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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.
| 451. |
A clock pendulum made of invar has a period of `2` s at `20^(@)C`. If the clock is used in a climate where average temperature is `40^(@)C`, what correction (in seconds ) may be necessary at the end of `10` days to the time given by clock? `(alpha_("invar") = 7 xx 10^(-7) "^(@)C, 1` "day" =`8.64xx10^(4)s`) |
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Answer» Correct Answer - 6 `T=2pisqrt((l)/(g)) implies (DeltaT)/(T) = (1)/(2) (Deltal)/(l) (1)/(2)(alphaDeltatheta) implies DeltaT = (T)/(2)(alphaDeltatheta)= ((10xx8.64 xx 10^(4))/(2)) (7xx10^(-7)) (20) = 6s` |
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| 452. |
A cylinder of cross-section area. A has two pistons of negligible mass separated by distances l loaded with spring of negligible mass. An ideal gas at temperature `T_(1)` is in the cylinder where the springs are relaxed. When the gas is heated by some means its temperature becomes `T_(2)` and the springs get compressed by `(l)/(2)` each . if `P_(0)` is atmospheric pressure and spring constant k=`(2P_(0)A)/(l)`, then find the ratio of `T_(2)` and `T_(1)` . |
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Answer» Correct Answer - 4 `(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2)` where `V_(1) = lA` and `V_(2) = 2lA` and `P_(1)=P_(0)` and `P_(2)= (kx)/(A) + P_(0) =2P_(0) implies (T_(2))/(T_(1)) = (P_(2)V_(2)/(P_(1)V_(1)) = 4` |
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| 453. |
A carbot freezer takes heat from water at `0^(@)C` inside it and rejects it to the room at a temperature of `27^(@)C`. The latent heat of ice is `336xx10^(3)Jkg^(-1)`. If 5kg of water at `0^(@)C` is converted into ice at `0^(@)C` by the freezer, then the energy consumed by the freezer is close to :A. `1.67 xx 10^(5) J`B. `1.68 xx 10^(6) J`C. `1.71 xx 10^(7) J`D. `1.51 xx 10^(5) J` |
| Answer» Correct Answer - A | |
| 454. |
A body cools from `50^(@)C` to `49.9^(@)C` in `5` sec. it cools from `40^(@)C` to `39.9^(@)C` in t sec. Assuming Newtons law of cooling to be valid and temperature of surrounds at `30^(@)C`, value of t//5 will be? |
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Answer» Correct Answer - 2 From `(theta_(1)-theta_(2))/(t) = k (theta_(1)+theta_(2)/(2)- theta_(0))` We have `(0.1)/(5) = "k"(39.9)/(2)` and `(0.1)/(t) =k(19.9)/(2) implies (t)/(5) = (39.9)/(19.9) = 2 implies t = 10 implies (t)/(5) = 2` |
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| 455. |
The ratio of work done by an ideal diatomic gas to the heat supplied by the gas in an isobatic process isA. `(2)/(3)`B. `(3)/(2)`C. `(2)/(5)`D. `(3)/(5)` |
| Answer» Correct Answer - C | |
| 456. |
Statement-`1`: On sudden expansion a gas cools. Statement -`2` : On sudden expansion , no heat is supplied to system and hence gas does work at the expense of its internal energy.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 | |
| 457. |
In figure shown, left arm of a `U-`tube is immersed in a hot water bath at temperature ` t^(@)C`, and right arm is immersed in a bath of melting ice, the height of manometric liquid in respective column is `h_(1)` and `h_(0)`. Determine the coefficient of expansion of the liquid. |
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Answer» The liquid is in hydrostatic equilibrium ` implies rho_(t) gh_(t) = rho_(0)gh_(0)` Where, ` rho_(t)` is density of liquid in hot bath, `rho_(0)` is density of liquid in cold bath. Volumes of a given mass M of liquid at temperature t and `0^(@)C` are ralated by `V_(t) = V_(0) (1+gammat) "Since" rho_(t)V_(t) = rho_(0) V_(0) implies rho_(t) = (rho_(0)V_(0))/(V_(t)) = (rho_(0))/((1+gammat))` Since ` h_(t) = (rho_(0)h_(0))/(rho_(t)) = h_(0)(1+gammat)` which on solving for ` gamma `, yields `gamma = ((h_(t)-h_(0)))/(h_(0)t) ` |
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| 458. |
Two moles of helium gas are taken over the cycle ABCDA, as shown in the P-T diagram The net work done on the gas in the cycle ABCDA is :A. `1076R`B. `1904R`C. ZeroD. `276R` |
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Answer» Correct Answer - D `W_(ABCDA) = W_(AB) + W_(DC) + W_(CD) + W_(DC)` =`muR(500-300) + 2.303muR(500)"log"(2)/(2) + muR(300-500) + 2.303muR(300)"log"(1)/(2) = 276R` Work done on the gas = `-276R` |
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| 459. |
A plane mirror is moving with velocity `4 (hat i) + 4 (hat j) + 8 (hat k)`. A point object in front of the mirror moves with a velocity `3 (hat i) + 4 (hat j) + 5 (hat k)`. Here, `hat k` is along the normal to the plane mirror and facing towards the object. The velocity of the image isA. `-3hati -4hatj + 5hatk`B. `3hati +4hatj + 11hatk`C. `-3hati -4hatj + 11hatk`D. `7hati +4hatj + 11hatk` |
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Answer» Correct Answer - B `vecV_(M) = 4hati + 5 hatj + 8hatk , vecV_(0) = 3hati + 4hatj + 5hatk` `because` Plane of the mirror is xy `therefore vecV_(I)= 3hati+4hatj` and for `veckv_(12)` =`2V_(MZ) - V_(OZ) = 2xx8 -5=11` `implies vecV_(I) = 3hati +4hatj + 11hatk` |
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| 460. |
For a compound microscope, the focal length of object lens and eye lens are `f_(o)` and `f_(e)` respectively, then magnification will be done by microscope whenA. `f_(o)=f_(e)`B. `f_(o) gt f_(e)`C. `f_(o) lt f_(e)`D. None of these |
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Answer» Correct Answer - C Magnification will be done by compound microscope only when `f_(o) lt f_(e)` |
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| 461. |
Which of the following is used in optical fibres?A. Total internal reflectionB. ScatteringC. DiffractionD. Refraction |
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Answer» Correct Answer - A The working of optical fibres is based on the phenomenon of total internal reflection, |
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| 462. |
The objective lens of a compound microscope produces magnification of 10. In order to get an overall magnification of 100 when image is formed at `25cm` from the eye, the focal length of the eye lens should beA. `4 cm`B. `10 cm`C. `(25)/(9)cm`D. `9 cm` |
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Answer» Correct Answer - C `m=m_(o)xxm_(e)m=m_(o)xx(1+(D)/f_(e))` `=10=10xx(1+(25)/(l_(e)))impliesf_(e)=(25)/(9)cm` |
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| 463. |
Large aperture of telescope are used forA. reduced spherical aberrationB. have high resolutionC. increase span of observationD. have low dispersion |
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Answer» Correct Answer - B The resolving power of an astronomical telescope is =`(D)/(1.22 lamda)` where D is the diameter of the objective lens . In order to improve the resolution of an astronomial telescope , one has to increase the diameter , i.e, linear aperture of the astronomial telescope |
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| 464. |
The image formed by an objective of a compound microscope isA. virtual and diminishedB. real and diminishedC. real and enlargedD. virtual and enlarged |
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Answer» Correct Answer - C The image formed by the objective of a compound microscope is real and enlarged as object lies between focus and centre of curvature . |
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| 465. |
To get three images of a single object, one should have two plane mirrors at an angle ofA. `60^(@)`B. `90^(@)`C. `120^(@)`D. `30^(@)` |
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Answer» Correct Answer - B The number of images formed when the angle between the mirror is `90^(@)` `N = (360)/(60) -1 = 4 -1 = 3` |
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| 466. |
AB is a man of height 2m and M is a mirror of length 0.5m and mass 0.1 kg. Initially top of mirror M and A are at the same level and the M starts falling freely always remaining vertical. If the level of the eyes of the manis 1.5 cm below his head. A, find the time after which the man sees the reflection of his feet. |
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Answer» Correct Answer - 0.318s |
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| 467. |
The solar constant for the earth is `sum`. The surface temperature of the sun is ` T K`. The sum subtends an angle `theta` at the earth :-A. `sum prop T^(4)`B. `sum prop T^(2)`C. `sum prop theta^(2)`D. `sum prop theta` |
| Answer» Correct Answer - A::C | |
| 468. |
The refractive indices of silicate fint glass for wavelength 400 nm and 700 nm are 1.66 aned 1.61 respectively. Find the minimum angles of deviation of an equailateral prism made of this glass for light of wavelength 400 nm and 700 nm. |
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Answer» The minumum angle of devision `delta_m` is given by `mu = (sin (A+delta_m)/(2))/(sin(A)/(2)) = (sin(30^@ + (delta_m)/(2)))/(sin [email protected])` `=2 sin(30^@ + (delta_m)/(2))` for 400 nm light, `1.66 = 2 sin (30^@ + delta_m//2)` or `sin(30^@ +delta_m//2) = 0.83` `or (30^@ + delta_m//2) = 56^@` For 700 nm light, `1.61 = 2 sin (30^@ + delta_m//2).` This gives `delta_m = 48^@` |
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| 469. |
There are two layers of water in a calorimeter, the lower one colder at temperature `30^(@)C` and the upper one hotter at temperature `60^(@)C`. The mass of lower water is m and volume is `V`. The mass of upper water is `2`m and volume is `2V`. The coefficient of volume expansion for water is `gamma`. When the temperature of the whole system becomes equal , the ratio of new volume to the old volume is . |
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Answer» Correct Answer - `1` mc`(T-30) + 2mc(T-60) = 0` `T= 50^(@)C` `DeltaV = V_(gamma)(T-30) + 2V_(gamma) (T-60) = 0` `DeltaV= 0` `V_(f) = V_(i)` |
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| 470. |
A plane mirror of circular shape with radius `r=20cm` is fixed to the ceiling .A bulb is to be placed on the axis of the mirror.A circular area of radius `R=1m` on the floor is to be illuminated after reflection of light from the mirror. The height of the room is `3m` What is maximum distance from the center of the mirror and the bulb so that the required area is illuminated? |
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Answer» Correct Answer - 75 cm |
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| 471. |
Statement-`1`: Equal amount of heat is supplied to two identical spheres `A` & `B` (see figure.) . The increment in temperature for sphere `A` is more than sphere `B`. Statement-`2` : Work done due to gravity on sphere `A` is positive while on sphere `B` is negative. 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 | |
| 472. |
An opaque cylindrical tank with an open top has a diameter of `3.00m` and is completely filled with water .When the setting sun reaches an angle of `37^(@)` above the horizon,sunlight ceases to illuminate any part of the bottom of the tank .How deep is the tank? |
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Answer» Correct Answer - 4m |
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| 473. |
Assertion `:` The setting sun apears to be red. Reason `:` Scattering of light is directly proportional to the wavelength .A. If both the assertion and reason are true and reason explains the assertion.B. If both the assertion and reason are true but reason does not explain the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - C The sun and its surroundings appear red during sunset or sunrise because of scattered light is inversely proportional to the fourth power fo wavelength of light, `i.e., Iprop(1)/(lambda^(4))` |
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| 474. |
A convex lens of focal length `12 cm` is made of glass of `mu=(3)/(2)`. What will be its focal length when immersed in liquid of `mu=(5)/(4)` ?A. `6 cm`B. `12 cm`C. `24 cm`D. `30 cm` |
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Answer» Correct Answer - D `(f_(I))/(f_(a))=((._(a)mu_(g)-1))/((._(I)mu_(g)-1))=(((3)/(2)-1))/(((3//2)/(5//4)-1))=(5)/(2)` `:. f_(I)=f_(a)((5)/(2))=(12xx5)/(2)=30cm` |
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| 475. |
Assertion `:` A double convex lens `(mu=1.5)` has focal length `10 cm`. When the lens is immersed in water `(mu=4//3)` its focal length becomes `40 cm`. Reason `:` `(1)/(f)=(mu_(1)-mu_(m))/(mu_(m))((1)/(R_(1))-(1)/(R_(2)))`A. If both the assertion and reason are true and reason explains the assertion.B. If both the assertion and reason are true but reason does not explain the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A Focal length of lens immersed in water is four times the focal length of lens in air. If means `f_(w)=4f_(a)=4xx10=40cm` |
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| 476. |
Assertion `:` A piece of red glass is heated till it glows in dark. The colour of glowing glass would be orange. Reason`:` Red and orange is complementary colours.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 The colour of glowing red glass in dark will be green as red and green are complimentary colours. |
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| 477. |
An object is kept at a distance of 16 cm from a thin lens and the image formed is real. If the object is kept at a distance of 6 cm from the same lens, the image formed is virtual. If the size of the image formed are equal, the focal length of the lens will be |
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Answer» Correct Answer - 11 cm For real image m = `(-f)/(f+16)` Virtual `m= (f)/(f+6)` `therefore (-f)/(f+16) = (f)/(f+16) implies f+16 = -f-6` `2f = -22 implies f = 11 cm` |
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| 478. |
Assertion `:` The colour of the green flower seen through red glass appears to be dark. Reason `:` Red glass transmits only red light.A. If both the assertion and reason are true and reason explains the assertion.B. If both the assertion and reason are true but reason does not explain the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A The red glass absorbs the radiations emitted by green flowers, so flower appears black. |
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| 479. |
An object is kept at a distance of 16 cm from a thin lens and the image formed is real. If the object is kept at a distance of 6 cm from the same lens, the image formed is virtual. If the size of the image formed are equal, the focal length of the lens will beA. 15cmB. 17cmC. 21cmD. 11cm |
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Answer» Correct Answer - d. Only convex lens can form a real well as virtual image. So, the given lens is a convex lens. Let f is the focal length of the lens and n the magniture of magnification. In the first case, when the image is real: `muv=-16` `v=+16n` So, applying `(1)/(v)-(1)/(u)=(1)/(f)` `(1)/(16)+(1)/(16)=(1)/(f)` or `1+(1)/(n)=(16)/(f)` In the second case, when image is virtual: `u=-6` `v=-6n` `f=+f` `:. (1)/(-6n)+(1)/(6)=(1)/(f)` `:.1-(1)/(n)=(6)/(f)` Adding (i) and (ii), we get `2=(22)/(f) or f=11cm` |
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| 480. |
Assertion `:` The focal length of the mirrorr is `f` and distance of the object from the focus is `u` , the magnification of the mirror is `f//u`. Reason : Magnification `=("Size of the image")/("Size of object" )`A. If both the assertion and reason are true and reason explains the assertion.B. If both the assertion and reason are true but reason does not explain the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A Magnification produced by mirrorr `m=(I)/(O)=(f)/(f-u)=(f)/(x) x` is distance from focus. |
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| 481. |
Light is incident normally on face AB of a prism as shown in Figure. A liquid of refractive index `mu` is placed on face AC of the prism. The prism is made of glass of refractive indes `3//2` . Find the limits of `mu` for which total internal reflection takes place on the face AC. |
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Answer» Critical angle between glass and liquid face is `sintheta_(c)=(3//2)/(mu)=(3)/(2mu)` Angle of incidence at face AC is `60^@` , i.e., `i=60^@` . For total internal reflection to take place, `i=theta_(c) or sini gt sintheta_(c)` or `sin60^@ gt (3)/(2mu)` or ` sqrt((3)/(2)) gt (3)/(2mu)` or `mu gt sqrt(3)` |
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| 482. |
Light is incident normally on face AB of a prism as shown in Figure. A liquid of refractive index `mu` is placed on face AC of the prism. The prism is made of glass of refractive indes `3//2` . Find the limits of `mu` for which total internal reflection takes place on the face AC. A. `mugt(3)/(4)`B. `mult(3sqrt(3))/(4)`C. `mugtsqrt(3)`D. `mult(sqrt(3))/(2)` |
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Answer» Correct Answer - c. Critical angle between glass and liquid face is `sintheta_(c)=(3//2)/(mu)=(3)/(2mu)` (i) Angle of incident at face AC is `60^(@)` i.e., `i=60^(@)` For total internal reflectino to take place, `i gt theta_(c)` or `sini gt sintheta_(c)` or `sin60^(@) gt (3)/(2mu)` or `(sqrt(3))/(2) gt (3r)/(2mu)` or `mu gt sqrt(3)` |
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| 483. |
A point object O is placed at a distance of 60 cm from a concave mirror of radius of curvature 80 cm. (a) At what distance from the concave mirror should a plane mirror be kept so that rays converge at O itself after getting reflected from the concave mirror and then from the plane mirror? (b) Will the position of the point where the rays meet change if they are first reflected from the plane mirror? |
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Answer» Correct Answer - (a) 90 cm (b) No. |
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| 484. |
A point object (O) is placed at the centre of curvature of a concave mirror. The mirror is rotated by a small angle q about its plole (P). Find the approximate distance between the object and its image. Focal length of the mirror is f. |
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Answer» Correct Answer - `4f theta` |
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| 485. |
A one eyed demon has a circular face of radius `a_(0) = 10 cm`. The eye is located at the centre of the face. At what distance from his face he must hold a convex mirror of 5 cm aperture diameter so as to see his complete face? Focal length of the mirror is 10 cm. |
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Answer» Correct Answer - 20 cm |
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| 486. |
Radius of a conductor increases uniformly from left end to right end as shown in fig. Material of the conductor is isotropic and its curved surface is thermally isolated from surrounding. Its ends are maintained at temperatures `T_(1)` and `T_(2) ( T_(1) gt T_(2))`: If, in steady state, heat flow rate is equal to H , then which of the following graphs is correctA. B. C. D. |
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Answer» Correct Answer - B Heat current flow rate is uniform everywhere . |
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| 487. |
Solids and liquids both expands on heating. The density of substance decreases on expanding according to the relation `rho_(2) = (rho_(1))/(1 + gamma(T_(2)- T_(1)))`, where , `rho_(1) rarr "density at" T_(1) , rho_(2) rarr "density at" T_(2) , gamma rarr` coefficient of volume expansion of substances. When a solid is submerged in a liquid , liquid exerts an upward force on solid which is equal to the weight of liquid displaced by submerged part of solid. Solid will float or sink depends on relative densities of solid and liquid . A cubical block of solid floats in a liquid with half ot its volume submerged in liquid as shown in figure (at temperature `T`) `alpha_(S) rarr` Coefficient of linear expansion of solid `gamma_(L) rarr "Coefficient of volume expansion of liquid"` `rho_(S) rarr "Density of solid at temperature" T` `rho_(L) rarr" Density of liquid at temperature" T` The relation between densities of solid and liquid at temperature `T` isA. `rho_(S) = 2rho_(L)`B. `rho_(S) = (1//2)rho_(L)`C. `rho_(S) = rho_(L)`D. `rho_(S) = (1//4)rho_(L)` |
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Answer» Correct Answer - B Let `M` be the mass of solid `therefore` Volume displaced = `(1xxM)/(2 xx rho_(S))` Thrust force =`rho_(L) xx (M)/(2rho_(S))g = Mg therefore rho_(1) = 2rho_(S)` |
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| 488. |
The temperature of an isotropic cubical solid of length `l_(0)`, density `rho_(0)` and coefficient of linear expansion `alpha` is increased by `20^(@)C`. Then at higher temperature , to a good approximation:-A. Length is `l_(0) (1+20alpha)`B. Total surface area is `l_(0)^(2) (1 +40alpha)`C. Total volume is `l_(0)^(3)(1+60alpha)`D. Density is `(rho_(0))/(1+ 60 alpha)` |
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Answer» Correct Answer - A::C::D Length `l = l_(0) (1+alphaDeltaT) = l_(0) (1+20alpha)` Area ` A =A_(0) (1+betaDeltaT) = 6l_(0)^(2)(1+40alpha)` Volume ` V=V_(0) (1+gammaDeltaT) = l_(0)^(3) ( 1+3alphaDeltaT) = l_(0)^(3) (1+60alpha)` Density `rho = (rho_(0))/(1+gammaDeltaT) = (rho_(0))/(1+60alpha)` |
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| 489. |
Solids and liquids both expands on heating. The density of substance decreases on expanding according to the relation `rho_(2) = (rho_(1))/(1 + gamma(T_(2)- T_(1)))`, where , `rho_(1) rarr "density at" T_(1) , rho_(2) rarr "density at" T_(2) , gamma rarr` coefficient of volume expansion of substances. When a solid is submerged in a liquid , liquid exerts an upward force on solid which is equal to the weight of liquid displaced by submerged part of solid. Solid will float or sink depends on relative densities of solid and liquid . A cubical block of solid floats in a liquid with half ot its volume submerged in liquid as shown in figure (at temperature `T`) `alpha_(S) rarr` Coefficient of linear expansion of solid `gamma_(L) rarr "Coefficient of volume expansion of liquid"` `rho_(S) rarr "Density of solid at temperature" T` `rho_(L) rarr" Density of liquid at temperature" T` If temperature of system increases, then fraction of solid submerged in liquidA. increaseB. decreaseC. remains the sameD. inadequate information |
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Answer» Correct Answer - D If `gamma_(liq) gt gamma_(solid)` `" "` Also `rho_(L) Vg = Mg` As `T uarr , rho_(L)darr , "so V displaced" uarr` Fraction of solid submerged should increased. |
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| 490. |
Solids and liquids both expands on heating. The density of substance decreases on expanding according to the relation `rho_(2) = (rho_(1))/(1 + gamma(T_(2)- T_(1)))`, where , `rho_(1) rarr "density at" T_(1) , rho_(2) rarr "density at" T_(2) , gamma rarr` coefficient of volume expansion of substances. When a solid is submerged in a liquid , liquid exerts an upward force on solid which is equal to the weight of liquid displaced by submerged part of solid. Solid will float or sink depends on relative densities of solid and liquid . A cubical block of solid floats in a liquid with half ot its volume submerged in liquid as shown in figure (at temperature `T`) `alpha_(S) rarr` Coefficient of linear expansion of solid `gamma_(L) rarr "Coefficient of volume expansion of liquid"` `rho_(S) rarr "Density of solid at temperature" T` `rho_(L) rarr" Density of liquid at temperature" T` Assume block does not expand on heating . The temperature at which the block just begins to sink in liquid isA. `T + (1)/(gamma_(L))`B. `T + (1)/((2gamma_(L)))`C. `T + (2)/((gamma_(L)))`D. `T + (gamma_(L))/(2)` |
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Answer» Correct Answer - A If volume change in solid is zero. Let at `T` , solid sinks `implies rho_(0)(V)/(2)g = Vrho_(S)g` (initially) Finally `rho_(T) Vg = Vrho_(S)g = (rho_(0))/(2)` `rho_(T) = (rho_(0))/((1+gamma_(L)DeltaT)) = (rho_(0))/(2) implies 2 = 1 + gamma_(L)DeltaT = DeltaT = (1)/(rho_(L))` `therefore` Temperature = `T + (1)/(gamma_(L))` |
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| 491. |
Pressure versus temperature graph of an ideal gas is shown in figure. Density of the gas at point A is `rho_(0)`. Density at B will be A. `(3)/(4)rho_(0)`B. `(3)/(2)rho_(0)`C. `(4)/(3)rho_(0)`D. `2rho_(0)` |
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Answer» Correct Answer - B Ideal gas equation `P = (rho)/(M)RT` For state `A` : `P_(0) = (rho_(0))/(M) RT_(0)` For state `B` `3P_(0) = (rho)/(M) R2T_(0) implies rho = (3)/(2)rho_(0)` |
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| 492. |
A small spherical monoatomic ideal gas bubble `(gamma= (5)/(3))` is trapped inside a liquid of density `rho_(l)` (see figure) . Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is `T_(0)`, the height of the liquid is H and the atmospheric pressure is `P_(0)` (Neglect surface tension). When the gas bubble is at a height y from the bottom , its temperature is :-A. `T_(0)((P_(0) + rho_(l)gH)/(P_(0) + rho_(l)gy))^(2//5)`B. `T_(0)((P_(0) + rho_(l)g(H-y))/(P_(0) + rho_(l)gH))^(2//5)`C. `T_(0)((P_(0) + rho_(l)gH)/(P_(0) + rho_(l)gy))^(3//5)`D. `T_(0)((P_(0) + rho_(l)g(H-y))/(P_(0) + rho_(l)gH))^(3//5)` |
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Answer» Correct Answer - B `T^(gamma)P^(gamma-1)` = constant `implies T_(2) = T_(1)((P_(2))/(P_(1))^((gamma-1)/(gamma))` =`T_(0)((P_(0) + rho_(l)g(H-y))/(P_(0) + rho_(l)gH))^(1-(3)/(5)) = T_(0) ((P_(0) + rho_(l)g(H-y))/(P_(0) + rho_(l)gH))^(2//5)` |
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| 493. |
When unit mass of water boils to become steam at `100^(@)C`, it absorbs Q amount of heat. The densities of water and steam at `100^(@)C` are `rho_(1)` and `rho_(2)` respectively and the atmospheric pressure is `P_(0)`. The increase in internal energy of the water isA. `Q`B. `Q +P_(0)((1)/(rho_(1)) - (1)/(rho_(2)))`C. `Q + P_(0)((1)/(rho_(2)) - (1)/(rho_(1)))`D. `Q - P_(0)((1)/(rho_(1)) + (1)/(rho_(2)))` |
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Answer» Correct Answer - B `Q = DeltaU +W implies +Q = DeltaU +P_(0)(V_(2)-V_(1))` ` implies Q = DeltaU + P_(0)((1)/(rho_(2))- (1)/(rho_(1))) implies DeltaU = Q + P_(0)((1)/(rho_(1))-(1)/(rho_(2)))` |
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| 494. |
If `H_(C) , H_(K)` and`H_(F)` are heat required to raise the temperature of one gram of water by one degree in Celcius , Kelvin and Fahrenheit temperature scales respectively then :-A. `H_(K)gtH_(C)gtH_(F)`B. `H_(F)gtH_(C) gtH_(K)`C. `H_(K) = H_(C) gt H_(F)`D. `H_(K) = H_(C) = H_(F)` |
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Answer» Correct Answer - C `H_(C) = msDeltatheta = ms (1)^(@)C` `H_(K) = msDeltatheta = ms (1)K = ms (1)^(@)C` `H_(F) = msDeltatheta = ms(1^(@)F) = ms (5//9)^(@)C` `therefore H_(C) = H_(K) gt H_(T)` |
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| 495. |
N molecules, each of mass m of gas `A and 2 N` molecules each of mass `2m` of gas `B` are containted in the same vessel which is maintained at temperature T. The mean square velocity of molecules of B type is denoted by `v^(2)` and the mean square velocity of A type is denoted by `(omega)^(2)`. the `omega^(2)//v^(2)` is:A. `1`B. `2`C. `1//3`D. `2//3` |
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Answer» Correct Answer - D `V_(A)= sqrt((3RT)/(m)) = (w)/(sqrt(3)) , V_(B) = sqrt((3RT)/(2m)) = v implies (w^(2))/(v^(2)) = (2)/(3)` |
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| 496. |
A vessel is partitioned in two equal halves by a fixed diathermic separator. Two different ideal gases ae filled in left (L) and right (R) halves the rms speed of the molecules in L part is equal to the mean speed of moleucles in the R equal to the ratio of the mass of a molecules in L part to that of a molecules in R part is A. `sqrt((3)/(2))`B. `sqrt(pi//4)`C. `sqrt(2//3)`D. `3pi//8` |
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Answer» Correct Answer - D `(V_("rms"))_(L) = (V_("avg"))R sqrt((3RT)/(M_(1))) = sqrt((8RT)/(piM_(2))) implies (M_(1))/(M_(2)) = (3pi)/(8)` |
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| 497. |
If mass-energy equivalence is taken into account , when water is cooled to form ice, the mass of water should :-(Note: The mass energy of an object is the energy equivalent of its mass , as given by `E= mc^(2)`, where m= mass of object & c = speed of light)A. increaseB. remain unchangedC. decreaseD. first increase then decrease |
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Answer» Correct Answer - C When water is cooled to form ice , the energy is released as heat so mass of water decreases. |
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| 498. |
A right angle prism`(45^(@) - 90^(@) - 45^(@))` of refractive index n has a plane of refractive index `n_(1)(n_(1) ltn)` cemented to its diagonal face. The assembly is in air. The ray is incident on AB. (i) Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle . (ii) Assuming n = 1.352 , calculate the angle of incidence at AB for which the refracted ray passes through the diagonal face undeviated. |
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Answer» Correct Answer - (i)`e = sin^(-1) (1)/(sqrt(2)) [(sqrt(n^(2) - n_(1)^(2)))- n_(1)]` (ii) `sin^(-1)(0.956)` (i) Given that `r_(2) = C implies r_(1) = 45^(@) - C` & sin e = n sin`(45^(@)-C)` sin e = n(`"sin"45^(@) "cos"45^(@) "sin"C`) `implies e = sin^(-1) (1)/(sqrt(2)) (sqrt(n^(2)-n_(1)^(2)) - n_(1))` (ii) to pass refracted ray undeviated `r_(2) = 0 implies r_(1) = 45^(@)` and sin e = n sin `45^(@)` `implies e = sin^(-1) ((1.352)/(sqrt(2))) = sin^(-1)(0.956)` |
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| 499. |
One kg of water at `373K` is converted into steam at the same temperature. The volume `1 cm^(3)` of water becomes `1671 cm^(3)` on boiling. Calculate the change in internal energy of the system , if heat of vaporisation is `540 cal g^(-1)`. Given standard atmospheric pressure `=1.013xx10^(5) Nm^(-2)`. |
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Answer» Volume of `1kg` of water = `1000cm^(3) = 10^(-3) m^(3)` , Volume of `1kg` of steam = `10^(3) xx 1671 cm^(3) = 1.671 m^(3)` Change in volume , ` DeltaV = (1.671-10^(-3))m^(3)= 1.670 m^(3)`, Pressure , `P = 1"atm". = 1.01 xx 10^(5) Nm^(-2)` In expansion work done , `W= P Delta V = 1.01 xx10^(5) xx 1.67J = (1.686xx15^(5))/(4.2) cal = 4.015xx10^(4) cal` But `DeltaU = Q-W`(first law of thermodynamics) or `DeltaU = (5.4xx 10^(5) - 0.04015 xx 10^(5)) cal = 4.9985 xx 10^(5) cal` |
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| 500. |
Ray optics is valid when characteristic dimensions areA. Of the same order as the wavelength of lightB. Much smaller than the wavelength of lightC. Of the order of one millimetreD. Much larger then the wavelength of light |
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Answer» Correct Answer - D Ray optics is valid when size of the objects is much larger than the order of wavelength of light . |
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