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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.
| 101. |
The specific heat of a metal at low temperature varies according to `S= (4//5)T^(3)` where `T` is the absolute temperature. Find the heat energy needed to raise unit mass of the metal from T = 1 K ` to `T= 2K`. |
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Answer» Correct Answer - 3 `Q= intmSdT = (mT^(4))/(5) implies (Q)/(m) = (15)/(5) = 3` |
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| 102. |
An object placed `10 cm` in front of a lens has an image `20 cm` behind the lens. What is the power of the lens`(` in dioptres `)` ?A. `1.5`B. `3.0`C. `-15.0`D. `+15.0` |
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Answer» Correct Answer - D `u=-10cm, v=20cm` `(1)/(f)=(1)/(v)-(1)/(u)=-(1)/(20)-((-1)/(10))=(3)/(20)impliesf=(20)/(3)cm` Now, `F=(100)/(f)=(100)/(20//3)=+15D` |
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| 103. |
A concave mirror forms an image of the sun at a distance of 12 cm from it.A. the radius of curvature of this mirror is 6cmB. to use it as a shaving mirro, it can be held at a distance of 8-10cm from the faceC. it an object is kept at a distance of 24cm from it, the image formed will be of the same size as the objectD. all the above alternative are correct. |
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Answer» Correct Answer - b.,c. Image of sun will be formed at the focus of mirro. Hence, focal length of mirro `=12cm`. To use it or shaving purpose, face should be between focus and pole, so that virtual image can be formed. |
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| 104. |
A concave lens of focal length `(1)/(3)m` forms a real, inverted image twice in size of the object. The distance of the object from the lens isA. `0.5 m`B. `0.166m`C. `0.33 m`D. `1m ` |
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Answer» Correct Answer - A `m=(m)/(f+u)` `-2=(1)/((3)/((1)/(3)+u))` `-(2)/(3)-2u=(1)/(3)` or `-2u =(1)/(3)+(2)/(3)=1` or`u=-(1)/(2)m=-0.5m` |
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| 105. |
A Concave lens of focal length `20 cm` produces an image half in size of the focal object. The distance of the real object isA. `20cm`B. `30 cm`C. `10 cm`D. `60 cm` |
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Answer» Correct Answer - A `m=(f)/(f+u)` `(1)/(2)=(-20)/(-20+u)` or `-20+u=-40` or `u=-40 +20` or `u=-20cm`. 10cm
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| 106. |
If in a planoconvex lens, the radius of curvature of the convex surface is `10cm` and the focal length is `30 cm`, the refractive index of the material of the lens will beA. `1.5`B. `1.66`C. `1.33`D. 3 |
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Answer» Correct Answer - C `(1)/(30)=(mu-1)((1)/(10))` or `mu-1=(1)/(3)` or `mu=1+(1)/(3)=(4)/(3)=1.33` |
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| 107. |
In which case does a light ray pass through the centre of a thin lens without deviation? |
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Answer» Correct Answer - when media on both sides of the lens is same |
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| 108. |
A cylindrical tube has a length of 60 cm. Three identical convex lenses, each of focal length `f = 10` cm are fixed inside the tube, one at each of the ends and one at the centre. One end of the tube is placed 10 cm away from a point source. The device casts an image of the object. How much does the image shift when the tube is moved away from the source by 10 cm. |
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Answer» Correct Answer - The image does not move |
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| 109. |
A convex lens of focal length 1.0m and a concave lens of focal length 0.25m are 0.75m apart. A parallel beam of light is incident on the convex lens. The beam emerging after refraction from both lenses isA. Parallel to principle axisB. ConvergentC. DivergentD. None of the above |
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Answer» Correct Answer - A Power of system `=(1)/(f_(1))+(1)/(f_(2))-(f)/(f_(1)f_(2))=(1)/(1)+(1)/(-0.25)-(0.75)/((1)(-0.25))` `=1-4+3=-3+3=0` Since power of the system is zero therefore, the incident parallel beam of light will remain parallel after emerging from the system. |
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| 110. |
A horizontal parallel beam of light passes through a vertical convex lens of focal length f. The optical centre of the lens is P. A small plane mirror is placed at point M inclined at `60°` to the axis of the lens. Distance `PM = f//2`. The mirror reflects the light passing through the lens and forms an image at point I. Find distance PI. |
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Answer» Correct Answer - `f//2` |
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| 111. |
A horizontal parallel beam of light passes though a vertical convex lens of focal length 40 cm. Behind the lens there is a plane mirror making an angle `theta` with the principal axis of the lens. The mirror intersects the principal axis at M. Distance between the optical centre of the lens and point M is `OM = 20 cm`. The light beam reflected by the mirror converges at a point P. Distance OP is 20 cm. Find `theta`. |
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Answer» Correct Answer - `60^(@)` |
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| 112. |
100g of ice at `0^(@)` is mixed with 100g of water at `100^(@)C`. What will be the final temperature (in K ) of the mixture? |
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Answer» Heat required by ice to raise its temperature to `100^(@)C`, `Q_(1) = m_(1)L_(1) + m_(1)c_(1)Deltatheta_(1)= 5xx80+ 5 xx 1xx100= 400+500 = 900"cal"` Heat given by the steam when condensed `Q_(2)= m_(2)L_(2)=5xx536=2680"cal"` As `Q_(2)gtQ_(1)`. This means that whole steam is not even condensed. Hence temperature of mixture will remain at `100^(@)C`. |
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| 113. |
540 g of ice at `0^(@)C` is mixed with 540 g of water at `80^(@)C`. The final temperature of the mixture isA. `0^(@)C`B. `40^(@)C`C. `80^(@)C`D. less than `0^(@)C` |
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Answer» Correct Answer - A Heat taken by ice to melt at `0^(@)C "is" Q_(2) = "m"L = 540xx80 = 43200"cal"` Heat given by water to cool upto `0^(@)C "is" Q_(2) = "ms"Deltatheta = 540 xx 1 x (80-0) = 43200"cal"` Hence heat given by water is just sufficient to melt the whole ice and final temperature of mixture is `0^(@)C`. |
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| 114. |
Two fines steel wires , fastened between the projectors of a heavy brass bar, are just taut when the whole system is at `0^(@)C`. What is the tensile stress in the steel wires the temperature of the system is raised by `200^(@)C`? (`alpha_("glass") = 2 xx 10^(-5) ^(@)C ^(-1), alpha_("steel") = 1.2 xx 10^(-5)"^(@)C^(-1), Y_("steel") = 200GNm^(-2)`) A. `3.2 Nm^(-2)`B. `3.2 xx 10^(8) Nm^(-2)`C. `32 xx 10^(8) Nm^(-2)`D. `0.48 Nm^(-2)` |
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Answer» Correct Answer - B Tensile Stress = `(Y_(steel))((Deltal)/(l)) = Y_(S) (alpha_(b)-alpha_(S))DeltaT` =` 200 xx 10^(9)(0.8 xx 10^(-5)) (200) = 3.2 xx 10^(8) Nm^(-2)` =`0.32 GNm^(-2)` |
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| 115. |
A fine steel wire of length `4`m is fixed rigidly in a heavy brass frame as ashown in figure . It is just taut at `20^(@)C`. The tensile stress developed in steel wire it whole system is heated to `120^(@)C` is :- (`Given alpha _("brass") = 1.8 xx 10^(-5) .^(@)C^(-1), alpha_("steel") = 1.2 xx 10^(-5) .^(@)C, Y_("steel") = 2 xx 10^(11) Nm^(-2), Y_("brass")= 1.7xx10^(7) Nm^(-2)`) A. `1.02 xx 10^(4) Nm^(-2)`B. `1.2 xx 10^(8) Nm^(-2)`C. `1.2 xx 10^(6) Nm^(-2)`D. `6 xx 10^(8) Nm^(-2)` |
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Answer» Correct Answer - B Stress = `Y` (strain) = `Y_(S)(alpha_(b)-alpha_(s))DeltaT= (2xx10^(11)) (0.6xx10^(-5)) (100) = 1.2 xx 10^(8) Nm^(-2)` |
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| 116. |
A parallel beam of light falls on a solid tranparent sphere. Q. If however thin beam is focussed at A, then find the refractive index of the sphere,A. 1.5B. 1.7C. `2.0`D. 2.5 |
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Answer» Correct Answer - c. `(mu)/(y)-(1)/(oo)=(mu-1)/(R)rArr(muR)/(mu-1)` If beam is focussed at A, than ` y=2R` `rArr 2R=(muR)/(mu-1)=mu=2` The above is valid for paraxial rays, hence beam should be thin. If `mugt2` , then `ylt2R` , hence beam can be focussed before A. For thin beam to be focussed at A, `y=R`. `rArr R=(muR)/(mu-1)rArr mu=oo` (not possible) |
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| 117. |
A parallel beam of light falls on a solid tranparent sphere. Q. For what value of refractive index `mu`, the thin beam can be focussed at centre of sphere.A. 1.5B. 2C. 2.5D. none of these |
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Answer» Correct Answer - d. `(mu)/(y)-(1)/(oo)=(mu-1)/(R)rArr(muR)/(mu-1)` If beam is focussed at A, than ` y=2R` `rArr 2R=(muR)/(mu-1)=mu=2` The above is valid for paraxial rays, hence beam should be thin. If `mugt2` , then `ylt2R` , hence beam can be focussed before A. For thin beam to be focussed at A, `y=R`. `rArr R=(muR)/(mu-1)rArr mu=oo` (not possible) |
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| 118. |
A ray of light traveling in a tranparent medium falls on a surface separating the medidum from air at an angle of incidence of `45^(@)` . The ray undergoes total internal refrlection. If n is the refractive index of the medium with respect to air, select the possible value (s) of n from the following:A. 1.3B. 1.4C. 1.5D. 1.6 |
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Answer» Correct Answer - c.,d. For total internal reflection to take place: Angle of incidence, `i gt` critical angle, `theta_(c) ["where" sintheta_(c)=(1)/(n)]` or `sin45^(@) gt (1)/(n)` or `(1)/(sqrt(2)) gt (1)/(n)` or `gt sqrt(2)` or n gt 1.414` Therefore, possible values of n can be 1.5 or 1.6 in the given options. |
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| 119. |
A student performed the experiment of determination of focal length of a concave mirror by `u-v` method using an optical bench of length 1.5 meter. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of `(u,v)` values recorded by the student (in cm) are: `(42,56),(48,48),(60,40),(66,33),(78,39)` . The data set (s) that cannot come from experiment and is (are) incorrectly recorded, is (are)A. `(42,56)`B. `(48,48)`C. `(66,33)`D. `(78,39)` |
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Answer» Correct Answer - c.,d. `(1)/(f)=(1)/(v)+(1)/(u)` (mirror formula) `f=-24cm` |
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| 120. |
A thermal insulated vessel contains some water at `0^(@)C`. The vessel is connected to a vaccum pump to pum out water vapour. This results in some water getting frozen. It is given latent heat of vaporization of water at `0^(@)C = 21 xx 10^(5) J//kg` and latent heat of freezing of water `=3.36 xx 10^(5) J//kg`. the maximum percentage amount of water vapour that will be solidified in this manner will be:A. `86.2%`B. `33.6%`C. `21%`D. `24.36%` |
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Answer» Correct Answer - A Let x = percentage of water solidified then heat lost = Heat gained `implies x xx 3.36 xx 10^(5) = (100 - x) xx 21 xx 10^(5)` `implies x= (100)/(1.16) = 86.2%` |
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| 121. |
The pressure of an ideal gas varies according to the law `P = P_(0) - AV^(2)`, where `P_(0)` and `A` are positive constants. Find the highest temperature that can be attained by the gasA. `(2P_(0))/(3R) (P_(0)/(3a))^(1//2)`B. `(3P_(0))/(2R) (P_(0)/(3a))^(1//2)`C. `(P_(0))/(R) (P_(0)/(3a))^(1//2)`D. `(P_(0))/(3R) (P_(0)/(3a))^(1//2)` |
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Answer» Correct Answer - A ` P= P_(0) -aV^(2)` From ideal gas equation `PV = nRT` `implies Rt = (P_(0)- aV^(2))V(n=1`) `implies T = ((P_(0)V - aV^(3))/(R)) implies (dT)/(dV) = 0 = ((P_(0) - 3aV^(2))/(R))` `implies V= sqrt((P_(0))/(3a))` and `(d^(2)T)/(dV^(2)) = -(6aV)/(R) (lt0)`. `therefore T_(max) = ((P_(0)-aV^(2))V)/(R) = (2P_(0))/(3R)sqrt((P_(0))/(3a))` |
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| 122. |
Infrared radiation is detected byA. spectrometerB. pyrometerC. nanometerD. photometer |
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Answer» Correct Answer - B Infrared radiations are detected by pyrometer. |
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| 123. |
Two rods of equal cross sections area are joined end the end as shown in figure. These are supported between two rigid vertical walls. Initially the rods are unstrained . If temperature of system is increased by `DeltaT` then junction will not shift if -A. `Y_(1)alpha_(1) = Y_(2)alpha_(2)`B. `Y_(1)alpha_(1) l_(1) = Y_(2)alpha_(2) l_(2)`C. `alpha_(1) = alpha_(2)`D. `Y_(2) alpha_(1) l_(2) = Y_(1)alpha_(2) l_(2)` |
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Answer» Correct Answer - B Stress developed at junction are same `Y_(1)alpha_(1)l_(1)DeltaT = Y_(2)alpha_(2)l_(2)DeltaT implies Y_(1)alpha_(1)l_(1) = Y_(2)alpha_(2)l_(2)` |
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| 124. |
Two rods of equal cross sections area are joined end the end as shown in figure. These are supported between two rigid vertical walls. Initially the rods are unstrained . If temperature of system is increased by `DeltaT` then shifting in junction if junction if `Y_(1)alpha_(1) gt Y_(2)alpha_(2)` is given by -A. `(l_(1)l_(2)(Y_(1)alpha_(2) - Y_(2)alpha_(1)))/(Y_(1)l_(1) + Y_(2)l_(2))`B. `(l_(1)l_(2)(Y_(1)alpha_(1) - Y_(2)alpha_(2)))/(Y_(1)l_(2) + Y_(2)l_(1))`C. `(l_(1)l_(2)(Y_(1)alpha_(1) - Y_(2)alpha_(2)))/(Y_(1)l_(1) + Y_(2)l_(2))`D. None of these |
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Answer» Correct Answer - B Let shifting in Junction be x towards right then `((Deltal)/(l))_(1) = (l_(1)alpha_(1)DeltaT-x)/(l_(1)) , ((Deltal)/(l))_(2) = (l_(2)alpha_(2)DeltaT + x)/(l_(2))` But `Y_(1) ((Deltal)/(l))_(1) = Y_(2)((Deltal)/(l))_(2)` So x = `(l_(1)l_(2)(Y_(1)alpha_(1) - Y_(2)alpha_(2))DeltaT)/(Y_(1)l_(2) + Y_(2)l_(1))` |
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| 125. |
Two rods of equal cross sections area are joined end the end as shown in figure. These are supported between two rigid vertical walls. Initially the rods are unstrained . If temperature of system is increased by `DeltaT` then thermal stress developed in first rod-A. is equal to thermal stress developed in second rodB. is greater than thermal stress developed in second rodC. is less than thermal stress developed in second rodD. None of these |
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Answer» Correct Answer - A As cross sectional area is same & equal and opposite force acting on both rods. So `F//A` = same |
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| 126. |
Three rods of equal length of same material are joined to form an equivalent triangle `ABC` as shown figure. Area of cross-section of rod `AB` is `S` of rod `BC` is `2S` and that of `AC` is `S` , then `{:(,"Column-I",,"Column-II"),((A),"Temperature of junction" B,(p),"Greater than" 50^(@)C),((B),"Heat current in" AB,(q),"Less than" 50^(@)C),((C),"Heat current in BC",(r),"Is equal to current in BC"),(,,(s),"Is " 2/3 "times heat current in AC"),(,,(t),"None"):}` |
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Answer» Let `R_(BC) = R` then `R_(AB) = R_(AC) = 2R` as `R = (l)/(kA)` `(100-T_(B))/(2R) = (T_(B)-0)/(R) implies T_(B) = 67.7 ^(@)C` `((DeltaQ)/(Deltat))_(AB) = ((DeltaQ)/(Deltat))_(BC)` and `((DeltaQ)/(Deltat))_(AB)= (2)/(3)((DeltaQ)/(Deltat))_(AC)` |
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| 127. |
A single ray traverses a glass plate (thickness = t) with plane surfaces that are parallel to each other. The emergent ray is parallel to the incident ray but suffers a lateral displacement d. Assuming that glass plate (refractive index `mu`) is placed in air, find the dependence of d on angle of incidence i. Plot the variation of d with i (changing from `0°` to `90°`) |
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Answer» Correct Answer - `d = t sin i[1-(cosi)/(sqrt(mu^(2)-sin^(2)i))]` |
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| 128. |
A point object `O` is placed on the principal axis of a convex lens of focal length `f=20cm` at a distance of 40 cm to the left of it. The diameter of the lens is 10. An eye is placed 60 cm to right of the lens and a distance `h` below the principal axis. The maximum value of `h` to see the image is |
| Answer» Correct Answer - B | |
| 129. |
Calorie is defined as the amount of heat required to raise temperature of 1 g of water by `1^@C` and it is defined under which of the following conditions?A. From `14.5^(@)C "to" 15.5 ^(@)C "at" 760 "mm of Hg"`B. From `98.5^(@)C "to" 99.5 ^(@)C "at" 760 "mm of Hg"`C. From `13.5^(@)C "to" 14.5 ^(@)C "at" 76 "mm of Hg"`D. From `3.5^(@)C "to" 4.5 ^(@)C "at" 76 "mm of Hg"` |
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Answer» Correct Answer - A `1` calorie is the heat required to raise the temp. fo `1` g of water from `14.5 "to" 15.5 ^(@)C "at" 760"mm of Hg"`. |
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| 130. |
The speed of sound in oxygen `(O_2)` at a certain temperature is `460ms^-1`. The speed of sound in helium (He) at the same temperature will be (assume both gases to be ideal)A. `460sqrt((200)/(21))ms^(-1)`B. `500sqrt((200)/(21))ms^(-1)`C. `650sqrt(2)ms^(-1)`D. `300sqrt(2)ms^(-1)` |
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Answer» Correct Answer - A `v=sqrt((gammaRT)/(M)) implies (v_(1))/(v_(2)) = sqrt((gamma_(1)M_(2))/(gamma_(2)M_(1))) = sqrt(((7)/(5)xx4)/((5)/(3)xx 32))` `implies (460)/(v_(2)) = sqrt((21)/(25xx8)) implies v_(2) = 460sqrt((200)/(21))m//s` |
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| 131. |
One kg of a diatomic gas is at pressure of `8xx10^4N//m^2`. The density of the gas is `4kg//m^3`. What is the energy of the gas due to its thermal motion?A. `6 xx 10^(4)J`B. `7 xx 10^(4)J`C. `3 xx 10^(4)J`D. `5 xx 10^(4)J` |
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Answer» Correct Answer - C Energy of the diatomic gas `(5)/(2)nRT = (5)/(2)PV = (5)/(2)xx 8 xx 10^(4) xx (1)/(4) = 5 xx 10^(4)J` |
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| 132. |
A beam of light composed of red and green ray is incident obliquely at a point on the face of rectangular glass slab. When coming out on the opposite parallel face, the red and green ray emerge formA. Two points propagating in two different directionsB. Two points propagating in two parallel directionsC. One point propagating in two different directionsD. One point propagating in the same directions |
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Answer» Correct Answer - B Angle of refraction will be different , due to which red and green emerge from different points and will be parallel. |
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| 133. |
A telescope has an objective lens of `10cm` diameter and is situated at a distance of one kilometre from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is `5000 Å`, of the order ofA. `0.5m`B. `5m`C. `5 mm`D. `5 cm` |
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Answer» Correct Answer - C Resolving limit of telescope is `theta prop (x)/(D)=(lambda)/(d)implies x=(lambdaD)/(d)` Given, `lambda=5000 Å=5000xx10^(-1)m` `D=1km=1000m` `d=10cm=0.1m` Hence, `x=(5000xx10^(-10)xx1000)/(0.1)` `=5xx10^(-3)m=5mm` |
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| 134. |
The focal lengths of the lenses of an astronomical telescope are `50cm` and `5cm`. The length of the telescope when the image is formed at the least distance of distinct vision isA. `45 cm`B. `55 cm`C. `(275)/(6)cm`D. `(325)/(6)cm` |
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Answer» Correct Answer - D Length of the telescope when final image is formed at least distance of distinct vision is `L=f_(o)+u_(e)=f_(o)+(f_(e)D)/(f_(e)+D)=50+(2xx25)/(5+25)=(325)/(6)cm` |
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| 135. |
In a terrestrial telescope, the focal length of objective is `90cm`, of inverting lens is `5 cm` and of eye lens is `6 cm`. If the final image is at `30cm`, then the magnification will be the final image is at `30cm, ` then the magnification will beA. 21B. 12C. 18D. 15 |
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Answer» Correct Answer - C `m=(f_(o))/(f_(e))(1+(f_(e))/(D))impliesm=(90)/(6)(1+(6)/(30))impliesm=18` |
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| 136. |
The thermal capacity of 40 g of aluminium (specific heat `=0.2 cal//gm^(@)C`)A. The molar mass of the gas is `40`gB. The molar mass of the gas cannot be determined from the data givenC. The number of degrees of freedom of the gas molecules is `6`D. The number of degrees of freedom of the gas molecules is `8` |
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Answer» Correct Answer - A::C `C_(P)-C_(V) =R, M (0.25 - 0.15) = 2[M`= molar mass] ` implies M = (2)/(0.05) = 40`g `(C_(P))/(C_(V)) = gamma = (0.2)/(0.15) = (4)/(3) =1 + (2)/(f)` `implies` f = degrees of freedom = `6` |
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| 137. |
One mole of an ideal monoatomaic gas is taken from `A` to `C` along the path `ABC` . The temperature of the gas at `A` is `T_(0)` . For the process `ABC` : A. Work done by the gas is `RT_(0)`B. Change in internal energy of the gas is `(11)/(2)RT_(0)`C. Heat absorbed by the gas is `(11)/(2)RT_(0)`D. Heat absorbed by the gas is `(13)/(2)RT_(0)` |
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Answer» Correct Answer - A::C Work done = Area of `ABC` with `V`- axis =`P_(0)(2V_(0)-V_(0)) + 0 = P_(0)V_(0) = nRT_(0) = RT_(0)` Change in internal energy = `nC_(V)DeltaT` ` 1 xx (3)/(2)R xx (4T_(0)-T_(0)) = (9)/(2)RT_(0)` `therefore` Heat absorbed = `(9)/(2) RT_(0) + RT_(0) = (11)/(2)RT_(0)` |
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| 138. |
A system `S` receives heat continuously from an electric heater of power `10 W`. The temperature of `S` becomes constant at `50^(@)C` when the surrounding temperature is `20^(@)C`. After the heater is switched off, `S` cools from `35.1^(@)C` to `34.9^(@)C` in `1 minute`. the heat capacity of `S` isA. `100J//"^(@)C`B. `300J//"^(@)C`C. `750J//"^(@)C`D. `1500J//"^(@)C` |
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Answer» Correct Answer - D Rate of cooling = ms`(-(d theta)/(dt)) = 4 sigma AT_(0)^(3) DeltaT` `implies 4sigmaA xx T_(0)^(3)(50-20) = 10` and `4sigmaA xx T_(0)^(3)(35-20) = "ms"(-(d theta)/(dt)) = ms((0.2)/(60))` `implies ms= (60)/(0.2) xx ((15 xx10)/(30)) = 1500J//^(@)C` |
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| 139. |
Prism angle of a prism is `10^(@)` ,Their refractive index for red & violet color is `1.51 & 1.52` respectively .Then dispersive power will be `(1) 0.5, (2)0.15, (3)0.019,(4) 0.032` |
| Answer» Dispersive power of prism `delta = ((mu_(v) - mu_(r))/(mu_(y) - 1)) "but" mu_(y)= (mu_(v)+ mu_(r))/(2) = (1.664-1.644)/(1.654-1)= 0.0305` | |
| 140. |
If the focal length of the objective lens is increased thenA. Magnifying power of the microscope will increase but that of telescope will decreaseB. Magnifying power of microscope and telescope both will increaseC. Magnifying power of microscope and telescope both will decreaseD. Magnifying power of microscope will decrease but that of telescope will increase |
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Answer» Correct Answer - D A microscope consists of lens of small focal lengths . A telescope consists of objective lens of large focal length. |
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| 141. |
The magnifying power of a microscope with an objective of `5 mm` focal length is 400. The length of its tube is `20cm.` Then the focal length of the eye`-` piece isA. `200 cm`B. `160 cm`C. `2.5 cm`D. `0.1 cm` |
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Answer» Correct Answer - C If nothing is said then it is considered that final image is formed at infinite and `moo((L_(oo)-f_(o)-f_(e)).D)/(f_(o)f_(e))~~(LD)/(f_(o)f_(e))` `implies400=(20xx25)/(0.5xxf_(e))impliesf_(e)=2.5cm.` |
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| 142. |
A man with normal near point (25 cm) reads a book with small print using a magnifying glass : a thin convex lens of focal length `5 cm`. (a) What are the closest and the farthest distances at which he can read the book when viewing through the magnifying glass ? (b) What is the maximum and the minimum angular magnifications (magnifying powers) possible using the above simple microscope ? |
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Answer» (a) As for normal eye far and near point are `oo` and `25` cm respectively, so for magnifier `v_(max) -oo "and" v_(min) = -25"cm"` . However , for a lens as `(1)/(v) - (1)/(u) = (1)/(f) implies u = (f)/((f//v)-1)` So u will be minimum when v= minimum = `-25 "cm i.e.," (u)_(min) = (5)/(-(5//25)-1) = -(25)/(6) = -4.17 "cm"` Ans u will be maximum when v= maximum =`oo` i.e., `u_(max) = (5)/(((5)/(oo)-1)) = -5` cm So the closest and farthest distance of the book from the magnifier (or eye) for clear viewing are `4.17 "cm and " 5` cm respectively. |
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| 143. |
A thin convex lens of focal length `5` cm is used as a simple microscope by a person with normal near point `(25"cm")` . What is the magnifying power of the microscope.? |
| Answer» Here , `f=5"cm" , D=25"cm" , M = ? MP = 1+ (D)/(f) = 1 + (25)/(5) = 6` | |
| 144. |
A parallel beam of monochromatic light is incident at one surface of a equilateral prism. Angle of incidence is `55^(@)` and angle of emergence is `46^(@)`. The angle of minimum deviation will beA. Less than `41^(@)`B. Equal to `41^(@)`C. More than `41^(@)`D. None of the above |
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Answer» Correct Answer - A By the hypothesis, we know that `i_(1)+i_(2)=A+deltaimplies55^(@)+46^(@)=60^(@)+deltaimplies41^(@)` But `delta_(m) lt delta`, so `delta_(m) lt 41^(@)` |
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| 145. |
A car is travelling at night along a highway shaped like a parabola with its vertex at the origin of the co-ordinate system. The car starts at a point 200 m West and 200 m North of the origin and travels in easterly direction. There is an animal standing 200 m East and 100 m North of the origin. At what point on the highway will the car’s headlight illuminate the animal? |
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Answer» Correct Answer - `X = 58.6 m; Y = 17.17 m` |
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| 146. |
The far point of a myopia eye is at `40 cm`. For removing this defect, the power of lens required will beA. `40D`B. `-4D`C. `-2.5 D`D. `0.25D` |
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Answer» Correct Answer - C For myopic eye `f=-`(defected far point ) `impliesf=-40cmimpliesP=(100)/(-40)=-2.5D` |
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| 147. |
A point source of light `S` is placed at the bottom of a vessel containing a liquid of refractive index `5//3`. A person is viewing the source from above the surface. There is an opaque disc of radius `1 cm` floating on the surface. The centre of disc lies vertically above the source `O`. The liquid from the vessel is gradually drained out through a tap. What is the maximum height of the liquid for which the source cannot be seen at all. |
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Answer» Correct Answer - 1.33 cm |
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| 148. |
A green light is incident from the water to the air - water interface at the critical angle `(theta)`. Select the correct statement.A. The spectrum of visible light whose frequency is more than that of various angles to the normalB. The entire spectrum of visible light will come out of the water at angle of `90 ^(@)` to the normalC. The entire spectrum of visible will come out of the water at angle of `90^(@)` to the normal.D. The spectrum of visible light whose frequency is less than that of green light will come out to the air medium. |
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Answer» Correct Answer - D Frequency of light (v) `gt` frequency of green light `(V_(G))` `mu` is also greater than `mu_(G)` and critical angle of light is less than green light therefore light will got total internal reflaction and not come out to the air . For frequency of light `(v) lt V_(G)` , light will not suffer T.I.R . Therefore light comes out to the air |
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| 149. |
If light travels a distance `x` in `t_(1)` sec in air and `10x` distance in `t_(2)` sec in a medium, the critical angle of the medium will beA. `tan^(-1)((t_(1))/(t_(2)))`B. `sin^(-1)((t_(1))/(t_(2)))`C. `sin^(-1)((10t_(1))/(t_(2)))`D. `tan^(-1)((10t_(1))/(t_(2)))` |
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Answer» Correct Answer - C From the formula `sin C=(1)/(._(1)mu_(2))impliessin C= ._(2)mu_(1)` `=(u_(1))/(u_(2))=(v_(2))/(v_(1))impliessin C=(10x//t_(2))/(x//t_(1))` |
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| 150. |
The critical angle for light going from medium X into medium Y is `theta` . The speed of light in medium X is v. The speed of light in medium Y isA. `v/(sin theta)`B. `v sin theta`C. `v cot theta`D. `v tan theta` |
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Answer» Correct Answer - A `n_(A) sin theta=n_(B) sin 90^(@)` `V=C/(n_(A))` `V_(B)=C/(n_(B))` `V_(B)=C/(n_(A) sin theta)=V/(sin theta)` |
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