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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. |
Copper of fixed volume V is drawn into a wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is Deltal. If Y represents the Young's modulus, then which of the following graphs is a straight line? |
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Answer» `Delta L` VERSUS V `Y = (FL)/(A Delta l) = (Fl)/((V/l)Delta l) = (Fl^2)/(V cdot Deltal)` `Delta l = (F cdot l^2)/(V cdot Y)` |
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| 2. |
The radius of gyration of a uniform disc about a line perpendicular to the disc is equal to its radius. Find the distance of the line from the centre of the disc. |
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Answer» Solution :RADIUS of the DISC `=R` Radius of gyration `=K` `I=I_(G )+Md^(2)` `MK^(2)=(MR^(2))/(2)+Md^(2)impliesK^(2)=(R^(2))/(2)+d^(2)` `R^(2)=(R^(2))/(2)+d^(2)` [`:.K=R]` `(R^(2))/(2)=d^(2)implies` DISTANCE `d=(R )/(SQRT(2))` |
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| 3. |
A particle which is constrained to move along x - axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x)=kx + ax^(3). Here, k and a are positive constants. For x ge 0, the functional form of the potential energy U (x) of the particle is |
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Answer»
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| 4. |
A weight mg is suspended from the middle of a rope whose ends are at the same level. The rope is no longer horizontal. What is the minimum tension required to completely straighten the rope is: |
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Answer» W/2 |
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| 5. |
If the length of a cylinder is measured to be 4.28 cm with an error of 0.01 cm. The percentage error in the measured length is nearly |
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Answer» `0.4%` |
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| 6. |
A particle starts from origin att = 0 with a velocity 5.0 hati m/s and moves in x-y plane under action of a force which produces a constant acceleration of (3.0i+2.0j) m//s^(2) , (a) What is the y-coordinate of the particle at the instant its x-coordinate is 84m?(b) what is the speed of the particle at this time? |
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Answer» Solution :The POSITION of the PARTICLE is given by `r(t) = v_(0) t + (1)/(2) a t^(2)` `= 5.0 hat(i) t + (1//2)(3.0 hat(i) + 2.0hat(j))t^(2)` `= (5.0 t + 1.5 t^(2))hat(i) + 1.0 t^(2) hat(j)` Therefore, `x(t) = 5.0 t + 1.5t^(2)` `y(t) = +1.0 t^(2)` Given `x(t) = 84m, t = ?` `5.0 t + 1.5t^(2) = 84 rArr t = 6s` At `t = 6 s, y = 1.0 (6)^(2) = 36.0 m` Now, the VELOCITY `v = (d r)/(d t) = (5.0 + 3.0 t)hat(i) + 2.0 t hat(j)` At `t = 6s, v = 23.0 hat(i) + 12.0 hat(j)` speed `= |v| = sqrt(23^(2) + 12^(2)) = 26ms^(-1)` |
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| 7. |
Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density). |
| Answer» SOLUTION :DECREASES, | |
| 8. |
If the distance travelled by a freely falling body in the last second of its journey is equal to the distance travelled in the first 2s, the time of descent of the body is |
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Answer» 5 s |
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| 9. |
A: The angle of contact of a liquid decrease with increase in temperature , R: With increase in temperature the surface tension of liquid decreases. |
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Answer» Both A and R are true and R is the correct explanation of A |
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| 10. |
A solid is completely immersed in a liquid. The force exerted by the liquid on the solid will a. Incrase if it is pushed deeper inside the liquid b. Change if its orientation in changed c. Decrease if it is taken partically out of the liquid d. Be in the vertically upward direction |
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Answer» a, and C are correct |
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| 11. |
It is easier to pull a lawn roller than to push it because pulling |
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Answer» INVOLVES sliding friction |
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| 12. |
A force P acts on a particle so as to accelerate it from rest to certain velocity. If P is replaced then by Q which decelerates it to rest. |
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Answer» <P>P must be EQUAL to Q |
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| 13. |
A boat takes 4 hr upstream and 2 hr down the stream for covering the same distance. The ratio of velocity of boat to the water in river is |
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Answer» `1:3` |
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| 14. |
(A): The position-time graph of a body may be a straight line parallel to time axis. (R) : It is possible that position of a body does not change with time |
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Answer» Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A) |
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| 15. |
A body is subjected under three concurrent forces and it is in equilibrium . The resultant of any two forces is |
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Answer» (a) coplanar with the third FORCE |
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| 16. |
In net force on a rigid body is zero. Then, match the following table. |
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Answer» |
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| 17. |
A body of mass m is at rest on a smooth horizontal plane. A force F = kt is applied on the body making an angle alpha with the horizontal. In the equation of force, t is time and k is a constant.At the instant the object loses contact with the plane, how far will it move along the plane and what will be its velocity? |
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Answer» Solution :As F = kt, force INCREASES with time. Vertical component of the force = Fsin`alpha`. The body loses contact with the plane when the vertical force equals the weight of the body. i.e., Fsin`alpha` = mg . Let the time after which the body loses contact be `t_(0)`. `therefore"" Fsin alpha = kt_(0) sin alpha = "mg"` `therefore""t_(0) = (mg)/(k sin alpha)` Horizontal component of the force F = Fcos`alpha` = kt cos`alpha` Hence horizontal acceleration = `("ktcos"alpha)/(m) = (dv_(x))/(dt)` `therefore"" dv_(x)= ("ktcos"alpha)/(m)cdot dt ` Integrating , `v_(x)= ("KCOS"alpha)/(m) (t^(2))/(2) + A ` `""` [where A = integration constant] As the body starts from rest, at t = 0, v = 0. INSERTING this in equation (1), A = 0. `therefore"" v_(x) = (kcos alpha)/(2m) t^(2) ` If the displacement along the plane is s, then `therefore""v_(x) = (ds)/(dt) = ("kcos"alpha)/(2m) t^(2) or, ds = ("kcos"alpha)/(2m)t^(2)dt ` Integrating again, s = `("kcos"alpha)/(6m) t^(3) + B` `""` [where B = integration constant ] As at the initial moment, the displacement is zero, i.e., at t = 0, s = 0. From equation (3), we get B = 0. Inserting the VALUE of B in equation (3), `s = ("kcos"alpha)/(6m)t^(3)` Hence, displacement along the plane in time `t_(0)`, `s_(0) = ("kcos"alpha)/(6m) t_(0)^(3) = ("kcos"alpha)/(6m) ((mg)/("ksin"alpha))^(3)` or,`s_(0) = (m^(2)g^(3)cos alpha)/( 6k^(2)sin^(3) alpha) ` and CORRESPONDING velocity, `v_(0) = ("kcos"alpha)/(2m) ((mg)/("ksin"alpha))^(2)` `therefore"" v_(0) = (mg^(2) cos alpha)/(2"k"sin^(2) alpha )` 
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| 18. |
Four spring connect with mass as shown in figure. Find frequency of S.H.M. |
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Answer» `(1)/(2pi) sqrt((K)/(4M))`  These two CONNECTED in series. So `K.=(k_(1)k_(2))/(k_(1)+k_(2))=k` Now all are in parallel so `k_(eq)=4k` so FREQ. `=(1)/(2pi) sqrt((4k)/(M))` |
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| 19. |
When a body is proiected from a level ground the ration of its speed in the vertical and horizonal direction is 4 : 3. If the velocity of projection is u. the time after which the raion of the velocities in the vertical and horizontal directions are reversed is |
| Answer» Answer :A | |
| 20. |
(A) : A disc released from the top of a fixed inclined plane rolls down without slipping if the inclined plane is rough only (R ) : Friction does no work in pure rolling |
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Answer» Both 'A' and 'R' and true and 'R' is the correct explantation of 'A' |
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| 21. |
A bird is tossing (flying to and fro) between two cars moving towards each other on a straight road. One car has speed of 27 km h^(-1)while the other has the speed of 18 km h^(-1). The bird statrs moving from first car towards the other and is moving with the speed of 36 km h^(-1)when the two cars were separated by 36 km. The total distance covered by the bird is |
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Answer» 28.8 KM |
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| 22. |
Consider the earth to be a homogeneous sphere. Scientist A goes deep down in a mine and scientist B goes high up in a balloon. The gravitational field measured by |
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Answer» A goes on DECREASING and that by B goes on INCREASING |
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| 23. |
If tension in the string is increased from 1 KN to 4 KN,other factors remaining unchanged, the frequency of the second harmonic will |
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Answer» be halved |
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| 24. |
A large vessel with a small hole at the bottom is filled with water and kerosene. The height of the water column is 20 cm and that of the kerosene is 25 cm. The velocity withwhich water flows out of the hole is (density of kerosene=0.8 gm/cc) (Neglect viscous force) |
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Answer» `2 ms^(-1)` |
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| 25. |
Calculate the energy of the (i) Moon orbiting the Earth and (ii) Earth orbiting the Sun. |
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Answer» SOLUTION :Gravitational CONSTANT`G = 6.67 xx10^(11) "Nm"^(2)kg^(-2)` Mass of the Monn `M_(m) = 7.35 xx10^(22)` kg Mass of the Sun `M_(5) = 2xx10^(30) kg` The distance between the Moonand the centre of the Earth ` R_(m ) 3.84 xx10^(6) km ` The distance betweenthe Earth and thecentre of the sun `R_(e) = 150 xx10^(6) km` (i) The energy of the Moonorbating the Earth `E_(m) = -(GM_(E )M_(m))/(2R_(m))` ` = (-6.67 xx10^(11) xx 6.02 xx10^(24) xx 7.35 xx10^(22))/(2xx3.84 xx10^(5) xx10^(3))` ` = (-295.1275)/(7.68) xx 10^(-9) xx10^(46)` `E_(m) = - 38.42 xx10^(27) ` Joule `E_(m) = -3.842 xx10^(28)` joule (or) J (ii) The energy of the Earth orbiting the Sun ` E_(m) = (-GM_(s) M_(E))/(2R_(E))` `= (-6.67 xx10^(-11) xx 2xx 10^(30) xx 6.02 xx10^(-24))/(2xx150 xx10^(6))` ` = (-80.3068 xx10^(43))/(3xx10^(8)) = - 26.7689 xx10^(35)` `E_(E ) = - 26.77 xx10^(35)` joule `E_(E ) = - 2.677 xx10^(36) ` joule (or) J |
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| 26. |
A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2m height further, the magnitude of the force is, (g= 10 ms^(2)) |
| Answer» ANSWER :C | |
| 27. |
A film of water is formed between two straight parallel wires each 10 cm long and at separation 0.5 cm. The work required to increase 1 mm distance between them is x (24 xx 10^(-7)) joule. Find x Surface tension of water = 72xx 10^(-3) N/m. |
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Answer» |
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| 28. |
Two wires of different material have same Lengths and areas of cross-section. What is the ratio of their increase in length when forces applied are the same ? (Y_(1)=0.90xx10^(11)Nm^(-2),Y_(2)=3.60xx10^(11)Nm^(-2).) |
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Answer» `4:1` |
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| 29. |
The displacement of a particle from its mean position (in meter) is given by Y=0.2sin(10pit+1.5pi) cos (10p t+1.5 pi). The motion of the particle is |
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Answer» PERIODIC but not SHM |
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| 30. |
What is meant by S.I. unit ? |
| Answer» SOLUTION :The S.I STANDARDS do not vary with TIME as they are based on the properties of colours. It is a rationalised system of UNITS. It is a coherent system of units. S.I system of units have permanence and reproduceability. | |
| 31. |
A shell of mass 2 kg moving at a rate of 4 m//s suddenly explodes into two equal fragments. The fragments go in directions inclined with the original line of motion with equal velocities, If the explosion imparts 48 J of translational kinetic energy to the fragments, find the velocity and direction of each fragment. |
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Answer» Solution :Conservation of linear momentum of along `X` and `Y` axes gives `"mu"=m_(1)v_(1) costheta_(1)+m_(2)v_(2)costheta_(2)` i.e. `2u=1xxv_(1)costheta_(1)+1xxv_(2)costheta_(2)` `implies v_(1)costheta_(1)+v_(2)costheta_(2)=2u`…………i and `0=m_(1)v_(2)sintheta+m_(2)v_(2)sintheta_(2)` As `m_(1)=m_(2)=1KG` `implies v_(1)sintheta_(1)=v_(2)sintheta_(2)` As `v_(1)=v_(2)=v`(say) `implies theta_(1)=theta_(2)=theta` (say) from eqn i `implies v costheta+vcostheta=2u` `implies v costheta =u=4`...............II  Also conservation of ENERGY gives `1/2mu^(2)+48=1/2(m_(1))v^(2)+1/2(m_(2))v^(2)` `implies 1/2xx2xx(4)^(2)+48=1/2xx1v^(2)+1/2xx1v^(2)` or `v^(2)=64` `implies v=8m//s` `costheta=u/v=4/8=1/2` `implies theta=60^(@)` |
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| 32. |
For a satellite escape velocity is 11.2 km/s from the earth. If the satellite is launched at an angle of 60° with vertical, then find the escape speed. |
| Answer» Solution :`IMPLIES` Escape SPEED `V_e=sqrt((2G)/R_e)`, does not DEPEND on the ANGLE of projection and hence escape velocity is unchanged so it is 11.2 m/s. | |
| 33. |
Write a short note on vectorproduct between two vectors. |
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Answer» SOLUTION :Torque `vectau=vecrxxvecF`. Where `vecF` is force and `vecr` is POSITION vector of a PARTICLE. (i) ANGULAR momentum `vecL=vecrxxvecp` where `VECP` is the linear momentum. (iii) Linear velocity `vecv=vecomegaxxvecr` where `vecomega` is angular velocity. |
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| 34. |
Explain with defination of angular position, angular displacement and angular speed for the motion of rigid body. |
Answer» Solution :In the figure a rigid body is rotating about a fixed axis Oz, normal to the plane of the figure.  All the particles of a rigid body move along CIRCULAR path and the centres of all these CIRCLES fall on a axis of rotation. Circular path is perpendicular to the axis. Suppose P is any particle of rigid body. This moves with radius (perpendicular distance from axis) r and about centre C of a circle. The linear velocity of particle at point P is `vecv` and is in the direction of tangent at point P. Suppose at `t=0` time particle is at P and at time `t+Deltat` it is on P. `therefore` ANGULAR displacement in `Deltat` time `=anglePCP` =`Deltatheta` In this time interval average angular velocity of particle `ltomegagt =(Deltatheta)/(Deltat)` As `Deltat` is small and when `Deltat` is near to zero then `(Deltatheta)/(Deltat)` is CALLED instateneous angular velocity `therefore` Instanteneous angular velocity is known as angular velocity also `therefore` Angular velocity `omega=underset(Deltatrarr0)lim(Deltatheta)/(Deltat)` `therefore omega=(d theta)/(dt)` Defination of angular position : The angle made by the line joining a particle with the centre of its circular path at a given time, with a given fixed REFERENCE line is called the angular position of the particle at that time. It is denoted by `theta` and it is scalar. Angular displacement : The change in angular position of a particle of rigid body is called angular displacement. It is denoted by `Deltatheta`. It is a vector. Angular velocity : time rate of change of angular position of a rigid body is known as angular velocity or instanteneous angular velocity. It is denoted by `omega` and it is vector. |
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| 35. |
The x-t plot of a car is shown in the figure. The average velocity is zero in the time interval between 0 and t then t=_____sec. |
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Answer» 2 |
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| 36. |
A given mass of gas at a pressure of 0.75 m of mercury is compressed isothermally until its volume is reduced to 3/4th of its volume. It is then made to expand adiabatically to a volume 1/5 times greater than its original volume. Calculate the final temperature if its initial temperature was 290 K.(Y=1.4) |
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Answer» <P> Solution :`P _(2) = (p_(1) V_(1)//V _(2)) = (0.75 xx 4//3) = 1 m of HG. T _(2)= T (V _(2) //V_(3)) ^( gamma -1) . V _(2) = (3//4) V _(1), V _(3) =[V _(1) + (V _(1) //5)]T_(2) = 290 xx (5//8) ^(0.4) = 240.3 K` |
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| 37. |
A steel meter scale is to be ruled so that millimetre intervals are accurate within about 5xx10^-5 m at a certain temperature. The maximum temperture variation allowable during the ruling is (coefficient of linear expansion of steel =10xx10^-6K^-1 |
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Answer» `2^(@) C` |
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| 38. |
The mass of the earth is 6.00 xx 10^24 kg and that of the moon is 7.40 xx 10^22kg. The constant of gravitation G = 6.67 xx 10^(-11) N-m^2kg^(-2). The potential energy of the system is - 7.79 xx 10^28J. The mean distance between the earth and moon is |
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Answer» `3.80 XX 10^(8)` m |
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| 39. |
For small and equal values of d and h, which is greater g_n or g_d ? Give reason for your answer. |
| Answer» SOLUTION :For SMALL and equal VALUES of d and h, change in acceleration due to gravity above earth (at h) is more. Decrease in .g. value of `g_h`is more than decrease in ‘g. value at `g_d`. There FORE `g_d GT g_h` | |
| 40. |
Can the value of a component ofa vector be greater than the value of the vector itself ? Discuss thecaseof rectangular components in this context. |
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Answer» Solution :By the LAW of parallelogram of VECTORS, TWO components of a vector form two adjacent SIDES of aparallelogram with the vector as the diagonal. Since both the adjacent sides of a parallelogram can be longer than the diagonal, the value of the components can be greater than that of the vector. But the rectangular components ofa vector, being two adjacent sides of a rectangle, cannot be greater than the value of the vector, because the DIAGONALOF a rectangle is always longer than any of the sides. |
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| 41. |
A particle moving with a velocity equal to 0.4 ms^(-1) is subjected to an acceleration of 0.15 ms^(-2) for 2 seconds in a direction at right angles to the direction of motion . The magnitude of the final velocity is |
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Answer» `0.3 MS^(-1)` |
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| 43. |
A gas undergoes on a change of state during which 100 J of heat is supplied to it and it does 20 J of work. The system is brought back to its original state through a process during which 20 J of heat is released by the gas. What is the work done by the gas in the second process ? |
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Answer» Solution :In a CYCLIC process `DELTA U=0` `:. Delta Q=Delta W` `(100-20)=20+W_(2) :. W_(2) =+60J` |
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| 44. |
The percentage errors in the measurement of mass and velocity of a body are 2%and 3%respectively. What is the percentage error in the estimation of its kinetic energy? |
| Answer» SOLUTION :KINETIC ENERGY, `K=1/2mv^2DeltaK/Kxx100=(DELTAM)/mxx100+2xx(DELTAV)/vxx100=2+2xx3=8%` | |
| 45. |
Schematic diagram of a device is shown below If T_1=100^@C T_2=0^@C.Q_1=4200J, find the value of Q_2 |
| Answer» SOLUTION :`T_2/T_1=Q_2/Q_1therefore 273/373=Q_2/4200 rarrQ_2=3075 J` | |
| 46. |
A stone is dropped from top of a tower 300m high and at the same time another is projected vertically upward with a velocity of 100 "ms"^(-1). Find when and where the two stones meet. |
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Answer» 2S, 200.9m |
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| 47. |
How will themomentum of a body change if its K.E. is doubled ? |
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Answer» |
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| 48. |
If man increases his speed to 6m//s in the same direction , then at what angle with the vertical will he observethe rain falling ? |
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Answer» `37^(@)`  We can see that `TAN alpha =(3)/(4) rArr alpha =37^(@)` HENCE option (a) is correct . |
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| 49. |
The end B of the rod AB which makes angle theta with the floor is being pulled with a constant velocity v_(0) as shown. The length of the rod is l. At the instant when theta=37^(@) |
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Answer» velocity of end A is `(4)/(3)v_(0)` downwards |
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