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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 person of mass 60 kg wants to lose 5kg by going up and down a 10m high stairs. Assume he burns twice as much fat while going up then coming down. If 1 kg of fat is burnt on expending 7000 kilo calories, how many times must he go up and down to reduce his weight by 5 kg? |
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Answer» SOLUTION :` 5xx7000xx10^(3) xx4.2J = 60xx15xx10xxN` `N=(21xx7xx10^(6))/9000 = 147/9xx10^(3) = 16.3xx10^(3)` times |
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| 2. |
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 = 10ms^(-2)) |
| Answer» Answer :A | |
| 3. |
Zeroth law of thermodynamics represents a) concept of temperature b) state of thermal equilibrium of a system c) that heat is a form of energy |
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Answer» a, B are CORRECT |
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| 4. |
A disc of mass m and radius R has a concentric hole of radius r. Find the M.I about a normal axis to its plane and passing through centre of disc. |
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Answer» Solution :Consider a ring of radius .X. and of thickness dx of the disc The M.I. of the ring about a normal AXIS through its centre is `dI=(m)/(pi(R^(2)-r^(2)))2pix.dx.x^(2)` The total M.I of the disc is obtained by integrating the above expression. `intdI=int_(r)^(R)(m)/(pi(R^(2)-r^(2)))2pix.dx.x^(2)` on solving `I=(m)/(2)(R^(2)+r^(2))` |
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| 5. |
A uniform string is vibrating with a fundamental frequency 'f. The new frequency, if radius and length both are doubled would be…….. |
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Answer» 2f `mu = (M)/(L) = (V RHO )/(L) =( AL rho )/(L ) = A rho = jpi r ^(2) rho ` Now fundamental frequency is, `f - (1)/(2L ) SQRT((T)/(mu)) = (1)/(2L ) sqrt ((T)/( pi r ^(2) rho ))` ` therefore f = (1)/(2Lr ) sqrt ((T)/(pi rho )) implies f prop (1)/(Lr )` `therefore (f _(1))/( f.) = (L.r.)/(Lr) implies (f)/(f.) = ((2L) (2r))/( Lr ) implies f._(1) = (f)/(4)` |
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| 6. |
The dimensions of a rectangular block measured with callipers having least count of 0.01 cm are 5 mm xx 10 mm xx 5 mm. The maximum percentage error in the measurement of the volume of the block is |
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Answer» `5%` `V= Lxx B xx H` the maximum relative ERROR in V is `(DeltaV)/(V)= (DeltaL)/(L)+ (DeltaB)/(B)+(DeltaH)/(H)` THEMAXIMUM percentage error in V is `(DeltaV)/(V) xx 100= (DeltaL)/(L)xx 100+ (DeltaB)/(B)xx100+ (DeltaH)/(H)xx100` `= (0.01)/(0.5)xx 100+ (0.01)/(1.0)xx 100+ (0.01)/(0.5)xx 100= 2%+ 1%+ 2%= 5%` |
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| 7. |
The relation between the escape velocity v_e and orbital velocity v_0 is given by |
| Answer» ANSWER :A | |
| 8. |
Fuel is consumed at the rate of 50 kg s^(-1) in a rocket. Find the thrust on the rocket if the velocity of the exhaust gases is 2 km s^(-1). Aslo calculate the velocity of the rocket at the instant, when its mass is reduced to 1/10th of its initial mass if its initial velocity is zero. (neglect gravity). |
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Answer» Solution :`(DM)/(dt)=50 kgs^(-1), u=2 km s^(-1)=2XX10^(3)ms^(-1)`, initial VELOCITY, `u_(0)=0` (i) The thrust on the rocket, `F=u(dm)/(dt)=2xx10^(3)xx50=1xx10^(5)N`. (II) The velocity of the rocket, `v=v_(0)+"u log"_(e )(m_(0))/(m)` `= 0+xx10^(3)log_(e )((m_(0))/(m_(0))xx10) " "(because m=(m_(0))/(10))` `= 2xx10^(3)log_(e )10=2xx10^(3)xx2.303 log_(10)10` `= 4.606xx10^(3)(m)/(s)`. |
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| 9. |
Principle of homogeneity of dimensios is the consistency test for any equation. If an equation fails this test, it is proved wrong. But if the equation passes thisconsistency test, it is not neacessarliy proved right. Why ? |
| Answer» Solution :The CONSISTENCY test for any equation is the principle of homogenetiy of dimensions, i.e., and equation is correct, when dimensions // powers of M, L, T on one side of the equation are equal to their respective values//powers on the other side of the equation. When power of even one of M, L, T fails to match, the equation in worng. However if an equation passes this consistancy test, it may still not be RIGHT. This is because mere dimensional CORRECTNESS of an equation does not ensure its physical correctness. For example, work = torque is DIMENSIONALLY correct, but not physically correct. Work is a scalar which is equivalent to energy. Torque is a vector, which REPRESENTS turning effect of a force. | |
| 10. |
A smooth washer impinges at a velocity v on a group of three smooth identical blocks resting on a smooth horizontal surface as shown in figure. The mass of each block is equal to the mass of the washer. The diameter of the washer and its height are equal to the edge of the block. Determine the velocity of blocks after the impact. |
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Answer» |
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| 11. |
Can a gas offer great resistance to bulk stress? |
| Answer» Solution :No. It can OFFER SMALL resistance to FORCES TENDING to decrease its VOLUME. | |
| 12. |
(A) : A solid sphere and hollow sphere when released from the top of a smooth and fixed inclined plane, reach the ground simultaneously.(R ) : Acceleration while sliding on a smooth inclined plane is independent of mass and dimension of the body. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 13. |
A wheel of 1 m radius completes half rotation on horizontal path. Magnitude of displacement form the point of contact with path initially is……… |
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Answer» `2pi`  If wheel completes 1 ROTATION, that HORIZONTAL distance is `2pi r.` Here, it completes half rotation. Hence distance covered on horizontal path `= pir = OP` and OQ =2R. From figure `PQ ^(2) =PO^(2) + OQ^(2)` `x^(2) = (pir) ^(2) + (2r)^(2) ` `x=1 sqrt (pi ^(2) +4) [because r =1m]` |
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| 14. |
The time period of a pendulum of infinite length on earth's surface is |
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Answer» INFINITY |
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| 15. |
Figures shows a smooth inclined plane of inclination theta fixed ina car. A sphere is set in pure rolling on the incline. For what value of a (the acceleration of car in horizontal direction) the sphere will continue pure rolling ? |
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Answer» `g costheta` |
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| 16. |
A uniform spherical sheel rolls down a fixed inclined plane without sliping. Find the ratio of rotational kinetic energy to translational kinetic energy as it reaches lowest point of the incline. |
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Answer» Solution :As there is no slip, `v = R omega` is valid and rotational kinetic energy `K_(R )=(1)/(2)I_(cm)omega^(2)=(1)/(2)mk^(2)omega^(2)` and translational kinetic energy `K_(T)=(1)/(2)MV^(2)=(1)/(2)MR^(2)omega^(2)` therefore `(K_(R ))/(K_(T))=(K^(2))/(R^(2))=(2)/(3)` |
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| 17. |
Convert the vector vecr=3hati+2hatj into a unit vector. |
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Answer» Here `=|r|=sqrt(3^(2)+2^(2))=sqrt(9+4)=sqrt13` `:.hatr=(vecr)/(|r|)=(3hati+2hatj)/(sqrt13)` |
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| 18. |
Three blocks of equal masses (each 3kg) are suspended by weightless strings as shown. If applied force is 100N, then T_(1) is equal to (g = 10m//s^(2)) |
| Answer» Solution :`T_(1)=F+mg` | |
| 19. |
The displacement of a particle varies according to relation x = 4 [Cos pi t + Sin Tpit ]. The amplitude of the particle is |
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Answer» -4 |
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| 20. |
A solid sphere and a hallow sphere of same mass and radius rotate about their diameter with same K.E. If torques of equal magnitude are applied on them, the ratio of their angular displacement before they are brought to rest is |
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Answer» `1:1` |
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| 21. |
Explain centrifugal force due to rotation of the earth |
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Answer» Solution :Earthspinsaboutitsownaxiswith anangularvelocity.Anyobjecton the SURFACEOF Earthexperiences a centrifugalforce. The centrifugalforceappearsto actexactlyin oppositedirectionfrom the axisofrotation . Wherer isperpendiculardistanceof the manfrom theaxisof ROTATION. byusingrightangletrianglethe distance `r= R COS theta` Here R =RADIUSOF tehEarthand `theta= `latitudeof theEarthwhere THEMAN isstanding |
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| 22. |
Taking that earth revolves round the sun in circular orbit of radius 15 xx 10^(10) m with a time period of lyear. The time taken by another planet which is at a distance of 540 xx 10^(10)m , to revolve round the sun in circular orbit once will be |
| Answer» Answer :A | |
| 23. |
A body of mass 1.0kg is suspended from a weightless spring having force constant 600 N//m. Another body of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of 3.0 m//"sec" and gets embedded in it. Find the amplitude of oscillation |
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Answer» Solution :By conservation of linear MOMENTUM in the collision `mv= (m+M)V` `implies V= (mv)/(m+M)= (0.5 xx 3)/((1+0.5))= 1m//"sec"` Now just after collision the SYSTEM will have `KE= (1)/(2)(m+M)V^(2)` at equilibrium POSITION. So after collosion by conservation of mechanical energy `KE_("max")= PE_("max")` `(1)/(2)(m+M)v^(2)= (1)/(2)KA^(2)` `implies A= Vsqrt((m+M)/(k))= 1sqrt((1.5)/(600))= (1)/(20)m= 5cm` |
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| 24. |
A satellite of mass m is moving in a circularorbit of radius r around earth (mass M) its angular momentum with respect to the centre of the orbit is L = (GMm^(x)r)^(1//y). Find the value of xy |
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| 25. |
A mass of 0.98 Kg suspended using a spring of constant K =300 Nm^(-1) is hit by a bullet of 20gm moving with a velocity 3m/s vertically. The bullet gest embedded and oscillates with the mass. The amplitude of oscillation will be |
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Answer» `0.15 CM` |
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| 26. |
A block has been placed on an inclined plane with the slope angle thetablock slides down the plane at constantspeed the coefficient of kinetic friction is equal to |
| Answer» Solution :`mu_(k )= tan theta` | |
| 27. |
An Idealrefrigeratorhas afreezerattemperature-12^(@)C . Thecoefficientofperformanceof theengineis 5.The temperatureof the air(towhichthe heatejected ) is |
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Answer» `50^(@)C` `5=(261)/(T_(H)+285)RARR""T_(H)=-232.8K=40.2^(@)C` |
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| 28. |
How can you distinguish a raw egg from a boiled egg by spinning it on a smooth table? |
| Answer» Solution :We spin the eggs on a table top. The egg that SPINS at lower rate is the raw egg. In the raw egg, the liquid matter inside it tries to GET away from the axis of ROTATION, therby increasing its moment of inertia. As `TAU = Ialpha`, `alpha` decreases, i.e.,the raw egg spins with less angularvelocity. The reverse is true for the boiled egg that rotates like a rigid body. | |
| 29. |
(a) Prove the theorem of perpendicular axes. (Hint : Square of the distance of a point (x, y) in the x-y plane from an axis through the origin perpendicular to the plane is x^(2)+y^(2)). (b) Prove the theorem of parallel axes. (Hint : If the centre of mass is chosen to be the origin Sigmam_(1)(r_(i)=0). |
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Answer» Solution :(a) PROOF of theorem of perpendicular axes : The moment of inertia of a planar body about an axis (z) perpendicular to its plane `(x-y)` is equal to the sum of its moment of inertia about TWO perpendicular axes (x and y) concurrent with perpendicular axes (x and y) concurrent with perpendicular axis and lying in the plane of the body. `therefore I_(z)=I_(x)+I_(y)` If planar body in yz plane then `I_(x)=I_(y)+I_(z)` If planar body in xz plane then `I_(y)=I_(x)+I_(y)` This theorem is APPLIED to planar body. Suppose planar body consists of many particles each of mass of m. Suppose the coordinate of P having mass m is `(x,y)` and distance r from origin.  Momentum of inertia of a particle about an axis OZ. `I=mr^(2)` If the moment of inertia of whole planar body about z-axis is `I_(2)` then `I_(z)=summr^(2)` `=summ(x^(2)+y^(2))` `therefore I_(z)=summx^(2)+summy^(2)....(1)` Now momentum of inertia of whole planar body about an axis x and y is `I_(x) and I_(y)` `I_(x)=summx^(2)andI_(y)=summy^(2)` `therefore I_(z)=I_(x)+I_(y)` from eqn. (1) it is proved. (b) Proof of theorem of parallel axes : The theorem of parallel axes states that the moment of inertia `I_(C )` of the body about a parallel axis passing through its centre of mass and the product of its mass M and the square of the distance d between the two parallel axes `therefore I=I_(C )+Md^(2)` where `I_(C )` = Moment of inertia about an axis passing through the centre of mass I = Moment of inertia about an axis passing through centre of mass parallel to any axis. M = Total mass of body `underset(i=l)overset(n)summ_(i)` d = perpendicular distance between two parallel axes  LL. axis passing through the centre of mass Total mass of SYSTEM of n particles `M=underset(i=l)overset(n)summ_(i)` Suppose any particle of body P is, LL. is `r_(i)` distance away axis The perpendicular distance of P from axis `ZZ.=OC+CP` `=d+r_(i)` Moment of inertia about an axis passing through centre of mass `I_(C)=underset(i=l)overset(n)summ_(i)r_(i)^(2)....(1)` If `I_(i)` is the moment of inertia of a POINT P about an axis ZZ. `I_(i)=m_(i)(OP)^(2)=m_(i)(r_(i)+d)^(2)` If moment of inertia of whole body about an axis ZZ. is I, then `I=underset(i=1)overset(n)sumI_(i)` `=underset(i=1)overset(n)summ_(i)(r_(i)+d)^(2)` `=underset(i=l)overset(n)summ_(i)(r_(i)^(2)+2r_(i)d+d^(2))` `=underset(i=1)overset(n)summ_(i)r_(i)^(2)+underset(i=1)overset(n)sum2m_(i)r_(i)d+underset(i=1)overset(n)summ_(i)d^(2)` `=I_(C)+0+Md^(2) [because underset(i=1)overset(n)summ_(i)r_(i)=0,underset(i=1)overset(n)summ_(i)=M]` `I=I_(C)+Md^(2)` proved. |
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| 30. |
An incompressible liquid is kept in a long conductying cylindrical cantainer, which is closed at its top by an airtight light piston. A cylinder of length 10 cm made of material of density 0.65g//cm^(3) floats with half-length submerged in the liquid as shown in the figure. Air trapped in the cylinder has density 1.30kg//m^(3) On placing extra weight on the piston, pressure of the air in the cylinder is increased to 100 times of the initial pressure.What can you conclude? (because Use Boyle's law of air i.e., P_(1)V_(1)=P_(2)V_(2) at constant temperature) |
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Answer» Cylinder moves downwards `becauserho_(m)Alg=rho_(a)A((l)/(2))g+rho_(w)A((l)/(2))g` `RARR rho_(m) = (rho_(a))/(2) + (rho_(w))/(2) rArr 0.65 = (0.0013)/(2) + (rho_(l))/(2)` `rArr rho_(l) = 1.2987 gm//c c` Determination of displacement (upwards) `100 rho_(a)x = 99 rho_(a) (5) + rho_(l)x` `rArr 100 xx 0.0013 x - 1.2987 x = 99 xx 0.00135 xx 5` `rArrx = 0.55 cm` |
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| 31. |
A body initially at rest is moving with uniform acceleration "a". It's velocity after n seconds is V. The displacement of the body in last 2 seconds is ……. |
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Answer» `(V (n -1 ))/(n )` `d _(n -2) = v _(0) (n -2) + 1/2 a (n-2) ^(2)` `= v _(0) n = 2 v _(0) + 1/2 an ^(2) - 2an + 2a …(2)` Subtracting equation (2) from equation (1), `d _(n-2) = 2v _(0) + 2an -2a` `v _(0) =0` `therefoer d _(n) - d _(n-2) = 2a (n-1)...(3)` Now, velocity is v in `n^(th)` second, `therefore `in `v = v _(0) + an, ` putting `v_(0) =0,` `v = an ` ` therefore a = (v)/(n) ...(4)` Putting value of equation (4) in equation (3), Distance covered in last 2 second, `d _(n) - d _(n-2) = 2a (n-1)` `= (2v (n-1))/(n)` Which is OPTION (B). |
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| 32. |
What is the number of significant figures in (3.20+4.80)xx10^(5) ? |
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Answer» 5 |
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| 33. |
A spring of force constant k is cut into two pieces,such that one piece is double the lenght of the other.Then,the long piece will have a force constant of |
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Answer» `2/3k` |
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| 34. |
Two particles A and B seperated by a distance (2R) are moving counter clockwise along the same circular path of radius (R) each with uniform speed (V). At time t = 0, A is given a tangential acceleration of magnitude a = (72V^(2))/(25piR)Then A collides B after certain |
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Answer» The time lapse for the TWO bodies collideis `(6piR)/(5V)` |
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| 35. |
What is the maximum speed of a particle executing SHM with amplitude of 3 cm and periodic time 6s ? |
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Answer» `(pi)/(2)` `= 3xx (2pi)/(6)= pi cm"/"s`. |
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| 36. |
A mass 'm' is released, with a horizontal speed v from the top of a smooth and fixed, hemispherical bowl, of radius r. the angle theta w.r.t. the vertical where it leaves contact with the bowl is |
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Answer» `sin^(-1)""((v^2)/(3rg)+(2)/(3))` |
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| 37. |
In a ballistics demonstration a police officer fires a bullet of mass 50.0 g with speed 200 ms^(-1) (see Table 6.2) on soft plywood of thickness 2.00 cm. The kinetic energy. What is the emergent speed of the bullet? |
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Answer» Solution :The initial kinetic energy of the bullet is `mv^(2)"/"2= 1000 J`. It has final kinetic energy of `0.1xx1000= 100 J`. If `v_(F)` is the EMERGENT speed of the bullet, `(1)/(2) mv_(f)^(2)= 100 J` `v_(f)= sqrt((2xx100 J)/(0.05 kg))` `= 63.2 ms^(-1)`. The speed is reduced by approximately 68% (not 90%). |
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| 38. |
A body starts with initial velocity u and moves with uniform acceleration.when the velocity has increased to 5u, the acceleration is reversed in direction, the magnitude remaining constant. Find its velocity when it returns to the starting point? |
| Answer» ANSWER :A | |
| 39. |
Ratio of average PE and average KE in SHM will be |
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Answer» 1:1 |
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| 40. |
An inventor claims to have developed an engine that during a certain time interval takes in 110 MJ of heat at 415 K, rejects 50 MJ of heat at 212 K while manages to do 16.7 kW of work. Do you agree with the inventor's claim ? |
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Answer» Solution :The CLAIMED EFFICIENCY `eta_c= W/(Q_H) = ( 16.7k Wh )/( 110 mJ)` butas ` 1kWh = 10 ^3 xx ( j xx S ) xx ( 60 xx 60 s)= 3.6MJ` `Soeta_C= ( 16.7xx 3.6)/( 110 )= 0.55= 55 %` whilemaximumpossibletheoeticalefficiency ` eta_("max")=1- (T_L )/( T_H)= (T_H -T_L)/(T_H)= ( 415 -212 )/( 415 ) = 0.49= 49 % ` Eqns. (1) and (2) it is CLEAR that claimed efficiency is greater than maximum possible theoretical efficiency, so inventor.s claim does not appear to be correct. |
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| 41. |
An object located at the origin at rest has mass M. It explodes into three pieces having masses M/4, M/3 and 5M/12. The pieces move on a horizontal frictionless plane (call it x-y plane). The fragment with mass M/4 moves with velocity 5.0 ms^(-1)at 37° to the x-axis. The fragment of mass M/3 moves wifh velocity 4.0 ms^(-1)in the direction making an angle 45° to the x-axis. a) What are the velocitity components of the third piece ? b) What can be said about the motion of centre of mass of the system after tile explosion ? |
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Answer» |
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| 42. |
A man thinks to remain in equilibrium by pushing in his hands and feet against two vertical parallel walls as the shown in the figure :(a) He must exert equal forces on both walls(b) The forces of friction at both walls must be equal(c ) The coefficients of friction between man and wall must be the at both ends(d) Friction must be present on both walls |
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Answer» A and B are CORRECT |
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| 43. |
20 g of steam at 100^(@)C is passed into 100 g of ice at 0^(@)C. Find the resultant temperature if specific latent heat of steam is 540 cal/g., specific latent heat of ice is 80 cal/g and specific heat of water is 1cal//g""^(@)C. |
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Answer» Solution :LET the resultant temperature of the mixture be `t^(@)C`. In the method of mixtures, heat lost by steam = heat gained by ICE `m_(s)L_(s)+m_(s)S(100-t)=m_(c)L_(c)+m_(c)s(t-0)` where `m_(s)` is the mass of steam, `L_(s)` is the latent heat of steam, s is the specific heat of water, `m_(c)` is the mass of ice and `L_(c)` is the latent heat of ice. Here, `m_(s)=20g,L_(s)=540cal//g,s` = `1cal//g""^(@)C,m_(c)=100G,L_(c)=80cal//g` `20xx540+20xx1(100-t)=100xx80+100xx1xxt` `10800+2000-20t=8000+100t` 120t = 4800 `t=40^(@)C`. |
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| 44. |
The gravitational force between two particles of masses m_1 and m_2 seperated by certain distance in air medium is F. If they are taken to vaccum and seperated by the same distance, then the gravitational force between them will be |
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Answer» GREATER than F |
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| 45. |
In an experiment of simple pendulum, the errors in the measurement of length of the pendulum (L) and time period (T) are 2% and 2% respectively. The maximum percentage error in the value of (L)/(T^(2)) is |
| Answer» SOLUTION :`6%` | |
| 46. |
A body is projected with a velocity 2"/"3rd of escape velocity from the surface of the earth is order to orbit the earth. Its orbital radius will be (R_e is radius of the earth) |
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Answer» `(9/8)R_(E )` |
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| 47. |
Simplify the following. Using Binomial theoremE=K[(1+(Deltatheta)/theta_(0))-1] |
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Answer» |
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| 48. |
What are thermal rediation ? What is the velocity of these radiation in vaccum. |
| Answer» SOLUTION :Heat ENERGY transmited by radiation is CALLED THERMAL radiation. The VELOCITY of thermal radiation in vaccum is `3 x 10^8 m/s`. | |
| 49. |
A body rolls down an inclined plane. If its K.E of rotational motion is 40% if its K.E. of translation, the body is |
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Answer» cylinder |
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| 50. |
A projectile has the maximum range of 500m. If the projectile is now thrown on an inclined plane of 30^(0) with the same speed, what is the distance covered by it along the inclined plane? |
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Answer» Solution :`R_(max)=(u^(2))/(G)"":.500=(u^(2))/(g)orusqrt(500g)` `V^(2)-u^(2)=2gs rArr 0-500g =2xx-(gsin30^(0))XXX` x=500m. |
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