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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. |
The diameter of one end of a tube is 2 cm and that of another end is 3 cm . Velocity and pressure of water at narrow end are 2ms^(-1)and1.5xx10^(5)Nm^(9-2) respectively . If the height differnce between narrow and broad ends is 2.5 m , find the velocity and pressure of water at the broad end .(Density of water is 1xx10^(3)kgm^(-3)). The narrow end is higher . |
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Answer» SOLUTION :The NARROW end of the flow tube : `d_(1)=2cm` `thereforer_(1)=1cm=1XX10^(-2)m` `v_(1)=2MS^(-1)` `P_(1)=1.5xx10^(5)Nm^(-2)` The broad end of the flow tube : `d_(2)=3cm` `thereforer_(2)=1.5cm=1.5xx10^(-2)m` `v_(2)=?,P_(2)=?` `A_(1)v_(1)=A_(2)v_(2)` `thereforev_(2)=(A_(1))/(A_(2))v_(1)=(pir_(1)^(2))/(pir_(2)^(2))*v_(1)` `thereforev_(2)=(r_(1)^(2))/(r_(2)^(2))*v_(1)=((1xx10^(-2))^(2))/((1.5xx10^(-2))^(2))xx2` `thereforev_(2)=089m//s` According to Bernoulli.s equation, `P_(1)+(1)/(2)rhov_(1)^(2)+rhogy_(1)=P_(2)+(1)/(2)rhov_(1)^(2)+rhogy_(2)` `thereforeP_(2)=P_(1)+(1)/(2)rho(v_(1)^(2)-v_(2)^(2))+rhog(y_(1)-y_(2))` `=(1.5xx10^(5))+(1)/(2)xx1xx10^(3)[(2)^(2)-(0.89)^(2)]+1xx10^(3)xx9.8xx2.5` `thereforeP_(2)=1.76xx10^(5)N//m^(2)` |
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| 2. |
In a standing wave experiment , a 1.2 - kg horizontal rope is fixed in place at its two ends ( x = 0 and x = 2.0 m) and made to oscillate up and down in the fundamental mode , at frequency of 5.0 Hz. At t = 0 , the point at x = 1.0 m has zero displacement and is moving upward in the positive direction of y - axis with a transverse velocity 3.14 m//s. What is the correct expression of the standing wave equation ? |
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Answer» `(0.1) sin ( pi//2) x sin (10 pi) t` `N = 5 Hz` `lambda = 2 l = 4 m` `V = n lambda = 5 XX 4 = 20 m//s` Using `v = sqrt((T)/(mu))` `T = 20^(2) xx 0.6 = 240 N` `((delta y)/( delta t))_(max) = 3.14 m//s` `( 2 A)omega = 3.14` Amplitude `2A = (3.14)/(2 xx (3.14) xx 5) = 0.1 m` EQUATION of standing wave is `y = (0.1) sin (pi)/(2) s sin (10 pi) t` |
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| 3. |
Aluminium wire of length 2.2 m and a Iron wire of length 1.6 m, both of diameter 3.0 mm, are connected end to end. When stretched by a load, the net elongation is found to be 0.70 mm. Obtain the load applied.Young's modulus of aluminium Y_(Al) =70 xx 10 ^(9) Nm ^(-2) Young's modulus of iron Y_(F)=120 xx 10 ^(9) Nm ^(-2) |
| Answer» SOLUTION :Appliec LOAD `F = 111.3N` | |
| 4. |
Moment of inertia does not depend on |
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Answer» mass distribution of BODY |
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| 5. |
A U-tube is supoorted with its limbs vertical and is partly filled with water. If internal diameter of the limbs are 1xx10^(-2)m and 1xx10^(-4) m respectively what will be the difference in heights of water columns in the two limbs (Surface tension of water is 0.07Nm^(-1)) |
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Answer» Solution :Surface TENSION `S=0.07Nm^(-1)` Densith `rho=1000 kgm^(-3),G=9,8ms^(-2)` Angle of contact `theta=0^(@)` RADIUS `r_(1)=0.5xx10^(-2)m, ` Radius `r_(2)=0.5xx10^(-4)m` Let `h_(1)` be the height of water in the limb of radius `r_(1)` Then `h_(1)=(2Scos theta)/(r_(1)rho g)=(2xx0.07xxcos 0^(@))/(0.5xx10^(-2)xx1000xx9.8)m` `=2.86xx10^(-3)m` similarly `h_(2)=2.86xx10^(-1)m` Difference in heights `=h_(2)-h_(1)=2.86xx10^(-1)m-2.86xx10^(-3)m` `=(0.286-0.00286)m=0.283m` |
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| 6. |
In a two dimensional motion,instantaneous speed v_(0) is a positive constant.Then which of the following are necessarily true? |
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Answer» The average velocity is not zero at any TIME. |
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| 7. |
A train standing at the outer signal of a railway station blows a whistle of frequency 400Hz in still air. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 ms^(-1). (b) recedes from the platform with a speed of 10ms^(-1) ? (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340ms^(-1). |
| Answer» SOLUTION :(i) (a) 412Hz, (B) 389Hz, (ii) `340 MS^(-1)` in each case | |
| 8. |
A bottle whose volume is 0.45 xx 10^(-3) m^(3) is sunk mouth downwards below the surface of water in a tank. How far must it be sunk so that 0.75 xx 10^(-4) m^(3)of water may run up into bottle ? The atmospheric pressure =10^(5) Nm^(-2) and density of water =1000 kg m^(-3) |
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Answer» |
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| 9. |
The relation between the slope of isothermal curve and slope of adiabatic curve is |
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Answer» SLOPE of ADIABATIC curve `= gamma` times slope of ISOTHERMAL curve |
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| 10. |
As a spherical wave propagates, ……….. . |
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Answer» the wave INTENSITY REMAINS CONSTANT |
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| 11. |
A satellite is revolving in a circular orbit of radius R. Ratio of KE and its PE is (only magnitude ) |
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Answer» `1:4` |
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| 12. |
Letthe speed of the planet attheperihelion pin be vPand the Sun-planet distance SP be r_(P). Relate {r_(P), v_(P)} to the corresponding quantities at the aphelion {r_(A), v_(A)}. Will the planet take equal times to traverse BAC and CPB ? |
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Answer» Solution : The MAGNITUDE of the angular MOMENTUM at P is `L_(p)` =`m_(p) r_(p) v_(p)`, since inspection tells us that rp and vp are mutually perpendicular. Similarly,` L_(A) = m_(p) r_(A) v_(A)`. From angular momentum conservation `m_(p) r_(p) v_(p) = m_(p) r_(A) v_(A)` or `(v_(p))/(v_(A)) = (r_(A))/(r_(p))` since `r_(A) gt r_(p), v_(p) gt v_(A).` the area SBAC bounded by the ellipse and the RADIUS vectors SB and SC is larger than SBPC in From Kepler’s second law, equal AREAS are swept in equal times. Hence the planet will take a longer time to traverse BAC than CPB. |
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| 13. |
Which of the following following expressions correspons to simple harmonicmotion along a straightline, where x is the displacement and a,b,c are positive constants? |
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Answer» `BX^(2)` |
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| 14. |
Derive an expression for the elastic energy stored per unit volume of a wire. |
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Answer» SOLUTION :When a body is stretched, work is done against the RESTORING force (internal force). This work done is stored in the body in the form of elastic energy. Consider a wire whose un-stretch length is L and area of cross section is A. Let a force produce an extension l and further assume that the elastic limit of the wire has not been exceeded and there is no loss in energy.Then , the work done by the force F is equal to the energy gained by the wire. The work done in stretching the wire by `dl, dW = F dl` The total work done in stretching the wire from 0 to l is `W = int_(0)^(l) F dl "" ....(1)` From Young.s modulus of elasticity, `Y = F/A xx L/l implies F = (YAl)/L "" ....(2)` SUBSTITUTING equation (2) in equation (1), we GET `W = int_(0)^(l) (YA l)/(L) dl` Since, l is the dummy variable in the INTERGRATION, we can change l to l. (not in limits), therefore `W = int_(0)^(l) (YAl^.)/(L) dl. = (YA)/L ((l.^2)/(2))_0^l = (YA)/L (l^2)/2 = 1/2 ((YAl)/(L)) l=1/2 Fl` `W = 1/2 Fl` = Elastic potential energy Energy per unit volume is called energy density `u = ("Elastic potential energy")/("Volume") = (1/2 Fl)/(AL)` `1/2 F/A l/L = 1/2 ("Stress" xx "Strain")`. |
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| 15. |
A lead bullet of mass 21 g travelling at a speed of 100m//s comes to rest in a wooden block . If no heat is taken away by the wood , the rise in temperature of the bullet in the wood nearly is (specified heat of lead =30 cal//kg^(@)C ) |
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Answer» Solution :Kinetic energy of the BULLET=heat gained by the bullet `1/2 mv^(2) = msDeltat` `Delta t =(v^(2))/(2S) = ((100^(2)))/(2xx0.03xx4.2xx1000) = 39.68^(@)C` |
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| 16. |
The oxygen molecule has a mass of 5.30 × 10^(-26) kg and a moment of inertia of 1.94×10^(-46) kg m^(2) about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is 500 m/s and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule. |
| Answer» SOLUTION :`6.75xx10^(12)" RAD s"^(-1)` | |
| 17. |
A ball of mass m is dropped from a heighth on a platform fixed at the top of a vertical spring, as shown in figure. The platform is depressed by a distance x. Then the spring constant is |
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Answer» `(MG)/((h + x))` Gain in elastic potential energy of the spring = `1/2 kx^2` According to the LAW of conservation of mechanical energy `mg(h + x) = 1/2 kx^2 " or " k = (2mg (h + x))/(x^2)` . |
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| 18. |
A body projected vertically upwards from the surface of the earth with a velocity equal to 1//3^(rd) of the escape velocity. The height reached by it if the radius of the earth is 6400 km is |
| Answer» ANSWER :D | |
| 19. |
A particle executes SHMof period 22 second and amplitude 10 cm. Find the distance it travels in 3.5 second starting from the point of zero displacement. |
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Answer» Solution :Amplitude of the particle = 10 cm=0.01 m, Time period `t=22s` and `phi_(0)=0` The angular velocity `OMEGA=(2pi)/T=(2xx22)/(7xx22)=2/7` rad/s The equation for displacement of the particle is `x=Asin (omega t+phi)=0.10sin(2/7t)` when `t=3.5s=7/2`s the distance it travels BECOMES equal to the displacement because the GIVES time is less than `T//4`. `:. x=0.10sin(2/7xx7/2)=0.10sin(1)` `=0.10sin (57^(@)17.) ( :. 1"radian"=57^(@)17.)` `=0.10xx0.8414=0.8414m` |
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| 20. |
A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J = –alphatheta , where J is the restoring couple and q the angle of twist). |
| Answer» Solution :Hint: The time period of a torsional PENDULUM is given by `T=2pisqrt(l/(ALPHA))`. where I is the moment of inertia about the axis of rotation. In our case `I=1/2MR^(2)`, where M is the mass of the DISK and R its radius. Substituting the given values, `alpha=2.0Nmrad^(-1)` | |
| 21. |
A sting breaks under a load of 4.8 kg A mass of 0.5 kg is attached to one end tothe string 2malong and is ratated in a horizontal circle caculate the greatest number of revolutions that the mass can make without breaking the string |
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Answer» Solution :Given`m=0.5 kg` `r=2 m` `g=9.8ms^(-2)` Themaximumtensionthat the stringcanwithstand `F=4.8 kg wt=4.8xx 9.8 N` LET themaximum NUMBER of revolutions persecond=v Now F = `mr epsilon^(2)= mr (2 pi v)^(2) = 4PI^(2) mrv^(2)` or `v^(2)=(F )/(4pi^(2)mr)` `=(4.8 xx 9.8)/( 4xx 9.8 xx 0.5 xx 2) =1.215` or `v = SQRT(1.215 )= 1.102 rps` `=1.102 xx 60 =66 .13 rpm` |
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| 22. |
A cyclist is riding on a cycle with speed v and while negotiating a circular path of radius r, he leans at an angle theta with the ground. Then, what will be the value of tantheta? |
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Answer» |
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| 23. |
A rectangular box lies on a rough inclined surface. The coefficient of friction between the surface and the box is mu. Let the mass of the box be m. (a) At what angle of inclination 0 of the plane to the horizontal will the box just start to slide down the plane ? (b) What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to a gt theta (c) What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed ? (d) What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration a ? |
Answer» SOLUTION :Considerdiagram SHOWNIN figure  (a ) For bxjuststartsto sliding downslope `F = mg sin theta` `N=m g COS theta` `(f )/(N )= tan theta` `MU = tan theta` `theta= tan ^(-1)(mu)` ( b)Whenangle ofinclinationincreasedto `alpha gttheta` Resultantforcebe `F_(1)` `F_(1)= mgsin alpha- f` (c ) Tokeepboxstationaryor movingwithuniformspeedupwardforceir requiredHerefrictionforcewouldbed indownwarddirection (d ) Whenbox isto bemovedwithaccelerationa upwardalong theplanenet forcebe `F_(3)` frictionwouldbe indownwarddirection `F_(3)= mg (sinalpha+ mu cos alpha )+ ma` |
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| 24. |
Springs of spring constants K, 2K, 4K, 8K ...... 2048 K are connected in series. A mass 'm' is attached to one end the system is allowed to oscillate. Its time period is approximately |
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Answer» `2pi SQRT((m)/(2K))` |
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| 25. |
The momentumof a body increased doubled .What is the percentage of increment in its kinetic energy ? |
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Answer» Solution :`K = 1/mv^(2) = (p^(2))/(2M) "" …(1)` `K = (p^(2))/(2m) ""….(2)` `p. =2p ` ` :. K = (4p^(2))/(2m)` ` :. K . = 4K ` Increase in KINETIC energy= `(DeltaK)/K xx100` `= (3K)/K XX 100` = 300 % |
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| 26. |
In the figure (i), (ii) and (iii) shown the objects, A,B and C are of same mass. String, spring and pulley are massless. C strikes B with velocity u in each case and stricks to it. The ratio of velocity of B in case (i) to (ii) to (iii) is- |
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Answer» |
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| 27. |
Among the following which are the examples of SHM?Oscillation of simple pendulum |
| Answer» SOLUTION :OSCILLATION of SIMPLE PENDULUM | |
| 28. |
When a man running on a levelground increases his speed by 4ms^(-2), if his kinetic energy increases by 125% his initial velocity is |
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Answer» `4 ms^(-1)` |
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| 29. |
A block of mass .m. is pulled by a force .F. on a level surface through a distance .S., then |
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Answer» The WORK done on the body against GRAVITATIONAL FORCE is zero |
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| 30. |
Explain various types of friction suggest a few methodsto reduce friction |
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Answer» Solution :There are two types of Friction: (i) Static Friction: Static friction is the force which opposes the initiation of motion of an object on the surface. The magnitude of static frictional force `f_s`lies between `0 LT= f_s lt= mu_s N` where ,`mu_s` - coefficient of static friction N - Normal force Kinetic friction: The frictional force exerted by the surface when an object slides as kinetic friction. Also called as SLIDING friction or DYNAMIC friction, `f_k = mu_ k N` where `mu_k`- the coefficient of kinetic friction N - Normal force exerted by the surface on the object Methods to reduce friction: Friction can be reduced (i) By using lubricants (ii) By using Ball bearings (iii) By polishing (IV) By streamlining |
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| 31. |
A body of mass m_1, projected vertically upwards with an initial velocity 'u' reaches a maximum height 'h'. Another body of mass m_2is projected along an inclined plane making an angle 30^@with the horizontal and with speed 'u'. The maximum distance travelled along the incline is |
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Answer» 2h |
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| 32. |
Two persons pull a rope by its two ends, exerting equal but opposite forces F. In another situation, one end of the rope is tied to a rigid support and a person pulls it with a force of 2F. In which case will the tension in the be greater ? |
Answer» SOLUTION :TENSION in the rope in the SECOND CASE will be greater i.e., T. = 2F, whereas the tension, in the FIRST case is T = F.
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| 33. |
An artificial satellite revolves around the earth, remaining close to the surface of the earth, Show that its time period is T = 2pi sqrt(R_e/g) |
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Answer» Solution :`IMPLIES` The necessary CENTRIPETAL force for satellite of mass m revolves around the earth is provided by GRAVITATIONAL force. `:.`Centripetal force = Gravitational force, Subsituting this in equation (1), `:. (mv^2)/R_e =mg ""...(1)` but `v= R_eomega` `:. v = R_e (2pi)/T"" [ :. omega =(2pi)/T]` `:.` Subsituting this in equation (1), `:. (R_e^2xx4pi^2)/(R_e.T^2)=g` `:. T^2= 4PI ^(2) R_e/g "" :. T = 2pi sqrt((R_e)/g)` |
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| 34. |
The length of a cyclinder is measured as 5 cm usinga verniercalipersof least count 0.1mm. The percentage error in the measured length is nerarly |
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Answer» `0.5%` |
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| 35. |
Statement 1: In process of photoelectric emission, all emitted electrons do not have same kinetic energy. Statement 2: If radiation falling on photosensitive surface of a metal consists of different wavelengths, then energy acquired by electrons absorbing photons of different wavelengths shall be different. |
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Answer» Statement-1 is true, Statement-2: is true, Statement-2 is a correct EXPLANATION for Statement-1. |
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| 36. |
Animals curl into a ball when it is very cold.Why? |
| Answer» Solution :Energy rediated is proprotional to the surface area of the body.They curl into a ball so as to DECREASE the surface area and thereby REDUCE the loss of heat by RADIATION. | |
| 37. |
The above question 26, the normal force acting, betweent : |
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Answer» wedge and incline PLANE is `Mg cos THETA` |
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| 38. |
What is the necessary condition for the formation of stationary waves ? |
| Answer» Solution :A stationary WAVE cant' be formed from two independent waves TRAVELLING in a MEDIUM in opposite directions. Acutually stationary wave in PRODUCED when a progressive wave and its reflected waves are SUPERPOSED. It can be produced in a finite medium which has its boundaries. | |
| 39. |
Statement I : The maximum speed at which a car can turn on level curve of radius 40 m, is 11 m/s , = 0.3 . Statement II : v = μRg = Rg v2 = 40 10 11 11 = 0.3. |
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Answer» If both STATEMENT- I and Statement- II are TRUE, and Statement - II is the CORRECT explanation of Statement– I. |
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| 40. |
The radius of gyration of a uniform rod of length L about an axis passing through its centre of mass is |
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Answer» `(L)/(SQRT(12))` |
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| 41. |
Two wheels of M.I. 3kg m^(2) and 5 kg m^(2) are rotating at a rate of 600 rpm and 800 rpm respectively in opposite directions. If are coupled so as to rotate with the same axis of rotation, the resultent speed of rotation (in rpm) |
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Answer» 275 |
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| 42. |
When a load of 20 kg is hung from a wire of cross sectional area 1mm^2, its elongated length is 600.5 cm. IF the load is withdrawn the wire shrinks by 0.5 cm in length. Calculate the young's modulus for the material of the wire. |
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| 43. |
The volume (V) of a moatomic gas varies with its temperature (T), as shown in the graph. The ratio of work done by the gas, to the heat absorbed by it, when it undergoes a change from state A to state B, is |
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Answer» `(2)/(5)` `:. dQ = nC_(P)dT`, where `C_(P)` is SPECIFIC heat at constant pressure. or `dQ = n((5)/(2)R)dT` Also, `DW = PdV = nRdT (because PV = NRT)` Required RATIO = `(dW)/(dQ) = ((nRdT)/(n(5)/(2)R)dT) = (2)/(5)` |
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| 44. |
A block of mass M is hanging over a smooth and light pulley through a light string. The other end of the strinig is pulled by a constant force F. The kinetic energy of the block increases by 20J n 1s. |
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Answer» the TENSION in the string is Mg |
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| 45. |
A diatomic ideal gas is used in a carnot engine as the working substance. If during the adiabatic expansion part of the cycle, the volume of the gas increases from V to 32 V, the efficiency of the engine is ……………… . |
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Answer» 0.25 `RARR""(T_(2))/(T_(1))=((V_(1))/(V_(2)))^(gamma-1)=((1)/(32))^((7)/(5)-1)` `(T_(2))/(T_(1))=((1)/(2^(5)))^((2)/(5))=(1)/(4)` `"EFFICIENCY"eta=1-(T_(2))/(T_(1))=1-(1)/(4)=0.75` |
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| 46. |
Four solids are shown in cross section. The sections have equal heights and equal maximum widths. They have the same mass. The one which has the largest rotational inertia about a perpendicular through the centre of mass is |
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Answer»
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| 47. |
Below are four equations expressing relationships between qualities. 'g' gravitational field strengths, r radius , rho a density, G the universal constant of gravitational, m a mass and T a time period. Which of the equations are true. (a) g T^(2) = 4pi r^(2)(b) (4)/(3) pi rho G r^(2) = g (c) g r^(2) = Gm(d) g = (4)/(3) pi rho Gr |
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Answer» only a & B are TRUE |
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| 48. |
Find ratio of radius of gyration of a disk and ring of same radii at their tangential axis in plane |
| Answer» SOLUTION :For DISC `MK^(2) = (5//4)MR^(2)`, For RING `MK^(2) = (3//2)MR^(2)`, RATIO = `sqrt(5)//sqrt(6)` | |
| 49. |
A body of mass 15kg is raised from certain depth. If the work done in raising it by 10m is 1620J, its velocity at this position is (g= 10 ms^(-2)) |
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Answer» `2 ms^(-1)` |
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| 50. |
Three particles of identical masses .m. are kept at the vertices of an equilateral triangle of each side length.a.. Find the gravitational force of attraction on any one of the particles. |
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Answer» SOLUTION : Let `F_A` and `F_B` be the forces due to the MASSES at A and B on the mass at C. Here `F_A=F_B` Hence the Resultant FORCE on the mass at C `F=2F_A cos theta//2 =2. (Gm^2)/a^2 cos 30^@` `=2xx(Gm^2)/a^2xxsqrt3/2 = sqrt3(Gm^2)/a^2` 
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