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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. |
Water flows through a horizontal tube of diameter 8cm and length 1km at the rate of 40 lt/s. Find the pressure difference required if n = 0.01 poise for water |
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Answer» `1.98xx10^(5) dyne-cm^(-2)` |
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| 2. |
Define atmospheric pressure what is its value ? |
| Answer» Solution :The PRESSURE EXERTED by the atmosphere at sea level is called atmospheric pressure. The atmospheric pressure at sea level is `1.013 XX 10^5 NM^(-2)` (or) Pa. | |
| 3. |
Why are mountain roads generally made winding upwards rather than going straight up ? |
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Answer» Solution :While going up a mountain the opposing force of friction ` F = mu R = mu mg COS 0 ` where `0` is angle of slope with HORIZONTAL To avoid skidding , `F` should be large `:. cos theta` should be large and hence `theta` must be small That is why mountain ROADS are generally made winding upwards. The road straight up would have LARGER slope . |
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| 4. |
Three measurements of the time for 20 oscillations of a pendulam give t_1=39.6s,t_2=39.9and t_3=39.5s. What is the precision in the measurements? What is the accuracy of the measurements? |
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Answer» SOLUTION :Precision is the least measurable value with that instrument in our case precision is `pm 0.1 sec`. Calculate of accuracy: Average value of measurements `=(39.6+39.9+39.5)/3=119/3=39.67` Error in each measurement = `Deltaa_1=39.67-39.6=0.07`, `Deltaa_2=39.9-39.67=0.23,` `Deltaa_3=39.67-39.5 =0.17` `Deltaa_(mean)=(0.07+0.23+0.17)/3=0.156` Precision `pm 1 sec`. In these measurements only two siginificant figures are believable. 3rd One is uncertain. Adjustment of `Deltaa_(mean)` upto given SIGNIFICANT FIGURE =0.156 adjusted to 0.2. So our value is accurate upto `pm0.2` So our result is 39.67 `pm` 0.2, when significant figures taken into ACCOUNT it is 39.7 `pm0.2 sec. |
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| 5. |
A small object of uniform density rolls up a curved surface with an initial velocity 'v'. It reaches upto a maximum height of (3v^(2))/(4g) with respect to the initial position . The object is |
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Answer» hollowsphere So by the LAW of conservation of mechanical energy , we have `(1)/(2) mv^(2) + (1)/(2) I omega^(2) = Mgh`  `(1)/(2) Mv^(2) + (1)/(2) I ((V)/(R))^(2) = Mg ((3v^2)/(4g)) (because omega = (v)/(R))` `(1)/(2) I (v^(2))/(R^(2)) = (3)/(4) Mv^(2) - (1)/(2) Mv^(2) , or I = (1)/(2) MR^(2)` Hence the object is disc. |
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| 6. |
A block of mass m is attached to the light spring of force constant k and released when it is in its natural length. Find amplitude of subsequent oscillations. |
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Answer» Solution :The maximum displacement of the spring in SUBSEQUENT motion will be `(2mg)/(k)` The equilibrium position of the SYSTEM will OCCUR at the extension of `(MG)/k` `therefore ` AMPLITUDE `=(2mg)/(k) - (mg)/(k) = (mg)/(k)` |
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| 7. |
If vec(P) = (k,2,3) and vec(Q) = (0,3,k) and vec(P) bot vec(Q) , then find the value of k . |
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Answer» <P> SOLUTION :`vec(P) bot vec(Q) rArr vec(P) . vec(Q) = 0 `` :. (k,2,3).(0,3,k) = 0 ` `:. 0+6 +3K = 0 ` ` :. 3k = -6` ` :. k = -2 ` |
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| 8. |
A student says that he had applied a force F=- r//x on a particle and the particle moved in simple harmonic motion. He refuses to tell whether r is a constant or not. Assume that he was worked only with positive x and no other force acted on the particle |
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Answer» As x increases k increases |
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| 9. |
Statement I: The viscosity of liquid increases with rise in temperature. Statement II: Viscosity ofaliquid is the property of the liquid by virtue of which it oppose the relative motion amongst its different layers. |
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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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| 10. |
A stationary object is divided into two fragment of mass 2M and 3M. Total energy of both fragment is E. Find energy of fragment with mass 2M. |
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Answer» `(2E)/(5)` `(M_(1)+M_(2) ) V=2Mv_(1)- 3M v_(2)` Where`M_(1) =2M,M_(2) = 3M V=0` `2Mv_(1) =3M_(2)` `v_(2) =(2)/(3) v_(1)` `v_(2)^(2)= (4)/(9) v_(1)^(2)` Totalenergy `E= (1)/(2) .(2M)v_(1)^(2) + (1)/(2) (3M)v_(2)^(2)` `=mv_(1)^(2) + (3)/(2)Mv_(2)^(2)` `=mv_(1)^(2) + (2)/(3) Mv_(1)^(2)` `E= (5)/(3) Mv_(1)^(2)` Kineticenergy of 2Mpart `E_(1) = (1)/(2) (2M) v_(1)^ (2)` Takingratio of (2)and (3) `(E_(1))/(E ) = ((1)/(2) (2M)v_(1)^(2))/((5)/(3)Mv_(1)^(2))` `E_(1) = (3)/(5)E` |
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| 11. |
When air is pumped into a cycle tyre the volume and pressure of the air in the tyre both are increased. What about Boyle's law in this case? |
| Answer» Solution :when there is pumped into a cycle tyre, more and more number of AIR MOLECULES are pumped in to a tire of the cycle . So number of molecules will be increased . Boyle's law is APPLICABLE only when number of molecules remains constant .In this case it does not OBEY Boyle's law . | |
| 12. |
Electron microscope works on the principle of |
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Answer» WAVE NATURE of light |
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| 13. |
How long does it take a brick to reach the ground if dropped from a height of 65 m? What will be its velocity just before it reaches the ground? |
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Answer» `2s,10m//s` |
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| 14. |
A pilot does not fall down when his aeroplane loops the loop. Why ? |
| Answer» SOLUTION :At the HIGHEST point the WEIGHT of the pilot is balanced by the CENTRIPETAL FORCE. | |
| 15. |
A column of Hg of 10cm length is contained in the middle of a narrow horizontal 1 m long tube which is closed at both ends. Both the halves of the tube contain air at a pressure 76 cm of Hg. By what distance will the column of Hg be displaced, if the tube is held vertical? (Assume temperature to be constant) |
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Answer» SOLUTION :If initially the length of air column on each side is L, ACCORDING to the given problem, `2L+10=100,i.e.,L=45cm""....(1)` Now if the tube is held vertical, the Hg column will be displaced downward by y such that `P_(B)+10=P_(A)""....(2)` Now applying Boyle.s law to air enclosed in side A, `P_(0)LA=P_(A)(L-y)A,i.e.,P_(A)=(LP_(0))/((L-y))""...(3)` while for air enclosed in side B, `P_(0)LA=P_(B)(L+y)A` i.e., `P_(B)=(LP_(0))/((L+y))" "...(4)` Substituting the value of `P_(A)andP_(B)` from equation (3) and (4) in (2), with L = 45 and `P_(0)` = 76 cm of Hg, we GET `(45xx76)/((45-y))-(45xx76)/((45+y))=10,ory^(2)+684y-(45)^(2)=0` or `y=[-684pmsqrt((684)^(2)+4(45)^(2))]//2` or `y=-342+345=3cm` 
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| 16. |
A coin is kept at distance of 10 cm from the centre of a circular turn table. If mu = 0.8, the frequency of rotation at which the coin just begins to slip is |
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Answer» 62.8 RPM |
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| 17. |
The diameter of a sphere is 1.23 m. Find its volume with due regard to the significant figures. |
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Answer» |
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| 18. |
A vector vec(A) when added to the vector vec(B)= 3hat(i) + 4hat(j) yields a resultant vector that is an in the positive y-direction and has magnitude equal to that of vec(B) find the magnitude of vec(A) |
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Answer» `SQRT10` |
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| 19. |
(A) : When external force acting on a body moving on a rough horizontal surface is doubled, the acceleration of the body is doubled.(R ) : The acceleration of a body moving on a rough horizontal surface is proportional to the resultant force acting on it. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct explanation of (A) |
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| 20. |
Which of the following relationships between the acceleration (a) and the displacement (x) of a particle involves simple harmonic motion? |
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Answer» a=0.7x |
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| 21. |
Check the correctness of the formula S = ut + 1/3 "at"^2 where S is the distance , u is velocity , a is acceleration and t is time. |
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Answer» Solution :DIMENSIONALLY, `LHS = [L]` `RHS = [LT^(-1) ] [T] + [LT^(-2) ] [T]^(2)` `= [ L] + [ L] = [L]` Since, the dimension of each TERM on both sides are same, the GIVEN equation is dimensionally correct . |
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| 22. |
For a particle, revolving in a circle with speed, the acceleration of the particle is (a) along the tangent |
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Answer» ALONG the tangent |
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| 23. |
At what temperature is rms speed of air molecules double of that at NTP? |
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Answer» `719^0` C |
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| 24. |
The escape velocity of a body from the surface of the earthis nearly |
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Answer» `7.98kms^(-1)` |
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| 25. |
If no external force acts on the system(a) Acleleration of centre of mass is zero(b) Velocity of centre of mass is zero(c ) Centre of mass may be at rest (or) moving with constant velocity(d) Velocity and momentum of centre of mass remains constant |
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Answer» a, B and C are TRUE |
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| 26. |
What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale ? |
| Answer» SOLUTION :`491.69`. | |
| 27. |
The value of temperature coefficient of resistance thermometer is |
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Answer» LOW and NEGATIVE |
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| 28. |
A police car moving at 22m/s chases a motorcyclist. The police man sounds his horn at 176 Hz, while both of them move towards a stationary siren of frequency 165 Hz. Calculate the speed of the motorcycle. If it is given that the motorcyclist does not observe any beats: |
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Answer» `33m//s` `176((330-v_(0))/(330-22))=165((330+v_(0))/(330))` `v_(0)=22m//sec` |
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| 29. |
If P = 106 kT, then the number of molecules per unit volume of the gas is [the letters have usual meanings.] |
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Answer» <P>`100 K` or `n/V = P/(kT) = (10^(6)kt)/(kT) = 10^(6)`. |
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| 30. |
The bulkmodulus for an incompresssible liquid is |
| Answer» Solution :`K= ("STRESS")/("STRAIN")=("stress")/(o)= OO` (Infinity ) | |
| 31. |
The velocity v of waves produced in water depends on their wavelength lambda, the density of water d, and acceleration due to gravity g. These quantities are related as (where k is a dimensionless constant) |
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Answer» `V^(2)= klambda^(-1)G^(-1)d^(-1)` |
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| 32. |
A man can swim in still water at a speed of 6 kmph and he has to cross the river and reach just opposite is point on the other bank. If the river is flowing at a speed of 3 kmph, and the width of the river is 2km, the time taken to cross the river is (in hours) |
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Answer» `(2)/(27)` |
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| 33. |
A man can swim in still water at a speed of 6 kmph and he has to cross the river and reach just opposite is point on the other bank. If the river is flowing at a speed of 3 kmph, he has to swim in the direction |
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Answer» `30^(@)` with the RIVER flow |
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| 34. |
An object reaches a maximum vertical height of 23.0 m when thrown vertically upward on the earth. How high would it travel on the moon where the acceleration due to gravity is about one sixth that on the earth ? Assume that initial velocity is the same. |
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Answer» 138m |
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| 35. |
The unit mass having r = 8 hat(i) - 4 hat(j) and v = 8 hat(i) + 4 hat(j) in its angular momentum is |
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Answer» 64 units in `-hat(k)` direction `L = m(r xx v)` For UNIT mass, `m=1` `:. |L|=[(8 hat(i)-4hat(j))xx(8hat(i)+4hat(j))]` ` rArrL=|{:(hat(i),hat(j),hat(k)),(8,-4,0),(8,4,0):}|` `RARR L=hat(i)(0-0)-hat(j)(0-0)+hat(k)(32+32)` `rArr L = 64 hat(k)` units. |
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| 36. |
Which of the following is true regarding projectile motion |
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Answer» HORIZONTAL VELOCITY is CONSTANT |
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| 37. |
A ring of radius 0.5 m and mass 10 kg is rotating about its diameter with angular velocity of 20 rad s^(-1). Its kinetic energy is |
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Answer» 10 J `E_(k)=(1)/(2)I omega^(2)rArr E_(k)=(1)/(4)mr^(2)omega^(2)=(1)/(4)xx10xx(0.5)^(2)xx(20)^(2)=250 J`. |
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| 38. |
Give the limitations and uses of turbulence. |
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Answer» Solution :Turbulence DISSIPATES kinetic ENERGY usually in the form of heat. Racing cars and planes are engineered to precision in order to minimum turbulence. On the other hand turbulence is SOMETIMES desirable . Turculence promotes mixing and increases the rates of transfer of mass, momentum and energy. The blades of a kitchen mixer induce TURBULENT flow and provide THICK milk shakes as well as beat eggs into a uniform texture. |
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| 39. |
A solid sphere of mass m andradius R rolls down from the top of a table. With how much angular speed will it touch the ground? |
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Answer» Solution :At position II the resultant of mg cos `theta ` and n supplies the necessary centripetal FORCE to roll on i.e., `momega^(2)R = mg cos theta -n` [ `omega` = angular SPEED of the sphere ] When n = 0, the sphere will not be in contact more with the table. Then `momega^(2)R = mg cos theta " ""or", cos theta = (omega^(2)R)/(g)` The centre of mass of thesphere is LOWERED down by R (1-cos `theta).` Let `I` be the moment of inertia of the sphere about the tan gent at the contact point between the table and the sphere. From theconservation law of mechanical energy, `mg .R (1-costheta) = (1)/(2)xxIomega^(2)= (7)/(10) mR^(2)omega^(2) [because I =(7)/(5) mR^(2)]` `:." " mgR(1-(omega^(2)R)/(g))=(7)/(10)mR^(2)omega^(2)` or, `mgR-momega^(2)R^(2)=(7)/(10)mR^(2)omega^(2)` `:. " " omega=sqrt((10g)/(17R))` Considering a FREE fall of thesphere as no torque acts on it this angular speed remains unaltered. It means that the sphere touches the ground with an angular speed `sqrt((10g)/(17R))`. 
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| 40. |
A bullet is fired into a huge wooden block . The bullet whilemoving insidethe blockloses half the velocity when it travles 6 cm inside the block . hOe far , then would it go inside the block ? And becomes rest . |
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Answer» Solution :Suppose the intial kinetic energy of bullet` = 1/2 mv_(0)^(2)` when the velocity becomes halved the KINETICENERGY= `1/2 ((mv_(0)^(2))/4)` According to WORK energy theorem , `W = DeltaK = K - K_(0)` ` :. F.d =1/2 ((mv_(0)^(2))/4)-1/2mv_(0)^(2)` ` :. F(6) =1/2 mv_(0)^(2) (1/4-1)` `F(6) =-3/8 mv_(0)^(2)` ` :. F = - 1/16 mv_(0)^(2)` If before coming to rest travels an additional DISTANCE d . Its kinetic energy becomes zero . ` :. Deltak =W` ` :. 0 -(mv_(0)^(2))/8 =[Fd] =-1/16 mv_(0)^(2) d` ` :. d = 2 cm ` |
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| 41. |
Two black bodies A and B emit radiations with peak intensities at wavelengths 4000 A and 8000 A respectively. Compare the total energy emitted per unit area per second by the two bodies. |
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Answer» |
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| 42. |
(A) : In an elastic collision of two bodies, the momentum and energy of each body is conserved.(R ) : If two bodies stick to each other, after colliding, the collision is said to be perfectly elastic. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A) |
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| 43. |
A particle of mass m, moving with velocity 'u' makes head on collision with identical stationary particle, which is elastic. The force of interaction increases to maximum and then becomes zero in time 'T' as shown in graph. What is the maximum value F_(0) of interaction force ? |
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Answer» SOLUTION :Force reaches maximum value in `(T)/(2)` time. The impulse of maximum force is, `= F_(0)XX(T)/(2)` SINCE, impulse = Change in MOMENTUM, mu = `F_(0).(T)/(2)`i.e., `F_(0)=(2m U)/(T)` |
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| 44. |
Is it possible for a cyclist to lean through an angle of 45^(@) with the vertical when he takes a turn? |
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Answer» |
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| 45. |
A pendulum that beats seconds on the surface of the earth were taken to a depth of (1//3)th the radius of the earth. What will be its time period of oscillation ? |
| Answer» Answer :A | |
| 46. |
Calculate the height at which the value of acceleration due to gravity becomes 50% of that at the surface of the earth. (Radius of the earth = 6400 km) |
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Answer» Solution :G at height , `g_h=g/(1+h/R)^2` In this problem , `g_h=50/100`g=0.5 g 0.5g= `g/(1+h/R)^2 rArr (1+h/R)^2 =g /(0.5g)=2` `1+h/R= SQRT2, h/R =sqrt2-1 rArr` the height , h=0.414 R =0.414 x 6400 = 2650 KM . |
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| 47. |
A mass M is moving with a constant velocity parallel to x axis. Its angular momentum w.r.t. origin |
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Answer» is zero |
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| 48. |
A particle is projected at an angle alpha with the horizontal.Show that if alpha is greater than cos^(-1)1/3=70.5^@, the distance of the projectile from the point of projection increases to a maximum above the horizontal plane. |
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Answer» |
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| 49. |
Three thin uniform rods each of mass M and length L are placed along the three axes of a cartesian coordinate system. The moment of inertia of the system aboutz-axis is |
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Answer» `(ML^(2))/(3)` Therefore , moment of inertia of the system about the z-axis = `(1)/(3) ML^(2) + (1)/(3) ML^(2) + 0 = (2)/(3) ML^(2)` 
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| 50. |
While reading the temperature of a body, only tube bulb is immersed in it. The percentage error committed while calculating the length of the mercury column corresponding to a temperature rise Delta T^(@) C. Ignoring the expansion of glass bulb will be: (volume expansivity of mercury is 182 xx 10^(-6)//°C and linear expansivity of glass is 8.3 xx10^(-6)//°C and linear expansivity of glass is 8.3 xx 10^(-6)//°C |
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Answer» `5%` |
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