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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. |
An athelete jumps at an angle of30^@ with a maximum speed of9.4 m//s. Find the range covered by him in the above jump. Suggest the angle by which |
| Answer» SOLUTION :`R=(mu^2sin2theta)/G=((9.4)^2sin60^@)/(9.8)=7.8m` RANGE will be MAXIMUM when `theta=45^@` | |
| 2. |
A pond has an ice layer of thickness 3 cm. If K of ice is 0.005 CGS units, surface tempera-ture of surroundings is -20^(0)C, density of ice is 0.9 gm/cc, the time taken for the thickness to increase by 1 cm is |
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Answer» 30 min |
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| 3. |
Obtain dimension of rho gv where, rho = density, g = acceleration, v = velocity |
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Answer» Solution :`[rhogv]=[rho][g][v]` `[ML^(-3)][LT^(-2)][LT^(-1)]` `-[ML^(-1)T^(-3)]` |
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| 4. |
United physical quantity |
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Answer» can have non-ZERO dimensions |
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| 5. |
The specific heat capacity of hydrogen at constant'pressure and constant volume are 14280 J kg ^(-1) K ^(-1) and 10110 J kg ^(-1) K ^(-1) respectively. Calculate the value of the universal gas constant. |
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Answer» <P> Solution :`C _(P) - C _(V) = R//M.14280 -10110= R//2 xx 10 ^(-3) , R = 8.34 mol ^(-1) K^(-1).` |
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| 6. |
F_(g), F_(e) and F_(n) represent the gravitational electromagnetic and nuclear forces respectively, then arrange them in increasing order of their strengths |
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Answer» `F_(N), F_(E), F_(G)` |
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| 7. |
Findmomentof inertia of a sector of mass M , radius R and of centralangle theta radians . |
Answer» SOLUTION : `I=(I_(DISC))/n = 1/2 ((Mxxn))/n R^(2) = 1/2 MR^(2)` |
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| 8. |
Two cylindrical vessels of the same type contain a liquid of density rho. The bases of both the vessels lie on the same horizontal plane. The depth of the liquid in the left vessel is h_(1) and that in the right vessel is h_(2). The area of cross-section of the base of each vessel is A. If the vessels are connected by a tube, then how much is the work done by gravity to equalize the levels of the liquid in the vessels (suppose h_(1)gth_(2))? |
Answer» SOLUTION :When the vessels are connected, the level of the liquid in the left-hand VESSEL will fall and that in the right-hand one will go up.  The initial difference in the liquid levels in the vessles = `h_(1)-h_(2)`. When the levels in both the vessels become the same, the level of liquid in the left-hand vessel will fall by `1/2(h_(1)-h_(2))` and that in the right-hand vessel will rise by `1/2(h_(1)-h_(2))`. MASS of this liquid = `1/2(h_(1)-h_(2))Arho`. The centre of gravity of this liquid moves up against gravity by `1/2(h_(1)-h_(2))`. `therefore` Increase in potential ENERGY of this liquid = work DONE against gravity `1/2(h_(1)-h_(2))Arhoxxgxx1/2(h_(1)-h_(2))=1/4Arhog(h_(1)-h_(2))^(2)`. |
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| 9. |
A mass 2kg is attached to the spring of spring constant 50 Nm^(-1). The block is pulled to a distance of 5 cm from its equilibrium position at x= 0 on a horizontal frictionless surface from rest at t=0. Write the expression for its displacement at anytime t. |
Answer» Solution :SPRING block system is shwon as in figure and it executes SHM with amplitude of 5 CM from mean position.  Here, spring CONSTANT `k= 50 N"/"m`, amplitude A= 5cm, mass attached m = 2 kg. ANGULAR frequency `OMEGA = sqrt((k)/(m))= sqrt((50)/(2))= 5 rad"/"s`. Displacement of block at time t, `y(t) = A sin (omega t+ phi)," where "phi=` initial phase at time t=0 `y(0)= A sin (phi)` `A= A sin phi""[therefore " at time "t=0, y= +A]` `therefore 1= sin phi` `therefore phi = (pi)/(2) rad` Required equation, `y(t) = A sin(omega t+phi)` `= 5sin (omega t+(pi)/(2))` `therefore y(t) = 5 cos omega t` `y(t)= 5 cos 5t` is a required equation where t is in second and y is in cm. |
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| 10. |
If two persons sitting on a swing instead of one, why the periodic time does not changed? |
| Answer» Solution :PERIODIC TIME of oscillation for a swing is `T= 2pi sqrt((l)/(G))`, here there is no mass term PRESENT. | |
| 11. |
A wire of mass 9.8 xx 10^(-3) kg per metre passes over a frictionless pulley fixed on the top of an inclined frictionless plane which makes an angle of 30^(@)with the horizontal Masses M_(1) and M_(2) are tied at the two ends of the wire The mass M_(1) rests on the plane and mass M_(2) hangs freely vertically downwards . The whole systemis in equilibrium Now a transverse wave propagates along the wire with a velocity of100 m//sIf g = 9.8 m//s^(2) calculate the valuse of masses M_(1) and M_(2) . |
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Answer» Solution :Here, ` m = 9.8 XX 10^(-3) kg//m` `theta = 30^(@) , G = 9.8 m//s^(2) ` ` upsilon = 100 m//s , M_(1) = ? M_(2) = ? ` The various forces acting on the system are shown in As the system of two masses is in equilibrium therefore , ` T = M_(1) g sin theta = M_(1) g sin 30^(@) = (M_(1) g )/(2) ` ` R = M_(1) g cos theta = M_(1) g cos 30^(@) = M_(1) g sqrt3/(2) ` Also ` T = M_(2) g` From (i) and (III) , ` T = (M_(1) g )/(2) = M_(2) g ` ` M_(1) = 2 M_(2) ` Now the velocity of transverse waves is ` upsilon= sqrt((T)/(m) ` ` T = upsilon^(2) xx m = (100)^(2) xx 9.8 xx 10^(-3) = 9.8 N ` From(iii) `M_2 = (T)/(g) = (98)/(9.8) = 10 kg ` From `M_(1) = 2 M_(2) = 2 xx 10 kg = 20 kg ` |
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| 12. |
Figure shows a rod of length l resting on a wall and the floor. Its lower end A is pulled towards left with a constant velocity v. Find the velocity of the other end B downward when the rod makes an angle theta with the horizontal. |
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Answer» Solution :In such type of problems, when velocity of one part of a body is GIVEN and that of other is required, we first FIND the relation between the two displacements, then differentiate them with respect to time. Here if the distance from the corner to the point A is x and up to B is y. Then `v = (dx)/(dt)` & `v_(B) = -(dy)/(dt)` (-sign DENOTES that y is decreasing) Further, `x^(2) + y^(2) = l^(2)` Differentiating with respect to time t `2x(dx)/(d t) + 2y(dy)/(d t) = theta ""x v = y v_(B)` `v_(B) = (v)(x)/(y) = v cot theta` |
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| 13. |
What is position vector? |
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Answer» Solution :It is vector which DENOTES the POSITION of a particle at any instant of TIME, with respect to some reference frame or coordinate system. The position vector `vecr` of the particle at a POINT P is given by `vecr=xhati+yhatj+zhatk` where x,y and Z are components of `vecr`. |
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| 14. |
In the above problems the accelration due to gravity is: |
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Answer» `-10ms^(2)` `g=-10m//s^(2)` |
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| 15. |
Explain Poisson's Ratio. |
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Answer» Solution :Poisson.s RATIO: The ratio fo LATERAL strain to ongitudinal is known as Possion.s ratio. In case of cylindrical rod if it has length l and radius r, after stretching, its radius decreases and length increases `therefore (Delta r)/( r ) =- (mu Delta l )/( l ) ""...(1)` Volume `V = pi r ^(2) l ""...(2)` FOr small CHANGE `(Delta V)/(V) = 2 (DELTAR )/(r) + (Delta l )/(l)` From equation (1), `(Delta V)/(V) = - 2mu (Delta l)/(l) + (Delta l)/(l) ""...(3)` `therefore (Delta V)/(V) = (Delta l)/(l) (1- 2mu ) = epsi _(1) (1- 2mu ) ""...(4)` Equation (4) shows that , for `Delta V gt 0` value of `mu` can not be increases from `0.5.` This equation is true for any cross-section of rod. |
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| 16. |
A composite bar has two segments of equal length L each. Both segments are made of same material but cross sectional area of segment OB is twice that of OA. The bar is kept on a smooth table with the joint at the origin of the co - ordinate system attached to the table. Temperature of the composite bar is uniformly raised by Delta theta. Calculate the x co-ordinate of the joint if coefficient of linear thermal expansion for the material is alpha^(@)C^(-1) |
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| 17. |
Two particles each of mass m, collide head on when their speeds are 2u and u. If they stick to each other on impact, when moving in opposite directions, their combined speed is |
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Answer» u |
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| 18. |
Suppose a machine consists of a cage at the end of one arm The arm is hinged at O as shown in figure such that the cage revolves along a vertical circle of radius r at constant linear speed v = sqrt(gr). The cage is so attached that the man of weight W, standing on a weighing machine inside the cage, remains always vertical. Then : |
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Answer» The reading of his weight on the MACHINE is equal to W at all positions |
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| 19. |
Figure shows the strain-stress curve for a given material. What are (a) Young's modulus and (b) approximate yield strength for this material ? |
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Answer» Solution :(a) From the given GRAPH for stress `150 xx 10 ^(6) Nm ^(-2)` corresponds strain is `0.002` `therefore` Young.s modulus ,= SLOPE of graph stress `to` strain `therefore` Young.s modulus `Y= ("Stress")/("Strain") =(150 xx 10 ^(6))/(0.002)` `therefore Y=7.5 xx 10 ^(8) Nm ^(-2)` (b) From graphs the YIELD strength of a material is the maximum stress, `= 300 xx 10 ^(6) Nm ^(-2)` `= 3 xx 10 ^(8) Nm ^(-2)` REALLY the yield strength is slightly less than `3 xx 10 ^(8) N m ^(-2)` |
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| 20. |
Explain why is hot soup more testy than the colder one ? |
| Answer» SOLUTION :The surface tesion of hot SOUP is LESS than that of COLDAND hence hot soup SPREADS over a larger area of the tongue . This makes hot soup more tasty than the cold one . | |
| 21. |
If specific heat of a substance is infinite, it means |
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Answer» HEAT is given out |
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| 22. |
A spherical ball of radius 3.0xx10^(-4)m and density 10^(4) kg//m^(2) falls freely under gravity through a distance h before entering a tank of water. If after entering the water the velocity of the ball does not change, find h in km. Viscosity of water is 9.8xx10^(-6)N^(-s)//m^(2). |
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Answer» `1.65` |
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| 23. |
The time period of a small satellite of mass .m. which describe a circular oribt of radius .r. very close to a spherical planet of mass M and of density rho = 1 kg m^(-3) is : |
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Answer» `(2pir^(3//2))/(SQRT(GM))` |
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| 24. |
The quanity of heat which can rise the temperature of x gm of a substance through t_(1)""^@C can rise the temperature of y of water through t_(2)""^@C is same . The ratio of specific heats of the substances is |
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Answer» `yt_(1)//xt_(2)` |
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| 25. |
On what factor(s) coefficient of linear expansion of a metal depend? Initial length, initial temperature, scaleof length or scale of temperature ? |
| Answer» SOLUTION :On SCALE of TEMPERATURE | |
| 26. |
Consider a Carnot's cycle operating between T_1 = 500 K and T_2 = 300 K producing 1 kJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs. |
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Answer» SOLUTION :Temperature of source `T_1` = 500 K Temperature of sink `T_2` = 300 K WORK done per cycle W = 1 kJ = 1000 J Heat transferred to the engine per cycle `Q_1` = ? Efficiency of a carnot engine , `eta=1-T_2/T_1` `=1-300/500` `=200/500` `=2/5` Now, `eta=W/Q_1` `THEREFORE Q_1=W/ eta=1000/2xx5` = 2500 J As `Q_1-Q_2=W` `therefore Q_2=Q_1-W`=2500 J -1000 J = 1500 J |
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| 27. |
Consider a sisc rotating in the horizontal plane with a constant angular speedomega about its centre (O). The disc has a shade region on one side of the diameter and an unshaded region on the other side as shown in the Fig. 2 (CF) . 37, When pebbles (P) and (Q) ar simultaneously proected at an angle towards. (R ). The velocity of projection is in they-z plane and is same for both pebbles with respect to the disc . Assume that (i) they lang back on the disc beforec teh disc has completed /8 rotation, (ii) their range is less than half the disc radius, and (iii),omega remains constant throughout. Then . |
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Answer» (P) lands in the SHADED ragion and (Q) in the unshaded region ` T= time ofr ` 1/8 th` rotation of disc ` = ( 2 PI)/(omega) xx 1/8 = (pi0/( 4 omega)`  .GIVEN range of each pebble ` lt r //2`. Therefore , pebble (P) lands in unshaded region and pebble (Q) the shaded region. Theus option (b) is true. |
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| 28. |
If object starts moving with acceleration beta t, then what will be the velocity after 't' time ? |
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Answer» SOLUTION :`a = BETA t ` `therefore (DV)/(dt) = beta t ` `therefore dv = beta t dt` ` therefore int dv = beta int t dt ` `therefore v = (beta t ^(2))/(2)` |
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| 29. |
A body initially at 80^(@)C cools to 64^(@)C in 5 minutes and to 52^(@) C in 10 minutes.What will be the temprature after 15 minutes and what is the temperature of the surroundings ? |
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Answer» solving `theta_(0)=16^(@)C, (52-THETA)//5alpha[(52+theta)//2]-16`solvin`theta=43^(@)C` |
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| 30. |
One solid sphere A and another hollow sphere B are of same mass and same outer radii. Their moments of inertia about their diameters are respectively I_(A)and I_(B) such that where d_(A) and d_(B) are their densities . |
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Answer» `I_(A) = I_(B)` |
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| 31. |
A ball is dropped freely from rest. If it travels a distance 55m in the 6th second of its journey find the acceleration. |
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Answer» |
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| 32. |
A ball of mass 0.4 kg moving with a uniform speed of 2 ms. strikes normally a wall and rebounds. Assuming the collision to be elastic and the time of contact of the ball with the walls is 0.4 s, find the magnitude of force exerted on the ball. |
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| 33. |
A particle is executing simple harmonic motion with an amplitude of 2m. The difference in the magnitudes of its maximum acceleration and maximum velocity is 4. The time period of its oscillation and its velocity when it is 1m away from the mean position are respectively, |
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Answer» `2s, 2sqrt(3)MS^(-1)` |
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| 34. |
If rate of heat loss ((dQ)/(dt))of two bodies of different masses are same, then |
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Answer» their RATES of fall of TEMPERATURE `((d theta)/(dt))` will be same |
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| 35. |
A ball of mass m collides with the ground at an angle alpha with the vertical. If the collision lasts for time t, the average force exerted by the ground on the ball is : (e = coefficient of restitution between the ball and the ground) |
Answer» Solution :IMPULSE = CHANGE in LINEAR momentum  `thereforeFt = m (eu cos alpha+u cos alpha) or F = (MU cos alpha(1+e))/(t)` |
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| 36. |
A small block of mass m slides along a smooth track as shown in the fig. (i) If it starts from rest at P, what is the resultant force acting on it at Q? (ii) At what height above the bottom of the loop should the block be released so that the force it exerts against the track at the top of the loop equals its weight? |
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Answer» Solution :By CONSERVATION of mechanical energy between points P and Q mg. (5R) `= mg R+ (1)/(2) mv^(2)` i.e., `v= sqrt(8gR)` Now in case of circular MOTION, N (or T) `= (mv^(2))/(R ) + mg cos theta` And as at Q, `theta = 90^(@)`  `N= (mv^(2))/(R )= (m(8gR))/(R )= 8mg` So RESULTANT force on m at Q. `F= sqrt((8mg)^(2) + (mg)^(2))= (sqrt65) mg` (ii) At highest point `N= (mv^(2))/(R )- mg (" as " theta = 180^(@))` But according to GIVEN problem N= mg so `(mv^(2))/(R ) + mg + mg` i.e., `v= sqrt(2gR)` If for ACHIEVING it h. the height, by conservation of mchanical energy again `mgh. = (1)/(2) mv^(2) + mg (2R)` or `h. = 2R + (v^(2))/(2g) = 2R + (2gR)/(2g)= 3R` |
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| 37. |
A disc is given an initial angular velocity omega_(0) and placed on a rough horizontal surface as shown. The quantities which will not depend on the coefficient of friction is/are |
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Answer» The TIME until rolloing begins |
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| 38. |
A screw gauge gives , the following reading when used to measure the diameter of a wire main scale reading: 0 mm Circular scale reading : 52 divisions Given that 1 mm on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is |
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Answer» `0.52` CM |
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| 39. |
Calculate the angular speeds of the second, minute and hour hands of a 12 hour dial clock. |
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Answer» Solution :(1) `omega` of the second that `=(2pi)/(T)=(2xx3.142)/(60)=1.047xx10^(-1)rads^(-1)` (2) `omega` of the minute hand `=(2xx3.142)/(60xx60)=1.745xx10^(-3)rads^(-1)` (3)`omega` of the hour hand=`(2xx3.142)/(12x3600)=1.454xx10^(-4)RAD^(-1)` |
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| 40. |
An ideal diatomic gas for which molar heat capacity at constant volume is C_v= (5R)/(2)is enclosed in is a closed vessel of volume 8m^3 at a pressure of 8 xx 10^5//m^2and temperature 400K. 10^6Jof heat is transferred to the gas. The final equilibrium pressure of the gas is |
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Answer» `8 XX 10^5 N//m^2` |
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| 41. |
(A) : The rise in temperature and increase in internal energy takes place quickly, when the gas heated at constant volume. (R ): The gas does no work, when it is heated at constant volume. |
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Answer» Both (A) and(R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 42. |
A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on control panel. The lift starts moving upwards with an acceleration 1.0 m//s^2. If g = 10 ms^(-2), the tension in the supporting cable is |
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Answer» `86 00 N` massof lift`, m= 940kg ,a= 1 m//s^2, , g= 10 m//s^2` LetT BETHE tensionin thesupportingcable `thereforeT -(M +m ) g= ( M+m) a` `T=(M+m) (a +g)` `=( 940+ 60)( 1+10)= 11000 N` 
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| 43. |
In column-l, the projection of a particle moving along a circle (uniformly) in x-y plane with its centre of origin along x and y axes, while in Column-II, the description of particle's motion is given. It is given that particle's angular velocity is constant and equal to omega and the radius of circle is A, delta != 0, pi/2 or pi For this situation match the column-I with column-II {:("Column - I","Column - II"),("A) "x(t)=Asin(omegat+delta-(pi)/(2))","y(t)=Asin(omegat+delta),"P) Uniform circular motion (clockwise)"),("B) "x(t)=A sin (omegat+delta)","y(t)=A sin (omegat+delta+(pi)/(2)),"Q) Uniform circular motion(anti - clockwise)"),("C) "x(t)=Asin (omegat)","x(t)=A cos (omegat),"R) At = 0 particle is neither on x - axis nor on y - axis"),("D) "x(t)=A cos (omegat+delta)","y(t)=A cos (omegat+delta+(pi)/(2)),"S) At t =0 particle is either on x - axis on y - axis"):} |
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Answer» |
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| 44. |
A seconds pendulum is mounted in a rocket. Its period of oscillation decreases when the rocket |
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Answer» comes down with UNIFORM ACCELERATION |
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| 45. |
An ice skater spins with arms outstreched at 1.9 rps. Her moment of inertis at this instant is 1,33 kg m^(2). She pulls in her arms to increase her rate of spin. If the moment of inertia is 0.48 kg m^(2) after she pulls in her arms, what is her new rate of rotation? |
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Answer» |
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| 46. |
The angular velocity and the amplitude of a simple pendulum are omega and a, respectively. The ratio of its kinetic and potential energies at a displacement x from the mean position is…………… |
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Answer» `(X^(2)omega^(2))/(a^(2)-x^(2)omega^(2))` Potential energy `U=(1)/(2)momega^(2)x^(2)implies(K)/(U)=(a^(2)-x^(2))/(x^(2))` |
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| 47. |
A spherical soap bubble A of radius 2 cm is formed inside another bubble B of radius 4 cm. Show that the radius of a single soap bubble which maintance the same pressure difference as inside the smaller and outside the larger soap bubble is lesser than radius of both soap bubbles A and B. |
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Answer» Solution :From the EXCESS pressure INSIDE a soap bubble `DeltaP = (4T)/R` Here the two bubbles having the same pressure and temperature. So the radius of the combined bubbles. `1/R = 1/(R_1) + 1/(R_2)` `1/R = 1/2 + 1/4 = (2 + 1)/4 = 3/4` `R = 4/3 = 1.33` `:. "" R = 1.33 cm` |
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| 48. |
If a projectile is projected with initial velocity 98 m/s in a direction of 30^(@)above the horizontal. For this situation which one of the following is a correct statement ? |
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Answer» Time of flight is 100 s `g = 9 . 8 m//s^(2)` `theta = 30^(@)` Time of flight`t_(f) = (2u SIN theta)/(g)` `t_(f) = (2 xx 98 xx sin 30^(@))/(9 . 8 ` `= 2 xx(8 xx 1)/(9 . 8 xx 2) = 10 s` |
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| 49. |
Zero errors of an instrument introduces |
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Answer» SYSTEMATIC error |
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