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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.
| 13501. |
A train sounding a horn of frequency 1000 Hz crosses a station at a unifrom velocity of 50 kmh^(-1)Calculate the frequency of the whistle relative to a listener, standing on the platform, before and after the crossing of the station . (velocity of sound = 340 ms^(-1)) |
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Answer» |
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| 13502. |
The planes of two rigid discs are perpendicular to each other. They are rotating about their axes. If their angular velocities are 3 rad/sec and 4 rad/sec respectively, then the resultant angular velocity of the system would be |
| Answer» Answer :C | |
| 13503. |
At atmospheric pressure, 2 g of water having a volume of 2.00cm becomes 3342cm^3of steamwhen boiled. The latent heat of vaporization of water is 539 cal/g at 1 atm. What is the amount of hat added to the system? |
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Answer» 2156 CAL |
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| 13504. |
Modulus of rigidity of ideal liquids is |
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Answer» infinity |
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| 13505. |
A boat starts from rest one end of a bank of a river of width d flowing with velocityu. The boat is steered with constant acceleration a in a direction perpendicular to the bank . If point of start is origin, direction of bank is x axis and perpendicular to bank is y axis. The equation of trajectory of the boat is y=(2ax^(2))/(N u^(2)), where N is an integer. The value of N is |
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| 13506. |
Diatomic gas at pressure .P. and volume .V. is compressed adiabatically to 1/32 times the original volume. Then the final pressure is |
| Answer» Answer :C | |
| 13507. |
At what height from the surface of earth the gravitational potential and the value of g are -5.4 xx 10^(7) J kg^(-2) and 6.0 ms^(-2) respectively ? Take the radius of earth as 6400 km : |
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Answer» 2600 km |
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| 13508. |
Estimate the change in the density of water in ocean at a depth of 500 m below the surface. The density of water at the surface=1030 kgm^(-3) and the bulk modulus of water =2.2xx10^(9)Nm^(-2) |
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Answer» `2.363 kg//m^(3)` |
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| 13509. |
A body of mass 5 kg is acted upon by a net force F which varies with time t as shown in graph, then the net momentum in SI units gained by the body at the end of 10 seconds is |
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Answer» 0 |
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| 13510. |
When a body is whirled in a vertical circle such that its total mechanical energy is constant with critical velocities then tension in the string changes with time. Arrange the following in increasing order of tension when (i) String is horizontal (ii) String makes an angle 0^(0)with downward vertical (iii) String makes an angle of 180^(0) with downward vertical (iv) String makes an angle 120^(0) with down Ward vertical |
| Answer» Answer :A | |
| 13511. |
A circular disc of radius 7//pi cm is placed horizontally on the surface of water. Find the vertical force required to separated it from the water surface. (Surface tension of water =0.070N/m) |
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Answer» Solution :The downward force on he circumference of the CIRCULAR disc `=Sxx2 pi R` WHER R is the rdius of the circular disc The VERTICAL force REQUIRED to separate it from the water surface = Downward force `F=Sxx2piR` `S=0.070N/m,R=7/(pi)cm=7/(pi)xx10^(-2)m` `F=0.070 xx2xxpixx7/(pi)xx10^(-2)=9.8xx10^(-3)n` |
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| 13512. |
Two capillary tubes of length 15 cm and 5 cm and radii 0.06 cm and 0.02 cm respectively are connected in series. If the pressure difference across the end faces is equal to pressure of 14 cm high water column then find the pressure difference across the (i) first tube and (ii) second tube |
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Answer» `0., 13.5 CM`of WATER COLUMN |
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| 13513. |
Two refracting media are separated by a spherical interface as shown in the figure. |
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Answer» If `mu_(2) gt mu_(1)`, then there cannot be real IMAGE of real object. |
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| 13514. |
If an ion engine with an exhaust velocity of 10^(8)ms^(-1) could be constructed,what percentage of the initial mass would have to be fuel if the engine were to be used to power a rocket that could attain the escape velocity of 11.2kms^(-1)? Neglect gravity and air resistance. |
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| 13515. |
A liquid surface behaves like a stretched elastic membrane. Name the liquid property for the above behaviour. |
| Answer» SOLUTION :SURFACE TENSION | |
| 13516. |
A young man of mass 60 kg stands on the floor of a lift which is accelerating downwards at 1m//s^(2) then the reaction of the floor of the lift on the man is (Take g = 9.8 m//s^(2)). |
| Answer» Solution :`R=m(g-a)` | |
| 13517. |
Assertion : For small deformations, the stress and strain are proportional to each other. Reason: A class of solids called elastomers does not obey Hooke's law. |
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Answer» where k is theproportionality constantand is KNOWN as modulusof elecrticity. |
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| 13518. |
A gas is taken through the cycle A rarr B rarr C rarr A, as shown. What is the net work done by the gas ? |
| Answer» Answer :D | |
| 13519. |
In parallelogram law of vector, the direction of resultant vector is given by |
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Answer» <P>`tanalpha=(Qcos THETA)/(P+Qsin theta` |
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| 13520. |
A ball of mass of 150g and velocity 12 ms^(-1) coming towards a batsman is hit by him with a force in such a way that the ball moves with velocity 30 ms^(-1) in the direction opposite to its original one. The time of contact between the ball and the bat is 0.01 s. Then the force applied on it by the batsman i s ....... |
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Answer» Solution :`Delta VEC( p ) = mv_(2)- vec( mv )_(2)` `=m(v_(2) - v_(1))` `=0.15 |30hat(I ) - (-12 hat(j ))` `=0.15[42hat(j )]` `vec( F )= (Delta vec(p ))/(Delta t )` `=(6.3)/( 0 .01)` `|vec(F ) | =630 N` |
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| 13521. |
Let omegabe the angular velocity of the earth.s rotation about its axis. An object weighed by a spring balance gives the same reading at the equator as at a height h above the poles (h lt lt R). The value of h is |
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Answer» `(OMEGA^(2)R^(2))/G` |
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| 13522. |
State whether the viscosity of a liquid increases or decreases due to an increases in temperature. |
| Answer» SOLUTION :DECREASES | |
| 13523. |
A solid sphere of radius R made of material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the surfaceof the liquid. When a mass M is placed on the piston to compress the liquid , find the frictional charge in the radius of the sphere. |
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| 13524. |
The stress-strain graphs for two materials are as shown. |
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Answer» Material (ii) is more elastic than material (i) and HENCE material (ii) is more brittle. |
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| 13525. |
What happens to the potential energy when two protons are brought together? |
| Answer» SOLUTION :POTENTIAL energy of the system INCREASES because work has to be DONE on the system against the electrostatic FORCE of repulsion. | |
| 13526. |
A tube of 0.8 mm radius is dipped into a liquid with surface tension and density 6 xx 10^(-2) N/m and 900kg/m^3 respectively. If the tube is kept vertical, the height of liquid risen in it will be (g=10ms^(-2)) |
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Answer» 0.017m |
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| 13527. |
A block of mass m is placed on a surface with a vertical cross- section given by y = x^(3)/6. If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is |
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Answer» `(1)/(6)` m The surface y = `(X^(3))/(6)` is curved. At any point on it, tan`theta = (dy)/(dx) = (3X^(2))/(6) = (x^(2))/(2)or, (x^(2))/(2) = 0.5 ` `therefore` x = 1 So, maximum height, `y_(m) = (1^(3))/(6) = (1)/(6) m` |
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| 13528. |
An ideal gas undergoing a change of state from A to B through four different paths I, II,III and IV as shown in the P-V diagram that lead to the same change of state tyhen the change in internal energy is |
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Answer» same in I and II but not in III and IV |
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| 13529. |
Aeroplanes are made to run on runway before take - off because it …………………. . |
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Answer» DECREASES friction |
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| 13530. |
The centripetal force required by a 1000 kg car that takes a turn of radius 50 m at a speed of 36 kmph is |
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Answer» 1000 N |
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| 13531. |
Soft iron is used in many electrical machines for: |
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Answer» LOW HYSTERESIS LOSS and low permeability |
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| 13532. |
Which of the following is not possible in a cyclic process ? |
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Answer» WORK DONE by the system is positive |
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| 13533. |
In a horizontal pipe a liquid of density d enters with velocity v and leaves with velocity 2v. The pressure difference between the ends of the pipe is |
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Answer» `3xV^(2)` |
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| 13534. |
Two disc of moments of inertia I_(1)andI_(2) about their respective axes (normal to the disc and passing through the centre), and rotating with angular speeds omega_(1)andomega_(2) are brought into contact face to face with their axes of rotation coincident. (a) What is the angular speed of the two-disc system? (b) Show that the kinetic energy of the combined system is less than the sum of the initial kinetic energies of the two discs. How do you account for this loss in energy? Take omega_(1)neomega_(2). |
Answer» Solution : Let `I_(1)andI_(2)` are the moment of inertia of the discs and their angular SPEED are `omega_(1)andomega_(2)`. When they brought into CONTACT the system of two discs having moment of inertia `I_(1)+I_(2)`. The angular speed of combined discs is `omega` Angular moment before collision `=I_(1)omega_(1)+I_(2)omega_(2)` Angular MOMENTUM after collision = `(I_(1)+I_(2))omega` From the law of conservation of angular momentum `I_(1)omega_(1)+I_(2)omega_(2)=(I_(1)+I_(2))omega` `therefore omega=(I_(1)omega+I_(2)omega_(2))/(I_(1)+I_(2))` (a) Rotational kinetic energy before collision `K_(i)=(1)/(2)I_(1)omega_(1)^(2)+(1)/(2)I_(2)omega_(2)^(2)....(1)` Rotational kinetic energy after collision `K_(f)=(1)/(2)(I_(1)+I_(2))omega^(2)` Putting the value of `omega`, `K_(f)=(1)/(2)(I_(1)+I_(2))((I_(1)omega_(1)+I_(2)omega_(2))/(I_(1)+I_(2)))^(2)` `=(1)/(2)((I_(1)omega_(1)+I_(2)omega_(2))^(2))/(I_(1)+I_(2))...(2)` Decrease in rotational kinetic energy = `K_(i)-K_(f)` `=(1)/(2)I_(1)omega_(1)^(2)+(1)/(2)I_(2)omega_(2)^(2)-(1)/(2)((I_(1)omega_(1)+I_(2)omega_(2))^(2))/(I_(1)+I_(2))` `=(1)/(2)[I_(1)omega_(1)^(2)+I_(2)omega_(2)^(2)-((I_(1)^(2)omega_(1)^(2)+I_(2)^(2)omega_(2)^(2)+2I_(1)I_(2)omega_(1)omega_(2)))/(I_(1)+I_(2))]` `=(1)/(2)([(I_(1)^(2)omega_(1)^(2)+I_(2)^(2)omega_(2)^(2)+I_(1)I_(2)omega_(1)^(2)+I_(1)I_(2)omega_(2)^(2)),(-I_(1)omega_(1)^(2)-I_(2)^(2)omega_(2)^(2)-2I_(1)I_(2)omega_(1)omega_(2))])/(I_(1)+I_(2))` `=(1)/(2)((I_(1)I_(2)(omega_(1)^(2)+omega_(2)^(2)-2omega_(1)omega_(2)))/(I_(1)+I_(2)))` `=(1)/(2)[(I_(1)I_(2)(omega_(1)-omega_(2))^(2))/(I_(1)+I_(2))]` `=(I_(1)I_(2)(omega_(1)-omega_(2))^(2))/(2(I_(1)+I_(2)))` It is positive `therefore K_(i)-K_(f)gt0` `therefore K_(i)gtK_(f)` Thus, when two discs are combined, the loss of rotational kinetic energy is due to the friction between the discs but the torque produced due to friction is initerial, hence the angular momentum remains UNCHANGED. |
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| 13535. |
The position of a particle is given by r=3.0that(i)+2.0t^(2)hat(j)+5.0hat(k). Where t is in seconds and the coefficients have the proper units for r to be in metres. (a) Find v(t) and a(t) of the particle. (b) Find the magnitude and direction of v(t) at t = 1.0s. |
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Answer» Solution :`v(t)(dr)/(dt)=(d)/(t)(3.0that(i)+2.0t^(2)hat(j)+5.0hat(k)) = 3.0hat(i)+4.0that(j)` ` a(t)=(dv)/(dt)=+4.0hat(j)` `a = 4.0 m s^(-2)` along y DIRECTION (B) `At t=1.0s, v=3.0hat(i)+4.0hat(j)` It.s magnitude is `v=sqrt(3^(2)+4^(2))=5.0ms^(-1)` and direction is `THETA = TAN^(-1)((v_(y))/(v_(x)))=tan^(-1)((4)/(3))=53^(0)` with x-axis. |
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| 13536. |
A ball of mass 'm' moving with a speed 2v strikes a heavy wall elastically, which is moved with a velocity 'v'. The work done by the heavy wall on the ball is |
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Answer» `(1)/(2) MV^(2)` |
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| 13537. |
A uniform rod has mass M and length 4L. It rests in equilibrium with one end on a rough horizontal surface at A. At point B, at a distance 3L from A, it is supported by a fixed smooth roller. The rod just remains in equilibrium when theta = 30^(@) (a) Find the normal force applied by the horizontal surface on the rod at point A. (b) Find the coefficient of friction between the rod and the surface. |
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Answer» (b) `(1)/(sqrt3)` |
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| 13538. |
A car weighing 100kg is on a slope that makes an angle 30^(@). With the horizontal. The component of car.s weight parallel to the slope is (g = 10 ms^(-2)) |
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Answer» 500N |
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| 13539. |
A body of mass M is attached to the lower end of a metal wire, whose upper end is fixed. The elongation of the wire is l. a) Loss in gravitational potential energy of M is Mg l b) The elastic potential energy stored in the wire is Mgl c) The elastic potential energy stored in the wire is 1//2 Mgl d) Heat produced is 1//2 Mgl |
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Answer» only a, B are true |
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| 13540. |
A cricket ball of mass 150g is moving with a velocity of 12 m/s and is hit by a bat so that the ball is turned back with a velocity ishit by a bat so that the ball is turned back with a velocity of 20 m/s. The force of blow acts for 0.01 s on the ball. What is the averate force exerted by the bat on the ball? |
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Answer» 240N Impulse = CHANGE in momentum `therefore F xx t= m(v-u)` `therefore F xx 10^(-2)= 150xx10^(-3)[ 20-(-12)]` `therefore F = (32xx15xx10^(-2))/(10^(-2)) = 480N` |
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| 13541. |
Non-conservative force is |
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Answer» FRICTIONAL force |
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| 13542. |
Find the position of centre of mass of carbon monoxide (CO) relative to carbon atom if the distance between the centre of the carbon and oxygen atoms is 1.130 xx 10^(-10). |
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| 13543. |
Where will a person hear maximum sound, at node or antinode? |
| Answer» Solution :Perception of SOUND is DUE to PRESSURE variations that is MAXIMUM at NODES. | |
| 13544. |
Under isothermal conditions two soap bubbles of radii a and b coalesce to form a single bubble of radius c. If external pressure is P_O, the surface tension is equal to: |
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Answer» `( P_O ( a^(3) + b^(3) -C^(3) ) )/( 4 ( a^(2) + b^(2) + c^(2) )` |
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| 13545. |
(A): Velocity- time graph for an object in uniform motion along a straight path is a straight line parallel to the time axis. (R) : in uniform motion of an object velocity increases as the square of time elapsed. |
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Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) |
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| 13546. |
Density of a substance at 0^(@)C is 10.6 gm/c.c and at 100^(@) C is 10 gm/c.c. coefficient of linear expansion of solid is |
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Answer» 0.0006`//^(@)C ` |
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| 13547. |
A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the sun and the earth in terms of the new unit, if light takes 8 min and 20 s to cover this distance? |
| Answer» Solution :Speed of light, C = 1 NEW UNIT per second Time, t = 8 min 20 s = 500 s Distance between the sun and earth `=ct=1xx500=500` new unit of length. | |
| 13548. |
Doppler effect in sound is due to …………… |
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Answer» MOTION of source |
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| 13549. |
Water is used in car radiators as coolant because of |
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Answer» Its density is more |
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| 13550. |
A body executing S.H.M has a maximum acceleration equal to 48m//sec^(2) and maximum velocity equal to 12m//sec. The amplitude of S.H.M is |
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Answer» 3 MG |
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