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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.
| 15451. |
What is meant by acceleration due to gravity? Is is a scalar or a vector? |
| Answer» SOLUTION :The ACCELERATION produced in a freely falling BODY under the gravitational PULL of the earth. It is a VECTOR having direction towards the center of the earth. | |
| 15452. |
A body is performing linear SHM. If the acceleration and the corresponding velocity of the body are a and v respectively, which of the following graphs is/are correct? |
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| 15453. |
A hollow cylinder of inner radius 3 cm and outer radius 5 cm and a solid cylinder of radius 2 cm are subjected to the same force. If they are made of same material and of same length, then the ratio of their elongations is |
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Answer» `1:1` |
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| 15454. |
"Water can be boiled in a paper cup". Explain how? |
| Answer» Solution :Conductivity of WATER is GREATER than that of paper. So water TAKES AWAY the heat. | |
| 15455. |
Time of flight of projectile depend only on vertical component of initial velocity. |
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| 15456. |
What are the theoritical units of Poission.s ratio? What are the practical limits of poission ratio ? |
| Answer» SOLUTION :Theoritical LIMITS of poission.s RATIO are -1 to +0.5. Practical limits of poission.s ratio are 0 to 0.5. | |
| 15457. |
A lead bullet hits a target at a velocity of 300 m cdot s^(-1). If it stops completely on hitting, determine the increase in temperature of the bullet. Assume that 50% of the heat produced is retained by the bullet. Specific heat of lead = 0.03 cal cdot g^(-1) cdot^(@)C^(-1). |
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| 15458. |
When is the tension maximum in the spring of a simple pendulum ? |
| Answer» SOLUTION :TENSION is MAXIMUM at the MEAN POSITION. | |
| 15459. |
An under water swimmer is at a depth of 12m below the surface of water. A bird is at a height of 18m from the surface of water, directly above his eyes. For the swimmer the bird appears to be a distance from the surface of water equal to (mu_W=4//3) |
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Answer» 24m |
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| 15460. |
A string is tied on a sonometer second end is hanging downward through a pulley with tension T. The velocity of the transverse wave produced is proportional to |
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Answer» `1/(sqrtT)` |
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| 15461. |
Define and illustrate the following terms.Equal vectors. |
Answer» SOLUTION :Two vectors `vec(A) and vec(B)`are said to be EQUAL when they have equal magnitude and same direction .
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| 15462. |
In a dark room with ambient temperature T_(0), a black body is kept at a temperature T. Keeping the temperatre of the black body constant (at T) sunrays are allowed to fall on the black body through a hole in the roof of the dark room. Assuimng that here is no charge in the ambient temperature of room, which of following statement is//are correct (i) The quantity of radiation absorbed by the black body in unit time will increase (ii) Since emissivity = absorptivity, hence, the equantity increase (iii) Black body radiation emitted by black body in unit time in the visible spectrum (iv) The reflected energy in unit time by the black body remains same |
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Answer» (i).(ii) |
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| 15463. |
A binary star system consists of two stars of masses m and 2 m. The separation between the centres is L. They revolve about the centre of mass of the system under their mutual gravitational interaction force. If K_(L) be the kinetic energy of lightbody and K_(H) be the kinetic energy of heavy body, then K_(L) + K_(H) is equal to |
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Answer» `(mL^(2)omega^(2))/(3)` |
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| 15464. |
A binary star system consists of two stars of masses m and 2 m. The separation between the centres is L. They revolve about the centre of mass of the system under their mutual gravitational interaction force.The system revolves with an angular velocity omega about the centre of mass given by |
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Answer» `sqrt((GM)/(L^(3)))` |
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| 15465. |
A binary star system consists of two stars of masses m and 2 m. The separation between the centres is L. They revolve about the centre of mass of the system under their mutual gravitational interaction force.Heavier star revolves in an orbit of radius |
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Answer» L |
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| 15466. |
8g oxygen, 14 g nitrogen and 22 g carbon dioxide are mixed in a container of volume 4 l. Find out the pressure of the gas mixture at 27^(@)C. Given R = 8.315 J cdot mol^(-1) cdot K^(-1). |
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Answer» Solution :If n = number of MOLES in a gas, then PV = nRT. So, `p = (nRT)/(V) = m/V(RT)/(V)`, where m = mass of the gas and M = molecular weight. Then the pressure DUE to oxygen, nitrogen and carbon dioxide gases, respectively, are `p_(1) = 8/32 (RT)/V cdot p_(2) = 14/28 (RT)/V cdot p_(3) = 22/44 (RT)/V` `:.` NET pressure of the gas mixture is `p = p_(1) + p_(2) + p_(3)` `=(8/32 + 14/28 + 22/44) RT/V = (1/4 + 1/2 + 1/2) (RT)/V = 5/4 (RT)/V` `=5/4 xx (8.315 xx 300)/(4 xx 10^(-3))` [Here, `T = 27^(@)C = 300K, V = 4 L = 4 xx 10^(-3) ^(3)]` `= 7.795 xx 10^(5) N cdot m^(-2)`. |
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| 15467. |
A man, standing on a truck running at a uniformspeed of 14.7m/s , throws a ball in such a way that after some time the ball returns exactly tohis hands. In that time, the truck moves through a distance of 58.8 m. Find out the velocity andangleof projection of the ball relative to (i) the truck, |
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| 15468. |
A man, standing on a truck running at a uniformspeed of 14.7m/s , throws a ball in such a way that after some time the ball returns exactly tohis hands. In that time, the truck moves through a distance of 58.8 m. Find out the velocity andangleof projection of the ball relative to (ii) the ground. |
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| 15469. |
Which of the following number has least number of significant figures |
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Answer» 0.8076 |
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| 15470. |
What is gradient? What is its dimension ? |
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Answer» Solution :In an HORIZONTAL streamline flow, the rate of change of velocity with DISTANCE `(du/dx)` in a direction PERPENDICULAR to the flow of the LIQUID is called the velocity gradient. Dimension of velocity gradient `(du/dx)=(LT^-1)/(L)=T^-1`. |
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| 15471. |
A body of mass m is projected vertically upwards with velocity v from earth.s surface and attains height ..h... Then (v_(e) is escape velocity ) |
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Answer» If `V = v_(e )` then `h = OO` |
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| 15472. |
Two particles were projected simu ltaneously in horizontal plane with same velocity u perpendicu lar to each other. The time after which their velocities makes angle 60^(@) with each other is k(u)/(g). Find the value of k |
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Answer» `vecv_(1)=uhati-gt hatk` For 2nd particle `vecv_(2)=uhatj-gt hatk` Now, `vecv_(1).vecv_(2)=v_(1)v_(2)cos theta` `g^(2)t^(2)=(U^(2)+g^(2))cos theta` `cos theta=(g^(2)t^(2))/((u^(2)+g^(2)t^(2)))` For `theta=60^(@)` `g^(2)t^(2)=u^(2)` LTB u=gt |
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| 15473. |
Explain surface tension on the basis of molecular theory. |
Answer» Solution :Let us consider three different molecules A, B and C in a given liquid a shown in Figure. Let a molecule 'A' be considered well inside the liquid within the sphere of influence. Since this molecule interacts with all other molecules in all directions, the net force experienced by A is ZERO. Now consider a molecule 'B' in which three-fourth lies below the liquid surface and one-fourth on the air. Since B has more molecules towards its lower side than the upper side, it experiences a net force in the downward direction. In a similar way, if another molecule 'C' is selected on the liquid surface (i.e., upper half in air and lower half in liquid), it experiences a maximum downward force due to the availability of more number of liquid molecules on the lower part. Hence it is obvious that all molecules of the liquid that FALLS within the molecules range inside the liquid interact with the molecule and hence experience a downward force. When any molecule is brought towards the surface from the interior of the liquid, work is done against the COHESIVE force among the molecules of the surface. This work is stored as potential energy in molecules. Hence the molecules on the surface will have greater potential energy than that of molecules in the interior of the liquid. But for a system to be under stable equilibrium, its potential energy (or surface energy) must be a minimum. Hence, in ORDER to maintain stable equilibrium, a liquid always TENDS to have a minimum number of molecules. In other words, the liquids tends to occupy a minimum surface area. This behaviour of the liquid gives rise to surface tension. |
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| 15474. |
What is the name of the temperature measuring instrument? |
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| 15475. |
On what does the efficiency of a heat engine depends ? |
| Answer» SOLUTION :it DEPENDS only on the TEMPERATURE of the SOURCE and the SINK | |
| 15476. |
A bullet emerges out from a wooden plank with 75% of its initial kinetic energy. The number of additional planks required to stop the bullet is |
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Answer» 2 |
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| 15477. |
Under what conditions is the average velocity equal to the instantaneous velocity? |
| Answer» SOLUTION :When a body moves with constant VELOCITY (UNIFORM motion), its AVERAGE velocity over any time interval is same as its instantaneous velocity. | |
| 15478. |
A shell projected from a level ground has a range R, if it did not explode. At the highest point, the shell explodes into two fragments having masses in the ratio 1:3, with each fragment moving horizontally immediately after the explosion. If the lighter fragment falls at a distance R from the point of projection, behind the point of projection, the distance at which the other fragments falls from the point of projection is |
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Answer» 2 R |
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| 15479. |
Two soap bubbles of radii r_1and r_2 combine to form a bigger bubble under isothermal conditions. The radius of bigger bubble will be |
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Answer» `(r_1r_2)/( r_2 - r_1)` |
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| 15480. |
In an experiment for finding the coefficient of expansion of air at constant pressure, the volume of air at 30^(@)C was 75 c.c. When the air was heated to 98^(@)C, keeping the pressure constant, it occupied 92 c.c. Calculate the volume coefficient of air. |
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Answer» Solution :Initial volume `V_(1)=75` c.c , Initial TEMPERATURE `t_(1)=30^(@)C` , Final volume `V_(2)=92` c.c, Final temperature `t_(2)=98^(@)C` Volume coefficient of air, `alpha=(V_(2)-V_(1))/(V_(1)t_(2)-V_(2)t_(1))=(92-75)/(75 times 98-92 times 30)=17/(7350-2760)=17/4590=0.003704l^(@)C` |
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| 15481. |
Two particles undergo SHM along parallel lines with the same time period (T) and equal amplitudes. At particular instant, one particle is at its extreme position while the other is at its mean position. The move in the same direction. They will cross each other after a further time. |
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Answer» `T//2` |
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| 15482. |
A uniform ladder of mass 10kg leans against a smooth vertical wall making an angle 53^(0) with it. The other end rests on a rough horizontal floor. Find the normal force and the frictional force that the floor exerts on the ladder. |
Answer» Solution : The LADDER is in equilibrium `:.N_(1)=f` and `N_(2)=W` TAKING torque about .B. `N_(1)(AO)=W(CB)`. or `N_(1)(AB)cos53=W((AB)/(2))sin53` or `N_(1)=(2)/(3)W` and `N_(2)=W=10xx9.8=98N` The FRICTIONAL force is `f=N_(1)=(2)/(3)W=65N` |
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| 15483. |
A system consists of two cubes of masses m_(1) and m_(2) respectively, connected by a spring a force constant k. Find the force (F) in newton that should be applied to the upper cube for which the lower one just lifts after theforce is removed. (take m_(1)=0.1kg, m_(2)=0.2kg, g=10m//s^(2)) |
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| 15484. |
A force acts on an object of mass 2 kg at rest, for 0.5 s. After the force stops acting, the object travels a distance of 5 m in 2 s. Hence the magnitude of the force will be … |
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Answer» SOLUTION :Herevelocityremainsconstant `v= (d )/( t) = (5 )/(2 ) = 2.5ms^(-1)` Nowinitialvelocity`v_(0)= 0 ms^(-1)` andforceacts on it As perNewtonssecondlaw of motion FORCEF `=(DELTA p )/(Delta t) = m(v-v_(0))/( Delta t )` `= 2(2.5 - 0)/(0.5)` `F= 10 N` |
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| 15485. |
Some quantity of heat is given to a substance.The quantity which determines the temperature of the body is |
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Answer» Mass |
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| 15486. |
All protons in an atom remain packed in a small nucleus inspite of the electrostatic repulsive force among them. Why? |
| Answer» Solution :The PROTONS are HELD inside the NUCLEUS due to the strong nuclear force between them, which is stronger than the ELECTROSTATIC force of repulsion. | |
| 15487. |
A boy while catching a ball experiences impulse of 6 Ns. If mass of ball is 200 gram, then what will be speed of ball during catch |
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Answer» Solution :` F Delta t= Delta p = m [m_(2)- v_(1) ]` `v_(1) =(F Delta t )/(m ) ` `=(6)/(0.2)` `=30 m//s` |
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| 15488. |
What happen when a compressed gas pushed a piston out and expands ? |
| Answer» Solution :The SPEED of the MOLECULES DECREASES when a compressed gas pushed a pistonout.Hence there kintech energy DECREASE. So, the TEMPERATURE of gas decreases. | |
| 15489. |
When light of intensity 1W//m^2 and wavelength 5xx10^(-7)m is incident on a surface, it is completely absorbed by the surface. If 100 photons emit one electron and area of the surface is 1cm^2 ,then the photoelectric current will be |
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Answer» Solution :0.4 `mu` A (3) 4.0 mA (4) `mu` A Number of photon / second `N=(IA)/((hc)/(LAMBDA))` Number of photo electron emitted/second `=N/100` Photo electric CURRENT `=N/100xxe` |
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| 15490. |
A spherical rain drop, falling in a constant gravitational field , grows by absorption of moisture from the surroundings at a rate proportional to its surface area. If it starts with zero radius, find its acceleration . |
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Answer» <P> Solution :`(dm)/(dt)=K.4pir^(2)`Where K is a constant and .r. is the radius of the drop at any instant . But `m=4/3pir^(3)P=4/3pir^(3)` SINCE P density of water is `1 gram//c.c` `(dm)/(dt)=4/3pi.3r^(2)(dr)/(dt) =4pir^(2)(dr)/(dt)=K.4pir^(2)``:. K=(dr)/(dt)` `:. r=Kt `(since it STARS with zero radius) Since `m(dv)/(dt) +V(dm)/(dt) =mg` `4/3pir^(3)(dv)/(dt)+v.k4pir^(2)=4/3pir^(2)g, (dv)/(dt)+vK3/r=g` But`(dv)/(dt)=a, v=at` `:. a+at.K.3/(Kt)=g` becomes `4a=g, a`=ACCELERATION of drop=`g//4` |
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| 15491. |
The value of 2.2 + 4.08 + 3.125+ 6.3755 with due regard to significant places is |
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Answer» `15.78` CM |
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| 15492. |
(A) : Paper pins are made to have pointed end. (R ) : Because pointed pins have very small area due to which even for small applied force it exert large pressure on the surface. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT explanation of (A) |
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| 15493. |
We often have the experience of pumping air into bicycle tyre using hand pump. Consider the air inside the pump as a thermodynamic system having volume V at atmospheric pressure and room temperature, 27^(@)C. Assume that the nozzle of the tyre is blocked and you push the pump to a volume 1//4 of V. Calculate the final temperature of air in the pump? (For air, since the nozzle is blocked air will not flow into tyre and it can be treated as an adiabatic compression). Take gamma for air = 1.4 |
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Answer» SOLUTION :Here, the process is adiabatic compression. The volume is given and temperature is to be found. We can use the EQUATION. `T_(i)V_(i)^(gamma-1)` `T_(i)=300K(273+27^(@)C=300K)` `V_(i)=V&V_(f)=(V)/(4)` `T_(f)=T_(i)((V_(i))/(V_(f)))^(gamma-1)=300K xx4^(1.4-1)=300Kxx1.741` `T_(2)~~522K or 249^(@)C` This temperature is higher than the boiling point of WATER. So it is very dangerous to touch the nozzle of blocked pump when you pump air. |
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| 15494. |
A solid sphere is suspended from a wire and is made to execute torsional oscillations with a period of 2 second. The torque required to twist the wire through unit radian is 6 xx10^(-2)N/m. Calculate the moment of inertia of the sphere about the axis of rotation. |
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| 15495. |
A body tied to a thread is made to revolvein a horizontal plane with a definite velocity and the thread does notsnap. But when the body is made to revolve in a vertical plane, the thread snaps. How is it possible? |
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Answer» Solution :Let the lengthof the thread be l , mass FO the body m, velocity of revolution v and TENSION in the thread T when the body is rotated in the horizontal plane [Fig.1.53].  Required centripetal force for revolution, `(mv^2)/(l sin theta) = Tsin theta.........(1) ` and mg =`T cos theta` .........(2) From equations (1) and (2) we get, `sin^2 theta +cos^2theta=(mv^2)/(lT)+((mg)/T)^2` `or, T^2=(mv^2)/(l)T+m^2g^2` `or, T^2-(mv^2)/lT+m^2g^2=0` `THEREFORE T=((mv^2)/l pmsqrt(((mv^2)/l)^2)+4m^2g^2)/2` `because Tgt0 and (mv^2)/l lt sqrt(((mv^2)/l)^2+4m^2g^2)` `T=1/2 [(mv^2)/l+sqrt(((mv^2)/l)^2+4m^2g^2)]` `or, T=1/2 [(mv^2)/l+sqrt(((mv^2)/l+2mg)^2-2*(mv^2)/l*2mg)]........(3)` Now, `(mv^2)/l+2mggt sqrt(((mv^2)/l+2mg)^2-2*(mv^2)/l*2mg)` `therefore` From equation (3) ,it can be ineferred that `Tlt1/2[(mv^2)/l+((mv^2)/l+2mg)]` `or, T lt ((mv^2)/l+mg)........(4)` Again, when TH body is revolved in thevertical plane , tension in the threadat the lowest pointof the circular path, `T^.=(mv^2)/l+mg`.......(5) `therefore` From equation (4) and (5) we get, `Tlt T^.` So, tension in the thread at the lowest point in the second case is greater than that in the first case, and HENCE, in the first casethough the thread does not snap, in the second case the thread may snap. |
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| 15496. |
A body weight 700 g on the surface of Earth. How much it weight on the surface of planet whose mass is1/7 and radius is half that of the Earth. |
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Answer» <P> Solution :Acceleration due to gravityg = `(GM)/(R^(2))``M_(e) = M/7 and R_(e) = R/2` On the planet `g_(p).(G(M/7))/((R/2)^(2)) = 4/7((GM)/(R^(2))) = 4/7 g ` Hence weighton the planet `W_(p) = 700 xx 4/7` `W_(p) = 400` GRAMM |
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| 15497. |
An open pipe is in resonance in 2nd harmonic with frequency f_1. Now one end of the tube is closed and frequency is increased to f_2 such that the resonance again occurs in nth harmonic. Choose the correct option. |
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Answer» `n= 3f_(2)=(5)/(4)f_(1)` `f_(1)=(v)/(lambda)=(v)/(L) "" [ lambda=L]` In case of closed, pipe, frequency of `n^(TH)` harmonic, `f_(2)=(mv)/(4L)` where n is odd number. clearly `f_(2)=(n)/(4)f_(1)` If `n=3, f_(2) lt f_(1)` which is not acceptable In `n=5, f_(2)=(5)/(4)f_(1)` which is acceptable |
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| 15498. |
A solid cylinder of mass 'm' rolls without slipping down an inclined plane making an angle theta with the horizontal. The frictional force between the cylinder and the incline is |
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Answer» `"mg sin"theta` |
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| 15499. |
A pendulum clock gives correct time at the equator. Will it gain or lose time if it is taken to the poles ? Why? |
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Answer» Solution :When a pendulum clock showing CORRECT TIME at equator is taken to poles then it will gain time. Acceleration due to gravity at poles is high. Time period of pendulum `T = 2PI sqrt(l/g)` When g INCREASES T decerases. So DECREASES . So number of oscillations made in the given time increases hence clock gains time. |
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| 15500. |
Can a body have constant velocity and varying speed ? |
| Answer» SOLUTION :No. WHENEVER the SPEED CHANGES, VELOCITY changes. | |
