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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 phenomena are reversible ? |
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Answer» Water fall |
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| 2. |
This question has statement -I and statement - II of the fourchoices given after the statements , choose the one that best describes the two statements . Statement - I : A point particle of mass m movingwith speed v collides with stationary point particle of mass M . If the maximumenergy loss possible is given as f(1/2 mv^(2)) " then" F = (m/(M+m)) Statement-II : Maximumenergy loss occurs when the particles get stuck together as a result of the collision . |
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Answer» Statement - I is true , Statement - II is true Statement - II is a correct explanation of Statement -I `1/2 mv^(2)+1/2 M(o)^(2)-1/2 (m+M)(v)^(2)` = energy loses Where `v = (mv+M XXO)/(m+M)` ` :. V. =m/(m+M) v` `1/2 mv^(2)-1/2 (m+m)(m/(m+m)v)^(2)=` Energy loses ` :. 1/2 mv^(2)-1/2 mv^(2)xx(m/(m+m)v)^(2)` = Energy loses ` :. 1/2 mv^(2)[1-m/(m+m)]^(2)` = Energy loses ` :. 1/2 mv^(2)[1-m/(m+m)]^(2)` = Energy loses `:. f = M/(m+M)` here in statement - I `m/(m+M)` is given . ` :. ` Statement -I is wrong and energy loses BECOMES POSITIVE so maximum energy will be DISSIPATED and hence statement -II is correct . |
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| 3. |
The position of a particle is given by vecr = hati + 2hatj - hatk and its linear momentum is given by vecp= 3hati+4 hatj - 2hatk. Then its angular momentum, about the origin is perpendicular to |
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Answer» yz-plane `= hati (-4 + 4 ) - hatj (-2 + 3) + hatk (4 - 6) = 0 hati 0 1 hatj - 2hatk` `vecL` has components along-yaxis and -z axis but it has no component in the x-axis . The angular MOMENTUM is in yz plane i.e., perpendicular to x-axis . |
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| 4. |
What is the difference between velocity and average velocity? |
Answer» SOLUTION :
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| 5. |
A body of mass 60 g is kept on an inclined plane of inclination 30^(@) . If the downward resultant force is 0.1764 N, estimate the value of the coefficient of friction. |
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Answer» |
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| 6. |
Physical interpretation of the definition of torque : |
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Answer» Solution :As shown in the figure a force `vecF` is acting on a PARTICLE P of rigid body. Force `vecF` is taken in the plane perpendicular to the axis of rotation. The axis is perpendicular to the plane of the figure and passing through point O. The position vector of P w.r.t. the centre of its CIRCULAR path is `VECR`. The angle between `vecF and vecr` is `theta`. To examine the effectiveness of the force in produing rotational motion, `vecF` can be resolved into two mutually perpendicular components. (1) Component of `vecF` parallel to `vecr` is `Fcostheta` which does not produce rotational motion. (2) Component of `vecF` perpendicular to `vecr` is `Fsintheta`. Which can produce the rotational motion. If F and/or `theta` large, `vecF` will be more effective. Moreover if the position vector `vecr` of the point of application of the force `vecF` is large, the effectiveness of the force in producing rotational motion is more. Thus the quantity responsible for producing rotational motion is not only `vecF` but `rFsintheta`. If can be written in vector FORM as under. `vectau = vecr xx vecF` Torque is the measure of effectiveness of force in producing rotational motion. |
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| 7. |
A swimmer can swim at 3 sqrt(2) km*h^(-1) in still water. He swims across a river when the velocity of current is 3 km *h^(-1). (i) In what direction should he swimto cross by the shortest path ? |
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Answer» |
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| 8. |
n moles of a gas expands from volume V_(1) to V_(2) at constant temperature T. The work done by the gas is |
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Answer» `nRT[upsilon_(2)/upsilon_(1)]` |
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| 9. |
Distinguish between isobaric and isochoric process. |
| Answer» Solution :Isobaric process is a PHYSICAL CHANGE taking place at constant pressure. VOLUME and TEMPERATURE change. Isochoric process is a physical change taking place at constant volume. Pressure and temperature change. | |
| 10. |
A mercury drop of radius 1 cm is sprayed into 10^(6) drops of equal size. The energy expended in joules is (surface tension of mercury is 460 xx 10^(-3)Nm^(-1)) |
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Answer» `0.057` |
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| 11. |
If the force acting on a body is inversely proportional to its speed, the kinetic energy of the body is: |
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Answer» constant |
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| 12. |
A swimmer can swim at 3 sqrt(2) km*h^(-1) in still water. He swims across a river when the velocity of current is 3 km *h^(-1). (ii) Inwhat direction should he swimto cross the river in minimum time ? In that case how far will he be carried by the river current if the river is sqrt(2) km wide ? |
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Answer» |
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| 13. |
Passage-II : The mass of hydrogen molecule is 3.32 xx 10^(-27) kg. If 10^(23) hydrogen molecules strike a fixed wall of area 2cm^(2) at an angle 45^(@) to the normal, per second and rebound elastically with a speed of10^(3) m//s then The pressure exerted on the wall by the hydrogen gas is |
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Answer» `1.346 XX 10^(3) N//m^(2)` |
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| 14. |
Passage-II : The mass of hydrogen molecule is 3.32 xx 10^(-27) kg. If 10^(23) hydrogen molecules strike a fixed wall of area 2cm^(2) at an angle 45^(@) to the normal, per second and rebound elastically with a speed of10^(3) m//s then The total force acting on the wall normally is |
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Answer» 10 N |
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| 15. |
Passage-II : The mass of hydrogen molecule is 3.32 xx 10^(-27) kg. If 10^(23) hydrogen molecules strike a fixed wall of area 2cm^(2) at an angle 45^(@) to the normal, per second and rebound elastically with a speed of10^(3) m//s then The change in momentum of each molecule normal to the wall is |
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Answer» `4.692 XX 10^(-24) ` kgm/s |
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| 16. |
How do animals sense impending danger of hurricane? |
| Answer» SOLUTION :Some animals are believed to be sensitive to be low frequency sound waves emitted by harricanes. they can also detect the SLIGHT DROPS in air and water PRESSURE that signal a storm.sapproach. | |
| 17. |
A cylinder of mass M and radius R is resting on a horizontal (which is parallel to x-y plane) with its axis fixed about y-axis and free to rotate about its axis. The platform is given a motion in x-directiongiven by x - A cosomegat. There is no slipping between cylinder and plateform. |
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Answer» The MAXIMUM torque acting on the cylinder during its MOTION is `1/3Momega^(2)"AR"` |
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| 18. |
What are the characteristics of stationery waves? Give the laws of transverse vibrations in a stretched string. |
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Answer» Solution :1.Stationary waves are characterised.by the confinement of a wave disturbance between two rigid boundaries. This means, the wave does not move forward or backward in a medium (does not advance), it remains STEADY at its place. Therefore, they are called ·.stationary waves or standing waves" 2. Certain points in the region in whichthe wave exists have maximum amplitude, called as anti-nodes and at certain points the amplitude is minimum or zero, called as nodes 3. The distance between two consecutive nodes (or) anti-nodes is `lamda/2` 4.The distance between a node and its neighbouring anti-node is `lamda / 4` 5. The transfer of energy ALONG the standing wave is zero. Laws of transverse vibrations in stretched strings: There are three laws of transverse vibrations of stretched strings which are GIVEN as follows: (i)The law oflength: For a given wire with tension T (which is fixed) and mass per unit length `MU`(fixed) the frequency varies inversely with the vibrating length. Therefore, ` f prop 1/l rArr f = C/l` ` rArr l xx f= C` , where C is a constant (ii)The law of tension: For a given vibrating length l(fixed) and mass per unit length `mu`(fixed) the frequency varies directly with the square root of the tension T, ` f prop sqrtT` ` rArr f = A sqrtT` , where A is a constant (iii)The law of mass: For a given vibrating length l (fixed) and tension T (fixed) the frequency varies inversely with the square root of the mass per unit length `mu` , ` f prop (1)/(sqrt mu) ` , where B is a constant |
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| 19. |
What is the work done in blowing a soap bubble from a radius of 3 cm to 5cm ? Surface tension of the soap solution is 0.03Nm^(9-1). |
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Answer» `0.2pimJ` (There are TWO free SURFACE of soap bubble) `=8piTxx(R_(2)^(2)-R_(1)^(2))` `=8xxpixx0.03[25xx10^(-4)-9xx10^(-4)]` `=8xxpixx0.03xx16xx10^(-4)` `=0.4pixx10^(-3)J` `thereforeW=0.4pimJ` |
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| 20. |
Two shperical balls of mass 1 kg and 4 kg are separated by a distance of 12 cm. The distance from 1 kg at which the gravitational force on any mass becomes zero. |
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Answer» 4 cm |
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| 21. |
A ball of mass 1 kghangs in equlibrium from two stings OAand OBin the diagram calcualte the tensions in stings OAand OB take g=10ms^(-2) |
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Answer» Solution :Givenmassof the ball `=1kg` `g=10 ms^(-1)` Variousforces actingat the point O are atshownin thediagram .Thethree forcesare inequilibrium.usingLamis THEOREM `(T_(1))/(sin150^(@)) = (T_(2))/(SIN 120^(@))= (10)/(sin 90^(@))` and `T_(1) 10 sin 30^(@) = 10 xx 0.5=5N` and `T_(2) =10 sin 60^(@)=10 xx (SQRT(3))/(2)` `=5sqrt(3) N` |
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| 22. |
A student comes to school by a bicycle whose tire is filled with air at a pressure 240 kPa at 27^(@)C. She travels 8 km to reach the school and the temperature of the bicycle tire increases to 39^(@)C. What is the change in pressure in the tire when the student reaches school? |
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Answer» Solution :We can take air MOLECULES in the TIRE as an ideal gas. The NUMBER of molecules and the volume of tire remain constant. So the air molecules at `27^(@)`C SATISFIES the ideal gas equation `P_(1)V_(1)=NkT_(1)` and at `39^(@)C` it satisfies `P_(2)V_(2)=NkT_(2)` But we know `V_(1)=V_(2)=V` `(P_(1)V)/(P_(2)V)=(NkT_(1))/(NkT_(2))` `(P_(1))/(P_(2))=(T_(1))/(T_(2)), P_(2)=(T_(2))/(T_(1))P_(1)` `P_(2)=(312K)/(300K)xx240xx10^(3)=249.6kPa` |
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| 23. |
Three vectors A, B and C add upto zero. Find which is false. |
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Answer» `(vecA XX vecB) xx vecC` is not ZERO unless `vecB, vecC` are parallel. |
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| 24. |
Theyoung'smodulusofa materialis2xx10 ^(11) N//m^2and itselasticlimitis1xx10 ^(8) N//m^2fora wireof1mlengthof thismaterial, themaximumelongationachievable is |
| Answer» Answer :4 | |
| 25. |
What is uniform motion ? |
| Answer» Solution :When an object MOVING on a straight line covers EQUAL DISTANCE in equal intervals of TIME is called UNIFORM motion. | |
| 26. |
Projection of the vector 2 hat(i) + 3hat(j) + 2hat(k) on the vector hat(i) - 2hat(j) + 3hat(k) is |
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Answer» `(2)/(SQRT14)` |
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| 27. |
A metallic ring of radius r and cross-sectional area A is fitted into a wooden circular disc of radius R (R gt r). If the Young's modulus of the material of the ring is Y, the force with which the metal ring expands is |
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Answer» |
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| 28. |
A cylindrical rod having temperature T_(1) and T_(2) at its end. The rate of flow of heat is Q_(1)("Cal")/(s) If all linear dimensions are doubled keeping temperature constant, then the rate of flow of heat Q_(2) in ("Cal")/(s) will be. . . . . . . . |
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Answer» `(Q_(1))/(4)` `H_(2)=KA_(2)((T_(1)-T_(2))/(L_(2)))` `:.(H_(1))/(H_(2))=(A_(1))/(A_(2))((L_(2))/(L_(1)))` `=((pid_(1)^(2))/(4))/((pid_(2)^(2))/(4))*(L_(2))/(L_(1))` `=(d_(1)^(2))/(d_(2)^(2))xx(L_(2))/(L_(1))` `:.(Q_(1))/(Q_(2))=(d_(1)^(2))/((2d_(1))^(2))xx(2L_(1))/(L_(1))` `=1/4xx2` `=1/2` `:.Q_(2)=2Q_(1)` |
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| 29. |
A mass m is supported by a massless stringwound around a uniform hollow cylinder of mass m and radius R. If the string does not slip on the cylinder, with what acceleration will the mass fall on release ? |
| Answer» Answer :A | |
| 30. |
If for two vectors vec(A) and vec(B), vec(A) xx vec(B) = 0then the vectors are : |
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Answer» PERPENDICULAR to each other |
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| 31. |
A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km/h. The average speed of the man over the interval of time 0 to 40 min, is equal to |
| Answer» Answer :D | |
| 32. |
An inventor claims to have developed an engine that during a certain time interval takes in 110 MJ of heat at 415 K, rejects 50 MJ of heat at 212 K while manages to do 16.7 kWh of work. Do you agree with the inventor.s claim ? |
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Answer» SOLUTION :The claimed efficiency `eta_(C ) = (W)/(Q_(H))=(16.7kWh)/(110MJ)` But as `1kWh=10^(3)xx(J//s) xx (60 xx 60s) = 3.6MJ` So `eta_(c )=(16.7xx3.6)/(110) = 0.55 = 55% …………(1) ` while maximum POSSIBLE THEORETICAL efficiency, `eta_("max")=1-(T_(L))/(T_(H))=(T_(H)-T_(L))/(T_(H)) = (415-212)/(415)` `=0.49=49%""......(2)` From Eqns. (1) and (2) it is clear that claimed efficiency is greater than maximum possible theoretical efficiency, so inventor.s claim does not appear to be correct. |
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| 33. |
In case of a freely falling body, the ratio of kinetic energyat the end of the third second to increase in kinetic energy in th next three seconds is |
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Answer» `1:1` |
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| 34. |
To maintain a rotor at a uniform angular speed of 200 rad san engine needs to transmit a torque of 180 Nm. What is the power required by the engine? Assume that the engine is 100% cfficient |
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Answer» <P> SOLUTION :Here `omega=200 "rad" s^(-1), tau=180 NM``:.` Power `, P= tau omega=180xx200=36,000W36 kW` |
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| 36. |
Figure a represents the speed - time graph for a particle . Find the distance covered by the particle between t=10min "and"t=30min |
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Answer» Solution :We draw perpendicular lines from the 10-minute point and the 30-minute point to the time -axis (Figure b). The DISTANCE covered is EQUAL to area of the RECTANGLE ABCD. It s value is `AB.AD =(30min-10min)xx(10km//h)` =`20minxx10km//h` `h=(20)/(60)hxx10km//h` `=(10)/3km` |
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| 37. |
A Dropof waterof massm and densityrhoplacedbetweentwowellcleanedglassplates, thedistancebetweenwhichis d.The forceofattractionbetweenthe platesis( n T m)/( rho d ^(n))( T= surfacetension ) |
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Answer» |
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| 38. |
The position vectors of three particles of mass m_(1)= 1kg, m_(2)=2 kg and m_(3)=4 kg are vec(r )_(1)=(hat(i)+4hat(j)+hat(k))m, "" vec(r )_(2)=(hat(i)+hat(j)+hat(k))m, and vec(r )_(3)=(2hat(i)-hat(j)-2hat(k))m respectively. Find the position vector of their centre of mass. |
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Answer» SOLUTION :The POSITION vector of centre of mass of the three particles is given by `vec(r )_(c )=(m_(1)vec(r )_(1)+m_(2)vec(r )_(2)+m_(3) vec(r )_(3))/(m_(1)+m_(2)+m_(3))` `vec(r )_(c )=(1(hat(i)+4hat(j)+hat(K))+2(hat(i)+hat(j)+hat(k))+4(2HAT(i)-hat(j)-2hat(k)))/(1+2+4)` ` = ((11hat(i)+2hat(j)-5hat(k)))/(7)=(1)/(7)(11hat(i)+2hat(j)-5hat(k))m` |
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| 39. |
The potential energy of a certain spring when stretched through a distance S is 10 joule. The amount of work in joule) that must be done on this spring to stretch it through additional distance S will be |
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Answer» 30 `U. = 1/2 k (S + S)^2 = U (1/2 kS^2) = 40 J "" ` (using (i)) `:. W = U. - U = 40 J - 10 J = 30 J`. |
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| 40. |
The gas having average speed four times as that of SO_(2) (molecular mass 64) is |
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Answer» He (molecular mass 4) |
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| 41. |
A light vertical spring is stretched by 0.2 cm when a weight of 10 g is attached to its free end. The weight is further down by 1.0 cm and released. Compute the frequency and maximum velocity of load. |
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Answer» Solution :(i) Force CONSTANT of the spring `k=("Restoring Force")/("Increase in length")=(mg)/("Increase in length") ( :. F=mg)` `=(10^(-2)xx9.8)/(2.0xx10^(-3))=49Nm^(-1)` Frequency `v=1/(2pi)SQRT(k/m)=1/(2pi)sqrt(49/(10^(-2)))=1/(2pi)xx70` (II) The amplitude of motion = The distance through which the WEIGHT is further pulled down =1.0cm i.e. `A=1.0cmimplies` Maximum velocity `=AW=10^(-2)mxx1/70"rad"s^(-1)=1/7000ms^(-1)` |
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| 42. |
According to Kepler's second law, the radial vector to a planet from the Sun sweeps out equal areas in equal intervals of time. This law is a consequence of: |
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Answer» LINEAR MOMENTUM |
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| 43. |
A satellite is revolving round the earth in an elliptical orbit, then (a) gravitational force exerted by the earth is equal to centripetal force at same point only (b) power associated with gravitational force is zero at every point (c) workdone by gravitational force is zero in some small parts of the orbit (d) at some points, the magnitude of gravitational force is greater than that of centripetal force |
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Answer» only a and B are TRUE |
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| 44. |
(A) A beam of white light gives a spectrum on passing through a hollow prism. (R ) The speed of light inside the hollow prism is more than that of incident. |
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Answer» Both A and R are true and R is the correct explanation of A |
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| 45. |
A uniform solid rod of mass 40kg and length 10m rests against a vertical smooth wall making an angle of 30^(@) with the vertical. Find the force of friction (f) and normal reaction (R ) exerted by the groundd on the rod. |
Answer» SOLUTION :For ROTATORY equilbrium, `Fbar(AB)=mgbar(CD)` AC=10m,m=40kg `sin60^(@)=(AB)/(AC)"":.bar(AB)=bar(AC) sin60^(@)=10xx0.866=8.66m` `sin30^(@)=(BC)/(AC)"":.BC=10xx0.5=5m` CD=`(BC)/2=2.5m"":.F(8.66)=40xx9.8xx2.5` i.e `F=(40xx9.8xx2.5)/(8.66)=113.16N` Force of friction=113.16N |
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| 46. |
Select the incorrect statement from the following statements : (i) The radius of a rain drop is 10^(-6)m (ii) The mass of a neutron is 10^(27)kg (iii) The average distance of Pluto's orbit is 10^(12)m (iv) The diameter of the hydrogen atom is 10^(-10)m |
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Answer» STATEMENTS (i) and (ii) are incorrect |
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| 47. |
An ideal refrigerator has a freezer at a temperature of -13^(@) C. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) will be |
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Answer» `325^(@) C` |
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| 48. |
Efficiency of an engine is eta1 at T1 =200 degree C and T2= 0 degree C and eta2 at T1 = 0 degree C and T2 = -200 degreeC find the ratio of eta1 / eta2 |
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Answer» 1 |
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| 49. |
The activity of the force to produce rotational motion in a body is called as |
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Answer» ANGULAR momentum |
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| 50. |
Obtain an expression for the aparent frequency of sound when the source of sound and the listener are movingaway fromeach other |
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Answer» Solution :For a source moving AWAY from an observer ( ASSUMED to be at REST ) apparent frequency of SOUND `f = f_(0) ((v)/( v+ v_(s)))` thereby WAVELENGTH `lambda = (v+v_(s))/( f_(0))` For an observer movingaway from the source , apparentfrequency, `f'= ( v- v_(0))/(lambda ) `i.e. `f' = f_(0)"" ((v-v_(0)))/( ( v+ v_(s)))` Note `:` For `v_(0) lt lt v, v_(S) lt lt v ` `f' = f_(0) (1- ( v_(0))/(v)) ( 1- ( v_(s))/(v))` 
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