Saved Bookmarks
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. |
If L and R reperesnt respectively the inductance and resisitance the dimensions of L/R are |
|
Answer» `M^0L^0T^0 ` |
|
| 2. |
A point on the rim of a whell 3 m in diameter has linear velocity of 18 ms^(-1). The angular velocity of the wheel is |
|
Answer» `4 "RAD s"^(-1)` |
|
| 3. |
Illustrate the existence of surface tension with example. |
Answer» SOLUTION :(i) Needle floats on water surface:A greased needle of steel and a piece of BLOTTING paper are take and gently placed over the water surface. Blotting paper soaks water and soon sinks down but the needle keeps floatinng. The floating needle CAUSES a litte depression, the FORCES F, DUE to the surface tension of the curved surface are inclined as shown in Figure. Thevertical components of these two forces support the weight of the needle. Now add liquid soap to the water and stir it. It is found that the needle sinks. (ii) Take a plastic sheet and cut out a piece in the shape of a boat. Atapering and smooth front with a notch at the back is suggested. A piece of camphor is put into the notch of the boat. Gently release the boat on the surface of the water and we find that the boat is propelled forward when the camphor dissolves. The surface tension is lowered, as the camphor dissolves and produces a difference in surface tension in the water nearby the notch. This causes the water to flow away from the back of the boat, which moves the boat forward. |
|
| 4. |
An object A of mass .m. is located at a point .P. at distance .r. and .2r. from two planets B and C of masses .M. and .6M. respectively as shown in the figure. If theescape of the object A form point .P. due to the gravitational infulence of only planet B is 5ksm^(-1) , then the escape speed of the object A from point .P. due to the gravitational influence of both the planets is ......... kms^(-1) . |
|
Answer» 1 |
|
| 5. |
If a monoatomic gas at 17^@C is suddenly compressed to 1/8th of initial volume B, then the new temperature will be ……. |
|
Answer» 887 K `T_1V_1^(gamma-1)=T_V_2^(gamma-1)` `THEREFORE T_2=T_1(V_1/V_2)^(gamma-1)` but `T_1`=17+273 =290 K `=290xx(8)^(5/3-1) "" [because gamma=5/3]` `=290xx(2^3)^(2/3)` =290 x 4 `therefore T_2` = 1160 K |
|
| 6. |
A tuck moving at 36 km. h^(-1) has to be brought to rest by applying brakes in such a way that there is no relative displacement of the goods in the truck with respect to the truck floor. Coefficient of friction between the goods and the truck floor is 0.2. Find the minimum distance the truck must move before it comes to rest . |
|
Answer» |
|
| 7. |
If the weight of the body and upthrust on the body in a liquid are equal, then the body |
|
Answer» Sinks |
|
| 8. |
An artificial satellite orbiting the earth doesnot falldown, because the earth.s attraction |
|
Answer» is balanced by the attraction of the moon |
|
| 9. |
A car takes a turn with velocity 50 ms^(-1) on the circular road of radius of curvature 10 m. Calculate the centrifugal force experienced by a person of mass 60 kg inside the car? |
|
Answer» Solution :Given: MASS of a person = 60 kg Velocity of the car `= 50 MS^(-1)` Radius of curvature = 10 m CENTRIFUGAL force ` F = (mV^2)/(R ) = (60 xx 50^2)/(10) = 15.00 N` |
|
| 10. |
The mass of spaceship is 1000kg. It is to be launched .from the earth.s surface out into free space. The values of .g. and .R. (radicles of earth) are 10m//s^2 and 6400km respectively. The required energy for this work will be: |
|
Answer» `6.4 XX 10^(11)` joules |
|
| 11. |
Statement 1: The angular momentum of a body moving in a circle with constant velocity remains conserved about any point on the circumference of the circle. Statement 2: If the torque acting on the body is zero than its angular momentum is conserved. Which one of thefollowing options is correct? |
|
Answer» Statement 1 and 2 are true and statement 2 is a correct explanation for statement 1. |
|
| 12. |
A bullet fired vertically up from the ground reaches a point inits path after 3s and 13 s, both times being taken from the instant of firing the bullet. The velocity of projection of the bullet is (g=10ms^(-2)) |
|
Answer» `80MS^(-1)` |
|
| 13. |
A pendulum clock that keeps correct time on the earth is taken to the moon. It will run |
|
Answer» at correct RATE |
|
| 14. |
A 500 kg car, moving with a velocity of 36 km h^(-1) on a straight road undirectionally, doubles its velocity in one minute. The power delivered by the engine for doubling the velocity is |
|
Answer» 750 W According to work ENERGY theorem Work done = change in kinetic energy = `1/2 mv^2 - 1/2 m u^2` `= 1/2 xx 500 [20^2 - 10^2] = (500 xx 30 xx 10)/(2)` The power delivered by the engine is `P=(500 xx 30 xx 10)/(2 xx 60) W = 1250 W` |
|
| 15. |
A body falls towards earth in air. Will its total mechanical energy be conserved during the fall ? Justify. |
| Answer» Solution :No, because a small PART of its energy is used against AIR resistive force which is non-conservative force. In this CONDITION, gain in KE < loss in PE. | |
| 16. |
Which is the weakest force of all the basic forces in nature? |
| Answer» SOLUTION : AMONG all basic forces, gravitational force is the WEAKEST forces in NATURE. | |
| 17. |
A piece of wood floats in water kept in a beaker. If the beaker moves with a vertical acceleration a, the wood will. |
|
Answer» SINK deeper in the LIQUID if `a` is upward |
|
| 18. |
Sqrings of constants k, 2k , 8k, 16k, .. Are connected in series. The mass 'm' kg is attached to the lower end of the last spring and the system is allowed to vibrate. The frequency of oscillation is |
|
Answer» `1/(pi) SQRT(m/(2K))` |
|
| 19. |
The accuracy in. measurement depends on the limit or the resolution of the measuring instrument. A physical quantity P is related to four observables a, b, c and d as P=(a^3b^2)/sqrt(cd) percentage errors in measurements of a, b, c and d are 1%,3%,2% and 3% respectively. What is the percentage error in the measurement ofP? |
| Answer» Solution :`(DELTAP)/Pxx100=3xx(DELTAA)/axx100+2xx(DELTAB)/bxx100+1/2xx(DELTAC)/cxx100+(Deltad)/(d)xx100=3xx1+2xx3+1/2xx2+3=13%` | |
| 20. |
Explain an example in which heat transfer is necessary. |
| Answer» SOLUTION :In a nuclear reactor, heat transfer systems need to be installed so that the enormous energy PRODUCED by nuclear FISSION in the CORE transits out sufficiently fast, THUS preventing the core from overheating. | |
| 21. |
State whether the fundamental law of calorimetry is applicable in the following cases : (ii) a chemical reaction occurs between a solid and a liquid in a calorimeter |
| Answer» SOLUTION :During a CHEMICAL reaction, heat is EVOLVED or absorbed. If this heat is not taken into consideration the fundamental law of calorimetry cannot be applied | |
| 22. |
There is a sphere of radius R and mass m. It is rotating with angular speed omega about a horizontal axis perpendicular to the plane of paper. Now it is placed gently between two inclined surface as shown in figure. The coefficient of friction between the sphere and each incline surface is mu. The time after which the rotation of the sphere will stop (2omega(1+mu^(2)))/(bsqrt(2gmu)) whereb is |
|
Answer» |
|
| 23. |
Let theta_(1)andtheta_(2) be the temperature of two ends of a metal rod 'AB' of length 'L' in steady state (theta_(1)gttheta_(2)). If theta be the temperature of the section at a distance 'x' from the hot end A then find 'theta'. (assume steady state conditions) |
|
Answer» SOLUTION :TEMPERATURE gradient is constant. i.e., `(theta_(1)-theta)/x=(theta-theta_(2))/(L-x)` After solving, we get `theta=theta_(1)-((theta_(1)-theta_(2))/L)x`. |
|
| 24. |
For body thrown horizontally from the top of a tower, |
|
Answer» the TIME of FLIGHT depends both on h and v |
|
| 25. |
A bomb is thrown vertically upwards. At its topmost position it explodes into a number of fragments. What would be the locus of the centre of gravity of the bomb ? |
| Answer» Solution :FORCES developed due to an explosion are INTERNAL forces and therefore the motion of the centre of gravity does not get affected. Hence, the centre of gravity would COME down VERTICALLY like a freely FALLING body. | |
| 26. |
The rotaitonal kinetic energy E=1/2I omega^(2). Use this equation to get the dimensional formula for omega, where I is the moment of inertia and omega is the angular velocity. |
|
Answer» |
|
| 27. |
At highest point velocity of projectile is …... (theta = angle of projection). |
|
Answer» |
|
| 28. |
Mention the factors on which moment of inertia depends. |
|
Answer» Solution : MOMENT of inertia of a BODY depends (i)on the distribution of mass in the body. If heavy mass is away from the axis of ROTATION, then M.O.L. increases, (II)M.O.L. depends on axis of rotation. M.O.L. isminimum about the axis passing through centre of mass in that plane. If axis changes from that position in the same plane M.O.L. increases. `I=I_(G)+MR^(2)` |
|
| 29. |
A copper calorimeter contains 100g of water at 16°C. When 15g of ice is added to it, the resultant temperature of the mixture is 4°C. Water equivalent of the calorimeter is |
|
Answer» 10g |
|
| 30. |
Problesm with accuracy are due to errors'. Is the statement true? |
| Answer» SOLUTION :YES, it is TRUE. | |
| 31. |
A particle perfor uniform circular motion with an angular momentum L. If the frequency of its motion is doubled and its kinetic energy is halved , the angular momentum becomes |
|
Answer» 2L |
|
| 32. |
A room contains oxygen and hydrogen moleculein the ratio 3:1 . Thetemperature of the room is 27^(@)C .The molar mass ofO_(2) is 32 g mol^(-1) and for H_(2) 3 g mol^(-1) . The value of gas constant R is 8.32 J mol^(-1)K^(-1) calculate: (a) rms speed of oxygen and hydrogen molecule. (b) Average kinetic energy per oxygen molecule and per hydrogen molecule. (c) Ratio of average kinetic energy ofoxygen molecules and hydrogen molecules. |
|
Answer» SOLUTION :Absolute temperature `T= 27^@C = 27 + 273 = 300K` Gas CONSTANT `R = 8.32 J "mol"^(-1) K^(-1)` For oxygen MOLECULE: molar mass ` M =32 gm//"mol" = 32 = xx 10^(-3) kg "mol"^(-1)` rms speed ` v_(rms) =sqrt( (3RT)/(M) ) = sqrt((3 xx 8.32 xx 300)/(32 xx 10^(-3) ))= 483.73 ms^(-1) ~~ 484 ms^(-1)` For hydrogenmolecule: Molar mass M ` = 2 xx 10^(-3) kg "mol"^(-1)` rms speed ` v_(rms) = sqrt( (3RT)/(M) ) = sqrt( ( 3 xx 8.32 xx 300)/(2 xx 10^(-3) )) = 1934 ms^(-1) = 1.93 K ms^(-1)` Note that the rms speed is inversely proportional to `sqrtM`and the molar mass of oxygen is 16 TIMES higher than molar mass of hydrogen. It implies that the rms speed of hydrogen is 4 times GREATER than rms speed of oxygen at the same temperature. ` (1934)/(484) = 4` |
|
| 33. |
If there are n no. of molecules per unit volume and m is the mass of each, V_(x) is the x - component of velocity pressure can be written as, |
|
Answer» `P=nmV_(X)^(2)` |
|
| 34. |
A bal of mass 5 kg hangs from a spring of force constant N//m and oscillates with certain time period. If the ball is removed, the spring shortens by |
|
Answer» `(K/G)m` |
|
| 35. |
Calculate the work done when a water drop of diameter 2mm is split into 10^(9) identical water droplets. Surface tension of water is 7.4xx10^(-2)Nm^(-1). |
|
Answer» |
|
| 36. |
A stone is dropped from the top of a tower of height 125 m. The distance travelled by it during last second of its fall is (g=10ms^(-2)) |
| Answer» ANSWER :B | |
| 37. |
A stone of mass 1kgis whirled in a circular pathof radius 1m find out the tension in the string if the linear velocity is 10 m/s |
| Answer» Solution :`T =(mv^(2))/(R )= (1xx (10)^(2))/(1 )= 100N` | |
| 38. |
The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on 60^@latitude becomes zero. |
|
Answer» `3sqrt((g_p)/R)` |
|
| 39. |
Two objects of masses 4kg and 3kg are placed along X and Y axes respectively at a distance of 1m from the origin. An object of mass 1kg is kept at the origin of the co-ordinate system. Calculate the resultant gravitational force attraction on the object at the origin. |
|
Answer» SOLUTION :Force due to 4kg mass on 1 kg mass at the ORIGIN is `F_1=(Gxx4xx1)/1^2=4G` Force due to 3 kg mass on 1KG mass at the origin is `F_2=(Gxx3xx1)/1^2`=3G The resultant force is found by using parallelogram law of vectors `F=sqrt(F_1^2+F_2^2+2F_1F_2costheta)` `=sqrt((4G)^2+(3G)^2+2(4G)(3G) COS 90^@)`= 5G 
|
|
| 40. |
A body is launched from the earths surface at an angle alpha =30^(@) to horizontal at a speed V_(0) = sqrt((1.5 GM)/(R )) Neglect air resistance and earths rotation. The height to which the body will rise from the surface of earth is (R radius of earth M -mass of earth and G-universal gravitational constant) |
|
Answer» 1.323 R |
|
| 41. |
A ball is projected with velocity v_(0) = sqrt(gh) at an angle 'theta' to the horizontal towards a vertical smooth wall distant 'd' from the point of projection. After collision the ball returns to the point of projection. The coefficient of restitution is |
|
Answer» `(d)/(h sin THETA - d)` |
|
| 42. |
A helicopter of m ass 1000 kg rises with a vertical a c c ele ra tio n o f 15 ms-^(2) . The crew an d th e passengers weigh 300 kg. Give the magnitude and direction of the,(a) Force on th e floor by th e crew and passengers. (b) Action of the rotor of the helicopter on the surrounding air. (c) Force on the helicopter due to the surrounding air. |
|
Answer» Solution :(a) ‘Free BODY’:crew and passengers Force on the system by the floor = F UPWARDS, weight of system = mg downwards, ` thereforeF – mg = ma` ` F- 300xx 10 = 300 xx 15` ` F= 7.5 xx 10^3 N ` upward By the Third Law, force on the floor by the crew and passengers = `7.5 xx 10^3 N`downwards. (B) ‘Free body’ :helicopter plus the crew and passengers Force by air on the system = R upwards, weight of system = mg downwards ` therefore R – mg = ma` `R - 1300 xx 10=1300 xx 15` ` R= 3.25 xx 10^4 N `upwards By the Third Law, force (action) on the air by the helicopter = `3.25 xx 10^4 N ` downwards. (c ) `3.25 xx 10^4 N ` upwards |
|
| 43. |
In equation of motion S=ut + 1/2 kt^(2) , the dimensional formula for K is................. |
|
Answer» `[LT^(-1)]` |
|
| 44. |
F_(g), F_(e) and F_nrepresent the gravitational, electro-magnetic and nuclear forces respectively, then arrange them in increasing order of their strengths |
|
Answer» `F_(N),F_(E),F_(G)` |
|
| 45. |
Pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of containing vessel. This law was first formulated by |
|
Answer» Reynolds |
|
| 46. |
If v_1 and v_2 be the velocities at the end of focal chord of projectile path and u is the velocity at the vertex of the path, then |
|
Answer» `v_1^2 + v_2^2 = U^2` |
|
| 47. |
Find the potential energy of a system of four particles placed at the vertices of a square of side l. Also obtain the potential at the centre of the square. |
|
Answer» Solution : Consider FOUR masses each of mass m at the corners of a square of side l,We have four mass pairs at distance l and two diagonal pairs at distance`sqrt(2)`l hence, W(r)= - 4 `(Gm^(2))/(l) - 2 (Gm^(2))/(sqrt(2)l)` = - `(2.Gm^(2))/(l ) (2 + (1)/(sqrt(2))) = - 5.41 (Gm^(2))/(l)` The gravtational potential at the CENTRE of the square `(r = sqrt(2)l//2)` is U(r)= - `4 sqrt(2) (Gm)/(l)` . 
|
|
| 48. |
How can we classify variables and constants based on dimension. Give example for each type? |
|
Answer» Solution :Dimensional variables : Physical quantities, that possessdimensions and have variable values are called dimensional variables. EG: Length, velocity, and acceleration etc. Dimensionless variables : Physical quantities that have no dimensions, but have variable values are called dimensionless variables. Eg: Specific GRAVITY, strain, refractive index etc. Dimensional CONSTANT : Physical quantities that posses dimensions and have constant values are called dimensional constants. Eg : Gravitational constant, Planck'sconstant etc. Dimensionless Constant : Quantitie which have constant values and also have no dimensions are called dimensionless constants. |
|
| 49. |
Neglecting the density of air ,the terminal velocity obtained by a raindrop of radius 0.3 mm falling through air of viscosity 1.8xx10^(-5)Nsm^(-2) will be …. |
|
Answer» `10.9ms^(-1)` `eta=1.8xx10^(-5)Nms^(-2)` Terminal velocity `v_(t)=(?)` `v_(t)=(2)/(9)(r^(2)g(rho-rho_(o)))/(eta)` `rho` = DENSITY of water `=10^(3)kg//m^(3)` `rho_(o)`=neglecting air density `v_(t)=(2)/(9)((9xx10^(-8))(9.8)(10^(3)))/(1.8xx10^(-5))` `=10.88` `v_(t)=10.9m//s` |
|
| 50. |
For a two - body system in absence of external forces, the kinetic energy as measured from ground frame is K_(0) and from center of mass frame is K_(cm). Pick up the wrong statement(s). |
|
Answer» The kinetic ENERGY as measured from center of mass frame is least. |
|