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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. |
For a particle oscillating along x-axis according to equation x = A sin omega t Magnitude of its mean velocity over first 3/8 of its oscillation is |
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| 2. |
Derive an expression for the time period (T) of a simple pendulum which may depend upon the mass (m) of the bob, length (l) of the pendulum and acceleration due to gravity (g). |
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Answer» Solution :Let `T= km^(a) l^(b) g^(c )` where k is a dimensionless constant. Writing the equation in dimensional FORM, we have `[M^(0) L^(0) T^(1)]= [M]^(a) [L]^(b) [LT^(-2)]^(c )= [M^(a) L^(b+ c)T^(-2c)]` EQUATING exponents of M, L and T on both sides, we get a= 0, b+ c= 0, `-2c= 1`. SOLVING the eq. we get a= 0, b=1/2, c= `-1//2` Hence, `T= K sqrt((l)/(g))`, where k is constant. |
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| 3. |
An Object is falling freely from height h due to gravitational field. Match the following : {:("Column-I","Column-II"),("(1) Potential energy = kinetic energy",(a) at(2h)/3height),("(2) Potential energy = 2 (kinetic energy)","(b) constant at all points"),(,(c)ath/2 height):} |
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| 4. |
A ball of mass m moving with a velocity collides head on elastically with another of the same mass m but moving with a - in the opposite direction). After the collision |
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Answer» `(v^2)/(4g)` |
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| 5. |
Define power and give its unit. |
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Answer» SOLUTION :Power is DEFINED as the rate of work done or energy DELIVERED P = work done / time taken Its UNIT si WATT . |
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| 6. |
For particle moving on circular path with constant speed angle between instantaneous velocity and instantaneous acceleration will be …... . |
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| 7. |
vec(A) and vec(B) are two vectors given byvec(A)=2hat(i)+3hat(j) and vec(B)=hat(i)+hat(j). The magnitude of the component of vec(A) along vec(B) is |
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Answer» `(5)/(SQRT2)` |
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| 8. |
The spring balance A reads 2 kg with a block m suspended from it. A balance B reads 5 kg when a beaker filled with liquid is put on the pan of the balance. The two balances are now so arranged that the hanging mass is inside the liquid as shown in figure. In this situation |
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Answer» The balance A will READ more than 2 KG |
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| 9. |
SI unit of force is |
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Answer» Newton |
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| 10. |
Ratio of adiabatic bulk modulus to isothermalbulk modulus is |
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Answer» <P>`C_(p)-C_(V)` |
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| 11. |
What is the smalles mass meaured indirectly so far ? |
| Answer» SOLUTION :The SMALLEST MASS measured indirectly so far is that of an electron `(~~10^(-30)kg).` | |
| 12. |
A wedge in the form of equilateral triangle isplaced on a rough horizontal surface as shown in the figure. The value of coefficient of frinction, can the block topple without slipping muge(1)/(sqrt(x)) |
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| 13. |
A force is applied on a body of mass 0.9 kg that is at rest. The force is applied for a duration of 5s and as a result the body covers a distance of 250 m. Find the magnitude of the force. |
| Answer» Answer :B | |
| 14. |
The law of conservation of momentum can be derived from |
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Answer» NEWTON 's first law of MOTION |
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| 15. |
A body is floating in a liqud. Weight of the body is |
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Answer» More than the WEIGHT of the body |
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| 16. |
An astronaut orbiting the earth in a circular orbit 120 km above the surface of the earth gently drops a spoon out of the space-ship. It will |
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Answer» FALL vertically doen to earth |
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| 17. |
The resultant of two forcews whose magnitudes are in the ratio 3:5 is 28N. If the angle of their inclination is 600, then find the mgnitude of each force. |
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Answer» Solution :LET `F_(1) and F_(2)` be the two forces. Then `F_(1)=3"x",F_(2)=5"x", R=28N and theta = 60^(0)` `R=sqrt(F_(1)^(2)+F_(2)^(2)+2F_(1)F_(2)COSTHETA)` `rArr28=sqrt((3"x")^(2)+(5"x")^(2)+2(3"x")(5"x")cos60^(0))` `rArr28=sqrt(9"x"^(2)+25"x"^(2)+15"x"^(2)) =7"x"rArr"x"=(28)/(7)=4`, `:. F_(1)=3xx4=12N, F_(2)=5xx4=20N`. |
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| 18. |
A person can see clearly upto 1m. Find the nature and power of the lens which will enable him to see things at a distance of 3m. |
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Answer» CONCAVE `-0.66D` |
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| 19. |
Law of gravitation is not applicable if (A) Velocity of moving objects are comparable to velocity of light (B) Gravitational field between objects whose masses are greater than the mass of sun. |
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Answer» A is TRUE, B is FALSE |
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| 20. |
A length of path ACB is 1500 m and the length of the path ADB is 2100 m. Two particles start from point A simultaneously around the track ACBDA. One of them travels the track in clockwise sense and other in anti-clockwise sense and other in anti-clockwise with their respective constant speeds. After 12 s from the start, the first time they meet at the point B. After minimum time (in s) in which they meet first at point B, again will they meet at the point B is time t_min = (12)^(x) s.The value of x is |
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Answer» 2 `therefore``(3600n_1)/v_1 = (3600n_2)/v_2` = t [total path length `= 1500 + 2100 = 3600`] or `n_1/n_2 = v_1/v_2 = (1500/12)/(2100/12) = 1500/2100 = 5/7` `therefore t_"min" = 3600 XX 5/v_1` `3600 xx 5/1500 xx 12 = 144 s = (12)^(x)` `therefore` x = 2 |
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| 21. |
The bar shown in the figure is made of a single piece of material. It is fixed at one end. It consists of two segments of equal lengh L//2 each but different cross sectional area A and 2A. Find the ratio of total elongation in the bar to the elongation produced in thicker segment under the action of an axial force F. Consider the shape of the joint to remain circular. (Given :Y is Young's modulus. |
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Answer» `=/_\L_2=(F(L//2))/(2AY)=(FL)/(4AY)` REQUIRED RATIO `=(/_\L_1+/_\L_2)/(/_\L_2)=(/_\L_1)/(/_\L_2)+1=2+1=3` |
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| 22. |
An ice ball has a metal piece embedded into it. The temperature of the ball is -theta^(@)C and it contains mass M of ice. When placed in a large tub containing water at 0^(@)C, it sinks. Assume that the water in immediate contact with the ice ball freezes and thereby size of the ball grows. What is the maximum possible mass of the metal piece so that the ball can eventually begin to float. Densities of ice, water and metal are sigma, rho and d respectively. Specific heat capacity of ice is s and its specific latent heat is L. Neglect the heat capacity of the metal piece. |
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| 23. |
A blackened platinum wire when gradually heated,first appears dull red,then blue and finally white.Explain. |
| Answer» SOLUTION :According to Wien.s displacement law,the wavelength of maximum emission SHIFTS to lower wavelength region when the temperature is increased.So, the wire turns to DULL RED, them to blue when it is heated.At very high temperature,it emits all radiations strongly and appears white. | |
| 24. |
If vecr and vecF are parallel or anti-parallel, then the torque is |
| Answer» ANSWER :A | |
| 25. |
Two cliff of heights 120 m and 100.4m are separated by a horizontal distance of 16m. If a car has to reach from the first cliff to the second, the horizontal velocity of the car should be |
| Answer» Answer :D | |
| 26. |
The critical velocity v of a body depends on the coefficient of viscosity eta the density d and radius of the drop r. If K is a dimensionless constant then v is equal to |
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Answer» `(K eta d)/(R )` |
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| 27. |
Which of the following is the evidence to show that there must be a force acting on earth and directed towards the sun? |
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Answer» Deviation of falling BODIES towards EAST |
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| 28. |
State and prove Archimedes principle. |
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Answer» Solution :Archimedes PRINCIPLE, It states that when a BODY is partially or wholly immersed in a fluid, it experiences an upward thrust equal to the WEIGHT of the fluid displaced by it and its opthrust acts through the centre of gravity of the liquid displaced. Proof: Consider a body of height "h" lying inside of destiny `rho` , at a depth bx below the freesurface of the liquid. AREA of cross section of the body is "a" . The forces on the sides of the body cancel out. Pressure at trhe upper face of the body, `P_(1),=xrhog`. Pressure of the LOWER face of the body,`P_(2)=(x+h)rhog`. Thrust acting on the upper of the body is `F_(1)=P_(1)a=xrhoga` acting vertically downwards, Thrust acting on the lower face of the body is `F_(2)=P_(2)a=(x+h)rhoga` The resultant force`(F_(2)-F_(1))` isacting on the body in the upward direction and is called upthrust (U). `U=F_(2)-F_(1)=(x+h)rhoga-xrhoa=ahrhog` But ah=V, Volume of the body=Volume of the liquid `U=Vrhog=Mg` Upthrust or buoyant force=Weight of liquid displaced . This proves the Archimedes principle. 
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| 29. |
Explainthe geometricalinterpretation of scalar product of two vectors . |
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Answer» SOLUTION :Suppose the scalarproduct of two vectros `vec(A) and vec(B)` is given as in the figure (a) `vec(A) . vec(B) = | hat(A) ||hat(B)| cos theta ` ` vec(A) . vec(B) = AB cos theta""` …(1) which is scalar .  where`theta `is the angle between `vec(A) and vec(B)` This product is shown in two ways. Method 1 : According to figure (b) , drawa perpendicular from the head of `vec(B)` on `vec(A)` resultingOM which is shownas in figureOM is the projection of`vec(B)` on to `vec(A)` or it is called the component of `vec(B )` inthe direction of `vec(A)` .  ` :.` O = component of `vec(B) ` along `vec(A)` ` = B cos theta ` ` :. vec(A) . vec(B ) = ABcos theta ` ` = A( B cos theta )` =A(OM) = Magnitude of `vec(A ) xx` component of `vec(A) ` along `vec(A)` Method 2 : According to figure (C ) , draw a perpendicular from the head of `vec(A)` on `vec(B)` resulting ON which is shownas in figure ON is the projection of `vec(A)` on a `vec(B )`or it is called component of `vec(A)` in the direction of `vec(B)`  ` :. ` ON = component of `vec(A )` along `vec(B)` ` = Acos theta ` ` :.vec(A).vec(B) = AB cos theta ` `= B(Acos theta)` = B(ON) = MAGNITUDEOF `vec(B) xx ` component of `vec(A) ` along `vec(B)` Hence , the scalar product of two vectors is equal to theproductof the magnitudeof first vectorwith theprojectionof secondvectorthe firstvector . |
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| 30. |
If distance travelled by particle is zero then its displacement is …... . |
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| 31. |
One source of sound moves towards stationary listener with a speed, (1)/(10)times speed of sound. Ratio of frequency of sound heard by listener to that emitted from source is ....... |
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Answer» `10/9`  We have, `(f _(L))/( V+ v _(L)) = (f _(S))/( v +v _(S))` `therefore (f _(L))/( f _(S))= (v + v _(L))/( v + v _(S)) = (v + 0)/( v - (v)/( 10 )) = (v)/( 9v//10 ) = (10)/(9)` |
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| 32. |
If veca"and "vecb are any two vectors such |veca+vecb|=|veca-vecb| find the angle between veca"and "vecb |
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| 33. |
Masses 1kg and 4kg are 6cm apart. The Point along the line joining the masses where the gravitational field due to them is zero is |
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Answer» 2CM from 1KG MASS |
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| 34. |
When water boils at a constant temperature of 100^@C, is there any change in the internal energy? |
| Answer» Solution :Yes, the INTERNAL energy of steam at `100^@C` is always greater then the internal energy of the same quantity of WATER at `100^@C`. | |
| 35. |
Water and caster oil are taken in two different bottles to the same level are shaken violently and kept on a table which will come to rest first? |
| Answer» SOLUTION :CASTOR OIL as it is more VISCOUS . | |
| 36. |
The velocity in SHM is ………………….. with acceleration, when particle is moving from extreme position to mean position. |
| Answer» SOLUTION :INPHASE | |
| 37. |
To what depth must a rubber ball be taken in deep sea so that its volume is decreased by 0.1% (The bulk modulus of rubber is 9.8 xx 10 ^(8) N//m^(2)and density of sea water is 10 ^(3) kgm^(-3)) |
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Answer» Solution :The percentage decrease in volume `(Delta V )/(V) XX 100 = 0.1` `therefore (Delta V)/(V) = (0.1)/(100)` Let the rubber BALL be taken up to depth h. `therefore` Bulk modulus `B = | (P )/( Delta V //V )| = (h rho g )/( Delta V //V)` `therefore h = ( B xx Delta V //V )/( rho g )` `therefore h = (9.8 xx 10^(8) xx 0.001)/( 10 ^(3) xx 9.8)` `therefore h = 100 m` |
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| 38. |
The displacement of a particle from its mean position (in metre) is given by y=0.2 sin (10pit+1.5pi) cos (10pit+0.5pi). The motion of particle is |
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Answer» PERIODIC but not S.H.M. |
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| 39. |
In planetary motion the areal velocity of position vector of a planet depends on angular velocity omega and the distance of the planet from sun (r ). If so the correct relation for areal velocity is |
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Answer» `(DA)/(DT) prop omega r` |
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| 40. |
A particle of mass m is located in a one dimensional potential field where potential energy of the particle has the form U (x) = (a)/(x^(2)) - (b)/(x)where a and b are positiveconstants. The position of equilibrium is: |
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Answer» `(B)/(2A)` |
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| 41. |
Two gases A and B having same pressure P, volume V , and Temperature T are mixed . If mixture has volume and temperature as V ans T respectively. Then pressure of mixture is : |
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Answer» 3P `P'=P_1+P_2=P+P=2P` |
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| 42. |
If the length of a wire increases by 1mm under 1kgwt. Then what will be the nature of the wire under 100kgwt.? |
| Answer» Solution :The WIRE will BREAK, because the STRETCHING force is INCREASED beyong breaking force. | |
| 43. |
Column - I describes some situations in which a small object moves. Column - II describes some characteristics of these motions. Match the situations in Column - I with the characteristics in Column - II {:("Column -I","Column -II"),("(A) The object moves on the axis under a conservative","(P) The object executes a simple"),("force in such a way that its ''speed'' and position''","harmonic motion."),("satisfy" v=c_(1)sqrt(c_(2)-x^(2))"where" c_(1) and c_(2) "are positive",),("constants.",),("(B) The object moves on the x-axis in such a way that","(Q) The object does not change its"),("its velocity and its displacement from the origin satisfy","direction"),("v=-kx, where k is a positive constant",),("(C ) The object is attached to one end of a mass-less","(R ) The kinetic energy of the object"),("spring of a given spring constant. The other end of the","keeps on decreasing"),("spring is attached to the ceilling of an elevator. Initially",),("everything is at rest. Teh elevator starts going upwards",),("with a constant acceleration a. The motion of the",),("object is observed from the elevator during the period",),("it maintains this acceleration",),("(D) The object is projected from the earth's surface","(S) The object can change its direction"),("vertically upwards with a speed" 2sqrt((GM_(e ))/(R_(e )))"where","only once".),(M_(e )"is the mass of the earth and" R_(e )"is the radius of the",),("earth Neglect forces from objects other than the earth.",):} |
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Answer» `v = omega sqrt(A^(2)-x^(2))` Where `omega` ands A are positive constants. The given above equation. HENCE the body executes. (b) If follows from equation v = -KX that v = 0 at x = 0, i.e. object comes to rest at `x = 0`. Thusof object starting from a NEGATIVE x valuescomes to rest x = 0 , its velocity (and hence kinetic energy) decrease with time. Hence correct choices are (q) are (r ). (c ) For an observer in the refrence frame of the elevator, a constant force acts on the object. This a pseudo force. Hence the motion of the object remains simple harmonic, which is choice (p). (d) Since the speed of the object is `sqrt(2)` time the escape velocity `(v_(e ) = sqrt(2GM_(e )//R_(e )))` the direction of motion of the object does not charge (since the force exerted by objects other than the earth are neglected) and the speed of the object keeps decreasing. Hence the correct choices are (q) and (r ). |
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| 44. |
Explain the meaning of heat and work with suitable examples. |
Answer» Solution :Meaning of heat: When an object at higher TEMPERATURE is placed in contact with ANOTHER object at lower temperature, there will be a spontaneous flow of energy from the object at higher temperature to the one at lower temperature. This energy is called heat. This process of energy transfer from higher temperature object to lower temperature object is called heating. Due to flow of heat sometimes the temperature of the body will increase or sometimes it may not increase.  Meaning of work: When you rub your hands against each other the temperature of the hands INCREASES. You have done some work on your hands by rubbing. The temperature of the hands increases due to this work. Now if you place your hands on the cheek, the temperature of the cheek increases. This is because the hands are at higher temperature than the cheek. In the above example, the temperature of hands is increased due to work and temperature of the cheek is increased due to heat transfer from the hands to the chin. It is shown in the Figure. By doing work on the system, the temperature in the system will increase and sometimes may not. Like heat, work is also not a quantity and through the work energy is transferred to system So we cannot use the WORD ..the object contains more work. or.less work.. Either the system can transfer energy to the surrounding by doing work on surrounding or the surrounding may transfer energy to the system by doing work on the system. For the transfer of energy from one hody to another body through the process of work, they need not be different temperatures. |
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| 45. |
Define the following. loss of kinetic energy in inelastic collision. |
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Answer» Solution :In PERFECTLY inelastic collision, the loss in kinetic energy during collision is transformed to another form of energy like sound, thermal, HEAT, light etc. Let `KE_(i)` be the TOTAL kinetic energy before collision and `KE_(f)` be the total kinetic energy after collision. Total kinetic energy before collision, `KE_(i)=(1)/(2)m_(1)u_(1)^(2)+(1)/(2)m_(2)u_(2)^(2)` ....(1) Total kinetic energy after collision, `KE_(f)=(1)/(2)(m_(1)+m_(2))v^(2)`....(2) Then the loss of kinetic energy is Loss of KE, `Delta Q=KE_(i)-KE_(f)` `=(1)/(2)(m_(1)+m_(2))v^(2)-((1)/(2)m_(1)u_(1)^(2)+(1)/(2)m_(2)u_(2)^(2))` ....(3) Substituting equation `v=(m_(1)u_(1)+m_(2)u_(2))/((m_(1)+m_(2)))` in equation (3), and on simplifying expand v by using the ALGEBRA `(a+b)^(2)=a^(2)+b^(2)+2ab`, we get Loss of KE, `Delta Q=(1)/(2)((m_(1)m_(2))/(m_(1)+m_(2)))(u_(1)-u_(2))^(2)`. |
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| 46. |
A ball of mass of 150g and velocity 12 m/s coming towards a batsman is hit by him with a force in such a way that the ball moves with velocity 20 m/s in the direction opposite to its original one. The time of contact between the ball and the bat is 0.01 s. Then the force applied on it by the batsman i s ..... |
| Answer» SOLUTION :480 N | |
| 47. |
A concave mirror of focal length 10 cm and a convex mirror of focal length 15 cm are placed facing each other 40 cm apart, A point object is placed between the mirrors, on their common axis and 15 cm from the concave mirror. Find the positions, nature of the image, and over all magnification produced by the successive reflections, first at concave mirror and then at convex mirror. |
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Answer» Solution :According to given problem, for concave mirror, `u=-15 cm, and f=-10 cm` So `1/v+1/(-15)=1/(-10)` i.e., `v=-30 cm` i.e., concave mirror will form real, inverted and enlarged image `I_1` of OBJECT O ata DISTANCE 30 cm from it, i.e., at a distance 40-30=10 cm from convex mirror.  FOr convex mirror the image, `I_1` will act as an object and so for it u=-10 cm and `f= pm 15 cm` `1/v+1/(-10)=1/15` i.e., v=+6 cm So final image `I_2` is formed at a distance 6 cm behind the convex mirror and is virtual as shown in FIGURE. Over all magnification `=m_1 times m_2=-2 times 6//10=-6//5`. Negativeindicates final image is virtual w.r.t given object. |
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| 48. |
Two liquids are at 40^(@)C and 30^(@)C. When they are mixed in equal masses, the temperature of the mixture is 36^(@)C. Ratio of their specific heats is |
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Answer» `3:2` |
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| 49. |
It is possible to project a particle with a given speed in two possible ways so that it has the same horizontal range .R.. The product of time taken by it in the two possible ways is |
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Answer» R/g |
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