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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.
| 37551. |
A particle of mass 1 kg moves from test along a straight line by the action of a force F which varies with the distance x as shown in the F-x graph. |
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Answer» MAXIMUM kinetic energy of the PARTICLE is 25 J |
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| 37552. |
An artificial satellite of mass 'm' is revolving around the earth, in a circular orbit of radius 'r'. If the mass of the earth is 'm' angular momentum of the satellite with respect to the centre of earth is (G-universal gravitational constant) |
| Answer» Answer :A | |
| 37553. |
A stone is allowed to fall from the top of a tower 300m height and at the same time another stone is projected vertically up from the ground with a velocity 100ms^(-1). Find when and where the two stones meet? |
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Answer» Solution :Suppose the two stones meet at a HEIGHT x from ground after t SECONDS. `x=100t-(1)/(2)g t^(2)…..(1)` `300-x=0+(1)/(2)g t^(2)…..(2)` Solve 1, 2 `t=3` SEC, `x=255.9m` |
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| 37554. |
Resonance is an example of : |
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Answer» TUNING of |
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| 37555. |
Calculate the excess of pressure inside an air bubble of radius 1 mm formed just below the free surface of water. Given surface tension of water 72xx10^-3Nm^-1. |
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Answer» <P> SOLUTION :EXCESS PRESSURE P = 2s/R= `(2xx72xx10^-3)/(1xx10^-3)=144Nm^-2` |
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| 37556. |
If the ratio of the orbital distance of two planets (d_1)/(d_2)=2, what is the ratio of gravitational field experienced by these two planets? |
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Answer» SOLUTION :Given:`(d_1)/(d_2)=2` FORMULA:Gravitational FIELD, `E=(GM)/(d^2)` `(E_2)/(E_1)=(((GM)/(d_(2)^(2))))/(((GM)/(d_(1)^(2))))=((d_1)/(d_2))^(2)=(2)^(2)` `(E_2)/(E_1)=4` `E_(2)=4E_(1)` |
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| 37557. |
A bomb, while falling vertically downward explodes, just when it attains the velocity u, into two fragments of mass ratio 2 : 1. If the heavier of the two masses starts moving vertically downwards with a velocity 2u, find the direction and magnitude of velocity of the lighter mass at that moment. |
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Answer» |
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| 37558. |
According to Newton's law of cooling, the rate of cooling of a body is proportional to (Delta theta)^(n) , where Delta thetais difference of temperature of body and surroundings, and n is equal to |
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Answer» 1 |
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| 37559. |
When is the work done on the body said to be zero ? |
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Answer» Solution :The work done on the body be zero under following situationusingformula `W = fd COS theta ` (1) Work done is zerowhenthe force.Asfor example, a blockmovingon a smoothhorizontaltableis notactedupon by a horizontalforcebutmayundergoa largedisplacement . (2) There is no displacement of bodyeven the force is exerted on it . As for example, when one push hard againsta rigidbrick wall , the force exert on the wall does not work . (3) When force and displacement are mutually perpendicular .  For example, during uniform circularmotion , the displacementof the body is always along tangent of the circle andthe force actingon the bodyis alwaystowardsthe centreof circleso forceand displacement is perpendicularto eachother . In this situation work doneby centripetal force is zero . (4) (a) When more than one force acts on body and the net force of all these forces is zero then work done is zero . For example: A CYLINDER is vertically tugwith string as shown in figure . Gravitational force of earth acting on the cylinder and hence tension in teh stringacting in upward directionbut here displacement of cylinder is zero so work done is also zero .  (b) A sphere is placed in a V - shaped ditch as shown in figure (C ). There are three forcesacting on it . Even though it doesn.t MOVE . So the displacement by every force is zero and hence total work is also zero . |
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| 37560. |
Statement 1 represents Assertion, statement 2represent Reasons .Statement I :In a non uniform circular motion, a body has one acceleration along the tangent to the circle and the other acceleration acts towards the centre of the circular motion . Setement 2: The magnitude and direction of the velocity of the bodychange with time in the non-uniform circular motion. |
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Answer» STATEMENT 1 is true and statement 2 is ture and explains statement 1 correctly. |
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| 37561. |
A disc of radius R=10 cm oscillates as a physical pendulum about an axis perpendicular to the plane of the disc at a distance r from its centre. If r=(R)/(4), the approximate period of oscillation is (Take g=10ms^(-2)) |
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Answer» 0.84 s  `T=2pisqrt((I)/(mgh))` Where I is the moment of inerrtia of the PENDULUM about an axiis through the pivot, m is the mass of the pendulum and h is the distance from the pivot to the centre of mass. In this case, a SOLID disc of R oscillates as a physical pendulum about an axiis perpendicular to the plane of the disc at a distance r from its centre. `thereforeI=(mR^(2))/(2)+mr^(2)=(mR^(2))/(2)+m((R)/(4))^(2)=(mR^(2))/(2)+(mR^(2))/(16)` `=(9mR^(2))/(16)""(becauser=(R)/(4))` Here, `R=10cm=0.1m,h=(R)/(4)` `thereforeT=2pisqrt(((9mR^(2))/(16))/((mgR)/(4)))=2pisqrt((9R)/(4g)` `=2pisqrt((9xx0.1)/(4xx10))=2pixx(3)/(2)xx(1)/(10)=0.94s` |
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| 37562. |
A particle is subjected to two S H Ms x_(1) = A_(1) sin wt and x_(2) = A_(2) sin (w t + pi//4). The resultant S H M will have an amplitude of |
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Answer» `(A_1 +A_2)/(2)` |
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| 37563. |
A wheel is rotating at the rate of 33 "rev min"^(-1). If it comes to stop in 20 s. Then, the angular retardation will be |
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Answer» `PI RAD s^(-2)` `=(2pixx33//60)/(20)=(11pi)/(200) "rad s"^(-2)`. |
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| 37564. |
Let the source propagate a sound waves whose intensity ata point (initially) be l. suppose we consider a case when the amplitude of the sound wave is doubled and the frequency is reduced to one-fourth. Calculate now the new intensity of sound at the same point? |
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Answer» Solution :Intensity of sound WAVE (old) = `I_(1)` AMPLITUDEOF sound wave `(A_(2)) = 2 A_(1)` FREQUENCY of the sound wave `f_(2) = (1)/(4) f_(1)` Intensity of sound wave `I_(2) = ?` `I_(1) PROP f_(1)^(2) A_(1)^(2) , I_(2) prop f_(2)^(2) A_(2)^(2)` `(I_(1))/(I_(2)) = (f_(1)^(2) * A_(1)^(2))/(f_(2)^(2) * A_(2)^(2)) = (f_(1)^(2) A_(1)^(2))/((1)/(16) f_(1)^(2) * 4 A_(1)^(2)) = (16)/(4) = 4` `I_(2) = (1)/(4) I_(1)` |
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| 37565. |
Find the derivative with respect to t, of the function x=A_(0)+A_(1)t+A_(2)t^(2) where A_(0),A_(1)andA_(2) are constants. |
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Answer» Solution :Note that here the independent variable is . and the dependent variable is .R.. The REQUIRED DERIVATIVE is `(dy)/(dx)=+A_(1)+2A_(2)t`. he second derivative is `(d^(2)X)/(dt^(2))=2A_(2)`. |
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| 37566. |
A synchronous satellite should be at a proper height moving |
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Answer» from WEST to EAST in equatorial PLANE |
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| 37567. |
The amount of work done by a moving body depends on the |
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Answer» MASS of body |
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| 37568. |
In order to double the frequnecy of the fundamental note emitted by a stratched string the length is reduced to 3/4 th of the original length andthe tension is changed. The factor by which the tension is to be changed is |
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Answer» `3/8` |
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| 37569. |
An electronand a proton are moving under the influence of mutual forces . In calculating the change in the kineticenergy of the system during motion , one ignores the magnetic force of one on another . This is , because |
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Answer» the TWO MAGNETIC FORCES are equal and opposite ,so they produce no net effect |
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| 37570. |
The earth is an approximate sphere. If the interior contained matter which is not of the same density everywhere, then on the surface of the earth, the acceleration due to gravity. |
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Answer» will be directed towards the centre but not the same everywhere |
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| 37571. |
Action and reaction are forces acting simultaneously and acting on same object. |
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Answer» |
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| 37572. |
A satellite orbits the earth at a distance equal to the radius of the earth above its surface. Find (i) its speed and (ii) its period |
| Answer» SOLUTION :`5592ms^(-1), 4HRS` | |
| 37573. |
A ball is thrown upwardsat a certain angle with the horizontal and it returns to the ground describing a parabolic path . Which of the following remains constant ? |
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Answer» MOMENTUM of the ball |
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| 37574. |
A body of mass 100 gm moving with an initial velocity of 4ms^(-1) collides with another body of mass 150 gm moving in opposite direction with velocity 6 ms^(-1). If the collision is perfectly inelastic, then after the collision, |
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Answer» the two bodies inter change their velocities |
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| 37575. |
Four identical balls A,B,C and D are placed in a line on a frictionless horizontal surface. A and D are moved with same speed .u. towards the middle as shown. Assuming elastic collisions, find the final velocities. |
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Answer» |
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| 37576. |
A transverse harmonic wave on a string is described by y(x,t)=3.0 sin (36t+0.018x +pi//4), where x and y are in cm and t is in s. The positive direction of x is from left to right. |
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Answer» The wave is travelling from right to left. (a As positive DIRECTION is from left to right and x positive, therefore, wave is travelling from right to left. (b) COMPARE the given equation with the standard form `y=rsin[(2pit)/(T)+(2pix)/(lambda0+phi)]` `(2PI)/(T)=36, (2pi)/(LAMBDA)=0.018` speed of wave, `upsilon=(lambda)/(T)=(36)/(0.018)=2000cm//s=20m//s` (c) Again, `T=(2pi)/(36)=(pi)/(18)` Frequency, `v=(1)/(V)=(18)/(pi)Hz=5.7Hz` |
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| 37577. |
A barometer kept in an elevator moving upwards with an acceleration 2ms^-2 reads 63.1 cm of mercury. What will be the possible air pressure inside the elevator? |
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Answer» SOLUTION :Pressure inside the elevator, = `(HRHO(g+a))/(rhog)=(63.1xx13.6xx10^2(9.8+2))/(13.6xx10^3xx9.8)` = 76 CM of Hg |
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| 37578. |
A person measures the time period of a simple pendulum inside a stationary lift and finds it to be 'T'. If the lift starts acceleration down wards with an acceleration g"/"3, the time period of the pendulum will be |
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Answer» `SQRT(3)T` |
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| 37579. |
The mean length of an object is 5 cm. Which of the following measurements is most accurate ? |
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Answer» `4.9 CM` |
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| 37580. |
If a simple pendulum is arranged in an artificial satellite its a) Time period becomes infinity b) Frequency becomes infinity c) Both time period and frequency become infinity d) It does not oscillate |
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Answer» a and d are CORRECT |
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| 37581. |
Explain why high & low tide are formed on seas. |
| Answer» Solution :The gravitational ATTRACTION of moon on sea water cause high tides. Tides at one PLACE cause low tides at another. Attraction by SUN also causes tides but only half of the MAGNITUDE. HENCE on new moon and fall moon days. When both effects add, tides are very high. | |
| 37582. |
A baloon of mass M is stationary in air. It has a ladder on which a man of mass m is standing. If the man starts climbing up the ladder with a velocity v relative to ladder, the velocity of balloon is |
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Answer» `(m)/(M)` v UPWARDS |
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| 37583. |
Find the time period of another pendulum of length equal to half of it at the same place. |
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Answer» Solution :The PERIOD of SIMPLE PENDULUM is given by `T= 2pisqrt((l)/(g)) ….(1) :. G= 4pi^(2)(l)/(T^(2))` i.e., `(l)/(T_(1)^(2))= (l_(2))/(T_(2)^(2)) implies (1.0)/(2^(2))= (0.50)/(T_(2)^(2))` or `T_(2)^(2)= (0.50)/(1.01) xx 4= 2 ` `:. T_(2)= sqrt(2)= 1.414s` |
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| 37584. |
Due to Doppler effect , the shift in wavelenght observed is0.1 Åfor A star producing a wavelength of 6000Å What is the velocity of recession of the star ? |
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Answer» |
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| 37585. |
The displacement of a periodically vibrating particle isy = 4 cos ^(2)((1)/(2)t)sin (1000t) .Calculate the number of harmonic waves that are superposed . |
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Answer» Solution :`y = 4 COS^(2) ((1)/(2)t) ` sin (1000t) `2 * 2 cos^(2) ((1)/(2)t)* ` sin (1000 t) = 2 (1 + cos t ) sin (1000 t ) = 2 sin (1000 t) + 2 sin (1000 t ) cos t = 2 sin (1000 t ) + sin (1000 t + t ) + sin ( 1000 t - t) = 2 sin (1000 t) + 1 sin (101 t ) + 1 sin (999 t ) ` = y_(1) + y_(2) + y_(3)` Here each of `y_(1),y_(2) and y_(3)`in the form of `a sin omegat` .Thus,each of them represents a harmonic wave . Hence, the number of SUPERPOSED harmonic WAVES = 3 . |
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| 37586. |
Consider the following statements and identify the correct statements(A) Frictional force always opposes motion(B) Static frictional force is always greater than kinetic frictional force for a given pair of surfaces.(C ) Frictional force is a non conservative force |
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Answer» All STATEMENTS are TRUE |
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| 37587. |
One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a spring of spring constant K. A mass m hangs from the free end of the spring. The area of cross-section and Young.s modulus of the wire are A and Y respectivley. Find the time period with which the mass m will oscillate if it is slightly pulled down and released? |
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Answer» Solution :`Y=(F)/(A),(L)/(x_(1)) rArr x_(1)=(FL)/(AY), F=Kx_(2) rArr x_(2)=(F)/(K)` `X=x_(1)+x_(2)=(F(LK+AY))/(KAY) rArr x= ma((LK+AY))/(KAY), (x)/(a) m=((LK+AY))/(KAY) rArr T=2pi SQRT((x)/(a))` `T=2pi sqrt((m(LK+AY))/(KAY))` |
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| 37588. |
According to kinetic theory of gases, the rms velocity of the gas molecules is directly proportional to ....... |
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Answer» `sqrtT` |
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| 37589. |
What is the condition for the vectors 2i+3j-4k and 3i-aj+bk to be parallel ? |
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Answer» `a=-9//2, b=-6` |
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| 37590. |
(A) When a body slides on rough surface, its momentum is not conserved.(B) When a ball falls from a height, momentum of earth ball system is conserved(C ) In case of explosion of flying bomb, momentum is conserved. |
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Answer» A, B are TRUE |
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| 37591. |
Which fall faster-big rain drops or small rain drops? |
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Answer» Solution :The TERMINAL velocity of rain drops is directly proportional to the sequence of radius of the drop. `V_(t)propr^(2)` HENCE big drops FALL faster. |
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| 37592. |
A refrigerator whose coefficient of performance is 5 extracts heat from the cooling compartment at the rate of 250 J per cycle. How mush electric energy is spend per cycle ? How much heat per cycle is discharged to the room ? |
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Answer» Solution :`beta = (Q_(2))/(W)` `:. W = (Q_(2))/(beta) = (250)/(5) = 50 J` `:.` Electric energy SPENT per cycle = 50 J Again, `Q_(1) = Q_(2) + W` or `Q_(1) = 250 + 50` or `Q_(1) = 300 J` `:.` HEAT energy discharged per cycle to room = 300 J |
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| 37593. |
Name and state the principle used .in solving the above problem. |
| Answer» SOLUTION :PRINCIPLE of HOMOGENEITY- of DIMENSIONS. | |
| 37594. |
Coefficient of restitution during the collision is changed to 1/2 keepting all other parameters unchanged. What is the velocity of the ball B after the collision ? |
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Answer» `(1)/(2)(3sqrt(3)hat(i) + 9hat(J)) m//s` |
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| 37595. |
In the quastion number 71, what is the difference in the masses of the pieces? |
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Answer» 0.02 g `=m_(2)-m_(1)=20.17g-20.15g` `=0.02`g (CORRECT upto two places of DECIMAL) |
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| 37596. |
The correct statement from the following is |
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Answer» A body having Zero velocity will not NECESSARILY have zero acceleration. |
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| 37597. |
A ray of light refracts from medium 1 into a thin layer of medium 2, crosses the layer and is incident at the critical angle on the interface between the medium 2 and 3 as shown in the figure. IF the angle of incidence of ray is theta the value of of theta is: |
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Answer» `sin^-1 (8/9)` |
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| 37598. |
A cylindrical container has cross sectional area of 0.20m^(2) and is open at the top. At the bottom, it has a small hole (A) kept closed by a cork. There is an air balloon tied to the bottom surface of the container. Volume of balloon is 2.2 litre. Now water is filled in the container and the balloon gets fully submerged. Volume of the balloon reduces to 2.0 litre.The cork is taken out to open the hole and at the same moment the whole container is dropped from a large height so as to fall under gravity. Assume that the container remains vertical. find the change in level of water inside the falling container 2 second after it starts falling. |
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Answer» |
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| 37599. |
A 0.4 kg body is rotated with a constant angular speed of 2 rps in a verticle circle of radius 1.2 m with the help of a string. The tension in the string when it is at the highest point is (g = 10 ms^(-2) and pi^(2) = 10) |
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Answer» 79.8 N |
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| 37600. |
A block of mass 1kg lies on a horizontal surface in the truck, the coefficient of friction between the block and the surface is 0.6. If the acceleration of the truck is 5ms^(-2) the frictional force acting on the block is |
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Answer» 2N |
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