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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. |
When two resistors of (600pm3%) ohm and (300pm 6%) ohm are connected in parallel, value of the equivalent resistance is |
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Answer» `200 "OHM" PM 1.5%` |
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| 2. |
Interal forces can change |
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Answer» the linear momentum but not the kinetic energy |
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| 3. |
The calorie is a unit of heat or energy and it equals about 4.2 J where 1J=1kgm^(2)s^(-2). Suppose we employ a system of units in which the unit of mass equals alpha kg, the unit of length equals beta m and the unit of time isgamma s. Show that the calorie has a magnitude of 4.2alpha^(-1) beta^(-2) gamma^(2) in terms of the new units. |
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Answer» Solution :Calorie is the of HEAT energy. DIMENSIONAL FORMULA of energy is `ML^(2)T^(-2)` `n_(1)(M_(1)L_(1)^(2)T_(1)^(-2))=n_(2)(M_(2)L_(2)^(2)T^(-2))` Given `n_(1)=4.2, n_(2)=?` `M_(1)=1kgL_(1)=1mT_(1)=1s` `M_(2)=alpha kg L_(2)=betamT_(2)=GAMMAS` `n_(2)=n_(1)((M_(1))/(M_(2)))((L_(1))/(L_(2)))^(2)((T_(1))/(T_(2)))^(-2)=4.2(1/(alpha))(1/(beta))^(2)(1/(gamma))^(-2)` `n_(2)=4.2alpha^(-1)beta^(-2)gamma^(2)` |
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| 4. |
If the displacement y of a particle is y =A sin (pt+qx) then dimensional formula of pq is |
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Answer» L |
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| 5. |
Mention the formulae for coefficient of performance of a refrigerator . |
| Answer» Solution :Coefficient of PERFORMANCE `beta` may be defined as the RATIO of AMOUNT of heat extracted from the cold reservior to the amount of work done by the PUMP in order to reject heat to the hot surrounding . | |
| 6. |
One end of a long metallic wire of length L, area of cross-section A and Young's modulus Y is tied to the ceiling. The other end is tied to a massless spring of force constant k and a mass m is hung from the free end of the spring. If m is slightly pulled down and released, then its time period of oscillation is |
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Answer» `2pi sqrt(m/k)` |
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| 7. |
For a particle in SHM, the K.E. at any instant is given by K = K_0 cos^2 omega t. The total energy of SHM is |
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Answer» `K_0` |
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| 9. |
A metal cube is weighing 4 N in air, 3 N in water and 2.5 N in a liquid |
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Answer» Relative DENSITY of liquid is smaller than that of CUBE. `=("Weight in air")/("loss of weight in WATER")` `=(4)/(4-3)=4` Relative density of liquid `=("loss of weight in liquid")/("loss of weight in water")` `=(4-2.5)/(4-3)=1.5` |
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| 10. |
When two waves interfere,does one alter the progress of the other? |
| Answer» Solution :When two waves interfere,the PROGRESS of NONE of them is altered.Rather they continue in their ORIGINAL direction. | |
| 11. |
A car of mass 1000 kg is moving at a speed 30 m/s. Brakes are applied to bring the car to rest. If the net retardation force is 5000 N, the car comes to stop after travelling d m in t s. Then |
| Answer» Answer :D | |
| 12. |
The diagram shows an estimeated force time graph for a baseball struck by a bat |
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Answer» SOLUTION :Formula (i)IMPULSE + AreaABC `=(1)/(2)xx 18000 xx (2.5 -1)` (ii) force= `("Impulse")/("time ")= (1.35 xx 10^(4))/( (2.5-1))` `=(13500)/(1.5) =900 N` (III)MAXIMUMFORCE= 1800 N |
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| 13. |
Derive an expression for Radius of gyration. |
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Answer» Solution :For bulk objects of regular shape with uniform mass distribution, the expression for moment of inertia about an axis involves their total mass and geometrical features like radius, length, breadth, which take care of the shape and the size of the objects. But, we need an expression for the moment of inertia which could take care of not only the mass, shape and size of objects, but also its orientation to the axis of rotation. Such an expression should be GENERAL so that it is applicable even for objects of IRREGULAR shape and non-uniform distribution of mass. The general expression for moment of inertia is given as, `I= MK^2` where, M is the total mass of the object and K is called the radius of gyration. The radius of gyration of an object is the perpendicular distance from the axis of rotation to an equivalent point mass, which would have the same mass as well as the same moment of inertia of the object. As the radius of gyration is distance, its unit is m. Its dimension is L. A rotating rigid body with respect to any axis, is considered to be made up of point masses `m_1, m_2 , m_3 ,....... mn` at perpendicular distances (or positions) `r_1 , r_2 , r_3.....r_n` respectively as shown in figure. The moment of inertia of that object can be written as, `I = sum m_1r_1^2 = m_1r_1^2 + m_2r_2^2 +m_3r_3^2 + ...... + m_n m_n^2` If we take all the n number of individual masses to be equal `m = m_1 = m_2 = m_3 = ....... = m_n` then `I = mr_1^2 + mr_2^2 + mr_3^2 + ....... + mr_n^2` `m = (r_1^2+ r_2^2 + r_3^2 + ..... + r_n^2)` ` =mn( (r_1^2 + r_2^2 + r_3^2 + ..... + r_n^2)/(n))` `I = MK^2` where, nm is the total mass M of the body and K is the radius of gyration. The expression for radius of gyration INDICATES that it is the root mean square (rms) distance of the particles of the body from the axis of rotation. In fact, the moment of inertia of any object could be EXPRESSED in the form, `I = MK^2` For example, let us take the moment of inertia of a uniform rod of mass M and length l . Its moment of inertia with respect to a perpendicular axis passing through the center of mass is, `I = 1/12 Ml^2` In terms of radius of gyration,`I = MK^2` Hence `MK^2 = 1/12 Ml^2` `K^2 =1/12 l^2` `K =1/12l " or"K = (1)/(2sqrt3) l " or" K = (0.289)l` |
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| 14. |
Friction can be reduced by |
| Answer» Solution : using ball bearings | |
| 15. |
By inserting a capillary tube upto a depth 1 in water, the rises to a height h. If the lower end of the capillary tube is closed inside water and the capillary is taken out and closed and opened, to what height the water will remain in the tube when 1gth |
| Answer» ANSWER :C | |
| 16. |
The efficiency of a heat engine if the temperature of source 227^(@)C and that of sink is 27^(@)C nearly |
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Answer» 0.4 |
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| 17. |
What is carnot's theorem . |
| Answer» Solution :WORKING between the same two temoeratures all REVERSIBLE ENGINES have same efficiency and no engine working between the same two TEMPERATURES can be more efficient than a carnot's reversible engine . | |
| 18. |
A glass vessel just holds 50gm of a liquid at 0^(@) C. If the coefficient of linear expansion of glass is 8 xx 10^(-6)//""^(@) C. The mass of the liquid it holds at 80^(@) C is [coefficient of absolute expansion of liquid 5 xx 10^(-4)//^(@) C ] |
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Answer» 46 gm |
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| 19. |
Sppose that a pointmass 'm'is movingunder a constantforce vecF = 2hati-hatj + hatk netweon .At someinstant , t=0, pointP(xm, ym, -1m) [m- metre ] is the instantaneous position of themass. We knowthattorque can beexpressed as thecross- product of positionvector and forcesvector, i.e., tau= vecr xx vecF . AtP, torquecan beexpessedastau= (-4hatj - 4 hatk)NmAt some other instant,t=3 sec, the pointmasshas anotherinstantaneouspositionQ(x_(1), y_(1), z_(1))suchthatthe displacementvectorsbetweenpoints P and Qand thegivenforce are mutually perpendicular. Also, x-component of torqure at Q is zero and yz-components are equal in magnitude and direction alongthe negativedirectionof therespectiveaxes. Usinga definitescale, if we constructa parallelogram with the positionvectorsof Q and thegivesforce vecF as itsadjacent sides , areaof thisparallelogramis 2sqrt(2)m^(2) . Areaof the given parallelogram , in fact , representsa physicalquantitywhose magnitude in SI systemcan beexpressed as 5times thegives are Answerthe following questions.Coordinates of P are : |
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Answer» `""= ("XI" + "yj"-hatk)xx (2i-j+hatk)` `""vectau_(p)=(y -1)i -(x + 2)j-(x +2y)hatk` Given`""vectau_(p) = - 4hatj - 4hatk` `therefore""x=2 and y=1` ALSO, `""vecr_(PQ)=(x_(1) -x) i+ (y_(1) - y)j +(z_(1) + 1)hatk` Given `""vecr_(PQ) botvecF` `therefore""vecr_(PQ) .vecF = 0` `therefore[(x_(1) - 2)hati + (y_(1) - 1) hatj + ( z_(1)+ 1)hatk].[2i - hatj+hatk]= 0......(i)` `""tau_(Q)= vecr_(QO) xxvecF= |{:("i","j","k"),(x_(1),y_(1),z_(1)),(2,-1,1):}|` `tau_(Q)= (y_(1) + z_(1))hati - j (x_(1) -2z_(1))+ k(-x_(1)- 2y_(1))` Given, `""y_(1) + z_(1) =0""...........(ii)` And,`""x_(1) -2z_(1) =x_(1) + 2y_(1)""......(iii)` Also,`""|vecr_(QO) xx vecF= 10sqrt(2)` `therefore (y_(1) + z_(1))^(2) + (x_(1) -2z_(1))^(2) + (x_(1) + 2y_(1))^(2) = 100 xx 2....(iv)` `therefore ""2(x_(1) - 2z_(1))^(2) = 100 xx 2` `therefore""x_(1) - 2z_(1) = 10` `therefore""vectau_(Q)= - 10 hatj - 10 hatk` `""W = 0, " as" vecr_(PQ)` is prependicular to `vecF` |
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| 20. |
Two bodies of masses 4kg and 16kg are at rest. The ratio of times for which the same force must act on them to produce the same kinetic energy in both of them is |
| Answer» ANSWER :B | |
| 21. |
Characteristic X-rays of K-series are obtained when the electron transition is from higher orbits to- |
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Answer» K-orbit |
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| 22. |
(A) : Absolute error is unitless and dimensionless. (R) : Errors are considered in measured quantities, not in the given constants. |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT explanation of (A) |
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| 23. |
Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms : [Hint : Assume the atoms to be 'tightly packed' in a solid or liquid phase, and use the known value of Avogadro's number. You should, however, not take the actual numbers you obtain for various atomic sizes too literally. Because of the crudeness of the tight packing approximation, the results only indicate that atomic sizes are in the range of a few Å]. |
| Answer» SOLUTION :CARBON[1.29 Å ], GOLD [1.59 Å], LIQUID Nitrogen [1.77 Å ], Lithium [ 1.73 Å ], Liquid FLUORINE[1.88 Å] | |
| 24. |
Inside a railway car a plumb bob is suspended from the roof and a helium filled balloon is tied by a string to the floor of the car. When the railway car accelerates to the right, then |
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Answer» both the plumb bob and BALLOON move to the LEFT |
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| 25. |
A student performs experiment with a simple pendulum and measure time period for 20 vibrations. If he measures time for 100 vibrations the error in the measurement oftime period will be reduced by a factor of |
| Answer» Answer :C | |
| 26. |
The speed with which a bullet can be fired is 150 ms^(-1). Calculate the greatest distance to which it can be projected and also the maximum height to which it would rise. |
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Answer» Solution :`R = (u^(2) sin 2 ALPHA)/(G) = ((150)^(2) sin 90^(@))/(9.8) = 2295.1m` `H_("max") = (u^(2) sin^(2) alpha)/(2g) = ((150)^(2) sin^(2) 45^(@))/(2 xx 9.8) = 573.97 m` |
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| 27. |
A particle executing SHM is an example of ......... |
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Answer» ACCELERATION of CONSTANT MAGNITUDE and direction |
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| 28. |
There is a long narrow and smooth groove in a horizontal table. Two identical blocks A and B each of mass m are placed inside the groove at some separation. An ideal spring is fixed to A as shown. Block A is given a velocity u to the right and it interacts with B through the spring. (a) What will be final state of motion of the two blocks ? (b) During their course of interaction what is the minimum kinetic energy of the system ? (c) The spring is removed and the two blocks are tied using a mass less string. Now A is set into motion with speed u. What will be the final state of motion of the two blocks in this case ? How much kinetic energy is lost by the system ? Where goes this energy ? |
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Answer» (b) `(m u^(2))/(4)` (c) Both will be travelling with velocity `(u)/(2)`. Loss in `KE=(m u^(2))/(4)` |
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| 29. |
A ball moving translationally collides elastically with another stationary ball of the same mass. At the moment of impact the angle between the straight line passing through the centres of the balls and the direction of the initial motion of the striking ball is equal to alpha = 45^(@). Assuming the ball to be smooth. Find the fraction eta. of the kinetic energy of the striking ball that turned into potential energy at the moment of the maximum deformation |
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Answer» 0.25 |
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| 30. |
For a given material the Youngs modulus is 2.4 times that of its rigidity modulus its poission's ratio is |
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Answer» 2.4 |
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| 31. |
An object falls from a bridge that is 45 m above the water.It falls deirectly into a small row-boat moving with constant velocity that was 12 m from the point of impact when the object was released .What was the speed of the boat? |
| Answer» Answer :D | |
| 32. |
A brass rod has length 0.2m area of cross section 1.0cm^2 and Young's modulus 10^11Nm^(-2)If it is compressed by 5kg. wt along its length then the change in its energy will be |
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Answer» An increase of`2.4 XX 10^(-5) J` |
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| 33. |
A person in an elevator accelerating upwards with an acceleration of 2 m s^(-2), tosses a coin vertically upwards with a speed of 20 m s^(-1). After how much time will the coin fall back into his hand ? ( take g=10 ms^(-2)) |
| Answer» SOLUTION :`3.33 s` | |
| 34. |
Mention some applications of viscosity in daily life. |
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Answer» Solution :Applications of viscosity are:(i) The oil used as a lubricant for heavy machinery parts should have a high viscous coefficient. To select a suitable lubricant, we should know its viscosity and how it varies with temperature [Note:As temperature increases, the viscosity of the liquid decreases]. Also, it helps to choose oils with low viscosity used in car ENGINER (light machinery). (ii) The highly viscous liquid is used to damp the motion of some instruments and is used as BRAKE oil in hydraulic brakes. (iii) Blood circulation through arteries and veins DEPENDS upon the viscosity of FLUIDS. (iv) Millikan conducted the oil drop experiment to determine the CHARGE of an electron. The knowledge of viscosity is used by him to determine the charge. |
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| 35. |
Two open organ pipes of lengths 50 cm and 50.5 cm produce 3 beats/s. Then the velocity of sound is . |
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Answer» 300 m/s |
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| 36. |
If work is done on system by non-conservative force, then potential energy increases. |
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Answer» |
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| 37. |
What is the gravitational field strength of a planet where the weight of a 60kg astronaut is 300N? |
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Answer» SOLUTION :Gravitational FIELD = `(F)/(m)` `=(300)/(60)` `=5 NKG^(-1)` |
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| 38. |
Four particles of masses 1 kg, 2 kg, 3 kg and 4 kg are placed at the four vertices A, B, C and D of a square side 1 m. Find the position of cejitre of mass of the system. |
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Answer» Solution :ASSUMING A as the origin, AB as x-axis and AD as y-axis we have Mass of FIRST particle, `m_(1)`= 1kg , Mass of SECOND particle, `m_(2)`= 2KG Mass of third particle, `m_3 = 3kg` , Mass of fourth particle, `m_4 = 4`kg Co - ordinates of first particle, (`x_(1),y_(1)`)= (0,0) Co - ordinates of second particle, `(x_2, y_2)` = (1,0) Co - ordinates of third particle `(x_3,y_3) = (1,1)` Co - ordinates of fourth particle `(x_4,y_4) = (0,1)` Co-ordinates of their CM are:  `x_(CM) =(m_(1)x_(1) + m_(2)x_(2) + m_(3)x_(3) + m_(4)x_(4))/(m_(1) + m_(2) +m_(3) + m_(4)) =((1)(0)+2(1) + 3(1) + 4(0))/(1+2+3+4)=0.5 m` Similarly, `y_(CM) =(m_(1)y_(1) + m_(2)y_(2) + m_(3)y_(3) + m_(4)y_(4))/(m_(1)+m_(2)+m_(3)+m_(4)) =(1(0) + 2(0) + 3(1) + 4(1))/(1+2+3+4) =0.7 m` `therefore` Co-ordinates of centre of mass `(x_(cm),y_(cm)) =(0.5 m, 0.7 m)` |
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| 39. |
Which one of the following statements is true? |
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Answer» A scalar quantity is the one that is conserved in a process. (b) The statement is false. It is because the TEMPERATURE (scalar quantity) can be negative. (c ) The statement is false. It is because gravitational potential energy (scalar quantitiy) vary from point to point is space. (d) The statement is true because the value of scalar does not change with ORIENTATIONOF axes. |
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| 40. |
The dimensions of Hubble's constant are : |
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Answer» `[M^(0)L^(0)T]` `=([LT^(-1)])/([L])=T^(-1)` `=M^(0)L^(0)T^(-1)` |
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| 41. |
Let vec(C ) = vec(A) + vec(B), |
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Answer» `|vec(C )|` is ALWAYS GREATER than `|vec(A)|` |
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| 42. |
If the kinetic energy of a body increases by 125% the percentage increase in its momentum is |
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Answer» 0.5 |
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| 43. |
gamma_(A) of liquid is 7/8 of gamma_( R) of liquid. alpha_(g) of vessel is |
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Answer» `( gamma_( R) )/( 8)` |
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| 44. |
In Internationan System of units, there are seven base quantities whose units are defined. Which physical quatity has a prefix with its unit? |
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Answer» MASS |
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| 45. |
M.I. of a solid sphere about its diameter is 64" kg m"^(2). If that sphere is recast into 8 identical small spheres, then M.I. of such small sphere about its diameter is |
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Answer» `"8 kg m"^(2)` |
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| 46. |
A solid cuboid with edges 1 cm, 2 cm and 3 cm is suspended in vacuum, its temperature falls from 100^(0)C to 90^(0)C in 100s. Another solid cuboid of same material and similar surface finish is kept in same enclosure. If sides of couboid are 2 cm, 4 cm and 6 cm, then time ttaken for this cuboid to cool from 100^(0)C to 90^(0)C will be |
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Answer» 100s |
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| 47. |
The linear momentum of a particle varies with time t as P = a + bt + ct^2. Then which of the following is correct ? |
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Answer»
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| 48. |
In a capillary tube,closed at one end,some air is enclosed by a mercury thread of length 10 cm. When the tube is kept horizontal the length of the air column is 17 cm. When it is held vertical with the open ends up,the length changes to 15 cm,what will be the length of the air column when the tube is held vertical with the open ends downwards? |
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Answer» Solution :Let atmospheric PRESSURE=pcmHg, cross sectional area of the tube =`a cm^2`. When the tube is horizontal pressure of the confined air `p_1=p cmHg` Volume of this confined air `V_1=17 a cm^3` When the tube is held vertical with the open ends up Pressure of the confined air ,`p_2(p+10) cmHg` Volume of this confined air ,`V_2=15 a cm^3` ACCORDING to Boyle.s law `p TIMES 17A=(p+10) times 15a or, p=75` `therefore` Atmospheric pressure=75 cmHg Again, when the tube is held vertical with the open and downwards, Pressure of the confined air , `p_3=(p-10) cmHg` volume of this confined air ,`V_3= h a cm^3` [where h=lengthe of the air column] According to Boyle.s law `p_1V_1=p_3V_3or,ptimes17a=(p-10)timesha` or,`h=(17p)/(p-10)=(17times75)/65=19.6 cm`. |
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| 49. |
A calorimeter of heat capacity 83.72J K^(-1) contains 0.48 Kg of water at 35^(0)C. How much mass of ice at 0^(0)C should be added to decrease the temperature of the calorimeter to 20^(0)C. |
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Answer» Solution :Heat CAPACITY of the calorimeter `= 83.72 J K^(-1)` , Fall in TEMPERATURE of the calorimeter `= (35^(0)C - 20^(0)C) = 15^(0)C` Heat lost by the calorimeter `= 83.72 xx 15 = 1255.8J` , Mass of water = 0.45 Kg Specific heat of water `= 4186 J Kg^(-1)K^(-1) ,` Fall in temperature of water `= 35^(0)C -20^(0)C = 15^(0)C` Heat lost by water `= 0.48 xx 4186 xx 15 = 30139.2J ,` LET .m. be the mass of ice added, Heat gained by ice during melting `= mL = m xx 0.335 xx 10^(6)J` Heat gained by ice (water) during rise in temperature `= m xx 4186 xx (20 - 0) = m xx 83720J` `therefore` Heat gained by the ice = Heat lost by the calorimeter and water `m xx (0.335 xx 10^(6) + 83720)` = 1255.8 +30139.2 = 31395 = , m[335000+83720] = 31395 `implies m = (31395)/(418720)=0.07498 Kg.` |
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| 50. |
A solid sphere of mass m and radius R rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation ((E_("sphere"))/(E_("cylinder"))) will be = .............. |
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Answer» Solution :Moment of inertia of sphere `I_(S)=(2)/(5)MR^(2)` Moment of inertia of cylinder `I_(C )=(MR^(2))/(2)` Rotational KINETIC energy of body `E=(1)/(2)Iomega^(2)` Here rotational kinetic energy of sphere `E_(S)=(1)/(2)I_(S)omega_(S)^(2)` `=(1)/(2)XX(2)/(5)MR^(2)omega_(S)^(2)` Rotational kinetic energy of cylinder `E_(C)=(1)/(2)I_(C)omega_(C)^(2)` `=(1)/(2)xx(MR^(2))/(2)xx(2omega_(S))^(2)` `=(1)/(2)xx(MR^(2))/(2)xx4omega_(S)^(2)` `therefore (E_(S))/(E_(C))=(1)/(5)` `therefore E_("sphere"):E_("cylinder")=1:5` |
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