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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. |
A duster weighs 0.5 N. It is pressed against a vertical board with a horizontal force of 11 N. If the co-efficient of friction is 0.5 the minimum force that must be applied on the duster parallel to the board to move it upwards is |
| Answer» ANSWER :C | |
| 2. |
The platinum wire has resistance of 2 ohms at the temperature of melting ice, 2.778 ohms at temperature of boiling water under standard pressure and 2.54 ohms at the boiling point of a liquid. Find the platinum scale temperature (t_(p)) and the corresponding correct temperature (t). Assume delta=1.5 for pure platinum. |
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Answer» <P> |
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| 3. |
The displacement of a particle executing SHM is given by x=0.10sin ((2pit)/T+phi) where all the quantities are in SI units. The time period of vibration is 3s. At t=0, the particle is displaced by 0.05 m. Find (i) the initial phase (ii) the phase angle corresponding to 0.0866 m and (ii) the phase difference between any two positons of the particle 2s apart. |
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Answer» Solution :For the particle in SHM `x=0.10 sin ((2pit)/T+PHI)` `=0.10sin((2pi)/3t+phi)( :. T=3s)` (i) At `t=0, x=0.05m` `0.05=0.10sin ((2pi)/3xx0+phi)=0.10sin phi` `sin phi =1/2impliesphi=(pi)/6`radian or `30^(@)` (ii) Substituting `x=0.0866`m in `x=0.10sin ((2pi)/3t + phi)` `0.0866=0.10sin((2pi)/3t+phi)` `sin((2pi)/3t+phi)=0.0866/0.10=0.866` `:.` Phase ANGLE `((2pi)/3t+phi)=(pi)/3` radian or `60^(@)`. (iii) Phase angle at `t=t_(1)` is `phi_(1)=((2pi)/3xxt_(1)+phi)` and at `t=t_(2)` is `phi_(2)=((2pi)/3xxt_(2)+phi)` Phase difference `=phi_(2)-phi_(1)` `=((2pi)/3t_(2)+phi)-((2pi)/3t_(1)+phi)=(2pi)/3(t_(2)-t_(1))` `=(2pi)/3x2-(4pi)/3` radian `( :. t_(2)-t_(1)=2)` |
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| 4. |
Choose the false statement. 1^(st) Law of thermo dynamics |
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Answer» does not tell US about the efficiency with which HEAT can be converted in to work. |
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| 5. |
In a harmonium the intermediate notes between a note and its octave form |
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Answer» An ARITHMETIC PROGRESSION |
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| 6. |
(A) : If body or a system has neither translational nor rotational motion then it is in complete mechanical equilibrium (R) : A body or system is said to be in mechanical equilibrium when net force and net torque acting on the body are zero |
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Answer» Both 'A' and 'R' are true 'R' is the CORRECT EXPLANATION of 'A' |
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| 7. |
A ring of mass m and radius R is placed on a rough inclined plane so that it rolls without slipping . Match the following table. |
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Answer» <P> |
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| 8. |
The units of universal gas constant is (a) J mol^(-1) K^(-1)(b) J kg^(-1) k^(-1) (c) Litre atm mol^(-1) k^(-1)(d) cal mol^(-1) k^(-1) |
| Answer» ANSWER :C | |
| 9. |
Consider the following statements and choosethe correct answer. a) if heat is added to a system, its temperaturemust always increase.b) if positive work is done by a system inthermodynamic process,its volume must increase. |
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Answer» both a and B are true |
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| 10. |
Derive expressions for displacement, velocity and acceleration of a particle executing simple harmonic motion. |
Answer» Solution : Consdier the position of the particle P on a circle of radius A at some instant of TIME t as shown in Figure. Its displacement y at that instant of time t can be derive as follows In `DeltaOPN`, `"sin" theta=(ON)/(OP)` `rArr ON=OP sin theta ""...(1)` But `theta= omega t, ON=y and OP =A` `y=A sin omega t ""..(2)` The displacement y takes maximum value (which is EQUAL to A) when `sin omega t=1`. This maximum displacement from the mean position is known as AMPLITUDE (A) of the vibrating particle. for simple harmonic motion, the amplitude is constant. Velocity : The rate of change of displacement is velocity. TAKING derivative of equation (2) with RESPECT to time, we get `v=(dv)/(dt)=(d)/(dt) (A sin omega t)` For circular motion (of constant radius), amplitude A is a constant. Angular velocity `omega` is a constant for uniform circular motion. `:.v=(dy)/(dt) A omegac t ......(3)` Using trigonometry identity, `sin^(2)omegat+cos^(2)omega t=1` `rArr cos omega t= sqrt(1-sin^(2)omega t)` We get, `v=A omega sqrt(1-sin^(2) omegat)` Form equation (2), `"sin" omega t=(y)/(A)` `v= A omega sqrt(1-((y)/(A))^(2))"....(4)` From equation (4), when the displacement `y=0`, the velocity `v= omega A` (maximum) and for the maximum displacementy=A, the velocity v=0 (minimum). Acceleration : The rate of change of velocity is acceleration. `a=(dv)/(dt)=(d)/(dt) (A omega cos omegat)` `a=- omega^(2)A sin omegat=- omega^(2)y...(5)` `:. a=(d^(2)y)/(d^(2))=- omega^(2)y ""....(6)` |
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| 11. |
A bullet is fired gun. The force on bullet is F=600-2xx10^(5)t newton. The force reduces to zero just when bullet leaves barrel. Find he impulse imparted to bullet |
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Answer» Solution :`F=600-2xx10^(5)t` F becomes zero as soon as the bullet leaves the barrel.`0 = 600-2xx10^(5)t, 6=2xx10^(5)t` `t=3XX10^(-3)s` Impusle `= int_(0)^(t)Fdt` `= int_(0)^(t)(600-2xx10^(5)t)dt=[600t-cancel(2)xx10^(5)(t^(2))/(cancel(2))]^(3xx10^(-3))` `= 600xx3xx10^(-3)-10^(5)xx9xx10^(-6)=0.9 NS` |
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| 12. |
What is the relation between angle offriction and angle of repose |
| Answer» SOLUTION :`mu_(s)`is GREATERTHAN `mu_(K)` | |
| 13. |
The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression, Delta V of water at the bottom of the ocean, given that the bulk modulus of water is2.2 xx 10 ^(9) Nm ^(-2) ("Take" g = 10 ms ^(-2)) |
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Answer» Solution :`h = 3000 m` Density of water `rho = 10 ^(3) KGM ^(-3)` Density of water `rho = 10^(3) kgm ^(-3)` Gravitational acceleration `g = 10 ms ^(-2)` `therefore` The pressure exerted on the bottom of ocean, `P = h rho g` `= 3 xx 10 ^(3) xx 10 ^(3) xx 10` `("FORCE")/("Area") = 3 xx 10 ^(7) Nm ^(-2)` Now bulk modulus, `B= ("Stress")/("VOLUME STRAIN") = ("Stress")/(Delta V //V)` `therefore (Delta V)/(V) = ("Stress")/(B) ` `= (3 xx 10 ^(7))/( 2.2 xx 10 ^(9)) [because` Stress = Pressure P] `= 1.36 xx 10 ^(-2)` `therefore` Volume strain `= (Delta V)/(V) xx 100 %` `= 1. 36 xx 10 ^(-2) xx 10 0 % = 1.36 %` |
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| 14. |
A wheel of radius .r. rolls without slipping with a speed V on a horizontal road. When it is at point .A. on the road, a small lump of mud separtes from the wheel at its highest point B and drops at point .C on the road. The distance AC will be |
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Answer» `vsqrt(r//g)` |
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| 15. |
The rectangular components of a vector lying in xy plane are (n+1) and 1. If co-ordinate system is turned by 60^(@). They are n & 3 respectively the value of .n. |
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Answer» 2 |
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| 16. |
If vector vecA =4 hati -6 hatJ +2hatk and vecB =6hati +8 hatj +mhatk are m utually perpendicular, m = .... |
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Answer» 24 `(4hat(I )- 6hat(j )+ 2hat(K )) (hat(I )= 8hat(I )+ MHAT(k )) = 0` `2 m=24` `m=12 ` |
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| 17. |
When a big drop of water is formed from m small drops of water, the energy loss is 3E where E is energy of bigger drop. If R is radius of bigger drop and r is radius of smaller drop then number of smaller drops(n) is |
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Answer» `(4R)/(r )` |
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| 18. |
The radius of reference circle is equal to the……….of the oscillator. |
| Answer» SOLUTION :AMPLITUDE. | |
| 19. |
In an experiment the valno of refractive index of glass was found to be 1.54, 1.53, 1.44,1.54, 1.56 and 1.45 m in successive measurement. Calculate (i) the mean vlaue of refractive index (ii) absolute error of each measurement (iii) mean absolute error (iv) relative error and (v) percentage error. |
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Answer» |
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| 20. |
A cylindrical drum open at the top contains 30 litres of water. Il drains out through a small opening at the bottom. 10 litres of water comes out in time t, the next 10 litres lp further time 12 and the last 10 litres in further time t3. Then |
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Answer» `t_(1)=t_(2)=t_(3)` |
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| 21. |
An ideal spring of spring constant k, is suspended from the ceiling of a room and a blok of mass m is fastened to its lower end. If the block is released when the spring is un-stretched, then then maximum extension in the spring is : |
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Answer» `4(Mg)/(k)` `MGX+(-(1)/(2)kx^(2))=0` `Mgx=(1)/(2)kx^(2)` `Mg=(kx)/(2)implies therefore X=(2Mg)/(k)` |
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| 22. |
To increase the length of an electric string of radius 3.5 mm by 1/20 the of its initial length, within its elastic limit, a 10N force is required, Calculate the Young's modulus for the material of the string. |
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Answer» SOLUTION :We KNOW that , `Y=(FL)/(pir^2l)` Here, `F=10N,l=1/20Land r=3.5mm=0.0035m.` `thereforeY=(10timesL)/(3.14times(0.0035)^2times1/20)=5.2 TIMES 10^6 N.m^-2` |
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| 23. |
A block of mas m is attached to two unstretched sprigs of spring constants k_(1) and k_(2) as shown in figure.The block is displaced towards right through a distance x and is released. Find the speed of the block as it passes through the mean positioni? |
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Answer» `xsqrt((k_(1)+k_(2))/m)` |
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| 24. |
Which of the following are vector quantities |
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Answer» Temperature |
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| 25. |
The percentage errors in quantities P, Q, R and S are 0.5%, 1%, 3% and 1.5% respectively in A = (P^2 Q^2)/(sqrt(Rs))the measurement of a physical quantity A = (P^2Q^2)/(sqrt(Rs)). The maximum percentage error in the value of A will be |
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Answer» 0.06 |
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| 26. |
A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms^(2) . Calculate the initial thrust (force) of the blast. |
Answer» Solution :Letrocketmovesin upwarddirectionwith accelerationaweigthandinitialthrustT onthe rocket  `:. T= mg +ma` `=2 xx10^(4)(9.8+5)` `T =29 .6 xx10^(4) N` Whichis THRUST actingon rocket T = `3.0xx 15^(5) N` |
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| 27. |
Two indentical ball A and B are release from lthe jpositions shoen in figure. They collied elastically on horizontal portion MN. The ratio of the heights attained by A and B after collision will be: (Neglect friction) |
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Answer» `1:4` |
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| 28. |
For a conservative system in one dimension , the force field can be derived fork is determined with a sonometer using …………. . |
| Answer» Answer :A | |
| 29. |
Awheel of radius R=2m performs pure rolling on a rough horizontal surface with speed v=10 m/s . In the figure shown, angle theta is angular position of point P on wheel reaches the maximum height from ground. Find the value of "sec"theta (take g=10 m//s^(2)). |
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Answer» |
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| 30. |
A driver takes 0.20 second to apply the brakes (reaction time). If he is driving car at a speed of 54kmh^(-1) and the brakes cause a deceleration of 6.0ms^(-1)? Find the distance travelled by car after he sees the need to put the brakes. |
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Answer» SOLUTION :(DISTANCE COVERED during 0.20 s) + (distance covered until REST) `=(15xx0.25)+[18.75]=21.75` |
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| 31. |
Explain : "By using significant digit we can prevent unnecessary long calculation. |
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Answer» Solution :In significant number with increase in number of significant digit accuracy increase. Measurement of physical QUANTITY INVOLVES addition, subtraction, multiplication and division which result into more number of significant digit. Result of calculation should be in CONSISTENCY of input value of measurement. For example, mass of object is m = 4.237 g and volume `V = 2.52 CM^(3)` Then density `RHO=(m)/(V)` `:. rho=(4.237)/(2.51)` `=1.68804780876 g cm^(-3)` Which is unnecessary long. Hence, it should be round off. Hence, practical value of density willbe `1.69 g cm^(-3)` |
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| 32. |
Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. Then the moment of inertia of the system about an axis along one of the sides of the square is |
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Answer» `(4)/(5) Ma^(2) + 2 Mb^(2)`  ABCD is a SQUARE of side b . FOUR spheres , each of mass M and radius a are PLACED at the four corners of the square as shown in the figure . We calculate moment of inertia of the system about AD . Moment of inertia of each of the spheres A and D about AD = `(2)/(5) Ma^(2)` Moment of inertia of each of the spheres B and C about AD = `((2)/(5) Ma^(2) + Mb^(2))` , using THEOREM of parallel axes. `therefore` Total moment of inertia of the system about AD ` = 2[(2)/(5) Ma^(2) + ((2)/(5) Ma^(2) + Mb^(2))] = (8)/(5) Ma^(2) + 2 Mb^(2)` |
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| 33. |
Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity (upsilon_(0)) and the braking capacity, or deceleration, -a that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of upsilon_(0) and a. |
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Answer» Solution :Suppose, vehicle stops after covering `d_(s)` distance when brakes are applied. Its initial velocity is `v_(0)` From equation `v ^(2) - v _(0) ^(2) = 2ax, v = 0 , v _(0) = v _(0), a =- a and X = d _(s)` `therefore 0- v _(0) ^(2) =- 2a d _(s)` ` therefore d _(s) = (v_(0) ^(2))/(2a) ` Note: Stopping distance is DIRECTLY proporational to the square of the initial speed of vehicle. If the speed of the vehicle is DOUBLED, then stopping distance will BECOME four times for given retardation. |
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| 34. |
Each of the two strings of length 51.6 cm and 49.1 cm are tensed separately by 20 N force. Mass per unit length of both the strings is same and equal to 1 g/m. When both the strings vibrate simultaneously the number of beats is ...... |
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Answer» 7 `f =(1)/( 2L ) sqrt ((T)/( MU ))` `therefore f _(1) = (1)/( 2 (0.516)) sqrt ((20)/( 0.001)) = 144 Hz` `therefore` Beat frequency `f _(2) - f _(1) = 144 - 137 =7 Hz` |
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| 35. |
Two bodies of masses 5kg and 4kg are tied to a string as shown. If the table and pulley are smooth, then acceleration of 5kg mass will be |
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Answer» `19.5m//s^(2)` |
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| 36. |
If a pendulum clock is shifted from earth to the space of moon. Then it |
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Answer» LOSES time |
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| 37. |
The longitudinal strain in a metal bar is 0.05. If the Poission.s ratio for the metal is 0.25, the lateral strain is |
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Answer» 0.0125 |
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| 38. |
A stone is projected horizontally with a velocity 9.8ms^(-1) from a tower of height 100 m. Its velocity one second after projection is |
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Answer» `9.8 MS^(-1)` |
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| 39. |
How will you determine the period of oscillation of a mass without the use of a clock? You are given a weightless spring, a metre scale and a known mass. |
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Answer» SOLUTION :The mass is made to oscillate. The period of oscillation `T = 2pisqrt(m/K) `, But `k = (MG)/x,x/g=m/k` `T = 2PI sqrt(x/g)` Hence, period can be found by measuring extension x PRODUCED by the mass using a metre scale. |
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| 40. |
n moles of a monoatomic gas is carried round the reversible rectangular cycle ABCDA as shown in the diagram. The temperature at A is T_0. The thermodynamics efficiency of the cycle____ |
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Answer» `15%` At A temperature =`T_0`, Volume =`V_0` , Pressure = `P_0` At A, `P_0V_0 =nRT_0` (cold) At B, `2P_0V_0=nRT` (HIGH) =T.=`2T_0` Maximum thermal efficiency of the reversible engine of this carnot.s cycle . `=eta=1-T_0/T=1-T_0/(2T_0)=1/2`=50% |
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| 41. |
Heat and Thermal capacity differ in the dimensions of |
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Answer» MASS |
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| 42. |
''A body with larger reflectivity is a poor emitter''. Why? |
| Answer» Solution :The reflectivity of the body is LARGE, so it is a poor absorber. ACCORDING to Kirchhoff.s law a poor absorber is a poor EMITTER. | |
| 43. |
A vessel of volume 2V contains 4 moles of a gas at certain temperature. Another vessel of volume 3V contains 3 moles of the same gas at same temperature then the ratio of the pressures is |
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Answer» `2:1` |
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| 44. |
A bus starts from rest with a constant acceleration of 5m//s^(2). At the same time a car travelling with a constant velocity 50 m/s overtakes and passes the bus. How fast is the bus travelling when they are side by side? |
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Answer» `10m//s` |
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| 45. |
When an ideal diatomic gas is heated at constant pressure, its internal energy increased by 50calthen the work done by the gas is |
| Answer» Answer :D | |
| 46. |
Velocity is defined as the rate of change of displacement. When does the average veiocity become equal to the instantaneous velocity? |
| Answer» SOLUTION :In UNIFORM MOTION | |
| 47. |
A cubical block is floating in a liquid with half of its volume immersed in the liquid. When the whole system accelerates upwards with acceleration of g/3, the fraction of volume immersed I the liquid is 1/N. Then N is equal to______ |
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Answer» |
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| 48. |
Which of the following paris is a correct pair |
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Answer» NEWTON 's second law and law of conservation of MOMENTUM |
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| 49. |
The angle of contact at the interface of water glassis 0^(@) ethylalcohol glassis 0^(@) mercury glass is 140^(@)and methyliodide glass is 30^(@)A glass capillary is put in a trough containing one of these four liquids . It is observed that the meniscus is convex.the liquid inthe trough is |
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Answer» water  If angleof contact `thetagt90^(@)` ,then meniscus will be CONVEX .Hence liquid of `theta=140^(@)`is mercury. |
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| 50. |
A particle starting from initial velocity v_0 moves on straight line with constant acceleration equation of distance travelled during n^"th" second will be …... . |
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Answer» SOLUTION :For given particle, DISTANCE TRAVELLED in `n^(th)` seconde is, d = (distance travelled in n seconds) - (distance travelled in (n-1) second) `= (v_(0) n + 1//2 an ^(2)) -(v_(0) (n -1) +1/2 a (n-1) ^(2))` (From equation `d = v _(0) t + 1/2 at ^(2))` `=( v _(0) n + 1/2 an ^(2)) - (v _(0) n - v_(0) + a//2 (n ^(2) -2n +1))` `=(v_(0) n + 1/2an ^(2) -v _(0) n + v_(0) -1/2 an ^(2) + an - (a)/(2))` `= v _(0) + a _(n) - (a)/(2)` `therefore d = v _(0) + a/2 (2n -1)` |
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