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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 long cylinder is fitted with a top on its wall at its lower end. Keeping the top closed, the cylinder is filled with water. Now the cylinder is made to float by placing it over a cork and then the tap is opened. State what will happen. |
| Answer» Solution :DUE to the lateral pressure of water, it will be seen that a jet of water COMES out through the opening of the TAP. If the opening of the tap is in the horizontal direction, then the water-jet will come out horizontally. Due to an equal but OPPOSITE reaction of this thrust, the cylinder (floating on the cork) will MOVE in the opposite direction. | |
| 2. |
A car of mass m starts from rest and acquires a velocity along east, v - vhat(i) (v gt 0) in two seconds. Assuming the car moves with uniform acceleration, the force exerted on the car is |
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Answer» `(mv)/(2)` eastward and is EXERTED by the car engine. massof the car =m velocityalongeast`vec( V)= v hat(j )` `vhat(j ) = 0 + a (2)` ForceF = ma Force act on car in `(mv)/(2)` in east directionforceactingon systemis frictionalfroce `= (mv)/(2)` force by engingwill be interalforce |
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| 3. |
Find out the location of centre of mass of a uniform rod. |
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Answer» Solution :Suppose a rod of mass M and length L is lying along the X-axis with its ONE END at x=0 and the other at x = L Mass per unit length of the rod =`M/L` Hence, mass of the element PQ of length dx situated at a DISTANCE V from the origin is DM =`M/L` dx The coordinates of the element PQ are (x, 0,0). Therefore, x -coordinate of CM of the rod will be: `x_(CM) = (int_(0)^(L)xdm)/(intdm) =(int_(0)^(L)(x)(M/Ldx))/M = I/L int_(0)^(L) xdx = L/2, y_(cm) =(int ydm)/(int dm)=0` |
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| 5. |
A plank of mass 3m is placed on a rough inclined plane and a man of mass m walks down the board. If the coefficient of friction between the board and inclined plane is mu = 0.5, the minimum acceleration of does not slide is: |
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Answer» `8 m//s^(2)` |
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| 6. |
If the unit of power is million erg/minute, the unit of force is 1000 dyne and the unit of length is 5/3 cm then the unit of time is in second). |
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Answer» 10 |
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| 7. |
A block of mass m moving with speed v collides with another block of mass 2m at rest. The lighter block comes to rest after the collision. The coefficient of restitution is |
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Answer» `(1)/(2)` |
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| 8. |
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 byX_(cm)=(1)/(M) sum_(i)x_(i) (C) x-coordinate of a rigidbody of continuous mass distribution represented by |
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Answer» a and B are TRUE |
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| 9. |
Two atomic clocks are allowed to run for 100 years and they differ by 0.1s only. What will be the accuracy of standard atomic clock in measuring 1s? |
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Answer» SOLUTION :Given,time, t= 100 years `=(100xx365+24)` days (24 days ADDED DUE to 24 leap years in 100 years) t= 36524 days `= 36524xx86400s` Also, `""trianglet=0.1s` Fractional error in masurment of time `(trianglet)/(t)=(0.1)/(36524xx86400)=0.32xx10^(-10)` Accuracy shown by atomic clock is `0.32xx10^(-10)` part in 1s. |
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| 10. |
Whichoneamong a solid, liquidgasof samemassand atthe sametemperaturehas thegreatestinternall energy. Whichone hasleastand why ? |
| Answer» SOLUTION :A gashasgreatestvalueof INTERNAL energy.Beinga negativepotentialenergypotentialenergyof itsmoleculesis smallest.internalenergyof SOLIDIS maximumbecausenegativepotentialenergyof itsmoleculesis MAXIMUM. | |
| 11. |
Four pendulums A, B, C and D are suspended from the same elastic support as shown in figure. A and C are of the same length, while B is smaller than A and D is larger them A. If A is given a transverse displacement, |
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Answer» D will vibrate with maximum amplitude Since length of PENDULUM A and C is same and amplitude of A and C are maximum. Here we have assessed that support is completely elastic and hence no damped oscillation. |
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| 12. |
A body released from certain height reaches the surface of the earth after 't' seconds. If the same body is released from the same height on the moon, it reaches the surface after (g_("moon")= g_(e )"/"6) |
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Answer» `(t)/(6)S` |
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| 13. |
If the velocity of light is taken as the unit of velocity and year as the unit of time find the unit of length? |
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Answer» SOLUTION :Dimensional formula for velocity is `v=[LT^(-1)]` `n_(1)[L_(1)T_(1)^(-1)]=n_(2)[L_(2)T_(2)^(-1)]` `L_(2)=((n_(1))/(n_(2)))((T_(1))/(T_(2)))^(-1)L_(1)`..i Here `n_(1)` is the numerial value of velocity in the system where unit of length is `L_(1)` and that TIME is `T_(1)` and `n_(2)` is the numerical value of velocity in the system where unit of length is `L_(2)` and that of time is `T_(2)`. `n_(1)=3xx10^(8)""L_(1)=1m, ""T_(1)=1s` `n_(2)=1""L_(2)=?""T_(2)=1"year"=365.25xx24xx60xx60s` Substituting these values in equation (i) `L_(2)=((3xx10^(8))/1)(1/(365.25xx24xx60xx60))^(-1)`(1) `=3xx10^(8)xx365.25xx24xx60xx60=9.468xx10^(15)m` This unit of length is called a light year. |
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| 14. |
ML^(-1) T^(-2) represents |
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Answer» Stress |
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| 15. |
In which of the following the final image is erect? (A) simple microscope (B) Compound microscope ( C) Astronomical microscope (D) Galilean telescope |
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Answer» (A), (C ) and (D) |
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| 16. |
Can you visualize the growth and development of useful scientific tool ? |
| Answer» Solution :Yes, For EXAMPLE, E.O. Lawarance was the FIRST to bulid an operation cyclotron for the study of subatomic particles in the year 1930. This cyclotron had a circumferece of about 13 inches. In contrast, the latest cyclotron developed is a super conducting super collider (SSC), which has a cirucfrence of about 52 miles. The machine has cost about 10 billion dollars and can give particles 100 million times more energy. The researches from SSC have LED us to new methods of treating cancer and AIDS , and development of more advanced computers. | |
| 17. |
If the angle between two vectors of equal magnitude P is theta, the magnitude of the difference of the vectors is |
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Answer» `2P cos""(THETA)/(2)` |
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| 18. |
Velocity - time graph of a moving object is shown below. What is the acceleration of the object? Also draw displacement - time graph for the motion of the object? |
Answer» SOLUTION :The given graph shows that the velocity of the OBJECT is CONSTANT. That  is, the velocity of the object is not changing, so the ACCELERATION of the object is zero. Since the acceleration of an object is given by `a=("Change in velocity")/("Time taken")`  Displacement - time graph for the motion of the object is shown in the figure above. |
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| 19. |
A body of mass m is drawn up a smooth inclined plane (Figure 5.21) of height h and length l by means of a string passing over the vertex of the plane. From the other end of the string hangs a mass m'. Show that in order that m may just reach the top of the plane, m' msut be deteched after m has moved through a distance (m+m')/(m')(hl)/(h+l) |
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Answer» |
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| 20. |
A particle is executing SHM on a straight line A and B are two points at which its velocity is zero. The particle is crossing certain point X (AX < XB) at successive intervals of 1. 2s and 3. 6s with a speed of 4 m/s. The ratio (AX)/(XB) is: |
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Answer» `SQRT(2)/(sqrt(2)+1)` |
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| 21. |
Planets move round the sun due to |
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Answer» MUTUAL attraction and REPULSION between sun and planet |
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| 22. |
The period of oscillation of a simple pendulum of length L suspended from the ceiling of a vehicle which moves with out friction down in an inclined plane of inclination alpha is given by |
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Answer» `2pisqrt(L/(G cos ALPHA))` |
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| 23. |
A 1 kg ball is rotated in a vertical circle by using a string of length 0.1 m. If the tension in the string at the lowest point is 29.4 N, its angular velocity at that position is |
| Answer» Answer :B | |
| 24. |
Consider a system of two particles kept in two different orientation as shown. Let U be the potential energy of the system associated only with the force of interaction between the particles. If U_(1) and U_(2) correspond to case 1 and case 3 respectively then |
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Answer» `U_(1)=U_(2)` |
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| 25. |
A body is thrown with velocity (4i+3j) metre per second. Its maximum height is (g=10ms^(-2)) |
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Answer» 2.5 m |
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| 26. |
The required percentage increase in the energy of the earth.s satellite to shift it from an orbit of radius r to (3r)/2 |
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Answer» `16.67%` |
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| 27. |
(a) It is known that density rho of air decrease with height y(in metres ) as rho =rho_(0)e^(-y//y_(0)) where rho_(0)=1.25kgm^(-3) is the density at see level , and y_(0) is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmospere remiansa constant (isothermal conditions). Also assume that the value of g remains constant. (b) A large He balloon of volume 1425m^(3) is used to lift a payload of 400g. Assume that the balloon maintains constant radius as it rises. How high does it rise? [take y_(0)=800m and rho_(He) = 0.18kg//m^(3)]. |
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Answer» Solution :We know that the rate of decrease of density `rho` of air with height y is directly proportional to density `rho` i.e., `-(d rho)/(dy) prop rho or (dp)/(dy) =-krho` where `K` is a constant of proportional . Here -ve SIGN shows that `rho` decreases as y increase. `:. (d rho)/(rho)=-K dy` Intergrating it within constions , as `y` changes from `0` to `y` , density change from `rho_(0)` to `rho`, we have `int_(rho_0)^(rho) (d rho)/(rho) = -int_(0)^(y) K dy` `[ log_(e)rho]_(rho_0)^(rho) = -Ky` or `log_(e)rho-log_(e)rho_(0)=-Ky` or `log(rho)/(rho_0) = -Ky` or `(rho)/(rho_0) = e^(-Ky)` or `rho=rho_(0)e^(-Ky)` Here `K` is a constant . suppose `y_(0)` is a constant such that `K=1//y_(0)`, then `rho=rho_(0)e^(-y//y_(0))`. (b) The balloon will rise to a height, where its density becomes equal to the air at that height. Density of balloon, `rho = (MASS)/(VOLUME) = ("PAY load" + "mass of He")/("volume") = ((400+1425xx0.18)kg)/(1425m^(3)) = (656.5)/(1425) kg//m^(3)` As, `rho = rho_(0)e^(-y//y_(0)) :. (656.5)/(1425) = 1.25e^(-y//8000)` or,`e^(-y//8000) = (656.5)/(1425xx1.25) or e^(y//8000) = (1425xx1.25)/(656.5) = 2.7132` Taking log on both the sides `(y)/(8000) = log_(e)2.7132=2.3026 log_(10)2.7132 = 2.3026xx0.4335 ~~1` `y=8000xx1 = 8000m = 8km`. |
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| 28. |
The phenomenon of Brownian movement may be taken as evidence of |
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Answer» kinetic THEORY of MATTER |
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| 29. |
A mass .m. is attachedwith rod asshown in figture. This mass is slightlystretchedand releasedthen find the time period. (Y is Young.smodulus of rod, A is cross sectionalarea of rod, .l. is its length). |
Answer» SOLUTION : 
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| 30. |
A body oscillates with SHM according to the equation (in SI units), x = 5 cos [2pi t + pi//4]. At t = 1.5 s, calculate the (a) displacement, (b) speed and (c) acceleration of the body. |
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Answer» Solution :The ANGULAR frequency `OMEGA` of the body `= 2PI s^(-1)` and its time period T = 1 s. At t = 1.5 s (a) DISPLACEMENT = (5.0 m) cos `[(2pis^(-1)xx1.5s+pi//4]` `=(5.0 m) cos [(3pi + pi//4)]` `=05.0xx0.707m` `=-5.0xx0.707m` `=3.535m` (b) Using Eq. (14.9), the speed of the body `=-(5.0m)(2pis^(-1))sin[(2pis^(-1))sin[(2pis^(-1))xx1.5s+pi//4]` `=-(5.0m)(2pis^(-1))sin[(3pi+pi//4)]` `=10pixx0.707ms^(-1)` `=22ms^(-1)` (c) Using Eq.(14.10), the acceleration of the body `=-(2pis^(-1))^(2)XX` displacement `=-(2pis^(-1))^(2)xx(-3.535m)` `=140ms^(-2)` |
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| 31. |
What do you mean by Rigid body? |
| Answer» SOLUTION :A rigid body is the oine which maintains its definite and FIXED shape EVEN when an external FORCE acts on it. | |
| 32. |
All trignometric functions are periodic but only sine or cosine functions are used to define SHM. Why? |
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Answer» Pythgoras |
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| 33. |
A particle of mass 10 g moves along a circle of radius 6.4 cm with constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal 8xx10^(-4)J by the end of the second revolution after the beginning of the motion? |
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Answer» `0.15m//s^(2)` `r=6.4cm` `(1)/(2)MV^(2)=8xx10^(-4)J` `THEREFORE v^(2)=(16xx10^(-4))/(m)` `=(16xx10^(-4))/(0.01)` `therefore v^(2)=16xx10^(-2)(m//s)^(2)` `2a_(T)S=16xx10^(-2)` Distance covered in second REVOLUTION `S=2pir+2pir` `a_(T)=(16xx10^(-2))/(2S)` `=(16xx10^(-2))/(2xx4pir)` `=(2xx10^(-2))/(3.14xx6.4xx10^(-2))` `=(2)/(3.14xx6.4)` `=0.099~~0.1m//s^(2)` |
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| 34. |
Two sitar strings A and B playing the note 'Ga' are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz. What is the of B ? |
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Answer» Solution :Here original frequency of A is `f _(A) = 324 Hz` `IMPLIES f _(B) = 318 Hz OR f _(B) = 330 Hz` `(because f _(A)-f _(B) = pm 6 Hz)` After reducing trnsino in A, its frequency is suppose `f._(A)` where `f ._(A) lt f _(A) (because f prop sqrtT).` Now, we have3 beats per second between A and B. Hence original frequency of B must be 318 Hz (because if it is 330 Hz then when frequency of A will decrease from 324 Hz, its DIFFERENCE with 330 Hz will be more than 6 Hz and so more than 6 beats will be produced per second which WOULD be wrong as per the STATEMENT). So,` f _(B) = 318 Hz` |
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| 35. |
Two wheels of M.I, 3kg m^(2) and 5kg m^(2) are rotating at the rate of 600 rpm and 800 rpm respectively in the same direction. If the two are coupled so as to rotate with the same axis of rotation, the resultent speed of rotation will be (in rpm) |
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Answer» 725 |
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| 36. |
When is the workdoneby a force is positive and negative? |
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Answer» becomes CONCAVE towards A |
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| 37. |
Two particles of same mass are projected simultaneously with same speed 20 ms^(-1) from the top of a tower of height 20m. One is projected vertically upwards and other projected horizontally. The maximum height attained by centre of mass from the ground will be (g = 10 ms^(-2)) |
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Answer» `10 SQRT(2)m` |
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| 38. |
A 400 kg satellite is in a circular orbit of radius 2R_Eabout the Earth. How much energy is required to transfer it to a circular orbit of radius 4R_E ? |
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Answer» Solution :Initial `E_(i) = (-GM_Em)/(4R_(E)` While finally `E_(F) =-(GM_(E)m)/(8R_E)` The CHANGE in the total energy is `DeltaE=E_(f) - E_(i)` `DeltaE=(gmR_E)/8=(9.81xx400xx6.37xx10^6)/8=3.13xx10^9J` |
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| 39. |
A particle is subjected to SHM as given by equations x_1 = A_1 sin omega t and x_2 = A_2 sin (omega t + pi//3) .The maximum acceleration and amplitude of the resultant motion are a_("max") and A, respectively, then |
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Answer» `a_("MAX") = omega^2 SQRT(A_1^2 + A_2^2 + A_1A_2)` |
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| 40. |
While catching the ball, the cricketer takes his hand in backward direction |
| Answer» SOLUTION :Whiletakinghishandin backwarddirectionbetweenball and batthesebydecreaseforceactingon HAND . Hencehe cancatchballeasily. | |
| 41. |
Sketch y = x - a for different values of a . |
Answer» Solution : This implies , when INCREASING the value of a ,the line SHIFTS towards right SIDE . For a = vt , y = x - vt satisfies the differential equation . Though this function satisfies the differential equation , it is not FINITE for all VALUES of x and t . hence , it does not represent a wave. |
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| 42. |
A glass bulb can withstand a pressure of 2 atm. It constains air at a temperature of 303 K and at one atmospheric pressure. What is the temperature at which the bulb bursts ? (Neglect the expansion of the bulb) gamma = 1.4 |
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Answer» SOLUTION :`P _(1) = 1 atm, T _(1) = 303 K P _(2) = 2 atm` `((P _(2))/( P _(1))) ^( GAMMA -1) = ((T _(2))/(T_(1))) ^(gamma)` `(2) ^( 1.4-1) = ((T_(2))/( 303)) ^(1//4)` `0.4 log 2 = 1. 4 log T_(2) - 1.4 log 303` `T_(2) = 369 K` |
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| 43. |
One end of an ideal spring is fixed to a wall at origin 'O' and axis of spring is parallel to x-axis. A block of mass m = kg is attached to free end of the spring and it is performing SHM. Equation of position of the block in co-ordinate system shown in figure is x = 10+35 m (10 t) Here 't' is in second and 'x' in cm. Another block of mass M = 3kg moving towards the origin with velocity 30 cm/sec collides with the block per forming SHM at t= O and gets stick to it calculate (a) New amplitude of oscillations, (b) New equation for position of the combined body (c) Loss of energy during collision [Neglect friction] |
| Answer» Solution :(a) 3 CM ,(B) x=10 - 3 sin 5T , (c) 0.135 J | |
| 44. |
Compared to that of solids and liquids,value of the coefficient of volume expansion of gases |
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Answer» is same |
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| 45. |
A tunnel is dug along the diameter of the earth. The gravitational force on particle of mass m placed in the tunnel at a distance 'x' from the centre. |
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Answer» Solution :MASS of earth `M = (4)/(3)pi R^(3) rho` Mass of IMAGINARY sphere, having radius x, `M^(1) = (4)/(3)pi x^(3) rho` (or) `(M^(1))/(M) = (x^(3))/(R^(3))` `therefore` GRAVITATIONAL force on .m. due to `M_(1)` `F = (GM^(1)m)/(x^(2))` (or) `F = (Gm)/(x^(2))((x^(3))/(R^(3))M) = (GMmx)/(R^(3))` 
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| 46. |
A normal eye is not able to see objects closer than 25cm because |
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Answer» the focal length of the eye is 25CM |
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| 47. |
The distance between the centres of carbon and oxyen atoms in the carbonmonoxid gas molecule is 1.13Å. Locate the centre of amss of the gas molecule relative to the carbon atom. |
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Answer» Solution :CENTRE of MASS `x_(CM) =(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))` `=(13xx0+16xx1.13)/(12+16)` `=0.6457Å` |
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| 48. |
The density of a material in CGS system of units is '4 g/cm^3'. In a system of units in which unit of length is 10 cm and unit of mass is 100 g. The value of density of material will be |
| Answer» ANSWER :C | |
| 49. |
The temperature vs time graph for two materials A and B has been plotted when the energy is given to both at the same rate. Initially both the materials are in solid state (below their melting point ) |
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Answer» `L_("fA")=L_("fB")` |
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| 50. |
U(r)=A/(r^(2))-B/r, where is the distance from the centre of the force and A and B are positive constants. |
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Answer» The equilibrium distance is GIVEN by2A/B |
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