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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. |
Four identical blocks of length L are arranged one over the other as shown. The maximum distance of the uppermost block from the edge of the lowermost block is x such that no block tumbles. Then x is: |
| Answer» Answer :C | |
| 2. |
A solid cylinder rolls without slipping down an inclined plane of height h. The velocity of the cylinder when it reaches the bouom is |
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Answer» `sqrt((2gh)/(3))` According to conservation of mechanical energy , `Mgh = (1)/(2) MV^(2) + (1)/(2) I OMEGA^(2) = (1)/(2) Mv^(2) + (1)/(2) ((1)/(2) MR^(2)) omega^(2)` = `(1)/(2) Mv^(2) + (1)/(4) MR^(2) ((v)/(R))^(2) = (1)/(2) Mv^(2) + (1)/(4) Mv^(2) = (3)/(4) Mv^(2)` or `v^(2) = (4)/(3) gh = v = sqrt((4)/(3) gh)` |
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| 3. |
Prove that V= sqrt( gamma CT) from the ideal gas equation for speed of sound is gases. |
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Answer» Solution :Let us consider an IDEAL GAS whose equation of state is `PV= nRT "" ...(1)` where P-pressure, V- volume, T-temperature, n-number of mole and R-universal gas CONTANT. For a GIVEN mass of molecules equation (1) can be written as `(PV)/(T)` constant `""...(2)` for a fixed mass m, density of the gas inversely varies with volume i.e., `rho prop (1)/(V),V=(m)/(rho) ""....(3)` Substituting equation (3) in equation (2), we get `(P)/(rho)=cT ""...(3)` where c proportionality is constant. `:. v=sqrt((gamma P)/(rho))= sqrt(gamma cT)` |
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| 4. |
A calorimeter contains 10 g of water at 20^(@)C. The temperature falls to 15^(@)C in 10 min. When calorimeter contains 20 g of water at 20^(@)C, it takes 15 min for the temperature to become 15^(@)C. Find the water equivalent of the calorimeter. |
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Answer» Solution :(i) LET water equivalent be `x` `(10 + x) s (Delta THETA)/(Delta t) = - k' (bartheta - theta_(0))` (b) `(10 + x) s ((20 - 15))/(10) = - k' ((20 + 15)/(2) - theta_(0))` `(20 + x) s ((20 - 15))/(10) = - k' ((20 + 15)/(2) - theta_(0))` `(10 + x)/(10) = (20 + x)/(15)` `150 + 15 x = 200 + 10 x` `5x = 50 IMPLIES x = 10 g` |
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| 5. |
From a complete ring of mass M and radius R, an are of making 30^(0) at centre is removed. What is the moment of inertia of the incomplete ring about an axis passing through the centre of the ring and bot^(er)to the plane of the ring |
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Answer» Solution :mass of in complete ring `=M-(M)/(2PI)xx(pi)/(16)=(11M)/(12)` MOI of inertia of incomplete ring `I=[((11m)/(12))]R^(2)` `I=(11)/(12)MR^(2)` |
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| 6. |
Choosethe correctalternative: |
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Answer» For a general rotationalmotion , angularmomentumL and angularvelocity`omega`NEED not beparallel |
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| 7. |
A mass is whirled in a circular path with constant angular velocity and its angular momentum is L.If the string is now halved keeping the angular velocity the same, the angular momentum is |
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Answer» Solution :We know that angular momentum `L=Mr^(2) omega` Here, MI and ca are constants `:. L prop R^(2)` If `(r)/(2)` becomes angular momentum becomes `(1)/(4)` th of its INITIAL value. |
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| 8. |
A star , 2 .5 times the mass of sum , collapse to a size of12 km with a speed of1.2 rps . T the object placed on thesurface remains stuck to the surface due to gravity ? [ mass of sun =2xx10^(30)kg] |
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Answer» SOLUTION :`g=(GM)/(R^(2))=(6.67xx10^(-11)xx25xx2xx10^(30))/(12xx10^(3))^(2)` `g=0.2316xx10^(13)ms^(-2)` `omega^(2) R = (1.2xx2pi)^(2)(1.2xx10^(7))=6.8xx10^(8)ms^(-2)` SINCE `g " GT"omega ^(2)R` , the object remains STUCK to its surface. |
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| 9. |
The speed of the wave represented by y =A sin (omega t - k x) is …………. . |
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Answer» `K// OMEGA` |
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| 10. |
The area under force time curve gives |
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Answer» WORK |
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| 11. |
Polar satellites |
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Answer» are sun-synchronous satellites. |
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| 12. |
In (P+(a)/(V^(2)))(V-b)=RT equation the, a,b will be ..... Where P= Pressure V= Volume and T= Temperature . |
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Answer» <P>`M^(1)L^(2)T^(-2)` `:.` Dimensional FORMULA of `(a)/(V^(2))=` Dimensional `:.` Dimensional formula of a= Dimensional formula of `PV^(2)` `=M^(1)L^(-1)T^(-2)(L^(3))^(3) "" [P]=[M^(1)L^(-1)T^(-2)]` `=M^(1)L^(-1)(T^(-2))(L^(6))[V]=L^(3)` `:. [a]=M^(1)L^(5)T^(-2)""...(i)` We have `V-b "" :. V=b` `:.` Dimensional formula of V= Dimensional formula of b `:. [b]=L^(3)....(ii)` `:. [a*b]=M^(1)T^(5)L^(-2)(L^(3))=M^(1)L^(8)T^(-2)` |
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| 13. |
Rain is falling vertically at 4 km *h^(-1) . What will be theapparent velocity and direction of the rain to a manmoving horizontally at 3 km *h^(-1) ? |
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Answer» |
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| 14. |
A motorcycle and a car start from rest from the same place at the same and travel in the same direction. The motorcycle acceleration at 1.0 m s^(-1) up to a speed of 36 km h^(-1) and the car at 0.5 m s^(-1) up to a speed of 54 km h^(-1). The time at which the car would overtake the motorcycle is |
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Answer» 20 s For motor cycle : Maximum speed attained = `36 km H^(-1) = 36 xx 5/18 = 10 m s^(-1)` Since its acceleration = `1.0 m s^(-2)`, the time `t_1` taken by it to attain the maximum speed is given by `v = u + at_1 implies 10 = 0 + 1.0 xx t_1 implies t_1 = 10 s`(sinceu = 0) The distance covered by motocycle in attaining the maximum speed is `S_1 = 0 + 1/2 at^(2)_1 = 1/2 xx 1.0 xx (10)^(2)` = 50 m The time during which the motorcycle moves with maximum speed is (t - 10) s. The distance covered by the motorcycle during the time is `S^(')_1 = 10 xx (t - 10) = (10t - 100)` m `therefore` Total distance travelled by motorcycle in time t is `S = S_1 + S^(')_1 = 50 + (10t - 100) = (10t - 50) m`...(i) For car : Maximum speed attained = `54 km h^(-1) = 54 xx 5/18 = 15 m s^(-1)` Since its acceleration = `0.5 m s^(-2)` The time taken by it to attain the maximum speed is given by 15 = `0 + 0.5 xx t_2` or `t_2 = 30 s`(since u = 0) The distance covered by the car in attaining the maximum speed is `S_2 = 0 + 1/2 at^(2)_2 = 1/2 xx 0.5 xx (30)^(2)` = 225 m The time during which the car moves with maximum speed is (t - 30) s. The distance covered by the car during this time is `S^(')_2 = 15 xx (t - 30) = (15t - 450)` m `therefore` Totoal distance travelled by car in time t is `S = S_2 + S^(')_2 = 225 + (15t - 450) = (15t - 225)`m...(II) From equations (i) and (ii), we GET `10t - 50 = 15t - 225` or `5t = 75` or t = 35 s |
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| 15. |
Two light strings of length 4 cm and 3 cm are tied to a bob of weight 500 gm. The free ends of the strings are tied to pegs in the same horizontal line and separated by 5 cm. The ratio of tension in the longer string to that in the shorter string is |
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Answer» `4:3` |
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| 16. |
Show that the horizontal vector and displacement vector in two dimensions . |
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Answer» SOLUTION :As horizontal range `=(u^(2) sin 2 THETA) /(G)` For `theta=45^(@), 2 theta =90^(@)` `THEREFORE` Horizontal range =`(u^(2) sin 90^(2))/(g)=(u^(2))/(g)` (MAXIMUM) |
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| 17. |
A block of mass 10 kg is in contact against the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be (g = 10 m//s^2) |
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Answer» `10 pi rad //s` Frictionalforce` ge `weight `mu N geMg` here ` ,N = Momega^2 r ` ` r=1 m, mu =0.1` forminimum` omegaM omega^2 r = MG` ` omega= sqrt((g)/( mu r)) = sqrt((10 ) /( 0.1 XX 1 )) = 10rad s^(-1)` 
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| 18. |
Two blocks of masses 2 kg and 4 kg are connected by a light string passing over a light smooth pulley clamped to the edge of a horizontal table. The 2 kg block is on the smooth horizontal table and the other block is hanging vertically, (i) Find the acceleration of the stem if it is released from rest, (ii) Find the tension in the string. |
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Answer» `6.54m//s^(2), 26 N` |
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| 19. |
A wooden block with a coin placed on its top floats in water as shown. After some time the coin falls into water.Then |
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Answer» L DECREASES and H INCREASES |
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| 20. |
Derive the time period of satellite orbiting the Earth. |
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Answer» Solution :Time period of the satellite: The distance covered by the satellite during one rotation in its orbit is equal to `2pi (R_(E )+H)` and time taken for it is the time period, T. Then, Speed , v =` ("Distancetravelled ")/("Time taken ")= (2pi(R_(E)+h))/T ""...(1)` Speed of the satellite, `V = sqrt((GM_(E))/((R_(E)+h)))` `sqrt((GM_(E))/((R_(E)+h))) = (2pi(R_(E)+h))/T` `T = (2pi)/(sqrt(GM_(E)) )(R_(E)+h)^(3/2)""...(2)` Squaring both sides of the equation (2) we get `T^(2) = (4pi^(2))/(GM_(E)) (R_(E)+h)^(3)`...(2) `(4pi^(2))/(GM_(E)) = `constantsay c `T^(2) =c(R_(E)+h)^(3)` Equation (3) IMPLIES that a satellite orbiting the Earth has the same relation between time and distance as that Kepler.s law of planetary motion. For a satellite orbiting near the surface of the Earth. h is negligible compared to the radius of the Earth.`R_(E)`Then, `T_(2) = (4pi^(2))/(GM_(E))R_(E)^(3)` `T^(2) = (4pi^(2))/(((GM_(E))/R_(E)^(2)))R_(E)RARR T^(2) = (4pi^(2))/g R_(E)` Since ` (GM_(E))/(R_(E)^(2)) = g ` `T = 2pisqrt(R_(E)/g)` By substituting the values of`R_(E) = 6.4 xx10^(6) ` m and g = `9.8 " m s"^(-2)`the orbital time period is obtained `T cong 85`minutes . |
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| 21. |
A long rod ABC of mass "m" and length "L" has two particles of masses "m" and "2m" attached to it as shown in the figure. The system is initially in the horizontal position. The work to be done to keep it vertical with A at the bottom is (g = acceleration due to gravity) |
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Answer» 2 MGL |
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| 22. |
Differentiate clearly between Thermal conduction and Thermal convection. |
Answer» SOLUTION :
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| 23. |
Consider a two particle system with particles having masses m_(1)andm_(2). If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position? |
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Answer» d `therefore (m_(1)d+m_(2)(-x))/(m_(1)+m_(2))=r_(cm)` `therefore m_(1)d-m_(2)x=0 [because r_(cm)=0]` `therefore m_(1)d=m_(2)x` `therefore x=(m_(1))/(m_(2))d` |
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| 24. |
What is musical scale ? |
| Answer» Solution :A series of NOTES whose fundamental FREQUENCIES have definite ratios and which produce a pleasing EFFECT on the EAR when in succession CONSTITUTE a musical scale. | |
| 25. |
Angle of friction between any two surfaces in contact is…… of force of limitting friction F and…..makes with….. |
| Answer» Solution :ANGLE which the RESULTANT , normal REACTION `R` , normal reaction . | |
| 26. |
A block of mass m is suspended by different springs of force constant shown in figure Let time period of oscillation in these four positions be T_1, T_2, T_3 and T_4. Then : |
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Answer» `T_(1)=T_(2)=T_(4)` |
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| 27. |
A body of mass 2 kg is moving with a velocity of vecu= 3hati +4hatj m// s. A steady force vecF=hati-2hatj N begins to act on it. After four seconds,the body will be moving along. |
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Answer» X-axis with a VELOCITY of `2 m//s` |
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| 28. |
A man of mass 60 kg is standing on a weighing machine kept in a box of mass 30 kg as shown in the diagram, If the man manages to keep box stationary find the reading of the weighing machine. |
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Answer» Solution :We know that NORMAL reaction = scale reading For MAN `T-Mg+R=M.0` For box : `T-R-mg=m.0` `CANCEL(T)-Mg+R=cancel(T)-R-mg` `2R=(M-m)g` `R=((60-30)xx10)/(2)=150N` 
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| 30. |
During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature the ratio (C_(p))/(C_(v)) for the gas is |
| Answer» Solution :For an adiabatic process `P prop T^((gamma)/(gamma-1))` | |
| 31. |
A liquid does not wet the surface of a solid if the angle of contact is |
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Answer» ZERO |
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| 32. |
Where are antinodes ? |
| Answer» SOLUTION :Antinodes are points of MAXIMUM displacement of the particles in a STATIONARY wave. | |
| 33. |
If the external forces acting on a system have zero resultant, the centre of mass |
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Answer» MAY MOVE but not accelerate |
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| 34. |
A stone is thrown horizontally with velocity g ms^(-1) from the top of a tower of height g metre. The velocity with which it hits the ground is (in ms^(-1)) |
| Answer» Answer :C | |
| 35. |
3400 has significant figures |
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Answer» 4 |
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| 36. |
The function (sin^(2)omegat) represents |
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Answer» A PERIODIC but not simple harmonic motion |
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| 37. |
The position vector of a moving particle at .t. sec is given by vecr = 3hat(i)+4t^(2)hat(j)-t^(3)hat(k). Its displacement during an interval of t = 1s to 3 sec is |
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Answer» `HAT(j) - hat(k)` |
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| 38. |
The concept " force causes motion" was given by |
| Answer» SOLUTION :Aristotle | |
| 39. |
Maximum wavelength of radiations from Sun and Moon are 25Å and 0.5xx10^(-6)m respectively ratio of their temperature is. . . .. . |
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Answer» `1:1000` `lamda_(m)T=` constant `:.(lamda_(m))_(s)T_(s)=(lamda_(m))_(m)T_(m)` `:.(T_(s))/(T_(m))=((lamda_(m))_(m))/((lamda_(m))_(s))=(0.5xx10^(-6))/(25xx10^(-10))` `=0.02xx10^(4)` `=200:1` |
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| 41. |
How will you measure the diameter of the Moon using parallax method? |
Answer» Solution :O - observation point on earth.  (i) In diagram, O is the observation point on the earth and d is the diameter of moon. An astronomical TELESCOPE held at O is focussed on moon, the image is observed into moon of a circular disc. (ii) `angleAOB=THETA` S - average distance between moon and the surface of earth. (III) As 'S' is very large COMPARED to the diameter, d of the moon, the diameter of the moon is considered as a circular arc of radius, S. `d=S times theta`. Hence d can be calculated, when 'S' is known and `theta` is MEASURED. |
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| 42. |
Figure showns two blocks P & Q kept in contact with the wall with the force F. The wall is smooth but the surfaces in between the blocks are rough, for the block to be at rest against the wall P is |
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Answer» F should be more than the weight of P & Q |
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| 43. |
A body of mass 100 gm moving with an initial velocity of 4ms^(-1) collides with another body of mass 150 gm moving in opposite direction with velocity 6ms^(-1). If the collision is perfectly inelastic, then after the collision, |
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Answer» The TWO bodies INTER change their velocities |
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| 44. |
A particle starts from point A moves along a straight line path with an acceleration given by a=p-qx where p, q are constant and x is distance from point A. the particle stops at point B. The maximum velocity of the particle is |
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Answer» <P>`(p)/(Q)` |
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| 45. |
(A) : It is harder to open and shut the door if we apply force near the hinge.(R ) : Torque due to a given force is maximum at hinge of the door. |
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Answer» Both (A) and (R ) are true and (R ) is the correct EXPLANATION of (A) |
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| 46. |
From a solid hemisphere of radius 'R' a cone of a base radius 'R' and height 'R' is removed as shown in the figure. The moment of inertia of the remainng body about an axis BB' in the plane of the base and passing through the centre 'O' is I_(0),I_(1) is the moment of inertia about AA' which is parallel to BB' and I_(2) is moment of inertia about an axis perpendicular to BB', and passing through 'O', then |
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Answer» `I_(1)=I_(0)` |
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| 47. |
A monkey of mass m clings to a rope over a fixed pulley . The opposite ends of the rope is tide to a weight of mass M lying on a horizontal plate neglecting friction, find the acceleration of both bodies relative to the plate and tension of the rope in the following case: (i)the monkey does not move with respect to rope.(ii)the monkey moves upwards with respect to rope with acceleration beta(iii) the monkey moves downwards with respect to the rope with acceleration beta |
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Answer» Solution :(i) If the monkey is at rest on the rope , then acceleratiion of both mass M and monkey are same , equal to a, (say) If T is tension in the rope , then the equation of motion of monkey is `mg-T=ma_(1)`….(1) and equation of motion of mass M is `T=Ma_(1)`....(2) Adding (1) and (2) mg`=(M+m)a_(1) `:. a_(1)=m/(M+m)G`....(3) (ii) When monkey moves upward with acceleration `beta`, the reaction m`beta` acts downward , THEREFORE EQUATIONS of motion now take the form (`a_(2)` is acceleration of mass, M RELATIVE to plate) `mg+mbeta-T=ma_(2)` (for monkey )....(4) `T=Ma_(2)`(for mass M)....(5) Adding , we get , `mg+mbeta=(M+m)a_(2)` `:. a_(2)=(m(g+beta))/(M+m)` `:.` The acceleration of mass relative to plate `a_(2)=(m(g+beta))/(M+m)`....(6) and acceleration of monkey M relative to plate `a_(2)=a_(2)-beta=(m(g+beta))/(M+m)-beta =(m(g+beta)-beta(M+m))/(M+m)=(mg+mbeta-mbeta-mbeta)/(M+m)=(Mg-Mbeta)/(M+m)` (iii) When monkey moves downward with acceleration `beta`, teh reaction `(R=mbeta)` acts upward, therefore the equations of motion take the form (`a_(3)` is acceleration of block relative to plate ) `mg-mbeta-T=ma_(3)`....(8) `T=Ma_(3)`....(9) Adding we get `mg-mbeta=(M+m)a_(3)` `:. a_(3)=(m(g-beta))/(M+m)` and acceleration of monkey relative to plate `a_(3)^(1)=a_(3)+beta=(m(g-beta))/(M+m)+beta=(mg-mbeta+Mbeta+mbeta)/(M+m)` `:. a_(3)^(1)=(mg+Mbeta)/(M+m)` |
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| 48. |
It istrue thatall bodies radiateenergy, whyare not to see all theobject in a dark room ? |
| Answer» Solution :At room temperature, enegyradiatedis primarilyin the infraredregion of ELECTROMAGNETICRADIATION . Our EYES areonlysensitiveto visiblelight . Our EYE can notdetect thisradiation .Hencewe can not see the OBJECTS in a DARKROOM . | |
| 49. |
(A) : 'Green houses' which are used to keep the plants fo warm atmosphere in winter are built with glass. (R) : Glass has the property of tranmitting shorter wavelength heat radiations through it while reflecting longer ones. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the CORRECT EXPLANATION of 'A' |
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| 50. |
A uniform solid spherical ball is rolling down a smooth inclined plane from a height h. The velocity attained by the ball when it reaches the bottom of the inclined plane is v. If the ball is now thrown vertically upwards with the same velocity v, the maximum height to which the ball will rise is |
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Answer» 5h/8  `mgh = (1)/(2)mv^(2) + (1)/(2) I omega^(2)` `= (1)/(2) mv^(2) + (1)/(2) ((2)/(5) m R^(2)) omega^(2)` (for solid sphere , `I = (2)/(5) mR^(2))` ` = (1)/(2) mv^(2) + (1)/(5) mv^(2) (because v = R omega)` `mgh = (7)/(10) mv^(2)` or `v = sqrt((10)/(7) GH)` Let H. be maximum height attained by the ball when it is thrown vertically upwards with velocity `v ( = sqrt((10)/(7) gh))` `therefore h. = (v^(2))/(2g) = (( sqrt((10)/(7) gh))^(2))/(2g) = (10)/(14) h = (5)/(7) h` |
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