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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. |
Which of the following is not an example of resonance? |
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Answer» TUNNING fork |
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| 2. |
A ball of density d, is dropped on a solid horizontal plane . It returns intime t_1 to thepoint from where it was droppedafter an elastic collision . From the same height, the ball is now dropped on a liquid surface ofdensity d_L (ii) if d=d_L how is the velocity of the ball in the liquid related tothe depth of the liquid? |
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Answer» |
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| 3. |
A ball of density d, is dropped on a solid horizontal plane . It returns intime t_1 to thepoint from where it was droppedafter an elastic collision . From the same height, the ball is now dropped on a liquid surface ofdensity d_L (i)if dlt d_L, the time required by the ball to return to the original height is t_2. Find the value of t_2 in terms of t_1 ,d and d_L. |
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Answer» |
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| 4. |
A body of mass m is attached to lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5s. Find the value of m is kg. |
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Answer» Solution :`T_(1)=3=2pisqrt((m)/(K))` `T_(2)=5=2pisqrt((m+1)/(k))` `(T_(1))/(T_(2))=(3)/(5)=sqrt((m)/(m+1))` `(9)/(25)=(m)/(m+1)implies9m+9=25m` `m=(9)/(16)kg` |
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| 5. |
A body constrained to move in Y direction, is subjected to a force given by vec(F) = (-2hat(i) + 15hat(j) + 6hat(k)) N. What is the work done by this force is moving the body through a distance of 1 m along Z axis? |
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Answer» 190 j |
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| 6. |
Which of the following is not a unit of time (A) par-sec (B) light-year (C ) micron (D) sec |
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Answer» only A |
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| 7. |
A plank of mass M and length L is placed at rest on a smooth horizontal surface. A small block of mass m is projected with a velocity v_0 from the left end of it as shown in figure. The coefficient of friction between the block and the plank is m, and its value is such that the block becomes stationary with respect to the plank before it reaches the other end. a. Find the time and common velocity when relative sliding between the block and the plank stops. b. Find the work done by the friction force on the block during the period it slides on the plank. Is the work positive or negative? c. Calculate the work done on the plank during the same period. Is the work positive or negative? d. Also, determine the net work done by friction, Is it positive or negative? |
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Answer» Solution :Displacement of block in `1s` `s=1/2xx10xx1^2=5m`  The WORK done by the APPLIED force in the first second, `W_F=vecF*vecs=F*scostheta=20xx5=100J` The work done by the WEIGHT of th block in the first |
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| 8. |
Two identical concentric rings each of mass M and radius R are placed perpendicular to each other. What is the moment of inertia about an axis passing through the center of this system and perpendicular to the plane of one of the rings? |
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Answer» `((3/2)MR^2)` |
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| 9. |
Can an object accelerate if its velocity is constant? |
| Answer» Solution :No, if the VELOCITY is constant, there is no a change in the velocity HENCE ACCELERATION is zero. | |
| 10. |
If the temperature of a Black body increases by 50% then amount of radiation emitted by it in a given time interval will |
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Answer» INCREASE by 800% |
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| 11. |
Explain thermal resistance. |
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Answer» Solution :In thermal steady state, heat current in a rod in thermal steady state, `H=(Q)/(t)=kA[(T_(1)-T_(2))/(L)]=([T_(1)-T_(2)])/(((L)/(kA)))` Comparing the above EQUATION with that of electric current `I=(V)/(R)=(V_(1)-V_(2))/(R)`is obtained from Ohm.s law. We can say that as we get electric current by applying potential difference, similarly we can have heat current between TWO ends we have heat current between two ends of conducting substance by establishing temperature difference. By analogy we can say that as the property of sucstance to OPPOSE the flow of electric current through it is called resistance R, similarly the TERM `(L)/(kA)=R_(H)` is a quantity opposing the flow of heat current can be called as thermal resistance `(R_(H))`. Moreover, mathematical comparison shows that the equation of thermal resistance is similar to that of electrical resistance , `R=rhol/A=l/(sigmaA)""` (Here `rho` is resistivity and `sigma` is conductivity) The equation of equivalent thermal resistance for series and parallel combination of heat conductors are similar to the equations of series and parallel combination of electric resistances. Substances, which are good conductors of electricity, are also good conductors of heat. |
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| 12. |
(A) An air bubble in jar of water shines brightly due to phenomenon of total internal reflection. ( R) Refraction of light is the phenomenon of change in the path of light, when it goes from one medium to another. |
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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 body moves with constant vellocity v in a straight line parallel to X-axis . The angular momentum with respect to the origin is |
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Answer» zero |
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| 14. |
A lead bullet just melts when stopped by an obstacle. Assuming that 2/7 th of the heat of absorbed by the obstacle. IF initial temperature is 27°C, melting point lead is 327°C, specific heat 0.03 cal/gm°C, latent heat of fusion of lead=6cal/gm and J=4.2J/cal, velocity of that bullet is |
| Answer» Answer :D | |
| 15. |
Energy required to move a body of mass.m. from orbit of radii 2R to 3R is: |
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Answer» `(GMM)/(2R^(2))` |
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| 16. |
Explain how lubricants help in reducing friction ? |
| Answer» SOLUTION :The lubricants form a uniform LAYER in BETWEENTHE two surfaces and this prevents the INTERLOCKING of the irregularities. | |
| 17. |
The first ball of mass m moving with a velocity u collides head on with the second ball of mass m at rest. If the coefficient of restitution is e, the ratio of final velocity of the second ball to the initial velocity of the first ball is |
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Answer» `(1 - E)/(1 +e)` `m u + 0 = mv_1 + mv_2` or `u = v_1 + v_2"" ………(i)` COEFFICIENT of restitution, `e = (v_2 - v_1)/(u_1 - u_2) ` But `u_1 - u_2 = u "" ( :. u_2 = 0)` `:. e = (v_2 - v_1)/(u)` or `eu = v_2 - v_1 "" .....(ii)` Addition equations (i) and (ii), we get `u (e + 1) = 2 v_2 " or " v_2 = (u(e + 1))/(2)` `:. ("Final VELOCITY of " 2^(nd) "ball")/("Initial velocity of " 1^(st) " ball) = (v_2)/u = (e + 1)/(2)`. |
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| 18. |
In the absence of air resistance, horizontal velocity of the projectile is |
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Answer» ALWAYS negative |
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| 19. |
A person cannot see distincly any object placed beyond 40cm from his eye. The power of the lens which will enable him to see distant stars clearly is. |
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Answer» <P> Solution :The person cannot see OBJECTS clearly beyond 0.4m, so his far point =0.4m DISTANCE of object=aHe should use lens which forms image of distant object at a distance of 40 cm infront of it. `-1/0.40-1/-infty=1/f=p IMPLIES P=(-10)/4=-2.5D` |
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| 20. |
Consider an ideal monoatomic gas of volume V at pressure P. The bulk modulus at constant temperature is |
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Answer» `(P)/(2)` |
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| 21. |
The dimensional formula for strain energy density is |
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Answer» `M^(1)L^(2)T^(-3)` |
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| 22. |
Match the following{:(,"List - I",,"List - II"),((a),"1HP",(e ),10^(7)ergs^(-1)),((b),"1KWH",(f),1.6xx10^(-19)J),((c ),"1eV",(g),36xx10^(5)J),((d),"1W",(h),"746W"):} |
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Answer» a-g, b-h, C-f, d-e |
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| 23. |
"Friction is a self adjusting force" Why ?- |
| Answer» SOLUTION :Applied FORCE always balances force of friction, till there is no relative motion between the TWO BODIES. | |
| 24. |
Dimensions of impulse are : |
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Answer» `[ML^(-2)T^(-1)]` `=[MLT^(-2)][T]` `=MLT^(-1)` |
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| 25. |
Two forces F_(1) and F_(2) are acting at a point, having resultant as F. If F_(2) is doubled F is also doubled. If F_(2) is reversed then also F is doubled. If F= 16 sqrtN, F_(1) = sqrt2N and F_(2) = n then find n |
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Answer» |
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| 26. |
A satellite of mass 'm' revolves around the earth of radius R at a height x from its surface. If 'g' is the acceleration due to gravity on the surface of the earth, than the orbital speed of the satellite is |
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Answer» `GX` |
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| 27. |
When two samples at different temperatures are mixed , the temperature of mixture can be |
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Answer» lesser than lower or greater than HIGHER temperature |
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| 28. |
Explain the term maximal static friction |
| Answer» SOLUTION :if theexternalforceis increasedaftera PARTICULAR limitthesurfacecannotprovidesufficientopposing fricitionalforceto balancetheexternalforceon THEOBJECT . Then theobjectstartsto SLIDE. | |
| 29. |
Which one of the following relationships between the acceleration a a and displacement x of a particle involve simple harmonic motion. |
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Answer» `a=0.7x` |
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| 30. |
If component of one vector in the direction of another vector is zero, then those two vectors |
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Answer» are PARALLEL to each other |
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| 31. |
Ratio of unit of Planck's constant and moment of inertia will give which physical quantity ? |
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Answer» time `E=hf` `h=(E)/(f)=(M^(1)L^(2)T^(-2))/(T^(-1))=M^(1)L^(2)T^(-1)` MOMENT of INERTIA, `1=Mr^(2)` `=KG m^(2)` `=M^(1)L^(2)T^(0)` `:.` Ratio `=(h)/(I)=(M^(1)L^(2)T^(-1))/(M^(1)L^(2)T^(0))` `=M^(0)L^(0)T^(-1)` Dimension of frequency |
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| 32. |
Which of the following options are correct? |
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Answer» Acceleration due to gravity decreases with increasing altitude at altitude,`g_h~~g(1-(2h)/R)` At depth d, `g_d = g(1- d/R)` In both cases, g decreases with increase in H and d. At latitude `PHI,g_0 = g - omega^(2) Rcos^(2) phi` In this case , `g_phi` increases with increase in `phi` , Also g is independent of mass. |
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| 33. |
Find the torque of a corce 7hati-3hatj-5hatkabout the origin which acts on a particle whose position vector is hati+hatj-hatk Given: vecF=7hati-3hatj-5hatk |
| Answer» Solution :`VEC TAU= vecr XX VECF|{:(hati, hatj, hatk),(1,1,-1),(7,-3,-5):}|` | |
| 34. |
(A) If you look out of the window of a fast moving train, the nearby trees, house etc. seem to move rapidly in a direction opposite to the train is motion , but the distant objects (hiss tops, the moon , the stars etc.) seem to be stationary . (R ) When a train moves rapidly rapidaly . On the other hand, in the case of far off objects, the line of sight changes its direction. |
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Answer» Both (A) and (R) are true and (R) is the correct EXPLANATION of (A) |
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| 35. |
Define latent heat. |
| Answer» Solution :LATENT HEAT is the AMOUNT of heat per unit mass transferred during change of STATE. | |
| 36. |
A body moving with a constant speed v in a circle of radius r. its angular acceleration is : |
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Answer» 0 |
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| 37. |
A small block of mass 'm' is projected horizontally with speed 'u' where friction coeffcient between block and plane is given by mu = cx where 'x' is displacement of the block on plane. Find the time of journey |
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Answer» |
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| 38. |
A small block of mass 'm' is projected horizontally with speed 'u' where friction coeffcient between block and place is given by mu = cx where 'x' is displacement of the block on plane. Find the displacemtn covered by the block before it comes to rest |
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Answer» |
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| 39. |
Find the fractional increase in the volume of a wire of circular cross section if its longitudinal strain is 2%. (sigma=0.4) |
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Answer» 0.004 |
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| 40. |
A conical glass capillary tube of length 0.1m has diameter 10^(-3) and 5 xx 10^(-4)respectively at its ends. When it is just immersed in a liquid at 0^@C with larger diameter in contact with liquid, the liquid rises to 8 xx 10^(-2) m in the tube. If another cylindrical glass capillary tube B is immersed in the same liquid at 0^@C, the liquid rises to 6 xx 10^(-2) m height. The rise of liquid in the tube B is only 5.5 xx 10^(-2) m when the liquid is at 50^@ C. Density of the liquid is (1//14)xx 10^(-2) Kg//m^3 and angle of contact is zero. Effect of temperature on the density of liquid and glass is negligible. The rate at which the surface tension changes with temperature considering the change to be linear is given by -1.4510^(-n)N//m^@ c What is the value of n ? |
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Answer» |
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| 41. |
In a forced oscillation of a particle the amplitude is maximum for a frequency w_(1) of the force, while the energy is maximum for a frequency w_(2) of the force, then |
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Answer» `w_(1)= w_(2)` |
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| 42. |
Which conservation law is expressed by the equation of continuity ? |
| Answer» SOLUTION :LAW of CONSERVATION of MASS | |
| 43. |
Among the following the error that can be eliminated is |
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Answer» SYSTEMATIC error |
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| 44. |
Four moles of a perfect gas heated to increases its temperature by 2^(@)C absorbes heat of 40 cal at constant volume. If the same gas is heated at constant pressure the amount of heat supplied is, (R=2" cal/mol K") |
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Answer» 28 cal |
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| 45. |
Consider the situation as shown in the figure. Find the maximum value of m_(1)//m_(2) so that the system remains at rest. What will happen if tan theta lt mu? |
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Answer» Solution :Since we are interested in MAXIMUM value of `m_(1)//m_(2)`, equilibrium is limiting `f = f_(max) = MU N` For `m_(1) : N_(1) = m_(1) g cos THETA ,f_(1) = mu N_(1) = mu m_(1) g cos theta` For `m_(2) : N_(2) = m_(2) g , f_(2) = mu N_(2) = mu m_(2) g` For the system to be at rest `m_(1) g sin theta LE (f_(1) + f_(2))` `m_(1) g sin theta le (mu m_(1) g cos theta + mu m_(2)g)` `m_(1) (sin theta - mu cos theta) le mu m_(2)` `(m_(1))/(m_(2)) le (mu)/((sin theta - mu cos theta))` `((m_(1))/(m_(2)))_(max) = (mu)/( sin theta - mu cos theta)` If `tan theta lt mu` [i.e. `theta lt tan^(-1) mu`] `(sin theta)/( cos theta) lt mu rArr sin theta lt mu cos theta` ` ( sin theta - mu cos theta)` lt 0 i.e. `((m_(1))/(m_(2)))` is negative i.e. the system will not slide for any value of `m_(1)//m_(2) theta lt tan ^(-1) mu`. 
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| 46. |
Ratio of force constants of two springs is 1 : 2 and the ratio of mechanical energy of them is 2 : 9, ratio of amplitudes of two bodies suspended at the end of spring is………. |
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Answer» SOLUTION :`(E_1)/(E_2)= ((1)/(2)k_(1)A_(1)^(2))/((1)/(2)k_(2)A_(2)^(2))` `therefore (2)/(9)= (1)/(2)XX ((A_1)/(A_2))^(2)` `therefore (2)/(9)xx (2)/(1) = ((A_1)/(A_2))^(2)` `therefore (4)/(9)= ((A_1)/(A_2))^(2)` `therefore (A_1)/(A_2)= (2)/(3)`. |
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| 47. |
What are geostationary and polar satellites? |
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Answer» Solution :Geo-stationary and polar SATELLITE: (i) The satellites orbiting the Earth have different time periods corresponding to different orbital radii. Orbital radius of a satellite if its time PERIOD is 24 hours is calculated below: Kepler's third law is used to find the radius of the orbit. `T^(2) = (4pi^2)/(GM_E) (R_(E) + H)^(3)` (R_(E) + h)^(3) = (GM_E T^2)/(4pi^2)` `R_(E) + h = ((GM_(E)T^(2))/(4pi^2))^(1//3)` (ii) Substituting for the time period (24 hrs = 86400 seconds), mass and radius of the Earth, h turns out to be 36,000 km. Such Satellites are called "geo-stationary satellites", They appear to be stationary when seen from Earth. (iii) Geo-stationary satellites is used for the purpose of telecommunicatio. Another type of satellite which is placed at a distance of 500 to 800 km from the surface of the Earth orbits the Earth from north to south direction. (iv)This Type of satellite which orbits Earth from North Pole to South Pole is called a polar satellite. (v) The time period of a polar satellite is nearly 100 minutes and the satellite completes many revolutions in a day. (VI) A polar satellite covers a small strip of area from pole to pote during one revoluation. In the next revoluation, it covers a different strip of area since the Earth would have moved by a small angle. (vii) In this way polar satellites cover the entire surface area of the Earth. |
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| 48. |
A ball is dropped from a heighth on to a floor. If, in each collision, its speed becomes e times of its striking value (a) Find the total change in momentum of the ball (b) Find the average force exerted by the ball on the floor. |
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Answer» Solution :(a) Change in MOMENTUM in I collision `= mv _(1) - (- mv _(0)) = m (v _(1) + v _(0))` Change in momentum in II collision `= m ( v _(2) +v _(1))` Charge in momentum in `n^(TH)` collision `= m ( v _(n) + v _(n-1))` Adding these all, total change in momentum of the ball is `DELTA p =m [ v _(0) + 2v _(1) +...+ 2v _(n-1) +v _(n)]` or `Delta p= m v _(0) [1 + 2 e + 2e ^(2) +......], [as v _(1) = ev _(0) , v _(2) = e ^(2) v _(0).......]` `Delta p = mv _(0) [1+ 2e ((1)/(1-e)) ]= m SQRT (2gh) [ (1 + e )/( 1-e )]...(1)` (b) Now as ` vecF = (vec(dp))/( dt) so, F _(av) = (Deltap)/( Delta T)` We know `Delta T = sqrt ((2h)/( g)) ((1+e)/( 1 -e ))` from `(1) Delta p =m sqrt (2gh) [ (1 +e)/( 1-e )]` `F _(av) = m sqrt (2gh) [(1+e)/(1-e)] xx sqrt ((g)/( 2h)) [ (1-e )/( 1 +e)] = MG ` |
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| 49. |
The moon moves around the earth in a circular orbit and the earth moves around the sun in elliptical orbit then.(a) For the earth - moon system, the gravitational attractions between them are the internal forces(b)The sun.s attraction on the earth and moon are external forces acting on the centre of mass of the system. |
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Answer» a is true, B is false |
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| 50. |
Three bodies, a ring, a solid cylinder and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity? |
Answer» Solution : Assume that conservation of energy of the rolling body. The potential energy lost by the body in rolling down the inclined PLANE is MGH. Where m is the mass of body, g is gravitational acceleration and h is height of slope. Equal kinetic energy gain by the body. The body start from the rest, the kinetic energy gained is equal to the final kinetic energy of the body. Hence final kinetic energy = `(1)/(2)Iomega^(2)+(1)/(2)mv^(2)` potential energy = final kinetic energy `mgh=(1)/(2)Iomega^(2)+(1)/(2)mv^(2)` `mgh=(1)/(2)mk^(2)omega^(2)+(1)/(2)mv^(2) [because I=mk^(2)]` `:. gh=(1)/(2)k^(2)xx(v^(2))/(R^(2))+(1)/(2)v^(2) [because omega=(v)/(R )]` `:. gh=(1)/(2)v^(2)[1+(k^(2))/(R^(2))]` `therefore v^(2)=(2gh)/(1+(k^(2))/(R^(2)))` `therefore v=[(2gh)/(1+(k^(2))/(R^(2)))]^((1)/(2))""...(1)` Hence, the velocity is independent of the mass of the rolling body (1) For a ring : `k=R :. k^(2)=R^(2)` From eqn. (1), `therefore v_("ring")=[(2gh)/(1+(R^(2))/(R^(2)))]^((1)/(2))` `=[(2gh)/(1+1)]^((1)/(2))=[(2gh)/(2)]^((1)/(2))` `=SQRT(gh)` (2) For solid sphere : `k=(R)/(sqrt(2))impliesk^(2)=(R^(2))/(2)` `:. v_("cylinder")=[(2gh)/(1+(R^(2))/(2R^(2)))]^((1)/(2))=[(2gh)/(1+(1)/(2)]]^((1)/(2))` `=[(4gh)/(3)]^((1)/(2))` `=sqrt((4gh)/(3))` (3) For solid sphere : `k=Rxxsqrt((2)/(5))impliesk^(2)=(2R^(2))/(5)` `:. v_("solid sphere")=[(2gh)/(1+(2R^(2))/(5R^(2)))]^((1)/(2))=[(2gh)/(1+(2)/(5))]^((1)/(2))` `=[(2gh)/((7)/(5))]^((1)/(2))` `=sqrt((10gh)/(7))` Hence, the sphere has the greatest and the ring has the least velocity of the centre of mass at the BOTTOM of the inclined plane. So solid sphere reached at first to the bottom of slope. |
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