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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.
| 1. |
Two rods A and B of different materials are welded together as shown in figure. Their thermal conductivities are `K_(1)` and `K_(2)`. The thermal conductivity of the composite rod will be A. `(3(K_(1) + K_(2)))/(2)`B. `K_(1) + K_(2)`C. `2 (K_(1) + K_(2))`D. `(K_(1) + k_(2))/(2)` |
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Answer» `(1)/(R_(eq)) = (1)/(R_(1)) + (1)/(R_(2))` `(K_(eq) . 2 A)/(L) = (K_(1) A)/(L) + (K_(2) A)/(L)` ` K_(eq) = (K_(1) + K_(2))/(2)` |
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| 2. |
A disc and a solid sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first ?A. DiskB. SphereC. Both reach at same timeD. Depends on their masses |
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Answer» `t prop (1)/(sin theta) sqrt((2 h)/(g)) sqrt(1 + k^(2)// R^(2))` `((k^(2))/(R^(2)))_("solid sphere") lt ((k^(2))/(R^(2)))_("Disc")` |
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| 3. |
In an electromagnetic wave in free space the root mean square value of the electric field is `E_(rms)=6 V//m`. The peak value of the magnetic field isA. `2.83xx10^(-8) T`B. `0.70xx10^(-8) T`C. `4.23xx10^(-8) T`D. `1.41xx10^(-8) T` |
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Answer» Correct Answer - A `E_(0)=sqrt(2)E_(rms)=6sqrt(2) V//m` `E_(0)=CB_(0)` `B_(0)=(E_(0))/C=(6sqrt(2))/(3xx10^(8))=2.83xx10^(-8) T` |
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| 4. |
A pendulum is hung the roof of a sufficiently high huilding and is moving freely to and fro like a simple harmonic oscillator .The acceleration of the bob of the pendulum is `20m//s^(2)` at a distance of `5m` from the meanposition .The time period of oscillation isA. `2 s`B. `1 s`C. `pi s`D. `2 pi s` |
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Answer» Correct Answer - C For `SHM, a = - omega^(2) x` `omega = sqrt((a)/(x)) = sqrt((20)/(5)) = 2` `:. (2 pi)/(T)` = 2 implies T pi s` |
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| 5. |
A small sphere falls from rest in a viscous liquid. Due to friction, heat is produced. Find the relation between the rate of production of heat and the radius of the sphere at terminal velocity.A. `r^(5)`B. `r^(4)`C. `r^(2)`D. `r^(3)` |
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Answer» Correct Answer - A `P = F_(V) . V^_(T), v_(T) prop r^(2), F_(v) porp rv_(T)` `P prop r^(5)` |
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| 6. |
A `npn` transistor is connected in common emitter configuration in a given amplifier. A load resistance of `800 Omega`is connected in the collector circuit and the voltage drop across `0.96` and the input resistance of the circuit is `192 Omega`, the voltage gain and the power gain of the amplifier will respectively be :A. `4,3.84`B. `3.69,3.84`C. `4,4`D. `4,3.69` |
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Answer» Correct Answer - A `beta=0.96, R_(L)=800 Omega, R_(i n)=192 Omega` `A_(V)=beta(R_(L))/(R_(i n))=(0.96)(800/192)=4` `A_(p)=beta^(2) (R_(L))/(R_(i n))=3.84` |
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| 7. |
In a common emitter transistor amplifier the audio signal voltage across the collector is `3 V`. The resistance of collector is `3 kOmega`. If current gain is `100` and the base resistance is `2 kOmega`, the voltage and power gain of the amlifier is :A. `15` and `200`B. `150` and `15000`C. `20` and `2000`D. `200` and `1000` |
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Answer» Correct Answer - B `A_(V)=beta(R_(L))/(R_(i n))=100xx3/2=150` `A_(p)=beta^(2)(R_(L))/(R_(i n))=100xx150=15000` |
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| 8. |
When a certain metallic surface is illuminated with monochromatic light of wavelength `lamda`, the stopping potential for photoelectric current is `3V_0` and when the same surface is illuminated with light of wavelength `2lamda`, the stopping potential is `V_0`. The threshold wavelength of this surface for photoelectrice effect isA. `4 lambda`B. `(lambda)/4`C. `(lambda)/6`D. `6 lambda` |
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Answer» Correct Answer - A `(hc)/(lambda)=(hc)/(lambda_(0))+e. 3V_(0).....(i)` `(hc)/(2lambda)=(hc)/(lambda_(0))+e.V_(0)......(ii)xx3` `(i)-3(ii)` `(hc)/(lambda)-(3hc)/(2lambda)=(hc)/(lambda_(0))-(3hc)/(lambda_(0))` `-1/(2lambda)=-2/(lambda_(0)) implies lambda_(0)=4lambda` |
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| 9. |
The de - Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature `T ("kelvin")` and mass `m`, isA. `h/(sqrt(3mkT))`B. `(2h)/(sqrt(3mkT))`C. `(2h)/(sqrt(mkT))`D. `h/(sqrt(mkT))` |
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Answer» Correct Answer - A `lambda=h/(sqrt(2mk))=h/(sqrt(2m.3/2 kT))=h/(sqrt(3mkT))` |
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| 10. |
A mass `m` moves in a circles on a smooth horizontal plane with velocity `v_(0)` at a radius `R_(0)`. The mass is atteched to string which passes through a smooth hole in the plane as shown. The tension in string is increased gradually and finally `m` moves in a cricle of radius `(R_(0))/(2)`. the final value of the kinetic energy is A. `(1)/(4) mv_(0^(2)`B. `2 mv_(0)^(2)`C. `(1)/(2) mv_(0^(2)`D. `m_(0)^(2)` |
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Answer» Angular momentum conservation `mv_(0) R_(0) = mv (R_(0))/(2) implies v = 2 v_(0)` `K_(f) = (1)/(2) mv^(2) = (1)/(2) m (2 v_(0))^(2)` `= 2 mv_(0)^(2)` |
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| 11. |
A sample of `0.1 g` of water of `100^(@)C` and normal pressure `(1.013 xx 10^(5) N m^(-2))` requires 54 cal of heat energy to convert to steam at `100^(@)C`. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample isA. `42.2 J`B. `84.5 J`C. `208.7 J`D. `104. 3 J` |
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Answer» Correct Answer - C `Delta Q = Delta U + Delta W` `Delta = Delta Q - Delta W = 54 xx 4.2 - P Delta V` `= 226.8 - 1.013 xx 10^(5) xx (167.1 - 0.1) xx 10^(-6)` `= 226.8 - 16.9` `~= 208.7 J` |
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| 12. |
A bolck of mass `10 kg` is moving in x-direction with a constant speed of `10 m//s`. it is subjected to a retardeng force `F=-0.1 x J//m`. During its travel from `x =20m` to `x =30 m`. Its final kinetic energy will be .A. `450 J`B. `275 J`C. `250 J`D. `475 J` |
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Answer» `F = ma` `a = (F)/(m) = (0.1 x)/(10) = 0.01 x` `v (dvb)/(dx) = - 0.01 x` `int_(10)^(v) v dv = - 0.01 int_(20)^(30) x dx` `|(v^(2))/(2)|_(10)^(v) = - 0.01 |(x^(2))/(2)|_(20)^(30)` `v^(2) - 100 = - 0.01 [(30)^(2) - (20)^(2)] = - 5` `v^(2) = 95` Final `K.E. K_(f) = (1)/(2) mv^(2) = (1)/(2) xx 10 xx 95 = 475 J` |
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| 13. |
A paritcal of mass `10 g` moves along a circle of radius `6.4 cm` with a constant tangennitial acceleration. What is the magnitude of this acceleration . What is the magnitude of this acceleration if the kinetic energy of the partical becomes equal to `8 xx 10^(-4) J` by the end of the second revolution after the beginning of the motion?A. `0.2 m//s^(2)`B. `0.1 m//s^(2)`C. `0.15 m//s^(2)`D. `0.18 m//s^(2)` |
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Answer» `K = (1)/(2) mv^(2) implies 8 xx 10^(-4) = (1)/(2) xx 10 xx 10^(-3) v^(2)` `implies v^(2) = 0.16` `v^(2) = u^(2) + 2 a_(t) (2 xx 2 piR)` `0.16 = 2 a_(1) (4pi xx 6.4 pi 10^(-2))` `a_(t) = (10)/(32 pi) = 0.1 m//s^(2)` |
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| 14. |
The acceleration due to gravity at a height `1km` above the earth is the same as at a depth `d` below the surface of earth. Then :A. `d = 1 km`B. `d = (3)/(2) km`C. `d = 2 km`D. `d = (1)/(2) km` |
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Answer» `g_(h) = g_(d)` `g (1 - (2 h)/(Re)) = g (1 - (d)/(Re))` `d = 2 h 2 km` |
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| 15. |
Two discs of same moment of inertia rotating their regular axis passing through centre and perpendicular to the plane of disc with angular velocities `omega_(1)` and `omega_(2)`. They are brought into contact face to the face coinciding the axis of rotation. The expression for loss of enregy during this process is :A. `(1)/(4) I (omega_(1) - omega_(2))^(2)`B. `I (omega_(1) - omega_(2))^(2)`C. `(1)/(8)(omega_(1) - omega_(2))^(2)`D. `(1)/(2) I (omega_(1) + omega_(2))^(2)` |
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Answer» Loss of energy `Delta K - (1)/(2) (I . I)/(I + I) (omega_(1) - omega_(2))^(2)` ` = (1)/(4) I (oemga_(1) - omega_(2))^(2)` |
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| 16. |
A 250-turns recantagular coil of length 2.1 cm and width 1.25 cm carries a current of `85 muA` and subjected to magnetic field of strength `0.85 T`. Work done for rotating the coil by `180^(@)` against the torque isA. `4.55 muJ`B. `2.3 muJ`C. `1.15 muJ`D. `9.1 muJ` |
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Answer» Correct Answer - D `W=MB(cos theta_(1)-cos theta_(2))` `=MB(cos0-cos180^(@))=2MB` `=2NIAB` `=2xx250xx85xx10^(-6)xx2.1xx1.25xx10^(-4)xx0.85` `=9.1xx10^(-6)J= 9.1 muJ` |
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| 17. |
A long solenoid of diameter 0.1 m has `2 xx 10^(4)` turns per meter. At centre of the solenoid is 100 turns coil of radius 0.01 m placed with its axis coinciding with solenoid axis. The current in the solenoid reduce at a constant rate to 0A from 4 a in 0.05 s . If the resistance of the coil is `10 pi^(2) Omega`, the total charge flowing through the coil during this time isA. `16 muC`B. `32 muC`C. `16 pimuC`D. `32 pi muC` |
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Answer» Correct Answer - B `DeltaQ=(Deltaphi)/R=(Delta(NBA))/R` `=(Nmu_(0)nA(i_(1)-i_(2)))/R` `=(100xx4pixx10^(-7)xx2xx10^(4)xxpi(0.05)^(2)xx(4-0))/(10pi^(2))` `=32xx10^(-6) C=32 muC` |
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| 18. |
The x and y coordinates of the particle at any time are `x=5t-2t^(2) and y=10t` respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t=2s is:A. ` 5 m//s^(2)`B. ` -4 m//s^(2)`C. ` -8 m//s^(2)`D. 0 |
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Answer» `x = 5t - 2t^(2)` `v_(x) = (dx)/(dt) = 5 - 4 t` `a_(x) = (dv_(x))/(dt) = - 4` `y = 10 t` `v_(y) = (4s)/(dt) = 10` `a_(y) = (dv_(y))/(dt) = 0` `a = a_(x) = - 4 m//s^(2)` |
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| 19. |
The photoelectric threshold wavelength of silver is `3250 xx 10^(-10) m`. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength `2536 xx 10^(-10) m` is `(Given h = 4.14 xx 10^(6) ms^(-1) eVs` and `c = 3 xx 10^(8) ms^(-1))`A. `~~0.6xx10^(6) ms^(-1)`B. `~~61xx10^(3) ms^(-1)`C. `~~0.3xx10^(6) ms^(-1)`D. `~~6xx10^(5) ms^(-1)` |
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Answer» Correct Answer - A::D `phi=1242/325=3.82 eV` `E=1242/253.6=4.89 eV` `K_(max)=E-phi=1.077 eV` `1/2mv_(max)^(2)=1.077xx1.6xx10^(-19)` `v_(max)=((2xx1.077xx1.6xx10^(-19))/(9.1xx10^(-31)))^(1//2)=0.6xx10^(6)m//s` |
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| 20. |
The magnetic susceptibility is negative forA. diamagnetic material onlyB. paramagnetic matrial onlyC. ferromagnetic material onlyD. Paramagnetic and ferromagnetic materials |
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Answer» Correct Answer - A Magnetic susceptibility is negative for diamagnetic material |
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| 21. |
A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system isA. `15 RT`B. `9 RT`C. `11 RT`D. `4 RT` |
| Answer» `U = U_(O_(2)) + U_(Ar) = 2 xx (5)/(2) RT + 4 xx (3)/(2) RT = 11 RT` | |
| 22. |
A Carnot engine, having an efficiency of `eta=1//10` as heat engine, is used as a refrigerator. If the work done on the system is 10J, the amount of energy absorbed from the reservoir at lower temperature isA. `90 J`B. `99 J`C. `100 J`D. 1 J |
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Answer» `eta = 1 - (T_(2))/(T_(1)) implies (1)/(10) = 1 (T_(2))/(T_(1)) implies (T_(2))/(T_(1)) = (9)/(10)` `(Q_(2))/(Q_(1)) = (T_(2))/(T_(1)) = (9)/(10) implies Q_(1) = (10)/(9) Q_(2)` `W = Q_(1) - Q_(2)` `10 = (10)/(9) Q_(1) - Q_(2) = (1)/(9) Q_(2)` `Q_(2) = 90 J` |
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| 23. |
The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe isA. `12.5 cm`B. `16 cm`C. `8 cm`D. `13.2 cm` |
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Answer» Correct Answer - D `(v)/(2l_(1)) = 3 (v)/(4 l_(2))` Therefore, `l_(1) = (2)/(3) l_(2) = (40)/(3) ~= 13.2 cm` |
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| 24. |
In a diffraction pattern due to a single slit of width a, the first minimum is observed at an angle `30^(@)` when light of wavelength 5000 Å is incident on the slit. The first secondary minimum is observed at an angle ofA. `sin^(-1)(3/4)`B. `sin^(-1)(1/4)`C. `sin^(-1)(2/3)`D. `sin^(-1)(1/2)` |
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Answer» Correct Answer - A `a=sin 30^(@)=1implies a=2lambda` `1st` secondary maximum `a sin theta_(1)=(3lambda)/2 sin theta_(1)=(3lambda//2)/(2 lambda)=3/4` `theta_(1)=sin^(-1)(3//4)` |
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| 25. |
Figure shows a circuit that contains three identical resistors with resistance `R=0.9 Omega` each, two identical inductors with inductance `L=2.0 mH` each, and an ideal battery with emf `epsilon=18 V`. The current `i` through the battery just after the switch closed is.......: A. `0.2 A`B. `2 A`C. `0` ampereD. `2 mA` |
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Answer» Correct Answer - B if we ignore `C`, `i=E/R=18/9=2 A` If we incule `C` `i=E/(R//2)=4A` |
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| 26. |
What is the minimum velocity with which a body of mass `m` must enter a vertical loop of radius `R` so that it can complete the loop?A. `sqrt(5 gR)`B. `sqrt(gR)`C. `sqrt(2 gR)`D. `sqrt(3 gR)` |
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Answer» To complete loop, velocity at lowerst point `v ge sqrt(5 g R)` `v_(m i n) = sqrt(5 g R)` |
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| 27. |
A refrigerator works between `4^(@)C` and `30^(@)C`. It is required to remove `600 calories` of heat every second in order to keep the temperature of the refrigerator space constant.The power required is (Take `1calorie= 4.2 J`)A. `2.365 W`B. `23.65 W`C. `236.5 W`D. `2364 W` |
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Answer» `T_(2) = 4^(@)C = 277 K, T_(1) = 30^(@)C = 303 K` `Q_(2) = 600 cal` `(Q_(1))/(Q_(2)) = (T_(1))/(T_(2)) implies (Q_(2) + W)/(Q_(2)) = (T_(1))/(T_(2))` `W = 236.5 W` |
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| 28. |
The approximate depth of an ocean is `2700m`. The compressibility of water is `45.4xx10^(-11)Pa^-1` and density of water is `10^3(kg)/(m^3)`. What fractional compression of water will be obtained at the bottom of the ocean?A. `1.0 xx 10^(-2)`B. `1.3 xx 10^(-2)`C. `1.4 xx 10^(-2)`D. `0.8 xx 10^(-2)` |
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Answer» `B = (Delta P)/(Delta V // V) implies `compressibility `= (1)/(B) = (Delta V // V)/(Delta P)` `(Delta V)/(V) = (Delta P)/(B) = (rho gh)/(B)` `10^(3) xx 10 xx 2700 xx 45.4 xx 10^(-11)` `= 1.2 xx 10^(-2)` |
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| 29. |
The diagram below show regions of equipotential: A positive chrages is moved from `A` to `B` in each diagram. A. In all the four cases the work done is the sameB. Minimum work is required to move `q` in figure `(a)`C. Maximum work is required to move `q` in figure `(b)`D. Maximum work is required to move `q` in figure `(c)` |
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Answer» Correct Answer - A `DeltaW=qDeltaV=q(40-10)` |
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| 30. |
An electron of mass `m` and a photon have same energy `E`. The ratio of de - Broglie wavelengths associated with them is :A. `1/C(E/(2m))^(1/2)`B. `(E/(2m))^(1/2)`C. `C(2mE)^(1/3)`D. `1/(xC)((2m)/E)^(1/2)` |
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Answer» Correct Answer - A `lambda_(e)=h/(sqrt(2mE)), lambda_(p)=(hc)/E` `(lambda_(e))/(lambda_(p))=1/(sqrt(2mE)). E/C=1/C(E/(2m))^(1/2)` |
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| 31. |
Two astronauts are floating in gravitational free space after having lost contanct with their spaceship. The two will:A. Move toward each otherB. move away from each otherC. Will become staitonaryD. Keep floating at the same distance between them |
| Answer» Due to attractive gravitational force, astronauts more towards each other. | |
| 32. |
A U-tube with both ends open to the atmosphere is partially filled with water. Oil, which is immiscible with water. Is poured into one side until it stands at a distance of `10 mm` above the water level on the other side. Meanwhile the water rises by `65mm` from its original level (see diagram). The density of the oil is: A. `425 kg m^(-3)`B. `800 kg m^(-3)`C. `928 kg m^(-3)`D. `650 kg m^(-3)` |
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Answer» Pressure is same at same level `BC:` `P_(a) + rho_(0)g (10 + 65 + 65) = P_(a) rho_(w)g (65+ 65)` `rho_(o) = (13 xx 1000)/(14) = 928 kg//m^(3)` |
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| 33. |
A particle executies linear simple harmonic motion with an amplitude `3cm` .When the particle is at `2cm` from the mean position , the magnitude of its velocity is equal to that of acceleration .The its time period in seconds isA. `(sqrt5)/(2 pi)`B. `(4 pi)/(sqrt(5)`C. `(2 pi)/(sqrt3)`D. `(sqrt5)/(pi)` |
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Answer» `v = omega sqrt((A^(2) - x^(2)))` `omega^(2) x = omega sqrt(A^(2) - x^(2))` `omega = (sqrt(A^(2) - x^(2)))/(x) = (sqrt((3)^(2) - (2)^(2)))/(2) = (sqrt(5))/(2)` `(2 pi)/(T) = (sqrt5)/(2)` `T = (4 pi)/(sqrt5)` |
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| 34. |
A particle is performing harmonic motion if its velocity are `v_(1)` and `v_(2)` at the displecement from the mean position are `y_(1)` and `y_(2)` respectively then its time period isA. `2 pi sqrt((x_(2)^(2) - x_(1)^(2))/(V_(1)^(2) - V_(2)^(2)))`B. `2 pi sqrt((V_(1)^(2) + V_(2)^(2))/(x_(2)^(2) + x_(1)^(2)))`C. `2 pi sqrt((V_(1)^(2) - V_(2)^(2))/(x_(2)^(2) - x_(1)^(2)))`D. `2 pi sqrt((x_(1)^(2) - x_(2)^(2))/(V_(1)^(2) - V_(2)^(2)))` |
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Answer» `v = omega sqrt(a^(2) - x^(2)) implies v^(2) = omega^(2) (a^(2) - x^(2))` `V_(1)^(2) = omega^(2) (a^(2) - x_(2)^(2)) = omega^(2) a^(2) - omega^(2) x_(2)^(2)` `V_(1)^(2) - V_(2)^(2) = omega^(2) (x_(2)^(2) - x_(1)^(2))` `omega = sqrt((V_(1)^(2) - V_(2)^(2))/(x_(2)^(2) - x_(1)^(2))) implies (2 pi)/(T) = sqrt((V_(1)^(2) - V_(2)^(2))/(x_(2)^(2) - x_(1)^(2))` |
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| 35. |
The ratio of the specific heats `(C_(P))/(C_(upsilon)) = gamma` in terms of degrees of freedom is given byA. `(1 + (n)/(3))`B. `(1 + (2)/(n))`C. `(1 + (n)/(2))`D. `(1 + (1)/(n))` |
| Answer» `gamma = 1 + (2)/(n)` | |
| 36. |
If energy `(E )` , velocity `(V)` and time `(T)` are chosen as the fundamental quantities , the dimensions formula of surface tension will beA. `[EV^(-1) T^(-2)]`B. `[EV^(-2) T^(-2)]`C. `[E^(-2) V^(-1) T^(-3)]`D. `[EV^(-2) T^(-1)]` |
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Answer» Surface tension `S = fn (E, V, T)` `S prop E^(a) V^(b) T^(C )` `M^(1) L^(0) ^(-2) prop [ ML^(2) T^(-2)]^(-2) [LT^(-1)]^(b) [T]^(c )` `prop M^(a) L^(a) L^(2a + b) T^(-2a - b + c)` Comparing powers of `M, L` and `T` `M : a = 1` `L : 2a + b = 0 implies b = - 2a = - 2` `T : - 2a - b + c = - 2` `- 2 + 2 + c = - 2 implies c = - 2` `a = 1, b = - 2, c = - 2` `S prop EV^(2) T^(-2)` |
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| 37. |
The efficiency of an ideal heat engine working between the freezing point and boiling point of water, isA. `6.25%`B. `12.5%`C. `20 %`D. `26.8%` |
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Answer» Correct Answer - D `eta = 1 - (T_(2))/(T_(1)) = 1 - (273)/(373) = (100)/(373) = (100 xx 100)/(373) = 26.8%` |
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| 38. |
The volume `(V)` of a monatomic gas varies with its temperature `(T)` as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state `A` to state `B`,is A. `(1)/(3)`B. `(2)/(7)`C. `(2)/(3)`D. `(2)/(5)` |
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Answer» Correct Answer - D From graph , `(V)/(T) = (nR)/(P) =` cosntant or P= constant Isobaric process, `(Delta W)/(Delta Q) = (nR Delta T)/(n C_(P) Delta T) = (R )/(5 R // 2) = (2)/(5)` |
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| 39. |
A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to `v(x) = beta x^(-2 n)` where `beta` and `n` are constant and `x` is the position of the particle. The acceleration of the particle as a function of `x` is given by.A. `- 2 nb^(2) x^(-4n -1)`B. `- 2 nb^(2) x^(-2n +1)`C. `- 2 nb^(2) x^(-4n + 1)`D. `- 2 nb^(2) x^(-2n - 1)` |
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Answer» `v = bx^(-2xn)` `(dv)/(dx) = (b) (-2n) x^(-2n - 1)` `a = v (dv)/(dx) = bx^(-2n) (b) (-2n) x^(-2n - 1)` `= - 2nb^(2) x^(-4n - 1)` |
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| 40. |
One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in (figure). The change in internal energy of the gas during the transition is `(gamma=3//5)` A. `- 20 kJ`B. `20 J`C. `- 12 kJ`D. `20 kJ` |
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Answer» `T_(A) = (P_(A) V_(A))/(nR) , T_(B) = (P_(B) V_(B))/(nR)` `(Delta U)_(A rarr B) = nC_(V) Delta T = n. (5R)/(2) (T_(B) - T_(A))` `= (5 nR)/(2) ((P_(B) V_(B) - P_(A) V_(A))/(nR))` `= (5)/(2) (P_(B) V_(B) - P_(A) V_(A))` `= (5)/(2) (2 xx 6 - 5 xx 4) xx 10^(3) = - 20 kJ` |
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| 41. |
A small signal voltage `V(t)=V_(0)sin omegat` is applied across an ideal capacitor `C`:A. Current `I(t)`, lags voltage `V(t)` by `90^(@)`B. Over a full cycle the capacitor `C` does not consume any energy from the voltage sourceC. Current `I(t)` is in phase with voltage `V(t)`.D. Current `I(t)` leads voltage V(t) by `180^(@)` |
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Answer» Correct Answer - B In capacitor, energy is stored |
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| 42. |
A body of mass `1 kg` begins to move under the action of a time dependent force `vec F = (2 t hat I + 3 t^(2) hat j) N`, where `hat i` and `hat j` are unit vectors along x-and y-axes. What power will be developed by the force at the time `t` ?A. `(2t^(2) + 3 t^(3)) W`B. `(2t^(2) + 4 t^(4)) W`C. `(2t^(3) + 3 t^(4)) W`D. `(2t^(3) + 3 t^(5)) W` |
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Answer» `vec F = 2 t hat i + 3 t^(2) hat j` `vec a = (d vec v)/(dt) = 2 t hat i + 3 t^(2) hat j` `int_(0)^(vec v) d vec v = int_(0)^(t) (2t hat i + 3 t^(2) hat j) dt` `vec v = t^(2) hat i + t^(3) hat j` `rho = vec F * vec v = (2 t^(3) + 3 t^(5)) W` |
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| 43. |
The bulk modulus of a spherical object is `B` if it is subjected to uniform pressure `p`, the fractional decrease in radius is:A. `(B)/(3p)`B. `(3p)/(B)`C. `(p)/(3B)`D. `(p)/(B)` |
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Answer» `B = - (Delta P)/(Delta V//V)` `(Delta V)/(V) = -(P)/(B)` `V = (4)/(2) pi R^(3)` `(Delta V)/(V) = 3 (Delta R)/(R )` `(Delta R)/(R ) = - (p)/(3B)` |
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| 44. |
If `theta_1` and `theta_2` be the apparent angles of dip observed in two vertical planes at right angles to each other, then the true angle of dip `theta` is given byA. `tan^(2) theta=tan^(2)theta_(1)+tan^(2) theta_(2)`B. `cot^(2) theta=cot^(2) theta_(1)-cot^(2) theta_(2)`C. `tan^(2) theta=tan^(2)theta_(1)-tan^(2) theta_(2)`D. `cot^(2) theta=cot^(2) theta_(1)+cot^(2) theta_(2)` |
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Answer» Correct Answer - D `tan theta_(1)=(tan theta)/(cos alpha), tan theta_(2)=(tan theta)/(cos(90-alpha))=(tan theta)/(sin alpha)` `cos^(2)alpha+sin^(2)alpha=1` `cot^(2) theta_(1)+cot^(2) theta_(2)=cot^(2) theta` |
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| 45. |
Which one of the following statement eis incorrect ?A. Frictional force oppose the relative motion.B. Coefficient of sliding friction has dimensions length.C. Limiting value of static friction is directly proportional to normal reaction.D. Rolling frictin is smaller than sliding friction |
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Answer» Correct Answer - B Coefficient of friciton is dimensionless |
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| 46. |
A solid sphere is in rolling motion. In rolling motion a body prosseses translational kinetic energy `(K_(t))` as well as rotational kinetic energy `(K_(r))` simutaneously. The ratio `K_(t) : (K_(t) + K_(r))` for the sphere isA. `10 : 7`B. `2 : 5`C. `5 : 7`D. `7 : 10` |
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Answer» Correct Answer - C `(K_(t))/(K_(t) + K_(r)) = (1)/(1 + (k^(2))/(R^(2))) = (1)/(1 + (2)/(5)) = (5)/(7)` |
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| 47. |
There object, `A` : (a solid sphere), `B` : (a thin circular disk) and `C` : (a circular ring), each have the same mass `M` and radius `R`. They all spin with the same angular speed `omega` about their own symmetry axes. The amount of work `(W)` required ot bring them to rest, would satisfy the relationA. `W_(B) gt W_(A) gt W_(C )`B. `W_(B) gt W_(C ) gt W_(B )`C. `W_(A) gt W_(B) gt W_(C )`D. `W_(C ) gt W_(A) gt W_(A)` |
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Answer» Correct Answer - D Work = Loss in `K.E. = (1)/(2) I omega^(2)` `I_(A) = (2)/(5) MR^(2) = 0.4 MR^(2)` `I_(B) = (1)/(2) MR^(2) = 0.5 MR^(2)` `I_(C ) = MR^(2)` `I_(C ) gt I_(B) gt I_(A)` `W_(C ) gt W_(B) gt W_(A)` |
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| 48. |
The kinetic energies of a planet in an elliptical orbit about the Sun, at positions `A,B` and `C` are `K_(A),K_(B)` and `K_(C)` respectively. AC is the major axis and `SB` is perpendicular to `AC` at the position of the sun as shown in the figure. Then A. `K_(B) lt K_(A) lt K_(C )`B. `K_(B) gt K_(A) gt K_(C )`C. `K_(A)B gt K_(B) gt K_(C )`D. `K_(A) lt K_(B) lt K_(C )` |
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Answer» Correct Answer - C From angular momentum conservation, `v_(A) ` : maximum, `v_(c )` : minimum `:. K_(A) gt K_(B) gt K_(C )` |
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