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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.
| 151. |
An ideal monoatomic gas undergoes a cyclic process `ABCA` as shown in the figure. The ratio of heat absorbed during `AB` to the work done on the gas during `BC` id A. `(5)/(2ln2)`B. `(5)/(3)`C. `(5)/(4ln2)`D. `(5)/(6)` |
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Answer» Correct Answer - C `W_(AB) = (2V_(0) - V_(0))P_(0) = P_(0)V_(0)` [isobaric process] `W_(BC) = |nRT(2T_(0))ln"(V_(0))/(2V_(0))| = 2P_(0)V_(0)ln2` [Isothermal process] `therefore (Q_(AB))/(W_(BC)) = ((3)/(2)P_(0)V_(0)+ P_(0)V_(0))/(2P_(0)V_(0)ln2) = (5)/(4ln2)` |
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| 152. |
A certain prism is found to produce a minimum of`38^(@)`.It produces a deviation of `44^(@)`when the angle of incident is either`42^(@)`or`62^(@)` .What is the angle of incidence when it is undergoing minimum deviation?A. `45^(@)`B. `49^(@)`C. `40^(@)`D. `55^(@)` |
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Answer» Correct Answer - B `delta_("min")= 38^(@), delta = 44^(@), i =42^(@)` `implies e = 62^(@)` `delta` = i + e -A, 44 = 42+62-A `implies A = 60^(@)` `therefore delta_("min") = 2i-A , 38 = 2i-60` `implies i = 49^(@)` |
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| 153. |
Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same velocity V. The mass of the gas in A is `m_A,` and that in B is `m_B`. The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The changes in the pressure in A and B are found to be `DeltaP and 1.5 DeltaP` respectively. ThenA. `4m_(A) = 9m_(B)`B. `2m_(A) = 3m_(B)`C. `3m_(A) = 2m_(B)`D. `9m_(A) = 4m_(B)` |
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Answer» Correct Answer - C Initial conditions `P_(1)V = n_(1)RT, P_(2)V= n_(2)RT` Final condition `(P_(1) - DeltaP)2V = n_(1)RT` `(P_(2) - 1.5DeltaP)2V = n_(2)RT implies (n-_(1)RT)/(2V) = DeltaP` `(n_(2)RT)/(2V) = 1.5DeltaP implies (n_(1))/(n_(2)) = (1)/(1.5) = (2)/(3) implies (m_(A))/(m_(B)) = (2)/(3)` |
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| 154. |
An object `3 cm` tall is placed on the principal axis of a concave mirrorr of focal length `9 cm` at a distance of `12 cm` form is. What is the nature and size of the image ?A. real, `9cm`B. virtual, `9 cm`C. real , `1cm`D. virtual, `1 cm` |
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Answer» Correct Answer - A `m=(-9)/(-9-(-12))=(-9)/(3)` or `m=-3` or `(I)/(O)=-3`or `I=-3xx3cm` or ` I=-9cm` The negative sign indicates that the image is real. |
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| 155. |
A `2.0 cm` tall object is placed `15 cm` in front of concave mirrorr of focal length `10 cm`. The size and nature of the image will beA. `1.0 cm, `realB. `4 cm, `virtualC. `4 cm`, realD. none of these |
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Answer» Correct Answer - C Using to mirrorr formula, `(1)/(v)+(1)/(u)=(1)/(f)` `(1)/(v)=(1)/(f)-(1)/(u)=(1)/(-10)-(1)/(-15)=(1)/(-10)+(1)/(15)` or `v=-30 cm` The image is formed at a distance of `30 cm` from the mirrorr on the same side of the object. It is a real image. Lateral magnification of the concave mirrorr, `m=(-v)/(u)=(h_(1))/(h_(o))` `(-(30))/(-15)=(h_(1))/(2)impliesh_(1)=-4 cm.` Negative sign shows that image is inverted. The image is real, inverted, of size `4 cm` at a distance of `30 cm` in front of the mirrorr. |
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| 156. |
A converging system of convex lenses free from chromatic aberration and of focal length 2.5 cm is to be constructed by using a convex lens of focal length 2 cm and dispersive power 0.04 and another convex lens of dispersive power 0.03. What should be the focal length of the second lens and at what distance from the first lens should it be placed ? |
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Answer» Correct Answer - 3cm |
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| 157. |
The figure shows the P-V plot of an ideal gas taken through a cycle ABCDA. The part ABC is a semi-circle and CDA is half of an ellipse. Then, A. the process during the path `A rarr B ` is isothermalB. heat flows out of the gas during the path `B rarr C rarrD`C. work done during the path `A rarr B rarr C` is zeroD. positive work is done by the gas in the cycle `ABCDA` |
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Answer» Correct Answer - B::D (i) For isothermal process curve should be hyperbola. (ii) Work done & change in internal energy are both negative. (iii) For higher pressure volume is increasing and lower pressure volume is decreasing . |
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| 158. |
A container of fixed volume has a mixture of a one mole of hydrogen and one mole of helium in equilibrium at temperature T. Assuming the gasses are ideal, the correct statement (s) is (are)A. The average energy per mole of the gas mixture is `2RT`.B. The ratio of speed of sound in the gas mixture to that in helium gas is `sqrt(6//5)`.C. The ratio of the rms speed of helium atoms to that of hydrogen molecules is `1//2`.D. The ratio of the rms speed of helium atoms to that of hydrogen molecules is `1//sqrt(2)` |
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Answer» Correct Answer - A::B::D `C_(V("mix"))= ((1)/((3)/(2)R) + (1)((5)/(2)R))/(2) = 2R` `C_(P("mix")) = 3R " " gamma_("mix") = (3)/(2) implies f =4` Average energy/mole = `f(1)/(2)RT= 2RT` `((V_("sound"))"mixture")/((V_("sound"))"He") = (sqrt(RT)/(2))/(sqrt(5RT)/(12)) = sqrt(6)/(5)` |
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| 159. |
Assertion `:` The images formed by total internal reflections are much brighter than those formed by mirrorrs or lenses. Reason `:` There is no loss of intensity in total internal reflection.A. Statement-I is true , Statement -II is true, Statement-II is correct explanation for Statement-I.B. Statement - I is true, Statement -II is true: Statement -II is NOT a correct explanation for statement -I .C. Statement-I is true, Statement-II is false.D. statement-I is false, statement- II is true. |
| Answer» Correct Answer - A | |
| 160. |
Assertion `:` The images formed by total internal reflections are much brighter than those formed by mirrorrs or lenses. Reason `:` There is no loss of intensity in total internal reflection.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - A In total internal reflection, `100%` of incident light is reflected back into the same medium, and there is no loss of intensity, while in reflection from mirrorrs and refraction from lenses, there is always some loss of intensity. Therefore images formed by total internal reflection are much brighter than those formed by mirrorrs or lenses. |
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| 161. |
Assertion `:` The focal length of the lens does not change when red light is replaced by blue light. Reason`:` The focal length of lens does not depends on colour of light used.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason both are false. |
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Answer» Correct Answer - D Focal length of the lens depends upon its refractive index as `(1)/(f)prop(mu-1). Since `mu_(b)gtmu_(r)` so `f_(b)ltf_(r)` Therefore, the focal length of a lens decreases when red light is replaced by blue light. |
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| 162. |
A car is fitted with a convex side-view mirror of focal length 20 cm. A second car 2.8m behind the first car is overtaking the first car at a relative speed of 15 `m/s`. The speed of the image of the second car as seen in the mrror of the first one is:A. `5.79cm, 6.9 cm`B. `6.45 cm, 5.16 cm`C. `2.7 cm, 4.8 cm`D. `0.1 m, 0.3m` |
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Answer» Correct Answer - B `m=-(v)/(u)=-(600)/(31)xx(1)/(-600)=(1)/(31)` Breadth of image `=(1)/(31)xx200cm =6.45 cm` Height of image `=(1)/(31)xx160cm= 5.16cm`. |
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| 163. |
A car is fitted with a convex side-view mirror of focal length 20 cm. A second car 2.8m behind the first car is overtaking the first car at a relative speed of 15 `m/s`. The speed of the image of the second car as seen in the mrror of the first one is:A. `19.35 cm`B. `17.45 cm`C. `21.48cm`D. `15.49 cm` |
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Answer» Correct Answer - A `(1)/(v)+(1)/(-600)=(1)/(20)` or `(1)/(v)=(31)/(600)` or `v-(600)/(31)cm=19.35cm.` |
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| 164. |
When monochromatic red light is used instead of blue light in a convex lens, its focal length will :-A. Does not depend on colour of lightB. increase the squeeze of the lensC. DecreaseD. Remain same |
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Answer» Correct Answer - B Focal length of a lens is given by `(1)/(f) = (mu-1) ((1)/(R_(1)) - (1)/(R_(2)))` `(1)/(f) propmu prop(1)/(lambda) implies f prop lambda` `lambda_(R) gt lambda_(B) implies f_(R) gt f_(B)` Focal length will increase |
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| 165. |
A car is fitted with a convex side-view mirror of focal length 20 cm. A second car 2.8m behind the first car is overtaking the first car at a relative speed of 15 `m/s`. The speed of the image of the second car as seen in the mrror of the first one is:A. 10 m/sB. 15 m/sC. `(1)/(10) m//s`D. `(1)/(15) m//s` |
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Answer» Correct Answer - D Velocity of image = `-m^(2)V_(0) = - ((f)/(f-u))^(2) V_(0)` = `-((20)/(20 -(-280)))^(2) (15) = -(1)/(15) m//s` |
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| 166. |
A telescope has an objective of focal length `50 cm` and eye piece of focal length `5 cm`. The least distance of distinct vision is `25 cm`. The telescope is focussed for distinct vision on a scale `200 cm` away from the objective. Calculate (i) the separation between objective and eye piece (ii) the magnification produced. |
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Answer» Given `u=-200cm, f=50cm` For image `I_(1)` object formed by objective lens, `(1)/(f)=(1)/(v)-(1)/(u)` we have `(1)/(v)=(1)/(f)+(1)/(u)=(1)/(50)+(1)/(200)=(4-1)/(200)=(3)/(200)` `rArr v=+(200)/(3) cm` Also, magnification produced by objective lens `m_(0)=(v)/(u)=-(200//3)/(200)=(1)/(3)` Image `I_(1)` acts as an object for eye lens. Here, `v=-25cm, f=5 cm` `:. (1)/(f)=(1)/(v)-(1)/(u)` `rArr (1)/(u)=(1)/(v)-(1)/(f)=-(1)/(25)-(1)/(5)=-(1+5)/(25)` `:. u=-(25)/(6)cm` And magnification produced by eye lens, `m_(e)=(v)/(u)=(-25)/((-25//6))=6` a. The separation between objective and eyepiece `=|V|+|u|=(200)/(3)+(25)/(6)=(425)/(6)=70.73cm` b. Magnification produced, `m=m_(0)xxm_(e)=-(1)/(3)xx6=-2` The negative sign show that the final image is inverted. |
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| 167. |
With a simple microscope, if the final image is located at the least distance of distinct of distinct vision (i.e. D) from the eye placed close to the lens, then the magnifying power isA. `D/f`B. `1+D/f`C. `f/D`D. `fxxD` |
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Answer» Correct Answer - B `1+D/1` |
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| 168. |
An observer views his own image in a convex mirror of radius of curvature R. If the least distance of distinct vision for the observer is d, calculate the maximum possible magnification. |
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Answer» Correct Answer - `(R)/(sqrt(R^(2)+d^(2))+d)` |
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| 169. |
An equiconcave diverging lens of focal length F is cut into two equal halves. The two halves are turned around and joined with some liquid between them.The lens obtained is converging with focal length F. If the refractive index of the liquid is 3 then what is the refractive index of the lens? A. 4B. 2C. 5D. 1.5 |
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Answer» Correct Answer - b. For initial diverging lens, `(1)/(-F)=(mu-1)((2)/(-R))` `rArr (1)/(F)=((mu-1)2)/(R)` (i) For final combination: `(1)/(F) =(1)/(f_(1))+(1)/(f_(2))+(1)/(f_(3))` `rArr (2(mu-1))/(R)=(mu-1)[-(1)/(R)]` `+(mu_(liq)-1)(2)/(R)+(mu-1)[-(1)/(R)]` `rArr mu=(mu_(liq)+1)/(2)=(3+1)/(2)=2` |
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| 170. |
A parallel beam of light falls normally on the first face of a prism of small refracting angle. At the second refracting face it is partly reflected and partly transmitted. The reflected light strikes the first face again and emerges from it making an angle of `4°` with the reversed direction of incident beam. The deviation suffered by refracted ray is `1°` from original direction of incident ray. Find the refractive index of glass of the prism and the angle of the prism. |
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Answer» Correct Answer - `mu = 2, A = 1^(@)` |
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| 171. |
Figure shows a glass `(mu_(g) = 1.5)` vessel, partly filled with water `( mu_(w) = (4)/(3))` . A ray of light is incident normally on the water surface and passes straight through. The vessel is tilted slowly till angle `theta` such that the light ray is emergent grazing the lower surface of the glass. Find `theta`. |
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Answer» Correct Answer - `theta = sin^(-1) ((3)/(4))` |
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| 172. |
A convex lens made of glass `(mu_(g)=3//2)` has focal length f in air. The image of an object placed in front of it is inverted, real and magnified. Now the whole arrangement distance between object and lens. ThenA. the new focal length will becom 4fB. the new focal length will become `f//4`C. new image will be virtual and magnifiedD. new image will be real, inverted and smaller in size |
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Answer» Correct Answer - a.,c. `(1)/(f_(air))=((3)/(2)-1)((1)/(R_(1))-(1)/(R_(2)))` `(1)/(f_(water))=((3//2)/(4//3)-1)((1)/(R_(1))-(1)/(R_(2)))` From these two equations we get, `f_(water)=4f_(air)=4f` In air image was inverted, real and magnified. Therefore, object was lying between f and 2f. Now the focal length has changed to 4f. Therefore, the object now lies between pole and focus. Hence, the new image will be virtual and magnified. |
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| 173. |
A convex mirror of focal length f produced an image `(1//n)^(th)` of the size of the object. The distance of the object from the mirror isA. `(n-1)f`B. `((n-1)/(n))f`C. `((n+)/(n))f`D. `(n+1)f` |
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Answer» Correct Answer - A `m=+(1)/(n)=-(v)/(u)impliesv=-(u)/(n)` By using formula `(1)/(f)=(1)/(-u/n)+(1)/(u)` `impliesu=-(n-1)f` |
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| 174. |
A luminous object is placed 20 cm from surface of a convex mirror and a plane mirror is set so that virtual images formed in two mirrors coincide. If plane mirror is at a distance of 12cm from object. Then focal length of convexA. 20cmB. 15cmC. 10cmD. 5cm |
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Answer» Correct Answer - d. For convex mirror, `u+v=12xx2=24cm` (because for plane mirror, distance of object `=` distance of image). Also, here for the convex mirror `u=20cm`, therefore `v=4cm` . Hence, using `(1)/(v)+(1)/(u)=(1)/(f)` We find `f=5cm` |
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| 175. |
Consider the situation shown in figure. Water `(mu_(W) = (4)/(3))` is filled in a breaker upto a height of 10 cm . A plane mirror fixed at a height of 5 cm from the surface of water. Distance of image from the mirror after reflection from it of an object O at the bottom of the beaker is A. 15 cmB. 12.5 cmC. 7.5 cmD. 10 cm |
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Answer» Correct Answer - B Initially observer is mirror for image formation of object O. As O is kept at the bottom , then it sees somewhat above the bottom. `therefore` Distance = `5 + (10)/(mu) = 5 + (10xx3)/(4)` = 12.5 cm (in front ) Hence image from the mirror after reflection is also at `12.5` cm, |
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| 176. |
The n transparent slabs of refractive index1.5 each having thickness 1cm, 2 cm,…to n cm are arranged one over another. A point object is seen through this combineation with near perpendicular light. If the shift of object by the combination is 1 cm, then find the value of n. |
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Answer» `x= `total shift `x=x_(1)+x_(2)+…+x_(n)` `=t_(1)(1-(1)/(mu))+t_(2)(1-(1)/(mu))+...+t_(n)(1-(1)/(mu))` or `1cm =(1-(2)/(3))1+2+3+…+n` `1=(n(n+1))/(6)` `6=n(n+1)rArrn=2` |
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| 177. |
At what distance from a convex mirror of focal length 2.5 m should a body stand so that his image has a height equal to half the original height ? The principal exis is perpendicular to the height. |
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Answer» We have `m = -(upsilon)/(u) = (1)/(2)` or , `upsilon = (u)/(2).` Also, `(1)/(u) + (1)/(upsilon) = (1)/(f)` or , `(1)/(u) + (1)/(-u//2) = (1)/(2.5m)` or `- (1)/(u) = (1)/(2.5 m)` or u = - 2.5m. |
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| 178. |
Consider the situation shown in figure. Water `(mu=4//3)` is filled in a beaker upto a height of 10cm. A plane mirror is fixed at a height of 5 cm from the surface of water. Distance of image from the mirror after reflection from it of an object O at the bottom of the beaker is A. 15cmB. 12.5cmC. 7.5cmD. 10cm |
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Answer» Correct Answer - b. Distance of first image `(I_(1))` formed after refraction from the plane surface of water is `(10)/(4//3)=7.5cm` from water surface `(d_(app)=(d_(actual))/(mu))`. Now, distance of this image is `5+7.5=12.5cm` from the plane mirror. Therefore, distance of second image `(I_(2))` will also be equal to 12.5 cm from the mirror. |
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| 179. |
Consider the situation shown in figure. A plane mirror is fixed at a height h above the bottom of a beaker containing water (refractive index `mu`) up of a bottom formed by the mirror. |
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Answer» The bottom of the beker appears to be shifted up by a distance `/_ =(1-1/mu)d` Thus, the apparent distance of the bottom from the mirror is `h/_ =h-(1-1/mu)d=h-d+d/mu`. The image is formed behind the mirror at a distance `h-d+d/mu` |
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| 180. |
Consider the situation shown in Figure. A plane mirror is fixed at a height h above the bottom of a beaker containing water (refractive index `mu`) upto a height d. Find the position of the image of bottom formed by the mirror. |
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Answer» The bottom of the beaker appears to be shifted up by a distance `Deltat=(1-(1)/(mu))d` Thus, the apparent distance of the bottom from the mirror is `h-Deltat=h-[1-(1//mu)]^(d)=h-d+(d//mu).` The image is formed behind the mirror at a distance `h-d+(d//mu).` |
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| 181. |
A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300K. The ratio of the average rorational kinetic energy per `O_2` molecules to that per `N_2` molecules isA. `1:1`B. `1:2`C. `2:1`D. depends on the moment of inertia of the two molecules |
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Answer» Correct Answer - A Average rotational `KE = 2 xx (1)/(2) kT` (for diatomic gas) |
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| 182. |
The ratio of the speed of sound in nitrogen gas to that in helium gas, at 300K isA. `sqrt(((2)/(7)))`B. `sqrt(((1)/(7)))`C. `sqrt((3))/(5)`D. `sqrt(6)/(5)` |
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Answer» Correct Answer - C `because V_(sound) = sqrt((gammaRT)/(M_(W)))` `therefore (V_(N_(2)))/(V_(He)) = sqrt((gamma_(N_(2)))/(M_(N_(2)))xx (M_(He))/(gamma_(He))) = sqrt((7)/(5) xx (1)/(28) xx (4)/(1) xx (3)/(5))= sqrt(3)/(5)` |
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| 183. |
I is a ray incident on a plane mirror. Keeping the incident ray fixed, the mirror is rotated by an angle `theta` about an axis passing through A perpendicular to the plane of the Fig. show that the refracted ray rotates through an angle `2theta`. Does your answer differ if the mirror is rotated about an axis passing through B? |
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Answer» Correct Answer - No |
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| 184. |
In the figure (a) the light is incident at an angle `theta` (slightly greater than the critical angle). Now keeping the incident ray fixed a parallel slab of refractive index `n_3` is placed on surface AB.(fig b) A. total internal reflection occurs at AB for `n_(3)=n_(2)`B. total internal reflection occurs at AB for `n_(3) gt n_(1)`C. the ray will return back to the same medium for alll values of `n_(3)`D. total reflection occurs at CD for `n_(3) lt n_(1)` |
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Answer» Correct Answer - C for combination of `n_(1)` and `n_(2)` crtical angle is `theta` when be places parallel slab of refractive index `n_(3)(n_(3) lt n_(1))`. Then crtical angle decreases and T.I.R. takes place for any value of `n_(3)` the ray will be return back in the medium `n_(2)` and reflection takes place at the surface CD. |
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| 185. |
A thin converging lens forms a real image of an object located far away from the lens. The image is formed at A at a distance 4x from the lens and height of the image is h. A thin diverging lens of focal length x is placed at `B [PB = 2x]` and a converging lens of focal length 2x is placed at `C [PC = 3x]`. The principal axes of all lenses coincide. Find the height of the final image formed. |
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Answer» Correct Answer - 2h |
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| 186. |
A person is suffering from myopic defect. He is able to see clear objects placed at `15 cm`. What type and of what focal length of lens he should use to see clearly the object placed `60 cm` away ?A. Concave lens of `20 cm` focal lengthB. Convex lens of `20 cm` focal lengthC. Concave lens of `12 cm` focal lengthD. Convex lens of `12 cm` focal length |
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Answer» Correct Answer - A For viewing far objects, concave lenses are used and for concave lens `u=` wants to see `=-60cm,v=` can see `=-15 cm` so from `(1)/(f)=(1)/(v)-(1)/(u)impliesf=-20cm.` |
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| 187. |
A man stands on vertical tower of height 20 cm. Calculate the distance up to which he will be able to see on the surface of the earth. Neglect the height of the man. Take the radius of the earth `= 6400` km |
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Answer» Correct Answer - 16 km |
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| 188. |
Wavelength of light used in an optical instrument are `lambda_(1) = 4000 Å` and `lambda_(2) = 5000Å` then ratio of their respective resolving powers (corresponding to `lambda_(1)` and `lambda_(2)`) isA. `16 : 25`B. `9:1`C. `4:5`D. `5:4` |
| Answer» Correct Answer - D | |
| 189. |
A ray of light passes through four transparent media with refractive indices`mu_1`,`mu_2`,`mu_3` and `mu_4` as shown in the figure. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have A. `mu_(1)=mu_(2)`B. `mu_(2)=mu_(3)`C. `mu_(3)=mu_(4)`D. `mu_(4)=mu_(1)` |
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Answer» Correct Answer - D For successive refraction through different media `mu sin theta=` constant. Here as `theta` is same in two extreme media `, mu_(1)=mu_(2)` |
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| 190. |
If the central portion of a convex lens is wrapped in black paper as shown in figure A. No image will be formed by the remaining portion of the lensB. The full image be formed but it will be less brightC. The central portion of the image will be missingD. There will be two images each produced by one of the exposed portions of the lens. |
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Answer» Correct Answer - B Since aperture of lens reduces so brightness will reduce but there will be no effect on size of image |
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| 191. |
A convex lens of focal length 20cm is placed 10cm away from a second convex lens of focal length 25 cm. What wil be the location of the image of an object at 30 cm in front of the first lens? |
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Answer» Let the image from the first lens be formed at a distance y to the right of lens. Refraction at the first lens: Here, `u=-3cm, f=+20cm, v=+y` From the lens equation, we get `(1)/(v)-(1)/(u) =(1)/(f_(1))` or `(1)/(y)-(1)/(-30) =(1)/(20)or y=60cm` Refraction at the second lens: Here, `u=60-10=+50cm, f=+25cm` Let the image be at a distance z to the right of lens2. Therefore, `v=+z. ` Substituting in the lens equation, we get `(1)/(v)-(1)/(u) =(1)/(f_(2))` or`(1)/(z)=(1)/(25) +(1)/(50) or z=(50)/(3) cm` |
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| 192. |
A pin is placed 10cm in front of a convex lens of focal length 20cm, made of a material having refractive index 1.5 . The surface of lens farther away from the pin is silvered and has a radius of curvature 22cm. Determine the position of the final image. Is the image real or virtual? |
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Answer» As radius of curvature of silvered surface is 22cm, so `f_(M)=(R)/(2)=(-22)/(2)=-11cm=-0.11m` And hence, `P_(M)=-(1)/(f_(M))-(1)/(-0.11)=(1)/(0.11)D` Further as the focal length of lens is 20cm, i.e., 0.20m, its power will be given by `P_(L)=(1)/(f_(L))=(1)/(0.20)D` Now as in image formation, light after passing through the lens will be reflected back by the curved mirror through the lens again `P=P_(L)+P_(M)+P_(L)=2P_(L)+P_(M)` i.e., `p=(2)/(0.20)+(1)/(0.11)=(210)/(11)D` So the focal length of equivalent mirror `F=-(1)/(P)=-(11)/(210)m=-(110)/(21)cm` i.e., the silvered lens behaves as a concave mirror of focal lneght `(110//210)` cm. So for object at a distance 10cm in front of it, `(1)/(v)+(1)/(-10)=-(21)/(110)i.e., v=-11cm` i.e., image will be 11 cm in front of th esilvered lens and will be real as shown in the figure. |
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| 193. |
A convex lens of focal length 10cm is painted black at the middle portion as shown in figure . An object is placed at a distance 20cm from the lens. Then, A. only one image will be formed by the lensB. the distance between the two images formed by such a lens is 6mmC. the distance between the images is 4mmD. the distance between the images is 2mm |
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Answer» Correct Answer - a. Only one image will be formed by this lens system because optic axis of both the parts coincide. Two images would have formed if their optic axis were different. |
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| 194. |
A ray of light is incident on a glass slab at grazing incidence. The refractive index of the material of the slab is given by `mu=sqrt((1+y)` . If the thickness of the slab is `d=2m`, determing the equation of the trajectory of the ray inside the slab and the coordinates of the point where the ray exits from the slab. Take the origin to be at the point of entry of the ray. |
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Answer» We have `(dy)/(dx)=cot(theta_(y))=sqrt(mu_(y)^(2)-sin^(2)theta)/(sintheta)` Here, `mu=sqrt((1+y))` and `theta=90^(@)` Therefore, `(dy)/(dt)=y^(1//2)` Integratin with the boundary condition that `y=0` at `zx=0`, we get `y=x^(2)//4` to be the equation of the path of the ray through the slab, the ray will obviously exit at the point `(2sqrt2m,2m)`. |
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| 195. |
A point object is placed at a distance of 25 cm from a convex lens of focal length 20cm. If a glass slab of thickness t and refractive index 1.5 is inserted between the lens and the object, the image is formed at infinity. The thickness t isA. 10cmB. 5cmC. 20cmD. 15cm |
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Answer» Correct Answer - d. Image will be formed at infinity if object is placed at focus of the lens, i.e., at 20cm form the lens. Hence, shift `=25-20=(1-(1)/(mu))mu` or `5=(1-(1)/(1.5))t ` or `t=(5xx1.5)/(0.5)=15cm` |
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| 196. |
What is the angle of incidence for an equilateral prism of refractive index `sqrt(3)` so that the ray si parallel to the base inside the prism?A. `30^(@)`B. `45^(@)`C. `60^(@)`D. Either `30^(@)` or `60^(@)` |
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Answer» Correct Answer - c. `sqrt(3)=sin((60^(@)+delta_(m))/2)/(sin((60^(@))/(2)))` `sqrt(3)=sin((60^(@)+delta_(m))/(2))` `sin60^(@)=sin((60^(@)+delta_(m))/(2))` or `(60^(@)+delta_(m))/(2)=60^(@)` or `delta_(m)=60^(@)rArri=(A+delta_(m))/(2)=(60^(@)+60^(@))/(2)=60^(@)` |
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| 197. |
An equilateral prism deviates a ray through `40°` for two angles of incidence. The two incidence angels differ by `20°`. Find their values. |
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Answer» Correct Answer - `60^(@)` and `40^(@)` |
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| 198. |
A parallel glass slab of refractive index `sqrt3` is placed in contact with an equilateral prism of refractive index `sqrt2`. A ray is incident on left surface of slab as shown. The slab and prism combination is surrounded by air. The magnitude of minimum possible deviation of this ray by slab-prism combination is: A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `60^(@)-sin^(-1)sqrt((2)/(3))` |
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Answer» Correct Answer - A |
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| 199. |
An object 240 cm in front of a lens forms a sharp image on a screen 12 cm behind the lens. A glass slab 1 cm thick, having refractive index 1.50 is placed between the lens and the screen with its refracting faces perpendicular to the principal axis of the lens. (a) By how much distance the object must be moved so as to again cast a sharp image on the screen? (b) Another identical glass slab is interposed between the object and the lens. How much further the object shall be moved so as to form a sharp image on the screen. |
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Answer» Correct Answer - (a) 320 cm (b) `(1)/(3)cm` |
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| 200. |
A ray of light undergoes deviation of `30^(@)` when incident on a n equilateral prism of refractive index `sqrt(2)`. The and made by the ray inside the prism with the base of the prism is `"________________"`. |
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Answer» For minimum deviation, `mu=sin(A+(delta_(m))/(2))/(sin((A)/(2)))=("sin"((60+30)/2))/(sin(60/2))=(sin45^(@))/(sin30^(@))=sqrt(2)` Since `mu=sqrt(2)` (given for the prism), the condition is for minimum deviation. In this case, the ray iniside the prism becomes parallel to the base. Therefore, the angle made by the ray inside the prism with the base of the prism is zero |
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