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. |
Consider a geosynchronous communications satellite of mass m placed in an equitorial circular orbit of radius r_(0). These satellite have an "apogee engine" which provides the thrusts needed to reach the final orbit. Once an error by the ground controllers causes the apogee engine to be fired. the thrust happens to be directed the Earth and, despite the quick reaction of the ground crew to shut the engine off, an unwanted velocity variation Deltav is imparted on the satellite. we characterize this boost by the parameter beta=Deltav//v_(0). the duration of the engine burn is always negligible with respect to any other orbital times, so that it can be considered as instantaneous.[Hing : under the action of central forces obeying the inverse square law, bodies follow trajectories descirbed by ellipses, parabolas or hyperbolas. in the approzimation mlt ltM the gravitating mass M is at one of the focuses. where l is a positive constant named the semilatus rectum and epsilon is the eccentricity of the curve. in terms of constants of motion: l=(L^(2))/(GMm^(2)) and epsilon=(1+(2EL^(2))/(G^(2)M^(2)m^(3)))^(1//2) where G is the Newton constant, L is the modulus of the angular momentum of the orbiting mass, with respect to the origin, and E is its mechanical energy, with zero potential energy at infinity. suppose betalt1 Determine the distance r_(min) of the satellite to the earth centre |
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Answer» `(betar_(0))/2` |
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| 2. |
The resistance of ap platinum theromometer at 0^(@)C, 100^(@)C and 203^(@)C is found to tbe 3.5, 5.2 and 6.9 ohms respectively. Find the temperature at which the resistance of the thermometer is 9.4 ohms. |
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Answer» <P> |
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| 3. |
Consider a geosynchronous communications satellite of mass m placed in an equitorial circular orbit of radius r_(0). These satellite have an "apogee engine" which provides the thrusts needed to reach the final orbit. Once an error by the ground controllers causes the apogee engine to be fired. the thrust happens to be directed the Earth and, despite the quick reaction of the ground crew to shut the engine off, an unwanted velocity variation Deltav is imparted on the satellite. we characterize this boost by the parameter beta=Deltav//v_(0). the duration of the engine burn is always negligible with respect to any other orbital times, so that it can be considered as instantaneous.[Hing : under the action of central forces obeying the inverse square law, bodies follow trajectories descirbed by ellipses, parabolas or hyperbolas. in the approzimation mlt ltM the gravitating mass M is at one of the focuses. where l is a positive constant named the semilatus rectum and epsilon is the eccentricity of the curve. in terms of constants of motion: l=(L^(2))/(GMm^(2)) and epsilon=(1+(2EL^(2))/(G^(2)M^(2)m^(3)))^(1//2) where G is the Newton constant, L is the modulus of the angular momentum of the orbiting mass, with respect to the origin, and E is its mechanical energy, with zero potential energy at infinity. suppose betalt1 Determine the new time period of the satellite if (beta=1/4) |
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Answer» `~~20 HRS` |
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| 4. |
Two blocks A and B are connected to each other by a string and a spring, the string passes over a frictionless pulley as shwon in figure. Block B slides over the horizontal top surface of a stationary block C and block A slides along the vertical side of C, both with the same uniform speed. The coefficient of friction between the surfaces of blocks is 0.2. Force constant of the spring is 1960 N/m. If mass of block A is 2 kg calculate the mass of block B and the energy stored in the spring. |
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| 5. |
using components methods, add the following vectorsvec(A) = A_(x) hat(i) + A_(y)hat(j) + A_(z) hat(k)vec(B) = B_(x) hat(i) + B_(y) hat(j) + B_(z) hat(k) . |
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Answer» SOLUTION :The two vectors `vec(A) and vec(B)`can be expressed as in a Cartesian coordinate system . `vec(A) = A_(x) hat(i) +A_(y)hat(j) +A_(Z) hat(k)` `vec(B) = B_(x) hat(i) + B_(y) hat(j) + B_(z) hat(k)` Then the addition of two vectors is EQUIVALENT to adding their CORRESPONDING x , y and z components . `:. vec(A) + vec(B) = (A_(x) + B_(y)) hat(i) + (A_(y) +B_(y))hat(j) + (A_(z) + B_(z))hat(k)` |
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| 6. |
The diameter of a flywheel is 1 m. It has a mass of 20 kg. It is rotating about its axis with a speed of 120 rotations in one minute. Its angular momentum in kg m s' is |
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Answer» `13.4` `r = (d)/(2) = (1)/(2) m = 0.5` m `omega = 2 pi epsilon = (2PI XX 120)/(60) = 4pi rad s^(-1)` `THEREFORE L = I omega = (Mr^(2))/(2) omega = (20 xx (0.5)^(2) xx 4pi)/(2)` `=10 xx 0.25 xx 4 xx 3.14 = 31.4 kg m^(2) s^(-1)` |
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| 7. |
The equation of a transverse wave is given by y = 10 sin {pi (0.01 x - 2t)}where y and x are incm and t is in seconds. Its frequency is ……….. |
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Answer» `10 s^(-1)` we have `omega = 2pi` Therefore , Frequency `F = (2pi)/(omega) = 1 s^(-1)` |
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| 8. |
Two identical co-centric rings each of mass M and radius R are places perpendicularly. What is the moment of inertia of the system about axis of one of the rings? |
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Answer» `1/2MR^2` |
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| 9. |
110J of heat is supplied to a gaseous system whose internal energy increases by 40J . The amount of externalwork done is |
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Answer» 150J |
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| 10. |
The dimensions of (a)/(b)in the equation P= (a^(2)-t^(2))/(bx)where P ispressure,x is distance and t is time are |
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Answer» `[M^(2)LT^(-3)]` From `P=(a-t^(2))/(bx)= (t^(2))/(bx)` `:. [b]= ([t^(2)])/([Px])= ([T^(2)])/([ML^(-1)T^(-2)][L])= [M^(-1)L^(0)T^(4)]` `:. [(a)/(b)]= ([T^(2)])/([M^(-1)L^(0)T^(4)])= [ML^(0)T^(-2)]` |
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| 11. |
What is the name of the force of attraction between the molecules of two different substances ? |
| Answer» SOLUTION :ADHESIVE FORCE | |
| 12. |
Paraffin wax expands on melting. The melting point of wax will |
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Answer» Increase with PRESSURE |
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| 13. |
A large block of ice 5 m thick has a vertical hole drilled in it and is floating in a lake. the minimum length of the rope required to draw a bucket full of water through the hole is ( density of ice = 900 "kg/m"^(3) ) |
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Answer» 0.5 m |
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| 14. |
You measure two quantities as A = 1.0 m +-0.2 m, B = 2.0 m +- 0.2 m . We should report correct value for sqrt(AB) as |
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Answer» `1.4 m+-0.04 m` Let `Y= sqrt(AB)= sqrt((1.0)(2.0))=1.414 m` ROUNDING off to two significant digit Y=1.4 m `(DeltaY)/(Y)=(1)/(2)[(DELTAA)/(A)+(DeltaB)/(B)]=(1)/(2)[(0.2)/(1.0)+(0.2)/(2.0)]=(0.6)/(2xx2.0)` `DeltaY=(0.6Y)/(2xx2.0)=(0.61xx1.4)/(2xx2.0)=0.212` Rounding off to one significant digit`DeltaY=0.2 m` Thus `sqrt(AB)=r+-Deltar=1.4+-0.2 m` |
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| 15. |
Three force F,F and sqrt(2) F acting at a point are in equilibrium. Find the angles between the forces. |
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Answer» Solution :Fig. shows the THREE force. Let the angles between the forces be `alpha, BETA` and `gamma`, as shown in the figure. `:. " " alpha+beta+gamma = 360^(@)` or, `" "beta+gamma= 360^(@)-alpha` `:. "" sin (beta+gamma) = -SINALPHA` By Lami.s theorem, `(sqrt(2)F)/(sinalpha)=(F)/(sinbeta)=(F)/(singamma)` `:. " " beta=gamma` Also, `" "(F)/(sinbeta)=(sqrt2F)/(sinalpha)or, sinalpha= sqrt(2)sinbeta.` Now, `" sinalpha=-sin(beta+gamma)=-sin2beta=-2sinbetacosbeta` Hence, `-2 sinbeta cos beta= sqrt(2)sin beta` or,`"" cosbeta=-(1)/(sqrt(2))=cos 135^(@)` `:. " " beta = 135^(@)=gamma` and `" " a= 360^(@) -(135^(@)+135^(@)) = 90^(@)`.
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| 16. |
Two masses 1.65 kg and 3.30 kg are joined together by a massles rod parallel to an incline to 30^(@), over which both slide down, with the 1.65-kg mass trailing behind the 3.30-kg mass. The coefficient of friction between the first mass and the plane is 0.226 and that between the second mass and the plane is 0.113 Find (a) the tension in the rod, (b) the common acceleration of the two masses [Hint : Make free-body diagram of the two bodies and consider their motion along and perpendicular to the plane] |
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| 17. |
The ratio of SI unit of CGS unit of a plysical constant is 10^(7). That costant is |
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Answer» UNIVERSAL GAS constant |
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| 18. |
Given, vecomega=2hatkandvecr=2hati+2hatj. Find the linear velocity. |
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Answer» `4hati+4hatj` `VECV=vecomegaxxvecr` `therefore vecv=(2HATK)xx(2hati+2hatj)` `=(2hatkxx2hati)+(2hatkxx2hatj)` `=4hatj+(-4hatj)(because hatkxxhati=hatj,hatkxxhatj=-hati)` `therefore vecv=-4hatixx4hatj` |
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| 19. |
A ball of mass m kg hangs from a spring of constant k. The ball oscillates with a period of T seconds. If the ball is removed the spring is shortened by |
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Answer» Solution :Using `T=2pisqrt(m/K)`, we GET `1/k` `=(T^(2))/((2pi)^(2)m)impliesk=((4pi)^(2))/(T^(2))m`…………1 For equilibrium `kx=mgimpliesx=(MG)/k` ……….2 Substitute 1 in 2 `impliesx=(gT^(2))/(4pi^(2))` |
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| 20. |
A unidirectional force F varying with time t as shown in the Fig. acts on a body initially at rest for a short duration 2T. Then the velocity acquired by the body is |
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Answer» `(piF_(0)T)/(4m)` |
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| 21. |
Molar gas constant is the same for all gases because at the same temperature and pressure, equal volumes of all gases have the same |
| Answer» Answer :A | |
| 22. |
One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is |
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Answer» 0.25 |
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| 23. |
A small object of uniform density roll up a curve surface with an initial velocity v. It reaches up to a maximum height of (3v^(2))/(4g) with respect to the initial position. The object is |
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Answer» Ring |
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| 24. |
A mancan swim atthe rate of4 km *h^(-1) in still water. At what angle should the man set himself relative to the current of 2 km*h^(-1) to cross the river perpendicular to the banks ? |
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Answer» |
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| 25. |
A ball impings directly upon another ball at rest and is itself brough to rest by the impact. If half of initial kinetic energy is destroyed in the collision, the coefficient of restitution is |
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Answer» `0.3` |
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| 26. |
Sand is falling on a conveyor belt at the rate of 5kg s^(-10. The extra power required to move the belt with a velocity of 6ms^(-1) is, |
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Answer» 30W |
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| 27. |
Velocity of sound in O_(2)gas is 460 ms^(-1) at certain temperature. Velocity of sound in He gas at the same temperature would be ....... (Consider both the gases to be ideal). |
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Answer» `1419 MS ^(-1)` considering oxygen and helium gases as first and second gas RESPECTIVELY, `(v_(1))/(v _(2) ) = sqrt ((T _(1))/( T _(2)) xx ( M _(02))/( M _(01)) xx (gamma _(1))/( gamma _(2))) ""(because ` R is constant) `therefore (460)/(v _(2)) = sqrt(1 + (4)/( 32) xx (7//5)/(5//3))""(because T_(1) - T _(2))` `therefore v _(2) = 460 xx sqrt ((32)/(4) xx (5)/(3) xx (5)/(7)) = 1419 (m)/(s)` |
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| 28. |
The distance travelled by a falling body in the last second of its motion, to that in the last but one second is 7:5, the velocity with which body strikes the ground is |
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Answer» 19.6 m/s |
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| 29. |
A boy stretches the rubber cord of his catapult by 6 cm. IF 0.588 N of force is required for stretching the cord by 1 cm, then calculate the average force acting on the cord of the catapult. With what velocity will a stone of mass 8 g be projected under such condition? |
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| 30. |
Why athlete taking long jump run for some distance before taking jump |
| Answer» SOLUTION :Whenathleteis runninghismomentumincrease. In giventimewhenmomentumincreasemoreforceact.Dueto thislargeforceha cantakelongjump . | |
| 31. |
A body falls freely from a height .h. after two seconds if accelaration due to gravity is reversed the body |
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Answer» CONTINUES to fall down |
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| 32. |
Length AB in the figure shown is 5m. The body is released from A. Friction is sufficient for pure rolling to take place. Suppose we have four bodies ring disc, solid sphere end hollow sphere. The angle theta is now gradually increased. Which body will start slipping in least time. All the bodies have same mass and radius, coefficient of friction is also same |
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Answer» Ring |
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| 33. |
A log of wood of mass 120 kg floats in water. The weight that can be on the raft to make it just sink, should be( density of wood = 600 "kg/m"^(3) ) |
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Answer» 80 kg |
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| 34. |
Length AB in the figure shown is 5m, The body is released from A. Friction is sufficient for pure rolling to take place. The maximum time which anybody (which can roll) can take to reach the bottom is : |
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Answer» 8s |
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| 35. |
A spherical mirror, whose reflecting surface is curved inwards, that is,faces towards the centre of the sphere, is called a |
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Answer» Concave mirror `"Area of ABG F"=2xx4=8` `"Area of BGHC"=(1)/(2)(4+2.5)xx0.5=1.625` `"Area of CDEH"=2xx2.5=5` `"Total area under F - t graph = Change in momentum"` `RARR m(V-u)=18.625` `rArr v=(18.625)/(2)+5=14.25ms^(-1)` |
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| 36. |
A spiral spring is stretched by a force. What type of strain is produced in it? |
| Answer» SOLUTION :SHEAR STRAIN. | |
| 37. |
A monkey of mass 15kg is climbing on a rope with one end fixed to the ceiling. If it wishes to go up with an acceleration of 1m//s^2 , how much force should it apply to the rope ? If the rope is 5m long and the monkey starts from rest, how much time will it take to reach the ceiling ? |
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Answer» `120N , SQRT10 SEC ` |
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| 38. |
In which case heat can be added to a body without causing any rise in temperature? |
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Answer» Solution :During CHANGE of state HEAT energy can be ADDED to a substance but there will be no change in temperature. Eg. : When boiling water is CONVERTING into steam. |
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| 39. |
Define one kilogram. |
| Answer» SOLUTION :ONE kilogram is the mass of the PROTOTYPE cylinder of platinum iridium alloy (whose height is equal to its DIAMETER), preserved at the International Bureau of Weights and MEASURES at Serves, near Paris, France. | |
| 40. |
Find the maximum velocity of a SHM particle having amplitude 'A' and periodic time T. |
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Answer» `(4A)/(T)` |
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| 41. |
A thin straight tube of uniform bore is movable about its centre on a smooth horizontal table and it contains a uniform thin rod of the same length and mass whose centre of gravity is nearly at the middle point of the tube. If the system be set in motion with angular velocity omega, the angular velocity of the tube as the rod leaves it is omega//b where b is |
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| 42. |
If a ball is thrown up in a moving train, it comes back to the thrower's hands. Why? |
| Answer» Solution :Both during its upward and downward MOTION, the ball continues to MOVE INERTIA of motion with the same HORIZONTAL velocity as the train. In this period, the ball covers the same horizontal distance as the train and so it COMES back to the thrower.s hand. | |
| 43. |
The frequency upsilonof vibrations of uniform string of length l and stretched with a force F is given by upsilon = (p)/(2l)sqrt((F)/(m)) where p is the number of segments of the vibrating string and m is constant of the string. What is the dimensions of m? |
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Answer» `[ML^(-1)T^(-1)]` `upsilon^(2)= (p^(2))/(4l^(2))(F)/(m)` or `m= (p^(2)F)/(4l^(2)upsilon^(2))` `:.` dimensions of m `("dimensions of m")/("dimensions of" l^(2) xx "dimensions of" upsilon^(2))` (`:.` p is a DIMENSIONLESS QUANTITY) `=([MLT^(-2)])/([L^(2)][T^(-1)]^(2))= [ML^(-1)T^(0)]` |
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| 44. |
A body is projected vertically upwards with a velocity 'U'. It crosses a point in its journey at a height 'h' twice, just after 1 and 7 seconds. The valueof U inms^(-1) is [g=10ms^(-2)] |
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Answer» 50 |
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| 45. |
A steel tape Im long is correctly calibrated for a temperature of 27.0 ^(@) C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0^(@) C .What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0^(@) C ? Coefficient of linear expansion of steel = 1.20 xx 10^(-5) K^(-1). |
| Answer» Solution :Actual LENGTH of the ROD at `45.0^(@) C = (63.0 + 0.0136) cm = 63.0136 cm`. (However, we should say that CHANGE in length up to three significant figures is` 0.0136` cm, but the TOTAL length is` 63.0 `cm, up to three significant places Length of the same rodat `27.0^(@) C = 63.0 `cm. | |
| 46. |
A particle of mass m is travelling with a constant velocity vec(v)=v_(0)hat(i) along the line y=b,z=0 Let dA be the area swpt out by the position vector from origin to the particle in time dt and L the amgnitude of angular momentum of particle about origin at any time t. Then (dA)/(dt)=(L)/("xm") where 'x' is |
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Answer» |
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| 47. |
Radius of corved path is r and the coefficient of friction between the wheel of a car and that path is mu. What should be the maximum speed with which the car can take a turn without skidding in that curved path? |
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Answer» Solution :Weight of the car = mg , so normal force OFFERED by the path = mg. Hence. LIMITING frictional force = `MU`mg. [ `mu` = coefficient of friction between the wheel of the car and the path ] The limiting frictional force provides the centripetal force necessary to prevent SKIDDING. so, if the maximum speed of the car is v, then the condition for no skidding in that path is, ` (mv^(2))/(r) = mu mg " " or, " " v = sqrt(mu r g ).` |
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| 48. |
Match the items in List-I with those in |
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Answer» a-f,b-e,c-h,d-g |
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| 49. |
Define solar constant. |
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Answer» |
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| 50. |
If displacement of a body s = (200 pm 5) m and time taken by it t = (20 pm 0.2) s, then find the percentage error in the calculation of velocity. |
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Answer» Solution :Percentage error in MEASUREMENT of DISPLACEMENT `= (5)/(200) xx 100` Percentage error in measurement of time `= (0.2)/(20) xx 100` `:.` MAXIMUM permissible error `= 2.5 + 1 = 3.5%` |
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