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 particle is projected with velocity 'u' at an angle 'theta' with horizontal {:("Column-I","Column-II"),("A) Average vertical velocity in complete journey","P) " u sin theta),("B) Average horizontal velocity in half of the journey","Q) "u cos theta),("C) Average velocity in half of journey","R) zero"),("D) Average velocity in complete journey","S) "U sqrt((sin^(2) theta)/(4)+cos^(2)theta)),(,"T) " U sqrt((1+3 sin^(2) theta)/(4))),(,):} |
|
Answer» |
|
| 2. |
Sir C. V. Raman got Nobel prize for his experiment on |
|
Answer» DISPERSION of light |
|
| 3. |
What is the time period of ratation of the earth around its axis so that the objects at the equator becomes weightless? (g=9.8 m//s^2 , Radius of earth =6400 km) |
|
Answer» Solution :g at the equator is `g_0=g-g_0=g-Romega^2` If bodies are to become WEIGHTLESS at the equator, `g_0=0`. `0=g-Romega^2 rArr Romega^2 =g` `omega=sqrt(g/R)` Time PERIOD of roration, `T=(2PI)/omega=2pisqrt(R/g)` `R=6400xx10^3` m , `g=9.8 m//s^2` `T=2pisqrt((6.4xx10^6)/9.8)`=5078 s =84 MINUTE 38 s. |
|
| 4. |
Sand drops from a stationary hopper at the rule of 5kg s^(1) onto a conveyor belt moving with a constant speed of 2ms^(-1). What is the force required to keep the belt moving and what is the power delivered by the motor moving the belt? |
|
Answer» Solution :Impulse J= Ft`-mv, F= (mv)/(t)` `v=2 MS^(-1), (m)/(t)= 5kgs^(-1) rArr F= 10N` POWER `P= (W)/(t)= (FS)/(t) = Fv` `=10 xx 2 = 20` watt |
|
| 5. |
The initial angular velocity of a heavy flywheel rotating on its axis is w_0. Its angular velocity decreases due to friction. At the end of the first minute its angular velocity is 0.8 omega_(0). What is its angular velocity at the end of third minute, if the frictional force is constant ? |
|
Answer» |
|
| 6. |
When will the motion of a simple pendulum be simple harmonic? |
| Answer» Solution :When the angular displacement `theta` is very SMALL MEANS `sin theta approx theta`, the MOTION of simple pendulum be considered as simple HARMONIC motion. | |
| 7. |
By placing the prism in minimum deviation position, images of the spectrum. |
|
Answer» BECOMES inverted |
|
| 8. |
A train stops at two stations s distance apart and takes time t on the journeyfrom one station to the other. Its motion is first of uniform acceleration a and then immediately of uniform retardation b, then therelation between a, b, s, t is 1/b+ 1/b = (kt^(2))/(2s) . The value of 'k' is |
|
Answer» |
|
| 9. |
A spherical black body with a radius of 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doulbed, the power radiated by it in watt would be |
|
Answer» 225 |
|
| 10. |
(A) : Acceleration of a moving particle can change its direction without any change in direction of velocity. ( R) : If the direction of change in velocity vector changes, the direction of acceleration vector also changes. |
|
Answer» Both (A) and ( R) are ture and ( R) is the correct EXPLANATION of (A) |
|
| 11. |
A carnot freezer takes heat from water at 0°C inside it and rejects it to the room at a temperature of 27° . The latent heat of ice is 336 xx 10^(3) "JKg"^(-1). If 5 kg of water at 0°C is converted into ice a 0°C by the freezer, then the energy consumed by the freezer is close to : |
|
Answer» `1.51 XX 10^(5)` J |
|
| 12. |
The radius of a pipe decreases according to r=r_(o)e^(-propx) , where alpha=0.50m^(-1)andxis the distance of a cross - section from the first end (x=0) . Find the ratioof Reynolds number for two cross -sections lying at the distance of 2m from each other . (take e=2.718) |
|
Answer» Solution :Reynold number `N_(R)=(rhovD)/(eta)` `THEREFORE` For a given liquid `N_(R)propvD` `therefore((N_(R))_(1))/((N_(R))_(2))=(v_(1))/(v_(2))XX(D_(1))/(D_(2))`….(1) ACCORDING to the equation of continuity, `A_(1)v_(1)=A_(2)v_(2)` `thereforepir_(1)^(2)v_(1)=pir_(2)^(2)v_(2)` `therefore(v_(1))/(v_(2))=(r_(2)^(2))/(r_(1)^(2))=((D_(2))/(D_(1)))^(2)`....(2) From equation (1) and (2) , `((N_(R))_(1))/((N_(R))_(2))=(r_(o)e^(-propx2))/(r_(o)^(-propx1))=e^(-alpha(x_(2)-x_(1)))` `=e^(-(0.5)(2))` `=e^(-1)` `(1)/(2.718)` `therefore((N_(R))_(1))/((N_(R))_(2))=0.368` |
|
| 13. |
The work done to increase unit area of the liquid surface at same temperature is called. |
|
Answer» SURFACE tension |
|
| 14. |
Figure shows distance - time graphs of two objects A and B . Which object is moving with a greter speed when both are moving ? |
Answer» Solution :A linecorresponding to the object B MAKES a LARGER angle with the time -AXIS. Its SLOPE is , therefore , larger than the slope of the line CORRESPONDING to the object A . Thusm the speed of B is greater than that of A.
|
|
| 15. |
The power end a capillary tube of diameter 2.00 mm is dipped in water 8 cm below the surface in a beaker it so that an air bubble is formed at the end of the tube inside water. The excess pressure inside the bubble will be (in Pascal) (T = 73 dyne/cm) |
|
Answer» 146 |
|
| 16. |
How much heat energy in joules must besupplied to 14 gms of nitrogen at room temperature to raise its temperature by 40^(@)C at constant pressure ? (Mol. Wt. of N_(2) = 28 gm, R = constant) |
|
Answer» 50R |
|
| 17. |
A bullet moving at 20 m/sec. strikes a wooden plank and penetrates 4 cm before coming to stop. The time taken to stop is |
|
Answer» 0.08 sec |
|
| 18. |
A wall of dimensions 2.00 m by 3.50 m has asingle-panewindowof dimensions0.75 m by 1.20 m .Ifthe inside temperature is 20^(@)C and the outside temperatureis -10^(@)C, effective thermalresistanceof the opaque wall and windoware 2.10 m^(2) KW^(-1) and 0.21m^(2) K W^(-1) respectively. The heat flowthroughthe entirewall will be . |
|
Answer» `215 W` The temperature difference, `T_(H) - T_(C) = 30 K` Area of window, `A_(i) = (0.75) (1.20) = 0.90 m^(2)` Heat flow through window pane, `((dQ)/dt)_(1)= (T_(H) - T_(C))/(R_(1)) = (0.90)(30)/(0.21) = 128.6 W` The area of the wall, `A_(2) = (2.00)(3.50) - (0.75)(1.20) = 6.10 m^(2)` Heat flow through the wall, `((dQ)/dt)_(2) = (T_(H) - T_(C))/(R_(2)) = ((6.10)(30))/((2.10)) = 87 W` Net heat flo, `(dQ)/(dt) = 128.6 + 87 = 215.6 W` |
|
| 19. |
Calculate the value of the gas constant for one gram of hydrogen, given that the density of hydrogen at N.T.P is 0.00009 g//cm^(3) |
|
Answer» 4.12J/K-g |
|
| 20. |
A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. If the minimum force that can be applied on A so that both the blocks move together is 12N, the maximum force that can be applied on B for the blocks to move together will be: |
|
Answer» 25N |
|
| 21. |
A simple pendulum of length one meter has a bob of mass 100 g. It is displaced through an angle 60^(@) from the vertical and then released. Kinetic energy of the bob when it passes through the mean position (g=10 m//s^(2)) |
| Answer» ANSWER :D | |
| 22. |
Friction between any two surfacesin contact is the force that opposes the relative motion between them The force of limiting friction (F) between any two surfaces in contact is directly proportional to the normal reaction (R) between them F prop R or F = mu R where mu is coefficient of limiting friction If theta is angle of friction then mu = tan theta A horizontal force of 1.2kg is applied on a 1.5kg block which rests on a horizontal surface If the coefficient of friction is 0.3 force of friction is . |
|
Answer» `0.45kgf` |
|
| 23. |
Friction between any two surfacesin contact is the force that opposes the relative motion between them The force of limiting friction (F) between any two surfaces in contact is directly proportional to the normal reaction (R) between them Fprop R or F = mu R where mu is coefficient of limiting friction If theta is angle of friction then mu = tan theta The acceleration produce in the block in the above question is . |
|
Answer» `9.8ms^(-2)` `f=` APPLIED force - force of friction `=1.2 -0.45 = 0.75 kg f = 0.75 XX 9.8 N` `a = (f)/(m) = (0.75 xx 9.8)/(1.5) = 4.9ms^(-2)` . |
|
| 24. |
An artificial satellite circulating the earth is at a height 3400 km from the surface of earth. If the radius of the earth is 6400 km and g = 9.8ms^-2, calculate the orbital velocity of the satellite. |
| Answer» SOLUTION :`v_0` = `SQRT(GM)/(R_g+h)` = `sqrt(gR_g^2)/(R_g+h)` = `sqrt(9.8 XX(6400xx10^3))^2/sqrt((6400 + 3400) 10^3` = `6400ms^-1` | |
| 25. |
A carnot engine, having an efficiency of eta=1//10 as heat engine, is used as a refrigerator. If the work done on the system is 10J, the emount of energy absorbed from the reservoir at lower temperature is |
|
Answer» 100J |
|
| 26. |
Equal quantities of ice at 0^@ and steam at 100^@C are mixed up. What will be the resultant temperature? |
|
Answer» |
|
| 27. |
A ball of mass 100 g is suspended by a string of 40 cm long. Keeping the string always taut, the ball describes a horizontal circle of radius 10 cm. Find the angular speed of the ball. |
|
Answer» |
|
| 28. |
Pendulum clocks generally run fast in winter and show in summer.Why? |
| Answer» Solution :TIME period of a pendulum is given by In summer,I increases with the increases of temperature. So T increases and the clock RUNS slow.The reverse happens in winter.To MINIMISE this invar with very small coefficient of expansion is used to MAKE pendulum clocks. | |
| 29. |
In the figure below, two systems A and B are seperated by an insulating fixed wall and insulated from surrounding by insulated wall |
|
Answer» The PRESSURE of A nd B will change with time |
|
| 30. |
In demaped oscillatory motion a block of mass 200 g is suspended to a spring of force constant 90N//m in a medium and damping constant is 40g/s.Find a time period of oscillation b. time taken for its amplitude of oscillationto drop to half of its initial value (c) time taken for its mecahnical enrgy to drop to half of its initial value. |
|
Answer» SOLUTION :Mass `m =200g=0.2kg` force constant k=90N/m damping constnat b=40g/s=0.04kg/s `sqrt(km)=sqrt(90xx0.2)=sqrt(18)` kg/s Here `b lt lt sqrt(km)` a. time period `T=2pisqrt(m/k)=2pisqrt(0.2/90)=0.3s` b. AMPLITUDE `=AE^(-bt//2m)` LET amplitude is dropped to half of its initial value after the time `T_(1//2)` Amplitude `A.e^((-b.T_(1//2))/(2m))=A/2` `impliese^((-b.T_(1//2))/(2m))=1/2` Take natural lograithms on both sides `(-b.(T_(1//2)))/(2m)="In"(1/2)impliesT_(1//2)=(In(2))//(b//2m)` `=2.302xx0.3010xx2m//b` `T_(1//2)=0.693xx(2m)/b=0.693xx(2xx0.2)/(0.04)=6.93s` c. Let the energyis dropped to half of its initial value after a time `t_(1//2)`. Initial energy `E_(0)=1/2kA^(2)` At time `t_(1//2)`, energy `=1/2E_(0)` `=1/2kA^(2)e^((-b.t_(1//2))/m)=1/2(1/2kA^(2))` `e^((-b.t_(1//2))/m)=1/2impliest_(1//2)=In(2)xxm/b=0.693xxm/b` `t_(1//2)=0.693xx0.2/0.004=3.46s` |
|
| 31. |
Assume that earth is a spherical planet of uniform density rho, radius R_(e), Mass m_(e) and acceleration due to gravity g. Then the gravitational constant G can be written as (3g)/(4pi rho R_(e))(b) (gR_(e)^(2))/(M_(e)) (c ) (3g)/(4pi rho R_(e)^(2)) (d) (12g)/(4pi rho R_(e)) |
|
Answer» only a & B are TRUE |
|
| 32. |
Consider the following equation of Bernoulli theorem P+(1)/(2)rho v^(2)+ rhogh= K (constant) Which of the following quantity has same dimensions as that of (K)/(P)? |
|
Answer» Thrust K has the same dimensions as each one of the FACTORS on the LHS i.e `P, (1)/(2)rhov^(2)` and `rhogh` `:. [(K)/(P)]= [M^(0)L^(0)T^(0)]` Angle has no dimensions , i.e `[THETA]= [M^(0)L^(0)T^(0)]` Hence, `[(K)/(P)]= [theta]` |
|
| 33. |
Of the following thermometers the one which is most useful for the measurement of a rapidly varying temperature is a |
|
Answer» PLATINUM resistance thermometer |
|
| 34. |
A 100 kg piston encloses 32 g of oxygen gas temperature of 27^@Cin a vertical cylinder of base area of 4 cm^2 . The air pressure outside is 1 xx 10^5 Pa . The axis of the cylinder is vertical, and the piston can move in it without friction. How much heat is to be transferred to the gas to raise the piston by |
| Answer» Answer :A | |
| 35. |
The work done per unit time in rotational motion is given by |
|
Answer» `vecF.v` |
|
| 36. |
What is meant by periodic and non-periodic motion ? Give any two examples , for each motion ? |
|
Answer» Solution :Periodic motion: Any motion which repeats itself in a fixed TIME INTERVAL is known as periodic motion. Examples: Hands in pendulum clock, swing of a cradle. Non-Periodic motion: Any motion which does not repeat itself after a REGULAR interval of time is known as non-periodic motion. EXAMPLE: Occurrence of Earth quake, eruption of volcano. |
|
| 37. |
Equal force is applied on one heavy weight and one light weight body . Which body experiences more work by applied force ? |
| Answer» SOLUTION :LIGHT WEIGHT BODY . | |
| 38. |
List the condition for the destructive interference to take place. |
|
Answer» Solution :`COS varphi=-1` `rArr varphi= 3pi, 3pi, 5pi, =(2n-1) PI`, where `n=1,2,...` i.e., this is the phase difference in which two WAVES OVERLAP to give destructive interference. |
|
| 39. |
Calculate the period of oscillation of a body falling freely inside the tunnel created along the diameter of theearth. Given average density of the material of the Earth 5500kgm^(-3) and universal gravitational constant 6.67xx10^(-11)Nm^(2)kg^(-2). |
|
Answer» Solution :We KNOW that time PERIOD of oscillation of a body dropped in the tunnel, `T=sqrt((3pi)/(Grho))` `"i.e."T=sqrt((3xx3.142)/(6.67xx10^(-11)xx5500))=sqrt((0.2567)/(10^(-8)))=0.5069xx10^(4)s=5069s` `T="1 hr 24 min 29 sec"` |
|
| 40. |
What is the nature of the displacement-time graph of a body moving with constant velocity ? |
|
Answer» |
|
| 41. |
(A) : If a body is in translational equilibrium it need not to be in rotational equilibrium. (R ) : In translation equilibrium net force is zero. |
|
Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
|
| 42. |
Calculate the centripetal acceleration of the Earth which orbits around the Sun. The Sun to Earth distance is approximately 150 million km. (Assume the orbit of Earth to be circular) |
|
Answer» SOLUTION : The centripetal acceleration `a_c = (v^2)/(r )` V - velocity of Earth around the ORBIT r - radius of orbit or distance of Earth to Sun Velocity of Earth is written in terms of angular velocity (`omega` ) as By substituting in the centripetal acceleration formula, `a_c = (omega^2 r^2)/(r ) =omega^2 r` But` omega = (2pi)/(T )`where T is time for the Earth to orbit around the Sun, which is one year. T = 365 days =` 365 xx 24 xx 60 xx 60s` `T = 31 xx 10^7`s `omega= 2.02 xx 10^(-7 )`rad per SEC `a_c = (2.02 xx 10^(-7))^2 xx (150 xx10^8) ` ` a_c =6.12 xx 10^(-3)ms^(-2)` |
|
| 43. |
Water falling from a 100 m high fall is to be used for generating electric energy. If 1.8 xx 10^(5) kg of water falls per hour and half the gravitational potential energy can be converted into electric energy, how many 100 W lamps can be lit using the energy ? |
|
Answer» |
|
| 44. |
Can a body have energy without momentum? |
| Answer» Solution :Yes, there is an INTERNAL energy in a body due to the THERMAL AGITATION of the particles of the body, while the vector SUM of the momenta of the moving particles may be zero. | |
| 45. |
A bimetallic strip made up of brass and iron remains linear at 20^(@)C. When the temperature is decreased to 0^(@)C, the strip bends.Which material remains on the convex side of the bent strip? |
|
Answer» |
|
| 46. |
The total mechanical energy of a harmonic oscillator of A =1mand force constant200 Nm^-1 is 150j. Then |
|
Answer» <P>The minimum P E is Zero |
|
| 47. |
Consider the following two statements. (A) Linear momentum of the system remains constant. (B) Centre of mass of the system remains at rest. |
|
Answer» A IMPLIES B and B implies A |
|
| 48. |
In anadiabaticexpansion of the airthevolumeis increased by 4%what ispereentage changein pressure?(Forair y = 1.4) |
|
Answer» <P> Solution :From equation for adiabatic process,`PV^(GAMMA)="constant"` Using differentiation, we get `Pgamma V^(gamma-1)dV+dP.V^(gamma)=0` `(dP)/(P)=-gamma(dV)/(V )` Volume .V. is increased. By `4%` and `gamma=1.4` `(dP)/(P)xx100=-gamma((dV)/(V)xx100)=-1.4xx4=-5.6` PRESSURE is DECREASED. by `5.6%`. |
|
| 49. |
The gross radiation emitted by a perfectly black body is |
|
Answer» dependent on its TEMPERATURE |
|
| 50. |
Name the factors that decide the state of matter . |
| Answer» SOLUTION :INTERATOMIC FORCE and TEMPERATURE | |