Saved Bookmarks
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 motor of an engine is rotating about its axis with angular velocity of 120 r.p.m. it comes to rest in 10s, after being switched off. Assuming constant decleration, calculate the number of revolutions made by it before coming to rest. |
|
Answer» Solution :Here, `omega_(0)=120 ` r.p.m. = 2 r.p.s. `= 4pi rad s^(-1)` `omega = 0`and `t=10 s "" omega = omega_(0)+at` `0=4pi+axx10` or `a=-0.4 PI rad s^(-2)` Also, the angle covered by the MOTOR, `THETA = omega_(0)t+(1)/(2)at^(2)` `because theta = 4pi xx10+(1)/(2)xx(-0.4 pi)xx10^(2)=40 pi-20pi` `= 20 pi` rad Hence, the number of revolutions completed, `n =(theta)/(2PI)=(20pi)/(2pi)=10` |
|
| 2. |
An electron of mass m is involved in head-on collision with an atom of mass M , initially at rest . As a result of the collision a certain amount of energy E is stored internally in the atom . What is the minimum initial velocity that the electron must have ? [Hint : Since energy is stored in the atom , the collision is essentially inelastic . mv = mv' + Mv'' and (1)/(2) mv^(2) = E + (1)/(2) mv^(2) + (1)/(2) Mv''^(2) . Solve for v' and apply the condition for real values of v'.) |
|
Answer» |
|
| 3. |
A stone is tied at end of string and whirled in horizontal plane with circular path of radius 20 cm. If its acceleration is 980 cms^(-2), then what will be its angular speed ? |
|
Answer» Solution :Centripetalacceleration `a_(c )= (V^(2))/( r )` `=r epsilon^(2) ` `epsilon = sqrt((a)/(r )) = sqrt((960)/( 20))` `epsilon = sqrt(49) =7 (rad)/( s )` |
|
| 4. |
The angle of inclination of an inclined plane is 60^(@). Coefficient of friction between 10kg body on it and its surface is 0.2, g=10 ms^(-2). The acceleration of the body down the plane in ms^(-2)is |
|
Answer» 5.667 |
|
| 5. |
The volume V of a liquid crossing through a tube is related to the area of cross-section A, velocity v and times t as V alpha A^(a) v^(b) t^(c ) which of the following is correct |
|
Answer» `a NE B ne C` |
|
| 6. |
At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at - 20 ^(@)C ? (atomic mass of Ar = 39.9 u, of He = 4.0 u). |
| Answer» SOLUTION :`2.52 XX 10^(3) K` | |
| 7. |
A physical quantity Q is found to depend on quantities x,y,z obeying relation Q = (x^(2)y^(3))/(z^(1)) . The percentage errors in x, y, and z are 2%, 3% and 1% respectively. Find the percentage error in Q. |
|
Answer» SOLUTION :Let Q = `(x^(2)y^(3))/(Z)` It is given, `(Deltax)/(x) = 2%, (Deltay)/(y)= 3%, (Deltaz)/(z) = 1%` `(DeltaQ)/(Q) = 2((Deltax)/(x)) + 3((Deltay)/(y)) + 1 ((Deltaz)/(z)) = 2(2%) + 3(3%) +1(1%)` `(DeltaQ)/(Q) = 4% + 9% + 1% = 14%` |
|
| 8. |
1000 drops of a liquid each of diameter 4 mm coasesce to form a single large drop. If surface tension of liquid is 35 dyne cm^(-1). Calculate the energy evolvedby the system in the process. |
|
Answer» SOLUTION :NUMBER of drops ` n =1000 impliesn^(1//3)=10impliesn^(2//3)=100` SURFACE tension of liquid `S=35"dyne"cm^(-1)` RADIUS of each small drop `r=2mm=0.2cm` ENERGY evolved in merging `W=4pir^(2)S[n-n^(2//3)]` `impliesW=4xx22/7xx(2xx10^(-1))^(2)35[1000-100]` `impliesW=(88xx35xx4xx10^(-2))/6[900]` `=15840` ergs `~~1.58xx10^(-3)J` |
|
| 9. |
The value of universal gas.Constant is 8.3 J/moK, the mean kinetic energy of 32gm of oxygen at -73^(@)C will be |
|
Answer» 480 J |
|
| 10. |
Find the ratio of escape velocities from the earth and the moon. (Mass of the earth = 6.0 xx 10^(24) kg, Radius of the earth = 6400 km, Mass of the moon = 7.4 xx 10^(22) kg Radius of the moon = 1740 km.) |
|
Answer» Solution :`V_(e) prop SQRT((M)/(R))` `(V_(e1))/(V_(E2)) = sqrt(((6.0xx 10^(24))/(7.4 XX 10^(22)))((1740)/(6400))) = (4.695)/(1)` |
|
| 11. |
Two light trollwys of masses m_(1) and m_(2) (= 3m_(1)) are connected by a spring. The spring is compressed and bound by a cord. The cord is burnt through, the spring expands and the trolleys move off in opposite direction. Assume that the coefficient of friction is the same for both trollwys. |
|
Answer» The relationship between the velocties `v_(1)` and `v_(2)` of the trolleys is `v_(1)//v_(2) = 3` |
|
| 12. |
A track (ACB) is in the shape of an arc of a circle. It is held fixed in vertical plane with its radius OA horizontal. A small block is released on the inner surface of the track from point A. It slides without friction and leaves the track at B. What should be value of q so that the block travels the largest horizontal distance by the time it returns to the horizontal plane passing through B? |
|
Answer» |
|
| 13. |
A stone dropped froma cliff the ground in 4 sec. Height of the cliff is |
|
Answer» 39.2 m |
|
| 14. |
One gets less hurt when one jumps from a height on sand than on hand floor. Why ? |
|
Answer» Solution :When a man jumps on any surface, the surface in contact with the man gets a LITTLE depressed. This depression is more in the case of a sandy surface than of a hand floor. The initial potential energy of the man is the same in both cases. So the kinetic energy just before HITTING the ground is also the same. Suppose , the mass of the man = m and he jumps from a height h. The average reaction force of the ground on him is R. He stops after depressing the ground by x. so his kinetic energy is used up in WORK done against the NET reaction force R-mg. As kinetic energy just before hitting the ground = potential energy at height = h =mgh , `(R -mg )x =mghor, R =(1+ h/x) mg` As x is more for a sandy surface , R is less and hence, the man is less hurt. |
|
| 15. |
The unit vector parallel to the resultant of the vectors A = 4i+3j + 6k and B = -i + 3j-8k is |
|
Answer» `(1)/(7) (3 hat(i) + 6hat(J) - 2hat(K))` |
|
| 16. |
A stone is projected from level ground with speed u and ann at angle theta with horizontal. Somehow the acceleration due to gravity (g) becomes double (that is 2g) immediately after the stone reaches the maximum height and remains same thereafter. Assume direction of acceleration due to gravity always vertically downwards. Q. The horizontal range of particle is |
|
Answer» `(3)/(4)(U^(2)sin2theta)/(g)` For remaining half, the time of flight is `t'=sqrt((2H)/((2g)))=sqrt((u^(2)sin^(2)theta)/(2g^(2)))=(t)/(sqrt(2))` `:.` Total time of flight is `t+t',T=t(1+(1)/(sqrt(2)))` `T=(usintheta)/(g)=(1+(1)/(sqrt(2)))` So horizontal range is u `costhetaxxT=(u^(2)sin2theta)/(2g)(1+(1)/(sqrt(2)))` |
|
| 17. |
A bomb is dropped from an aeroplanewhen it is directly above a target at an height of 1000m .The aeroplane is moving horizontally with a speed of 500 kmh^(-1). By how much distance will the bomb miss the target ? |
|
Answer» Solution :As the aeroplane ismoving horizontally , the initial DOWNWARD VELOCITY of the bomb , `u_(y) = 0` Also, y = 1000 m `g = 9 . 8 ms^(-2) , t = ?` `:. 1000 = 0 + (1)/(2) xx 9 . 8t^(2)` ` = 4 . 9 t^(2)` (or)`t = sqrt((1000)/(4 . 9)) = (100)/(7) s` `{:("Horizontal"),("velocity of the"),("aeroplane"):}} = 500 kmh^(-1)` `= 500 xx (5)/(18) ms^(-1)` `= (1250)/(9) ms^(-1)` `{:("DISTANCE by"),("Which the bomb"),("misses the target"):}} = {{:("Horizontal distance covered"),("by the bomb before it"),("hits the ground"):}` = Horizontal velocity `xx`time `= (1250)/(9) xx (100)/(7) = 1984. 13 m ` |
|
| 18. |
Explain elasticity using intermolecular forces. |
| Answer» Solution :When a deforming force is applied in a solid, its atoms are pulled apart or BIND TOGETHER. When the deforming force is removed, interatomic forces of attraction or repulsion restore the atoms to their EQUILIBRIUM positions. If a body REGAINS its original shape and size after theremoval of deforming force, it is said to be elastic and the PROPERTY is called elasticity. | |
| 19. |
Distinguish between cohesive and adhesive forces. |
| Answer» Solution :The force between the like MOLECULES which holds the liquid TOGETHER is called .cohesive force.. When the liquid is in CONTACT with a solid, the molecules of the these solid and liquid will experience an ATTRACTIVE force which is called .adhesive force.. | |
| 20. |
Express a unit of electrical energy in terms of joule. |
|
Answer» Solution :1 ELECTRICAL unit = 1 kWh `=1xx(10^(3)W)xx(3600s)` 1 electrical unit = `3600xx10^(3)WS` 1 electrical unit = `3.6xx10^(6)J` `1kWh=3.6xx10^(6)J` |
|
| 21. |
A block of mass 500 g is at rest on a horizontal table. What steady force is required to give the block a velocity of 200 cms^(-1)in 4 s? |
|
Answer» SOLUTION :Use`F= ma ` ` a = (V - u)/(t ) = (200 - 0)/(4) = 50 cm//s^2` ` F = 500 xx 50 =25,000 `dyne |
|
| 22. |
For liquids, the graph between temperature and coefficient of viscosity is |
|
Answer» STRAIGHT line |
|
| 23. |
Five persons A, B, C, D & E are pulling a cart of mass 100kg on a smooth surface and cart is moving with acceleration 3 m//s^(2)in east direction. When person 'A' stops pulling, it moves with acceleration Im/s in the west direction. When person 'B' stops pulling, it moves with acceleration 24m//s^(-2)in the north direction. The magnitude of acceleration of the cart when only A & B pull the cart keeping their directions same as the old direction is |
|
Answer» ` 26 m//s^2 ` |
|
| 24. |
A small source of sound vibrating at frequenc 500 Hz is rotated in a circle of radius (100//pi)cmat a constant angular speed of 5.0 revolutions per second . The speed of sound in air is 330 m .s^(-1) . An observer (A) is situated at a great distance on a straight lineperpendicular to the plane of the circle , through its centre . Another observer (B) is at rest at a great distance from the centre of the circle byt nearly in the same plane . After something the source of sound comes to rest after reaching the centre of the circle . At that time , another observer (C) moves towards the source with a constant speed of 20 m .s^(-1) , along the radial line to the centreThe change in the frequency of the source heard by C will be |
|
Answer» 0.06 |
|
| 25. |
A small source of sound vibrating at frequenc 500 Hz is rotated in a circle of radius (100//pi)cmat a constant angular speed of 5.0 revolutions per second . The speed of sound in air is 330 m .s^(-1) . An observer (A) is situated at a great distance on a straight lineperpendicular to the plane of the circle , through its centre . Another observer (B) is at rest at a great distance from the centre of the circle byt nearly in the same plane . After something the source of sound comes to rest after reaching the centre of the circle . At that time , another observer (C) moves towards the source with a constant speed of 20 m .s^(-1) , along the radial line to the centreThe mininum and the maximum values of the apparent frequency heard by B will be |
|
Answer» 455 HZ and 535 Hz |
|
| 26. |
A small source of sound vibrating at frequenc 500 Hz is rotated in a circle of radius (100//pi)cmat a constant angular speed of 5.0 revolutions per second . The speed of sound in air is 330 m .s^(-1) . An observer (A) is situated at a great distance on a straight lineperpendicular to the plane of the circle , through its centre . Another observer (B) is at rest at a great distance from the centre of the circle byt nearly in the same plane . After something the source of sound comes to rest after reaching the centre of the circle . At that time , another observer (C) moves towards the source with a constant speed of 20 m .s^(-1) , along the radial line to the centreThe apparent frequency of the source heard by A will be |
|
Answer» GREATER then 500 Hz |
|
| 27. |
A particle of mass 0.1 kg executes SHM under a force F=-10x(N). Speed of particle at mean position is 6m/s. Find its amplitude of oscillations . |
|
Answer» SOLUTION :From conservationfo MECHANICAL ENERGY `1/2kA^(2)=1/2mv^(2)` or `A=vsqrt(m/k)=6sqrt(0.1/10)=6/10=0.6m` |
|
| 29. |
The minimum and maximum distance of a satellite from the centre of the earth 2R and 4R respectively. Where R is the radius of the earth and M is the mass of earth. Its maximum speed is |
|
Answer» `sqrt((GM)/(6R))` |
|
| 30. |
How will you calculate the length of the S period. |
|
Answer» Solution :According to Newton.s second LAW `F= (dp)/(dt)` i.e. `dp = F dt` INTEGRATE it over a time interval from `t_1 ` to `t_f` `int_i^f dp = int_(t_1)^(t_f)F dt` `P_f - P_i = int_(t_1)^(t_2) F. dt` `P_1 to` initial momentum of the object at `t_1` `P_f to ` Final momentum of the object at `t_f` `P_f - P_i = DELTAP=` CHANGE in momentum during the time interval `Deltat. int_(t_1)^(t_f) F dt` = J is called the impulse. If the force is constant over the time interval `Deltat`, then `int_(t_1)^(t_f) F dt = F int_(t_1)^(t_f) dt = F(t_f - t_i) = F Deltat` `J = F Deltat= Deltap` Hence the proof. |
|
| 31. |
A spring balance has a scale that reads 0 to 20 kg. The length of the scale is 10 cm. A body suspended from this balance, when displaced and released, oscillates, with period of (pis)/10. What is the mass of the body |
|
Answer» 4.9 kg |
|
| 32. |
Calculate the rate at which water flows thorugh a capillary tube of length 0.5 m with an internal radius of 1 mm. Coefficient of viscosity is 1.3 xx 10^(-3) kg//m^(-s). The pressure head is 20 cm of water . |
|
Answer» Solution :`P= (20-10^(-2) xx 1000 xx 9.8) Nm^(-2) , r = 10^(-3)m , l= 0.5 m` `ETA = 1.3 xx 10^(-3)Kg//m^(-s) ,` From Poiseuill.s formula `V = (PI Pr^4)/(8 etal)` `=(3.14xx20 xx 10^(-2) xx 1000 xx 9.8 xx (10^(-3))^4)/(8xx1.3 xx 0.5) = 1.2 xx 10^(-6) m^3 s^(-1)`. |
|
| 33. |
Inner and outer radii of a spool are r and R I, respectively. A thread is wound over its inner surface and spool is placed over a rough horizontal surface. Thread is pulled by a force F as shown in Fig. In case of pure rolling, which of the following statements are false? |
|
Answer» Thread unwinds, spool rotates anticlockwise and friction acts leftwards. About the instantaneous axis if rotation, MOMENT produced by `F` is clockwise. Therefore, the spool rotates clockwise. In that case, acceleration will be RIGHTWARD and thread will wind. If rotational motion of the spool is considered about its own, then the resultant moment on it must be clockwise. But moment produced by the force is anticlockwise and its magnitude is equal to `F_(1)//r`. Hence, moment produced by the friction (about its own axis) must be clockwise and its magnitude must be greater than `Fr`. It is possible only when friction acts leftwards. Therefore, Option (b) is correct |
|
| 34. |
A car accelerates from rest at a constant rate alpha for sometime after which it decelerates at a constant rate beta to come to rest.If the total time lapsed it t seconds,find (i) the maximum velocity reached and (ii) the total distance travelled. |
|
Answer» `((ALPHA^(2)+BETA^(2))/(alpha beta))t` |
|
| 35. |
A thin rotates about an axis passing through its centre and perpendicular to its plane with a constant angularvelocity .omega. I is the moment of inertia of that discand .L. is its angular momentum about the given axis. Then rotational kinetic energy of the disc .E. is(A) E alpha L^(2)(B) E alpha L^(-2)(C ) E alpha I(D) E alpha I^(-1) (if L constant) |
|
Answer» A & C are correct |
|
| 36. |
A vanis moving with a speed of 72 Kmph on a level road, where the coefficient of friction between tyres and road is 0.5. The minimum radius of curvature, the road must have, for safe driving of van is |
|
Answer» 80 m |
|
| 37. |
A car of mass 1200 kg is traveling around a circular path of radius 300 m with a constant speed of 15m/s. calculate its angular momentum. |
|
Answer» Solution :ANGULAR momentum, `L=rmv` `=300xx1200xx15` `L=54xx10^(5)kgm^(2)s^(-1)` |
|
| 38. |
Two rigid bodies have same angular momentum about their axes of symmetry. If same torque is applied about their axes, the ratio of times after which they will be stopped is |
|
Answer» `1:1` |
|
| 39. |
The acceleration due to gravity on moon is 1/6th of that on the earth. If the length of the seconds pendulum is 96 m on the earth, what is its length on the moon ? |
|
Answer» Solution :For SECONDS pendulum , time period is 2S , `T= 2pi sqrt(l/g)` for SIMPLE pendulum time remains CONSTANT in both the planets , `l/g `= constant `therefore (l_m)/(g_m) = (l_e)/(g_e) ` (or) `(l_m)/((g_3)/(6)) = (96)/(g_e) = lm = 16 ` CM. |
|
| 40. |
A cart of mass 500 kg is taken along a polished inclined surface to the top of a platform which is at a height of 2m above the ground level. Find the work done by the external force in moving the cart to the top. |
|
Answer» |
|
| 41. |
Can a vector be zeroif any of itscomponents is not zero? |
|
Answer» Solution :In TERMS of componentsin three dimestions, `vecA = hati A_(x) + hatj A_(y) + HATK A_(Z)`with `A = sqrt(A_(x)^(2)+ A_(y)^(2) + A_(z)^(2))` so if any of `A_(x),A_(y) and A_(z)`is not zero(or NULL)all itscomponentsmustvanish. |
|
| 42. |
An eraser weighing 2N is pressed against the black board with a force of 5N. If the coefficient of friction is 0.4. How much force parallel to the black board is required to slide the eraser upwards |
| Answer» ANSWER :C | |
| 43. |
Which of the following does not hold when two particles of masses m_(1) and m_(2) undergo elastic collision |
|
Answer» when `m_(1)=m_(2)` and `m_(2)` is stationary, there is maximum TRANSFER to kinetic energy in head an collision |
|
| 44. |
A uniform disc of mass m and radius R with a spring connected (spring constant k) to its centre stand in static equilibrium on a very rough incline plane. Now a constant moment of couple M_(0), is applied at the centre as shown such that the disc rolls down incline. The maximum distance the disc's centre of mass will go down as measured from initial position is |
|
Answer» `(M_(o))/(RK)+(2mgsintheta)/(3K)` |
|
| 45. |
(A) : Out of the measurements A= 10.00 and B = 10.000, B is more accurate. (R) : Percentage error in B is less than the percentage error in A. |
|
Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
|
| 46. |
In an experiment if the measured values of a physical quantity have a high degree of closeness to the true value. These measurements are said to be |
|
Answer» both PRECISE and ACCURATE |
|
| 47. |
An important milestone in the evolution of the universe just after the Big Bang is the Planck time t_(p), the value of which depends on three fundamental constants-speed of light in vacuum c, Gravitational constant G and Planck's constant h. Then, t_(p) prop |
|
Answer» `Ghc^(5)` `:. [(Gh)/(c^(5))]^(1//2)=[ T]` HENCE, `t_(p) prop ((Gh)/(c^(5)))^(1//2) :. [t_(p)]= [T]` |
|
| 48. |
The ratio of the time periods of two satellites at heights h_(1), h_(2) from the surface of the earth. |
|
Answer» `((h_(1))/(h_(2)))^(3//2)` |
|
| 49. |
How much heat energy is gained when 5kg of water at 20^(@)C is brought to its boiling point (Specific heat of water = 4.2 kJ kg^(-1^(@))C^(-1) |
|
Answer» Solution :`Q _(2) = mcd T + mL = 2000 xx 1 xx 25 + 2000 xx 80 = 2.1 xx 10 ^(5) cal. Q _(1) = Q _(2) T _(1) //T_(2)` `Q _(1) = 2.1 xx 10 ^(5) xx 298 // 258 = 2.42 xx 10 ^(5) cal. ` Energy needed `= Q _(1) - Q _(2) = 3.25 xx 10 ^(4) cal= 3.25 xx 10 ^(4) xx 4.2 //3600 = 37.92 Wh.` |
|