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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A stone projected vertically up with velocity v from the top of a tower reaches the ground with velocity 2v. The height of the tower is |
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Answer» `(V^(2))/(2G)` |
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| 2. |
Which of the following sets of forces acting simultaneously on a particle keep it in equilibrium ? |
| Answer» Answer :C | |
| 3. |
A man squating on the ground gets straight up and stand . The force of reactionof ground on the main during the process is |
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Answer» constant and equal to mg in magnitude RESULTANT force = firction + mg `F_(R) gt mg` When the man gets straight up in that case friction `approx ` mg |
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| 4. |
What mass must suspended from the free end of a steel wire of length 2m and diameter 1mm to stretch it by 1mm ? (Y=2xx10^(11)Nm^(-2)) |
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Answer» Solution :`r = 1/2 xx 10^(-3) m , L = 2m, Y = (Mg)/(PI r^2) L/e` `M = (Y(pi r^2)e)/(gL) = (2 xx 10^11 xx pi (1/2 xx 10^(-3))^2 xx 1 xx 10^(-3))/(9.8 xx 2)` ` = (pi xx 10^2 )/(9.8 xx 4) = (100pi)/(39.2) = 8.015 kg ` |
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| 5. |
To the free end of spring hanging from a rigid support, a block of mass m is hung and slowly allowed to come to its equilibrium position. Then, stretching in the spring is d. If the same block is attached to the same spring and allowed to fall suddenly, the amount of stretching is (force constant k) |
| Answer» Answer :B | |
| 6. |
A particle executes SHM of type x=Asinomegat. It takes time t_(1) from x=0 to x=(A)/(2) and t_(2) from x=(A)/(2) to x=A. The ratio t_(1):t_(2) will be |
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Answer» `1:1` or `t_(2)=(T)/(4)-t_(1)` At time, `t=t_(1)x=(A)/(2)` `THEREFORE (A)/(2)=Asinomegat_(1)oromegat_(1)=(pi)/(6)ort_(1)=(pi)/(6omega)` `therefore t_(2)=(T)/(4)-(pi)/(6omega)=(2PI)/(4omega)-(pi)/(6omega)=(2pi)/(6omega) (because T=(2pi)/(omega))` `therefore t_(1):t_(2)=1:2` 
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| 7. |
A pillar of height 10 m carries a load of 80 tonne.Calculate the contractions in length and in volume of the pillar. Area of cross section of the pillar is 100 cm^2, Poisson's ratio of the material =0.2 and Young 's modulus =1.96 times 10^11 N.m^-2. |
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Answer» |
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| 8. |
The angular momentum of a system of particles is conserved: |
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Answer» CENTRE of the circle |
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| 9. |
The moment of the froce, vecF=4hati+5hatj-6hatk at (2,0,-3), about the point (2,-2,-2) is given by |
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Answer» `-7hati-4hatj-8hatk` `VECR=(2,0,-3)-(2,-2,-2)=(0,2,-1)` `THEREFORE vectau=vecrxxvecF` `=(0,2,-1)xx(4,5,-6)` `=|(hati,hatj,hatk),(0,2,-1),(4,5,-6)|` `=hati(-12+5)-hatj(4)-hatk(-8)` `=-7hati-4hatj-8hatk` |
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| 10. |
What is human life expectancy ? |
| Answer» SOLUTION :About 100 YEARS which is of the ORDER of `10^9` SEC. | |
| 11. |
An elevator which can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ms. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power |
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Answer» Solution :Downward force on the ELEVATOR is : `F=mg+f=22000N` `therefore "power SUPPLIED by motor to BALANCE this force is"`: `p=F_(V)=44000W=(44000)/(746)=59hp` |
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| 12. |
Which of the following digits are significant? |
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Answer» ZERO digits |
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| 13. |
A mild steel wire of length 1.0 m and cross - sectional area 0.50xx10^(-2)" cm"^(2) is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid - point of the wire. Calculate the depression at the mid point. |
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Answer» |
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| 14. |
Show that the ratio of velocities of equal masses in an inelastic collision when one of masses is stationary is v_(1)/v_(2)=(1-e)/(1+e) |
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Answer» Solution :`e=("velocity of speration" ("after collision"))/("velocity of APPROACH" ("before collision"))` `=((v_(2)-v_(1)))/((u_(1)-u_(2)))=((v_(2)-v_(1)))/((u_(1)-0))=(v_(2)-v_(1))/u_(1)` implies `v_(2)-v_(1)=eu_(1)` From the law of conservation of linear momentum, `mu_(1)=mv_(1)+mv_(2)impliesu_(1)=v_(2)` Using the equation (2) for `u_(1) `in (1), we GET `v_(2)-v_(1)=e(v_91)+v_(2)` On simplification, we get `v_(1)/v_(2)=(1-e)/(1+e)` |
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| 15. |
Moment of inertia of a uniform solid cylinder about an axis passing through the center and along the axis of the cylinder is |
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Answer» `MR^(2)` |
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| 16. |
The adjacent sides of a parallelogram are vec(P)=2i-3j+k and vec(Q)=-2i+4j-k. What is the area of the parallelogram ? |
| Answer» Answer :C | |
| 17. |
Length of a simple pendulum is 2m and mass of its bob is 0.2 kg. If the tension in the string exceeds 4N, it will break. If the bob is whirled in horizontal plane, the maximum angle the string can make with vertical during rotation is |
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Answer» `30^@` |
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| 18. |
Angular momentum L of a body with moment of inertia I and angular velocity omega (rad)/s is equal to |
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Answer» `I/omega` |
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| 19. |
A person trying to lose weight (dieter ) lifts a 10 kg mass , one thousand times , to a height of0.5 m each time .Assume that the potential energy lost each time she lowers the mass is dissipated . (a) How much work does she do against the gravitational force ? (b) Fat supplies 3.8 xx10^(7) J of energy per kilogram which is converted to mechanical energy with a 20 % efficiency rate . How much fat will the dieter use up ? |
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Answer» SOLUTION :(a) Mass of the weight , m = 10 kg Height to which the person lifts the weight , h = 0.05 m Number of times the weight is lifted , n = 1000 ` :. ` Work done against gravitational force , = n (mgh) ` = 1000 xx10 xx 9.8 xx 0.5` ` = 49 xx10^(3) J = 49000 J ` (b) Energy equivalent of 1 kg of fat = `3.8 xx10^(7) J ` Efficiency rate = 20 % Mechanical energy supplied by the person.s body ` 20/100 xx3.8 xx10^(7) J ` ` = 1/5 xx3.8 xx10^(7) J ` Equivalent mass of fat lost by the dieter , ` m = 1/(1/5 xx3.8 xx10^(7)) xx49 xx10^(3)` ` m = 245/(3.8) xx10^(-4) = 0.0064 kg ` = `6.4 ` G |
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| 20. |
At critical temperature the surface tension of a liquid ………….. . |
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Answer» is ZERO |
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| 21. |
What happens when a metal spring dissolves in an acid ? |
| Answer» SOLUTION :The internal energy of the ACID INCREASES followed by INCREASE in temperature. | |
| 22. |
A motor vehicle travelled the first third of a distance s at a speed of v_(1)=10 kmph, the second third at a speed of v_(2)=20 kmph and the last third at a speed of v_(3)=60 kmp. Determine the mean speed of the vehicle over the entire distance s. |
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Answer» 15 kmph |
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| 23. |
A body of mass 1 kg accelerates uniformly from rest to a velocity of 1 m s^(-1) in 15 seconds. At what time the kinetic energy of the body will be 2/9 J ? |
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Answer» Solution :`a = (1 - 0)/(15) ms^(-2) = 1/15 ms^(-2)` ALSO, `1/2 xx 1 xx v^2 = 2/9 or v = 2/3 ms^(-1)` Now, `2/3 = 0 + 1/15 xx t " or " t = 10 s` |
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| 24. |
A solid sphere, disc solid cylinder all of the same mass and made of the same material are allowdd to roll down (from rest) on the inclined plane, then: |
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Answer» SOLID SPHER reaches the BOTTOM first |
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| 25. |
A block slides down an incline of angle 30^(@) with an acceleration g/4. Find the kinetic friction coefficient. |
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Answer» SOLUTION :`mu=1/(2SQRT3)` |
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| 26. |
Derive the expresssion for moment of inertia of a rod about its centre and perpendicular to the rod. |
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Answer» Solution :Let US consider a uniform rod of mass (M) and length (l) as shown in figure. Let us find an expression for moment of inertia of this rod about an AXIS that passes through the CENTER of mass and perpendicular to the rod: First an origin is to be fixed for the coordinate system so that it coincides with the center of mass, which is also the geometric center of the rod. The rod is now along the x axis. We take an infinitesimally small mass (dm) at a distance (x) from the origin. The moment of inertia (dl) of this mass (dm) about the axis is `dI = (dm) x^2`  As the mass isuniformly distributed , , the mass per unit length`(lamda)` of the rod is `lamda = M/l` The (dm) massof the infinitesimally small lengthas , `dm= lamda dx = M/l dx` The moment of inertia (I) o the entire rod can be found by integrating dI `I= INT dI = int (dm) x^2 in ( M/l dx) x^2` ` I = M/l int x^2 dx` As the mass is distributed on either side of the origin the limits for integration are taken from to `- l//2 " to " l//2` ` I = m/l int_(-l//2)^(l//2) x^2 dx= M/l [(x^3)/(3)]^(l//2)` ` I = M/l [ (l^3)/(24) - (- (l^3)/(24) )] = M/l [ (l^3)/(24) + (l^3)/(24) ]` ` I = M/l [2( (l^3)/(24) ) ]` ` I = 1/12 ml^2` |
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| 27. |
Statement I : Degree Fahrenheit is the smallest unit for measuring temperature. Statement II : Fahrenheit was the first temperature scale used for measuring temperature. |
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Answer» STATEMENT I is TRUE, statement II is true , statement II is a correct EXPLANATION for statement I. |
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| 28. |
A person of 60 Kg mass is in a lift which is coming down such that the man exerts a force of 150 N on the floor of the lift. Then the acceleration of the lift is (g=10 ms^(-2)) |
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Answer» `40.0 MS^(-2)` |
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| 29. |
A person stands in contact against a wall of cylindrical drum of radius r , rotating with an angular velocity omega. If is the coefficient of friction between the wall and the person, the minimum rotational speed which enables the person to remain stuck to the wall is |
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Answer» `SQRT((g//mu r)) ` |
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| 30. |
A body of mass 5kg moving along a straight line is accelerated from 4ms^(-1) " to " 8ms^(-1) with the application of a force of 10N in the same direction. Then, |
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Answer» Work DONE by the FORCE is 120J |
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| 31. |
Define elementary physical quantities . |
| Answer» SOLUTION :A physical quantity made by itself is called an elementary physical quantity.(As they are complete by themselves, they are also called FUNDAMENTAL or BASEC QUANTITIES). | |
| 32. |
Angle (in rad) made by the vector sqrt(3)hati + hatj with the x-axis is |
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Answer» `pi//6` |
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| 33. |
A simple pendulum of length l having bob of mass M is vibrating with amplitude a. Then the maximum tension in the string is |
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Answer» `MG` |
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| 34. |
Derive the kinematic equations of motion for constant acceleration . |
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Answer» Solution :Let US consider an object moving in a STRAIGHT line with uniform or constant acceleration 'a' Let u be the velocity of the object at time t = 0 Let v be velocity of the body at a later time t. Velocity - time relation :The acceleration of the body at any instant is given by ` = (dv)/(dt) or dv = a dt` Integrating both SIDES with the condition that as time CHANGES from 0 to t, the velocity changes from u to v. For the constant acceleration, `underset(u)overset(v)intdv = underset(0)overset(t)intadt = a underset(0)overset(t)int dt rArr [v]_(u)^(v)a [t]_(0)^(t)` `v - u = at (or) v = u + at "". . . (1)` Displacement-time relation : The velocity of the body is given by `v = (ds)/(dt) or ds = vdt` But , v = u + at, `:. `ds = (u + at) dt Let us assume that initially at time t=0, the particle started from the origin. At a later time t, the particle displacement is s. Further assuming that acceleration is time - independent, we get `underset(0)overset(z)int ds = underset(0) overset(t)int u dt = underset(0)overset(t) int at dt (or)` `:. s = ut +(1)/(2) at^(2)`. . . (2) Velocity- displacement . relation :The acceleration is given by the first derivative of velocity with respect to time . `a = (dv)/(dt) = (dv)/(ds)(ds)/(dt) = (dv)/(ds) v` Let is displacement traversed [since ds/dt = v] `:. a = (1)/(2) (dv^(2))/(ds)` or ds ` = (1)/(2a) d (v^(2))` When the velocity changes from `u^(2) "to"v^(2)` , displacement changes from 0 to s, integrating the above EQUATION, we get `underset(0)overset(s)int ds = underset(m) overset(v) int (1)/(2a) d(v^(2))` `:. s = (1)/(2a) (v^(2) - u^(2))` `rArr :. v^(2) = u^(2) + 2as "". . . (3)` Let us derive the displacement s in terms of initial velocity u and finla velocity v . From the equation (1) we can write, at = v - u Substitute this in equation (2), we get `s = ut (1)/(2) (v -u) t` `:. s = ((u +v)t)/(2)` |
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| 35. |
Which one out of the following statement is false ? |
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Answer» A body can have ZERO velocity and still be accelerated |
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| 36. |
A spherical glass ball of mass 1.34 xx 10^(-4) kg and diameter 4.4 xx 10^(-3) m takes 6-4s to fall steadily through a height of 0-381 m inside a large volume of oil of specific gravity 0.943. Calculate the coefficient of viscosity of oil. |
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Answer» |
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| 37. |
A disc of mass 100 g and radius 10 cm has a projection on its circumference. The mass of projection is negligible. A 20 g bit of putty moving tangential to the disc with a velocity of 5ms^(-1) strikes the projection and sticks to it. The angular velocity of disc is |
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Answer» `14.29rads^(-1)` |
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| 38. |
A car of mass .m. starts from rest and accelerates so that the instantaneous power delivered to the car has a constant magnitude p_(0). The instantaneous velocity of this car is proportinal to |
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Answer» <P>`t^(-1//2)` |
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| 39. |
Explain Carnot cycle |
| Answer» SOLUTION :A Carnot CYCLE consists of two ISOTHERMAL and two adiabatic processes. | |
| 40. |
Find the efficiency of the thermodynamics cycle shown in figure for an ideal diatomic gas. |
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Answer» Solution :Let n be the number of moles of the gas and the temperature be `T_(0)` in the state A. Now, work DONE during the CYCLE `W=1/2xx(2V_(0)-V_(0))(2P_(0)-P_(0))=1/2P_(0)V_(0)` For the heat `(DeltaQ_(1))` given during the process `A rarr B`, we have `DeltaQ_(1)=DeltaW_(AB)+DeltaU_(AB)` `DeltaW_(AB)`=area under the STRAIGHT line AB `=1/2(P_(0)+2P_(0))(2V_(0)-V_(0))=(3P_(0)V_(0))/2` APPLYING equation of state for the gas in the state A&B. `(P_(0)V_(0))/T_(0)=((2P_(0))(2V_(0)))/T_(B) rArr T_(B)=4T_(0)` ` :. U_(AB)=nC_(v)DeltaT=n[(5R)/2](4T_(0)-T_(0))=(15nRT_(0))/2=(15P_(0)V_(0))/2` ` :. DeltaQ_(1)=3/2P_(0)V_(0)+15/2P_(0)V_(0)=9P_(0)V_(0)` Obviously, the processes. `BrarrC and CrarrA` involve the abstraction of heat from the gas. `"Efficiency"=("Work done per cycle")/("Total heat supplie per cycle")` i.e., `eta=(1/2P_(0)V_(0))/(9P_(0)V_(0))=1/18` |
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| 41. |
A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is at 3 cm away from A going towards B. |
Answer» Solution :In the figure, the point A and B, 10 cm apart, are the the EXTREME positions of theparticle in SHM, and the point O is the MEAN position. The direction from A to B is positive, as indicated. The direction of velocity, displacement and acceleration from A to B is positive.  A 3 cm AWAY from A, that is, at D and going TOWARDS B, v is positive, acceleration and force, being directed towards O are also positive. (+, +, +). |
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| 42. |
(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2(MR^(2))/(5), where M is the mass of the sphere and R is the radius of the sphere. (b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be (MR^(2))/(4), find its moment of inertia about an axis normal to the disc and passing through a point on its edge. |
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Answer» Solution :(a) The moment of inertia of a sphere about it diameter = `(2)/(5)MR^(2)`. So sphere is solid sphere according to the theorem of parallel axes, the moment of inertia of about a axis to the tangent `I=I_(C)+Md^(2)` `=(2)/(5)MR^(2)+MR^(2)[because d=R]` `=(7)/(5)MR^(2)`  (b) Diameter of DISC are AB and CD. The moment of inertia of a disc about centre of mass and its diameter EF. `I_(C)=(MR^(2))/(4)` HENCE moment of inertia of disc about AB and `CDtoI_(EF)` `I_(EF)=I_(AB)+I_(CD)` `=(MR^(2))/(4)+(MR^(2))/(4)=(MR^(2))/(2)`  Now according to the theorem of parallel axes moment of inertia about axis DG. `I=I_(EF)+Md^(2)` `=(MR^(2))/(2)+MR^(2)` `=(3MR^(2))/(2)` |
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| 43. |
A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of 15 ms^(-1)Then, the frequency of sound that the observer hears in the echo reflected from the cliff is ....... (Take velocity of sound in air = 330 ms^(-1)) |
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Answer» 800 Hz  We have `(f _(L))/( v + v _(L)) = (f _(S))/( v + v ._(S))` `therefore (f _(L))/( 330 + 0) = (800)/( 330 - 15)` `therefore f _(L) = 800 xx (330)/(315) = 838.1Hz` |
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| 44. |
The sum of the given two numbers with regard to significant figures is |
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Answer» `4.55xx10^(-6)` |
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| 45. |
At which of the following temperatures the value of surface tension of water is minimun? |
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Answer» `4^0C` |
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| 46. |
A force F acing on a body depends on its displacement s as F alpha s^(-1"/"3). The power delivered by F will depend on displacement as |
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Answer» `S^(2/3)` |
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| 47. |
Consider the following statements for a paramagnetic substance kept in a magnetic field (a) if the magnetic field increases, the magnetisation increases (b) magnetisation increases |
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Answer» both (a) and (B) are TRUE |
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| 48. |
Masses 1 kg, 1.5 kg, 2 kg and M kg are situated at (2,1,1), (1,2,1), (2,-2,1) and (-1,4,3). If their centre of mass is situated at (1,1,3/2), the value of M iss |
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Answer» 1 kg |
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| 49. |
In pure rolling, the velocity of the point of the rolling object which comes in contact with the surface line |
| Answer» ANSWER :B | |