `(3)/(5sqrt(2))`D. `(3)/(5)`m |
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Answer» Correct Answer - C In this case only x will change so required length of mirror =`(1(1+2)"sin"45^(@))/(1+2xx2)` =`(3 xx (1)/(sqrt(2)))/(5) = (3)/(5sqrt(2))m` |
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| 96. |
A man of height H is standing in front of an inclined plane mirror at an angle `theta` from horizontal. The vertical separation between man and inclined plane is x. Man can see its complete image in length `(H(H+x)"sin"theta)/(H +2x)` of mirror . (Given : x = H = 1m and `theta = 45^(@)`) If man starts moving with velocity `sqrt(2)ms^(-1)` along inclined plane, find out length of mirror at t = 1s in which he can see his complete image,A. `(sqrt(2))/(3)m`B. `(2)/(3)`C. `(1)/(3)m`D. `(1)/(3sqrt(2))m` |
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Answer» Correct Answer - A In this case x, H and `theta` remains same so required length of mirror = `(H(H+x)"sin"theta")/(H + 2x)` =`(1(1+1)"sin"45^(@))/(1 + 2 xx1) = (sqrt(2))/(3)m` |
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| 97. |
The distance of an object from the pole of a concave mirror is equal to its radius of curvature. The image must be:A. realB. invertedC. same sizedD. erect |
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Answer» Correct Answer - A If object is at centre then image forms on the centre, and real and if object is virtual then image forms at `V=-R/3` (in both the cases image is real) |
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| 98. |
For a normal eye, the cornea of eye provides a converging power of `40 D` and the least converging power of the eye lens behind the cornea is `20 D`. Using this information, the distance between the retina and the cornea eye lens can be estimated to beA. `5 cm`B. `2.5cm`C. `1.67cm`D. `1.5 cm` |
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Answer» Correct Answer - C `P_(eff)=40D+20D=60D` `f=(100)/(P_(eff))` |
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| 99. |
A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by appropriately placingA. A convex mirrorr of suitable focal lengthB. A concave mirrorr of suitable focal lengthC. A concave lengs of suitable focal lengthD. A convex lens of suitable focal length less than `0.25 m` |
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Answer» Correct Answer - D Convex mirrorr and concave lens do not form real image. For concave mirrorr , `vgtu`, so image will be enlarged, hence only convex lens can be used for the purpose. |
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| 100. |
A beam of light of wavelength 600nm from a distance source falls on a single slit 1mm wide and a resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark frings on either side of central bright fringe isA. 1.2cmB. 1.2mmC. 2.4cmD. 2.4mm |
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Answer» Correct Answer - d. The distance between the first dark frigne on either side of the central bright fringe `=` width of central maximum `=(2Dlambda)/(a)` `=(2xx2xx600xx10^(-9))/(10^(-3))=2.4xx10^(-3)m=2.4mm.` (d) is the correct option. |
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