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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. |
All oscillatory motions are necessarily periodic motion in conservative fields but all perodic motion are not oscillatory. Simple pendulum is an example of oscillatory motion. |
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Answer» Both 'A' and 'R' are TURE and 'R' is the CORRECT explanation of 'A' |
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| 2. |
Two blocks having masses 8 kg and 16 kg are connected to the two ends of a light spring. The system is placed on a smooth horizontal floor. An inextensible string also connects B with ceiling as shown in figure at the initial moment. Initially the spring has its natural length.A constant horizontal force F is applied to the heavier block as shown. What is the maximum possible value of F so that lighter block doesn't loose contact with ground. (##NAR_NEET_PHY_XI_P2_C06_E11_012_Q01.png" width="80%"> |
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Answer» 10 N ` (##NAR_NEET_PHY_XI_P2_C06_E08_012_S01.png" width="80%"> `Kx - T COS THETA = 0` `Rightarrow T = Kx/(cos theta), T sin theta + N -mg = 0` When N = 0 then `T sin theta = mg Rightarrow Kx/(cos theta)sin theta = mg ` `x = mg /(K TAN theta) = (80)/(K XX(4/3)) = 60/K` If spring has to just extend till this value then from WORK energy theorem we get `Fx = 1/2Kx^(2) Rightarrow F = 30N` |
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| 3. |
A particle executing SHM has amplitude of 4cm, and its acceleration at a distance of 1 cm from the mean position is 3 cms^(-2). What will its velocity be when it is at a distance of 2 cm from its mean position ? |
| Answer» Answer :C | |
| 4. |
The ventilation of building is necessary for |
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Answer» ENTRY of OUTSIDE oxygen |
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| 5. |
When torque acting upon a system is zero, which of the following will remain constant? |
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Answer» force |
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| 6. |
The displacement (in metre) of a particle varies with time (in second) according to the equation y =- (2)/(3) t ^(2) + 16 t +2.How long does the particle take to come to rest ? |
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Answer» 12 s `y =- 2/3 t ^(2) + 16 t +2` `THEREFORE` Velocity `U = (dy)/(DT) = (d)/(dt) [- (2)/(3) t ^(2) + 16 t +2]` `therefore v = -(4)/(3) t + 16` When a particle is at rest, v =0 `therefore 0=- (4)/(3) t + 16` `therefore 4/3 t = 16` `therefore t = 12s.` |
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| 7. |
A porter moving vertically up the stairs with a suitcase on his head does work. Give reason. |
| Answer» SOLUTION :The porter lifts the suitcase vertically to the UPSTAIRS. Force has to be APPLIED on the suitcase against the force of gravitation. HENCE the porter does WORK. | |
| 8. |
The sprinking of water reduces slightly the temperature of a closed room because, |
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Answer» Temperature of WATER is less than that of the room. |
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| 9. |
Explain the construction and working of transformer. |
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Answer» Solution :Venturimeter : This device is USED to measure the rate of flow (or say flow speed) of the incompressible fluid flowing through a pipe. It works on the principle of Bernoulli.s theorem . It consists of two WIDER tubes A and A. (with cross sectional area A) CONNECTED by a narrow between the wide the narrow tubes. The manometer contains a LIQUID of density `.rho_m.`. Let `P_1` be the pressure of the fluid at the wider region of the tube A. Let us assume that the fluid of density `.rho.` flows from the pipe with speed `.v_1.` and into the narrow region, its speed increases to `.v_2.` . According to the Bernoulli.s equation, this increase in speed is accompanied by a decrease in the fluid pressure `P_2` at the narrow region of the tube B. Therefore, the pressure difference between the tube A and B is noted by measuring the height difference `(DeltaP = P_1 - P_2)` between the surfaces of the manometer liquid. From the equation of continuity, we can say that `Av_1 = av_2` which means that `v_2 = A/a v_1` Using Bernoulli.s equation , `P_1 + rho (v_1^2)/(2) = P_2 + rho (v_2^2)/(2) = P_(2) + rho 1/2 (A/a v_1)^(2)` From the above equation, the pressure difference `DeltaP = P_1 - P_2 = rho (v_1^2)/(2) ((A^2 - a^2))/(a^2)` Thus, the speed of flow of fluid at the wide end of the tube A `v_1^2 = (2 (DeltaP)a^2)/(rho (A^2 - a^2)) implies v_1 = sqrt((2 (Delta P)a^2)/(rho(A^2 - a^2)))` The volume of the liquid flowing out per second is `V = A v_1 = A sqrt((2(DeltaP)a^2)/(rho(A^2 - a^2))) = aA sqrt((2(DeltaP))/(rho(A^2 - a^2)))`. |
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| 10. |
A car starts from rest and travels with uniform acceleration alpha for some time and then with uniform retardation beta and comes to rest. If the total time of travel of the car is t, then what isthe maximum velocity attained by the car ? |
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Answer» |
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| 11. |
The unit of viscosity in the CGS system is poise (P) and that in SI is poiseuille(Pl). Which of the following statement is correct ? |
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Answer» 1 P = 1 Pl `THEREFORE 1Pl=10"GCM"^(-1)s^(-1)=10P` |
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| 12. |
State in the following cases, whether the motion is one ,two (or) three dimensional .A kite flying on a windy day . |
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Answer» |
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| 13. |
An astronaut tries to fill ink in an ink pen in an artificial satellite. The ink |
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Answer» Will be FILLED into the pen |
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| 14. |
A molecule moving along a straight line possesses x degree of freedom. Here, x refers to |
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Answer» one |
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| 15. |
A ladder AP of length 5m inclined to a vertical wall is slipping over a horizontal surface with velocity of 2m/s, when A is at a distance 3m from 0, the velocity of CM at this moment is: |
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Answer» Solution :From figure, `Y = sqrt(L^(2)-x^(2))` `(DY)/(DX) = x/sqrt(L^(2)-x^(2)) (dx)/(dt) =-(3 xx 2)/4 = -3/2 m//s` `V_(CM) = sqrt([1/2(dx)/(dt)]^(2) + [1/2(dy)/(dt)]^(2))` `=sqrt(1^(2) + (3/4)^(2)) = 1.25 m//s` 
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| 16. |
The order of magnitude of 45,71,100 = 4.571xx10^(6) metre is |
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Answer» 10 |
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| 17. |
A satellite is revolving round the earth in a circular orbit of radius .a. with velocity v_(0), A particle is projected from the satellite in forward direction with relative velocity v = (sqrt(5//4)-1)v_(0). Calculate, during subsequent motion of the particle, the maximum distance from earth.s centre |
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Answer» 2a/3 |
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| 18. |
A man rowing a boat up stream is at rest with respect to the bank. Is he doing work ? |
| Answer» Solution :No, the MAN applies force for rowing the BOAT UPSTREAM. But he is at rest with respect to the bank i.e., DISPLACEMENT is zero and work DONE is zero. | |
| 19. |
Statements : (a) Allgebric sum of moments of mass about centre of mass is equal to zero (b) x-co-ordinate of centre of mass of system of particles in a plane is represented by x_(cm)=(1)/(M)summ_(1)x_(1) (c ) x-co-ordinate of a rigidbody of continuous mass distribution represented by x_(cm)=(1)/(m)intxdm |
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Answer» a and B are true |
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| 20. |
(A): When two bodies of equal masses collide, their veloctities are always interchanged. (R ): Momentum is conserved in any collision |
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Answer» Both .A. and .R. are true and .R. is the CORRECT explanation of .A. |
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| 21. |
What isstaionarywave? Instationarywave , whatis thephasedifferencewave?In stationarywave, whatis thephasedifferentbetweentwoparticlesoftwoconsercutiveloops ? |
| Answer» SOLUTION :THEMOTION ofpartivclesintwoadjactantloops areoppositein DIRECTION. Hencephasedifference`= 180 ^(@)`. | |
| 22. |
Consider two containers A and B containingidentical gases at the same pressure, volume and temperature. The gas in container A is compressed to half of its original volume isothermally while the gas in container B is compressed to half of its original value adiabatically. The ratio of final pressure of gas in B to that of gas in A is |
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Answer» `2^(gamma-1)` |
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| 23. |
One mole of an ideal gas at an initial temperature of TK does 6R Joule of work adiabatically. If the ratio of specific heat of this gas at constant pressure and at constant volume is 5/3,the final temperature of gas will be ____ |
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Answer» (T-4)K `W=(P_1V_1 -P_2V_2)/(gamma-1)` `=(nRT_1-nRT_2)/(gamma-1) =(NR(T_1-T_2))/(gamma-1)` `THEREFORE 4=T-T_2` `therefore T_2=(T-4)` K |
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| 24. |
(1mum)/(1fm)=...... |
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Answer» |
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| 25. |
A body of weight 10 N attached to one end of a string of length 1m is rotated in a vertical plane. At the instant when the stringmakes an angle of 60^(@) with the downward vertical the speed of the bob is sqrt(9.8) ms^(-1). The tension in the string at that instant is equal to |
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Answer» 10 N |
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| 26. |
(A) : Dampled oscillation indicates loss of energy. (R) : The loss in damped oscillation may be due to friction, air resistance etc. |
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Answer» Both 'A' and 'R' are TURE and 'R' is the correct explanation of 'A' |
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| 27. |
Estimate the average thermal energy of helium atom at a temperature of 27^@C. (Boltzmann constant is 1.38xx10^(-23) JK^1) |
| Answer» SOLUTION :Average THERMAL ENERGY = `3/2k_BT = 3/2 XX 1.38 xx 10^(-23) xx 300 = 6.21xx10^(-21)J` | |
| 28. |
A stick is thrown in air. It lands a little away from the thrower. The locus of the path of the centre of mass of the stick will be a parabola |
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Answer» in all cases |
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| 29. |
A body of mass 1.0 kg is rotating on a circular path of diameter 2.0 m at the rate of 10 rotations in 31.4 s. The angular momentum of the body, in kg m^2/s, is |
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Answer» `1.0` `therefore` Radius `R = (D)/(2) = 1.0` m The moment of inertia of the body is `I = MR^(2) = (1.0 Kg) (1.0 m)^(2) = 1.0 kg m^(2)` The ANGULAR velocity of the body is `omega = 2 pi epsilon = 2 xx 3.14 xx (10)/(31.4)` = rad/s = 2 rad/s The angular momentum of the body is `L = I omega = (1.0 kg m^(2))` (2 rad/s) = 2 kg `m^(2)//s` |
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| 30. |
Does the P.E. of the spring decreases or increases when it si compressed or stretched ? |
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Answer» |
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| 31. |
A gas is equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity for example does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise independence is given by the so called law of atmospheres n_2=n_1 exp[-mg(h_2-h_1)//K_BT] where n_2, n_1 refer to number density at heights h_2 and h_1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column: n_2=n_1exp[-mgN_A(p-p')(h_2-h_1)//(pRT)] where p is the density of the suspended particle and p' that of the surrounding medium [N_A is Avogadro's number and R the universal gas constant ] |
| Answer» SOLUTION :USE Archimedes principle to FIND the apparent weight of the SUSPENDED PARTICLE. | |
| 32. |
If the temperature of a black body is raised from 300K to 600K by what factor the rate of emission shall increase ? |
| Answer» Solution :The TEMPERATURE of the BODY is doubled. So by Stefan.s law the ENERGY radiated will became 24 = 16 TIMES. | |
| 33. |
For a gas, the difference between the twospecific heats is 4150J Kg^(-1) K^(-1)and the ratio of specific heats is 1.4. What is the specific heat of the gas at constant volume in J Kg^(-1) K^(-1)? |
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Answer» 8475 |
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| 34. |
The centre of mass of three particles of masses 1kg, 2kg,3kg is at (2,2,2). The position of fourth mass of 4kg to be placed in the system so that new centre of mass is at (0,0,0) is |
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Answer» `(-3,-3,-3)` |
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| 35. |
The material of prism used for obtaining spectrum of heat radiations is |
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Answer» Flint glass |
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| 36. |
A planet revolving round the Sun in an elliptical orbit, whose semi-major axis is double that of its semi-minor axis. When the Sun is assumed to be at rest at the mid point of semi-major axis, planet takes 24 hours to travel through a path bod as shown in the figure. Then the time taken by the planet to travel along dab is................ |
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Answer» 744 MINUTES |
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| 37. |
Is angular momentum a vector or a scalar? If it is a vector give its direction |
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Answer» Solution :Angular momentum is a vector. Angular momentum `barL=m(barxxbarv)` The direction of angular momentum is perpendicular to the PLANE CONTAINING`barr` and `barv` i.e., if `barr` and `barv` are in X-y plane then angular momentum is along Z-axis. Its direction is given by RIGHT HAND screw rule. |
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| 38. |
A 0.5 kg block slides from a point A [fig.1.17] on a horizontal track with an initial speed of 3 m.s^-1 towards a weightless horizontal spring of length 1 m and of force constant 2 N.m^-1. The part AB of the track is frictionless and the part BC has coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distance AB and BD are 2 m and 2.14 m respectively. Find the total distance covered by the black before it comes to rest (g=10 m.s^-2) |
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Answer» Solution :Initial kinetic energy of the block `=1/2mv^2=1/2times0.5times(3)^2=2.25J`  The part AB of the horizontal TRACK is frictionless and HENCE during the PASSAGE of the block over this path no energy is LOSS in energy in the part BD =workdone against function in that path `=mumg times BD=0.2times0.5times10times2.14=2.14J` So, the kinetic energy of the blok when it reaches the point D=2.25-2.14=0.11 J Let us ASSUME that the block compresses the spring through an amount x. According to the principle of conservation of energy. work done against friction+work done in compressing the spring =0.11 or`mu mgx+1/2kx^2=0.11` or,`0.2times0.5times10timesx+1/2times2x^2=0.11` or,`x^2+x-0.11=0` or`(x+1.1)(x-0.1)=0` `therefore` x=0.1 m [as x cannot be -1.1m] `therefore` The distance covered by the block before it comes to rest =2+2.14+0.1=4.24m. |
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| 39. |
Give the expression for beat period. |
| Answer» Solution :BEAT PERIOD `t_(b) = (1)/(f_(b)) =((1)/( f_(1)-f_(2)))s` | |
| 40. |
A bus of mass 2400 kg is moving on a straight road with a speed of 30 km/h. A car of mass 1600 kg is following the bus with speed of 40 km/h. How fast is the centre of mass of the system of two vehicles moving? |
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Answer» `70km//h` `=(2400xx30+1600xx40)/(2400+1600)` `=(72000+64000)/(4000)` `=(136000)/(4000)` `=34km//h` |
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| 41. |
A wire of length 2 m with the area of cross section 10^(-6) m^(2) is used to suspend a load of 980 N. Calculate (i) The stress developed in the wire (ii) the strain and (iii) the energy stored . Given Y = 12 xx 10^(10) Nm^(-2). |
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Answer» SOLUTION :(i) Stress - `F/A = (980)/(10^(-6)) = 98 xx 10^(7) Nm^(-2)` (II) Strain = `("Stress")/(Y) = (98 xx 10^(7))/(12 xx 10^(10)) = 8.17 xx 10^(-3)` (no unit) (iii) Since, volume = `2 xx 10^(-6) m^(3)` ENERGY = `1/x ("stress" xx "strain") xx "volume" implies 1/2(98 xx 10^(7)) xx (8.17 xx 10^(-3)) xx 2 xx 10^(-6) = 8 "joule"` |
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| 42. |
On a large tray of mass M an ice cube of mass m, edge L is kept. If the ice melts completely, the centre of mass of the system come down by |
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Answer» `(ML)/(2(M+m))` |
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| 43. |
A block of mass 2m is attached to a string of length L = 5 m. The block is released when the string is horizontal , it picks up a particle of 2 m kept at rest at the lowest point. The maximum height attached by the combined mass is |
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Answer» `1 m` `2 MV = ( 2m + 2m)v'` `v' = (v)/(2) = sqrt((gL)/(2)) LT sqrt( 2 gL)` `0 = v'^(2) - 2gh rArr (gL)/(2) = 2gh` `h = (L)/(4) = (5)/(4) = 1.25 m`   
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| 44. |
A long steel wire of length 'L' is suspended from the ceiling of a room. A sphere of mass 'm' and radius 'r' is attached to the lower end of the wire. The height of the ceiling is (L+2r^1+1) . When the sphere is made to oscillate as a pendulum, its lowest point just touches the floor. The velocity of the sphere at the lowest point will be (L > > r^1 , 1 and r is the radius of the wire) |
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Answer» `SQRT((pi r^2 ye)/(m) - Lg)` |
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| 45. |
The moment of inertia of a fly wheel making 300 revolutions per minute is 0.3kgm^(2), find torque required to bring it to rest in 20s. |
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Answer» Solution :Torque `TAU=Ialpha=I((omega_(2)-omega_(1)))/(t)` `:.tau=(0.3xx(0-10pi))/(20)` `=-0.15pi=-0.471Nm` |
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| 46. |
A Carnot engine whose efficiency is 40%, receives heat at 500 K. If the efficiency is to be 50%, the source temperature for the same exhaust temperature is |
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Answer» 900 K where `T_(1)` and `T_(2)` are the temperature of SOURCE and sink RESPECTIVELY. `:. (T_(2))/(T_(1)) = 1 - eta = 1 - (40)/(100) = (60)/(100) = (3)/(5) (because eta = 40%)` or `T_(2) = (3)/(2)T_(1) = (3)/(5) xx 500 K = 300 K` ...(i) LET `T._(1)` be the temperature of the source for the same sink temperature. `:. (T_(2))/(T._(1)) = 1 - eta. = 1 - (50)/(100) = (1)/(2) "" (because eta. = 50%)` or `T._(1) = 2T_(2) = 2 xx 300 K = 600 K` (Using (i)) |
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| 47. |
Percentage errors in the measurement of mass and speed are 3% and 2% respectively. The error in the calculation of kinetic energy is ............ |
| Answer» SOLUTION :`7%` | |
| 48. |
Is a large brake on a bicycle more effective thana small one ? |
| Answer» Solution :The FORCE of friction is INDEPENDENT of area of contact between the surfaces. So far the same braking force LARGE BRAKE is not NECESSARY. | |
| 49. |
A given mass of air at one atmosphere and 273K is allowed to expand to four times its original volume adiabatically. Calculate the final temperature. |
| Answer» Solution :`T_(1) =T_(1) (V_(1)//V_(2)) ^(GAMMA -1) = 273 (1//4) 6(1.4 -1) = 156.8K` | |