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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. |
Density of a liquid .A. is 0.5 g/c.c and that of liquid .B. is g/c.c . Heat capacity of 8 litres of .A. is equal to that of 10 litres of .B.. Then their specific heats ratio is |
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Answer» `4:5` |
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| 2. |
According to Newton's law of cooling, the rate of cooling of a body is proportional to (Deltatheta)^(n),"where "Deltatheta is difference of temperature of body and surroundings, and n is equal to |
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Answer» TWO |
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| 3. |
Elastic potential energy possessed by a spring is |
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Answer» `1/2mv^(2)` |
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| 4. |
In which of the following cases it is certainly not possible for the diac to perform pure rolling. |
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Answer» A ROUGH inclined PLANE |
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| 5. |
Force acting on the particle moving with constant speed is |
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Answer» ALWAYS zero |
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| 7. |
A uniform rod of length 1 and mass m is suspended by two vertical inextensible string as shown in the fig. The tension in the left string when right string snaps, is |
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Answer» Solution :`mg-T=ma ""…(1)` `alpha=(tau)/(I)=(mg(L)/(2))/((ml^(2))/(3))=(3g)/(2I) ""…(2)` `a=(l)/(2) alpha ….(3)` solving EQUATION (1), (2) and (3) we GET `T=(mg)/(4)` 
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| 8. |
The maximum and minimum magnitude of the resultant of two vectors are 17 units and 7 units respectively. Then the magnitude of resultant of the vectors when they act perpendicular to reach other is |
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Answer» 14 |
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| 9. |
Ratio of distances covered by object falling freely under gravity in 1 ^(st),2^(nd)&3^(rd)second 18 ........ |
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Answer» `1:3:5` In `d _(n) = v _(0)t + a/2 (2n -1) , v_(0) =0,` `d_(n) = (a)/(2) (2n -1)` `THEREFORE ` For `n =1, d _(1) = g /2 (2XX 1-1) = g/2` For `n = 2, d_(2) = g/2 (2 xx 2-1) = (3g)/(2)` For `n =3, d _(3) = g/2) (2 xx 3-1) = (5G)/(2)` `therefore d _(1) : d_(2) :d _(3) = (g)/(2) : (3g)/(2) : (5g)/(2) = 1:3:5` |
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| 10. |
Statement I : Acceleration of a moving particle can change its direction without any change in the direction of velocity. Statement II : If the direction two particles moving with constant velocities always remains vector also changes. |
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Answer» Statement I is TRUE, statement II is true , statement II is a correct EXPLANATION for statement I. |
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| 11. |
Give the quantities for which the following are the dimensions: M^(1)L^(2) |
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Answer» |
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| 12. |
On the surface of earth, the escape velocity of a body is 11.2 km//s. If mass of earth is made half and radius becomes twice, then the escape velocity will become |
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Answer» 5.6 km/s `V_(e) = 11.2 km//s` `V_(e)` = Final escape velocity `= sqrt((2GM)/(2(2R))) = sqrt((1)/(4)(2GM)/(R ))` `V._(e) = (1)/(2)V_(e)` = 5.6 km/s |
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| 13. |
Three spherical balls of masses 1kg, 2kg and 3kg are placed at the corners of an equilateral triangle of side 1m. Find the magnitude of the gravitational force experted by the 2kg and 3kg masses on the 1kg mass. |
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Answer» Solution :Mass of the spherical ball at the corner .A. of the equilateral triangle = 1kg Mass of the spherical ball at the corner .B. of the equilateral triangle = 2kg Mass of the spherical ball at the comer .C. of the equilateral triangle = 3kg Side of equilateral triangle = 1m If .F. is the force of attraction between 1kg, 2kg masses then `F_1 = G XX (1 xx 2)/((1)^2) rArr F_1 = 2G` If `.F_2.` is the force of attraction between 1kg, 3kg masses then ` F_2 = G xx (1xx3)/((1)^2) rArr F_2 = 3G` The ANGLE between the forces `F_1` and `F_2`is `60^@`. If `.F_R.`is the resultant of these two forces, then according to parallelogram law of VECTORS we can write `F_R = sqrt(F_1^2 + F_2^2 +2F_1F_2 Cos theta ) rArr F_R = sqrt((2G)^2 + (3G)^2 +2 xx 2G xx 3G xx cos 60^@) rArr F_R = sqrt19 G` |
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| 14. |
A boat of mass 40 kg is floating on water. A boy of mass 10 kg on the boat moves by 8 m towards the shore. What is the distance by which the boat moves away from the shore? |
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Answer» |
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| 15. |
If x = a cos t is the displacement in time t then, acceleration is : |
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Answer» a COS t `a = (d)/(dt) (-a sin t) = - a cos t` |
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| 16. |
The gas thermometers are more sensitive than liquid thermometers because |
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Answer» GASES expand more than liquids |
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| 17. |
When the wind blows at a high speed, the proofof a house is blown off because: |
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Answer» the PRESSURE under the ROOF increases |
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| 18. |
A solid floats in liquid A with half its volume immersed and in liquid B with 2/3 of its volume immersed. The densities of the liquids A and B are in the ratio. |
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Answer» Solution :WEIGHT of liquid A DISPLACED is `W_A = 1/2 V rho_A G` Weight of liquid B displaced is `W_B = 2/3 rho_B g` According to Archimede.s principle `W_A = W_B` `1/2 V rho_A g = 2/3 V rho_B g` `(rho_A)/(rho_B) = 4/3` |
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| 19. |
The temperature of a wire is doubled. The Young's modulus of elasticity |
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Answer» will ALSO double |
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| 20. |
In C.G.S. system the magnitude of the force is 100 dyne. In another system where the fundamental physical quantities are kilogram, metre and minute, the magnitude of the force is (new units) |
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Answer» 0.036 |
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| 21. |
An oscillating simple pendulum is kept in a lift which is accelerating upwards. Will there be any change in the time period ? |
| Answer» Solution :When the lift ACCELERATES UPWARDS `g.=g+a`DUE to increase in the value of g, T decreases. | |
| 22. |
A simple pendulum is swinging in a vertical plane. Its heighest point above ground is 50cm. If the maximum velocity of the pendulum is 1.4 ms^(-1), its lowest position above ground is |
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Answer» 20cm |
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| 23. |
While walking on ice one should take small steps to avoid slipping. This is because smaller steps ensure |
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Answer» LARGER friction |
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| 24. |
A metal cube is placed in an empty vessel. When water is filled in the vessel so that the cube is completely immersed in the water, the force on the bottom of the vessel in contact with the cube |
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Answer» Will increase |
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| 25. |
What is the largest average velocity of blood flow in an artery of radius 2xx10^(-3)m if the flow must remain laminar > (b) What is the corresponding flow rate ? (Take vicosity of blood to be 2.084xx10^(-3)Pa s). |
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Answer» Solution :Density of blood `rho=1.06xx10^(3)KGM^(-3)` Radius of the artery `r=2xx10^(-3)` The COEFFICIENT of viscosity of blood, `eta=2.84xx10^(-3)`Pas For streamline flow `N_(R)=2000` (a) Suppose the largest AVERAGE velocity of blood is v, Reynold.s number `R_(e)=(RhovD)/(eta)` `thereforev=(R_(e)eta)/(rgoD)` `(2000xx2.084xx10^(-3))/(1.06xx10^(3)xx2xx2xx10^(-3))` `=983xx10^(-3)` `=0.983` `thereforev=0.98ms^(-1)` (b)Suppose flow ate of blood is V, `thereforeV=` arrea of cross section of artery `xx` velocity of blood `=pir^(2)xxv` `=3.14xx(2xx10^(-3))^(2)xx0.98` `=12.3xx10^(-6)` `=1.23xx10^(-5)m^(3)s^(-1)` |
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| 26. |
A uniform metal rod is suspended from one end. When it is displaced through 60^(@) from the vertical, the change in potential energy is .U., the change in potentialenergy when it is displaced further through 60^(@) is |
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Answer» U |
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| 27. |
A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal to - (K//r^(2)), where K is constant. What is the total energy of the particle ? |
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Answer» `(K)/( R)` |
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| 28. |
A block of metal weighing 2kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate of 1 kg//s and at a speed of 5 m//s. The initial acceleration of the block will be |
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Answer» `2.5m//s^(2)` |
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| 29. |
the frequency of oscillation of mass .m. suspended by a spring is n_(1). If the spring is made half, the same mass oscillates with the frequency n_(2). Then the value of n_(2) is |
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Answer» `n_(1)"/"2` |
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| 30. |
The value of acceleration due to gravity .g_p. on the surface of a planet with radius double that of earth and same mean density as that of the earth is (g_e to acceleration due to gravity on the surface of earth) |
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Answer» `g_p=2g_e` |
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| 31. |
A capillary tube of radius r and height h_(1) is connected to a broad tube of large height as shown in the figure. Water is poured into the broad tube – drop by drop. Drops fall at regular intervals. Plot the variation of height of water in both tubes with time. Initially the tube and capillary are empty. Neglect the volume of the connecting pipe. |
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Answer» |
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| 32. |
Two particles move parallel tox-axis about the origin with same amplitude 'a' and for omega . At a certain instant they are found at a distance a/3 from the origin on opposite sides but localities are in the same direction. What is the phase difference between the two? |
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Answer» `COS^(-1) 7/9` |
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| 33. |
Which of the following is meaningful ? |
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Answer» VECTOR/Vector |
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| 34. |
The motion of a particle along a straight line is described by the function s=6+4^(2)-t^(4) in SI units. Find the velocity, acceleration, at t=2s, and the average velocity during 3^(rd) seconds. |
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Answer» Solution :`s=6+4t^(2)-t^(4)` Velocity `=(ds)/(dt)=8t-4t^(3)" when "t=2` Velocity `=8xx2-4xx2^(3)` Velocity `=-16m//s` Acceleration `a=(d^(2)s)/(dt^(2))=8-12t^(2)" when "t=2` acc `=8-12xx2^(2)=-40` acc `=-40m//s^(2)` displacement in 2 seconds `s_(1)=6+4.2^(2)-2^(4)=6m` displacement in 3 seconds `s_(2)=6+4.3^(2)-3^(4)=-39m` displacement during `3^(rd)` SECOND `=s_(2)-s_(1)=-39-6=-45m` `therefore` AVERAGE velocity during `3^(rd)` second `=(pm45)/(1)=-45m//s` `-ve` SIGN indicates that the body is moving in opposite direction to the initial direction of motion. |
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| 35. |
The speed of a solid sphere after rolling down from rest without sliding on an inclined plane of vertical height h is |
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Answer» `sqrt((4)/(3)GH)` |
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| 36. |
A solid sphere, a hollow sphere and a disc, all having samc mass and radius, are placed at the top of an incline and released. The friction coefficients between the objects and the incline are same and not sufficient to allow pure rolling. Least time will be taken in reaching the bottom by |
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Answer» the SOLID SPHERE |
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| 37. |
The momentum of a body is doubled. By what percentage does its kinetic energy increase ? |
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Answer» <P> Solution :`K=(p^(2))/(2M)rArr (K_(1))/(K_(2))=(p_(1)^(2))/(p_(2)^(2))`Let `K_(1)=K, p_(1)=p` then `p_(2)=2p, K_(2)=?` `(K)/(K_(2))=(p^(2))/((2p)^(2)), K_(2)=4K`. % increase in kinetic energy = `("Increase in kinetic energy")/("Initial kinetic energy")XX100` `(K_(2)-K_(1))/(K_(1))xx100=(4K-K)/(K)xx100=300%` |
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| 38. |
What is the order of age of earth ? |
| Answer» Solution :AGE of earth is of the order is `10^(17)s.` | |
| 39. |
A vertical cylinder of height 100 cm contains air at a constant temperature and its top is closed by a frictionless pistion at atmospheric pressure (76 cm of Hg) as shown figure ( a). If mercury is slowly poured on the pistion, due to its weight air is compressed. Find the maximum height of the mercury column which can be put on the pistion. |
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Answer» Solution :When mercury is poured on the top of the piston, due to increase in pressure, the volume of air wiU decrease according to Boyle.s law. If FINAL mercury column of height X is poured on the PISTION then GAS pressure in equilibrium can be given as `P_(f) = (76 + x) ` CM of Hg As at1nospheric pressure is equivalent to the pressure due to a mercury column of height 76cm. If A be the area of cross secion of cylinder then we have according to Boyle.s law `P_(1) V_(1) = P_(2) V_(2) ` or `(76) (100 A) = (76 + x) (100 - x) A ` or `7600 = 7600 + 24 x - x^(2) " or " x = 24 ` cm . |
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| 40. |
A body of mass 100gm is performing SHM with a period of 4.5sec and amplitude 7cm. Velocity of the body at equilibrium position in cm/sec is approximately |
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Answer» 10 |
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| 42. |
A horizontal platform with an object placed on it is executing SHM in the vertical direction. The amplitude of oscillation is 3.92xx 10^(-3)m. What must be the least period of these oscillations, so that the object is not detached from the platform? |
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Answer» `0.1256 s` For `R= 0, mg= ma implies mg= m A omega^(2)` `implies omega = sqrt(g"/"A) = sqrt((9.8)/(3.92xx 10^(-3)))= 50` `implies T= (2pi)/(omega)= (2pi)/(50)= 0.1256s`. 
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| 43. |
Two particles of masses 4kg and 8kg are kept at x=-2m and x=4m respectively. Then the net gravitational force acting on a third particle of mass 1kg kept at the origin is |
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Answer» G |
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| 44. |
A body starts from rest and moves with uniform acceleration of 5ms^(-2) for 8 seconds. From that time the acceleration ceases. Find the distance covered in 12s starting from rest. |
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Answer» Solution :The VELOCITY after 8 seconds `v=0+5xx8=40m//s` Distance covered in 8 seconds `s_(0)=0+(1)/(2)xx5xx64=160n` After 8S the body moves with UNIFORM velocity and distance covered in 4s with uniform velocity is `S=vt=40xx4=160m` The distance covered in `12s=160+160=320m` |
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| 45. |
A stone is dropped from a hill of height 180m. Two seconds later another stone is dropped from a point P below the top of the hill. If the two stones reach the ground simultaneously, the height of P from the ground is |
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Answer» 100 m |
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| 46. |
Convert the unit of work done from MKS system to CGS system (i.e. joule ) |
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Answer» Solution :Dimensional formula of energy is `M^(1) L^(2) T^(2)`. Let `IJ = n "ergs" RARR n = J// "ergs"` `THEREFORE n_(2) = n_(1) [(M_1)/(M_2) ]^(1) [(L_1)/(L_2) ]^(2) [(T_1)/(T_2)]^(-2)` `rArr n = [ (1kg)/(1g) ] [(1 m)/( 1cm)]^(-2) [ (1s)/( 1s) ]^(-2) ((n_2)/(n_1) = n)` `rArr n= [ (1000g)/( 1g) ] [ (100 cm)/( 1cm ) ]^(2) [(1s)/( 1s) ]^(-2) ` `rArr n= [1000] xx [ 100]^(2) xx [1]^(2) = 10^(7)` `therefore 1 "joule" = 10^(7) "ergs"` |
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| 47. |
Consider three solid spheres, sphere (i) has radius r and mass m, sphere (ii) has radius r and mass 3 m, sphere (iii) has radius 3r and mass m, All can be placed at the same point on the same inclined plane, where they will roll without slipping to the bottom, If allowed to roll down the incline, then at the bottom of the incline |
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Answer» sphere (i) will have the largest speed For a sphere, `(I)/(mv^(2))=(2)/(5)` Hence, `a=(5)/(7)g sin theta=` constant, Hence, speed of all sphere is same at the BOTTOM. Sphere (ii) has the largest MASS. Hence, it will have the maximum kinetic energy. |
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| 48. |
On the arrangement shown in figure Pulleys are small light and springs are ideal K_1 , K_2 , K_3 and K_4 are force constant of the springs calculate period of small vertical oscillations of block of mass 'm' |
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Answer» Let T be the extra tension Then T = ma If `x_1 , x_2 , x_3 and x_4` are the extensions of springs. `2T = K_1 x_1 = K_2x_2 = K_3 x_3= K_4 x_4` `x=2(x_1 + x_2 + x_3+x_4)` `x= 4T (1/(K_1) + 1/(K_2) + 1/(K_3) + 1/(K_4))` from T = ma `4m(1/(K_1) + 1/(K_2) + 1/(K_3) + 1/(K_4)) a= x , omega = SQRT(a/x) and T = (2pi)/(omega)` |
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| 49. |
The vector product of two non-zero vectors will be minimum when theta is equal to |
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Answer» (a) `0^(@)` |
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