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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 train is moving with constant acceleration. When the ends of a train pass by a signal their speeds are u and v respectively. Calculate the speed of the midpoint of the train while passing the signal. |
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Answer» SOLUTION :Suppose length of a train is l. Now, speeds of both ends are different, so we can say that the speed changes from u to v in TRAVELLING DISTANCE l. Hence, using the equation of motion `v ^(2) - v_(0) ^(2) = 2ax,` putting `v=v, v _(0)=u and x = l,` we get `v^(2) -u ^(2) =2AL ""...(i)` Now, suppose the speed of the midoint of a train, while passing the signal is v.. This means that the speed changes from u to v. in travelling a distance `1/2.` `therefore v .^(2) -u ^(2) =2a ((1)/(2)) = al ...(ii)` Taking rato of (i) and (ii),` (v ^(2) - u ^(2))/(v^(.2) - u ^(2))` `therefore 2 (v^(.2) - u ^(2)) = v ^(2) -u ^(2)` `therefore v^(.2) - u ^(2) = (v ^(2) - u ^(2))/( 2)` `therefore v ^(.2) = (v ^(2)- u ^(2) )/(2) + u ^(2)` `therefore v ^(.2) = (v ^(2) + u^(2))/( 2)` `therefore v.= sqrt ((v ^(2) + u ^(2))/(2))` |
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| 2. |
A car moving with a speed of 50 km/hr can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr the minimum stopping distance is |
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Answer» 12 m |
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| 3. |
A particle of mass m is placed inside a spherical shell, away from its centre. The mass of the shell is M. |
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Answer» The PARTICLE will move towards the centre |
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| 4. |
If the distance between the earth and moon were doubled, then the gravitational force between them will be |
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Answer» Halved |
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| 5. |
Two perpendicular vectors oversettoA and oversettoBare of magnitudes 12 and 5 respectively. The magnitude of their vector sum is |
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Answer» 17 |
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| 6. |
Explain what is stress ? Write its unit and dimensional formula. |
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Answer» Solution :When forces are applied on a body in such a manner that the body is still in static equilibrium it is deformed to a small or large extent. Deformation depending upon the nature of the material of the body and the magnitude of the deforming force. The deformation may not be NOTICEABLE VISUALLY in many materials but it is there when a body is subjected to a deforming force, a restoring force is developed in the body. This restoring force is equal in magnitude but OPPOSITE in the DIRECTION to the applied force. The restoring force per UNIT area is known as stress. If F is the force applied and A is the area of cross section of the body, Magnitude of stress `=F/A` Its SI unit is `Nm^(-2)` or Pascal (Pa). Its dimensional formula is (`ML^(-1)T^(-2)`). |
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| 7. |
Where does the centre of buoyancy lie in a displaced liquid. |
| Answer» SOLUTION :at the CENTRE of GRAVITY | |
| 8. |
A person crossing a road with a certain velocity due north sees a car moving towards east The relative velocity of the car w.r.t the person is sqrt(2)times that fo the velocity of the persons. The angle made by the relative velocity with the east is |
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Answer» `30^(@)` |
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| 9. |
A rectangular metal slab of mass 33.333 g has its length 8.0 cm, breadth 5.0 cm and thickness 1mm. The mass is measured with accuracy up to 1 mg with a senstitive balance. The length and breadth are measured with a vernier calipers having a least count of 0.01 cm. The thickness is measured with a screwgauge of least count 0.01 mm. Calculate the percentage accuracy in density from above measurements. |
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Answer» 0.13 |
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| 10. |
If the earth were a homogeneous sphere of radius R and a straight hole bored in it through its centre, show that a particle dropped into the hole will execute shm and find out its period. |
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Answer» Solution :Acceleration due to gravity at the surface of the earth `g= (GM )/R^2` M-Mass of the earth and R-radius G-Gravitational constant If `rho`is the DENSITY of earth then `M = 4/3piR^2rho` `:. g = 4/3 piRrhoG ""...(1)` At a depth h below the surface of the earth `g.=4/3 pi(R-h)rhoG""...(2)` Dividing eq. (2) by eq. (1) `(g.)/g=(4/3pi(R-h)rhoG)/(4/3piRrhoG)` `g.=g""((R-h))/R` Let X be the distance of the particle from the centre of the earth. Then R-h = x Then acceleration `=(d^2x)/(dt^2) =g.` i.e.`(d^2x)/(dt^2) = -(gx)/R` Negative sign is put to show that acceleration acts towards the centre of the earth which is opposite to the direction of increasing x. `(d^2x)/(dt^2)+g/Rx=0` `omega^2=g/R,omega=sqrt(g/R)` Hence the particle DROPPED into the hole execute SHM Priod `T=(2pi)/omega=2pisqrt(R/g)` |
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| 11. |
Two blocks are connected over a massless pulley as shown in figure. The mass of block A is 10 kg and the coefficient of kinetic friction is 0.2 Block A sliders down the incline at constant speed. The mass of block B in kg is |
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Answer» 5.4 `m_(A)g sin 30^(@)=mu m_(A)g cos 30^(@)+m_(B)g` `therefore"" m_(B)=((m_(A))/(2))-((mu m_(A)sqrt(3))/(2))` `=10[(1)/(2)-0.2xx(sqrt(3))/(2)]=3.3 kg ` |
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| 12. |
A tank with vertical wall is mounted so that its base is at a height H above the horizontal ground.The tank is filled with water to a depth h. A hole is punched in the side wall of the tank at a depth X below the water surface. Tohave maximum range of the emerging stream, the value of X is |
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Answer» `(H+h)/4` |
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| 13. |
The centre of mass of a non uniform rod of length L whose mass per unit length lambda = (Kx^(2))/(L), where L is a constant and x is the distance from one end is |
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Answer» 3L/4 |
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| 14. |
From a high tower at time t=0, one stone is dropped from rest and simultaneously another stone is projected vertically up with an initial velocity the graph between distance betweeen the particles and time before either hits the ground is : |
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Answer»
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| 15. |
A ball is thrown down with an initial velocity v_0 from a height of 10m. The ball lost 50% of its energy on coming in contact with the ground , and then bouncing back up tothe same height. Find v_0 of the ball. If theinitial velocity was upward instead of downward, up to what height would the ball have bounced ? |
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Answer» |
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| 16. |
A bullet is dropped from some height, when another bullet is fired horizontally from the same height. They will hit the ground |
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Answer» DEPENDS UPON MASS of bullet |
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| 17. |
Cooking gas containers are kept in a lorry moving with uniform speed.The temperature of the gas molecules inside will |
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Answer» INCREASE |
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| 18. |
A mercury manometer consists of two unequal arms of equal cross-section and length 100 cm and 50 cm. The two open ends are sealed with air in the tube at a pressure of 80 cm of mercury. Some amount of mercury is now introduced in the manometer through the stopcock connected to it. If mercury rises in the shorter tube to a length 10 cm in steady state, find the length of the mercury column risen in the longer tube. |
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Answer» Solution :For shorter ARM , INITIAL volume of air = A `times` 50 cm When MERCURY is introduced in the tube, the volume of air `=A times 40cm^(3)` Let `P_(1)` be the pressure of the air in shorter arm, then by PV = constant, we have `80 times (A times50)=P_(1) times (A times 40)` `THEREFORE P_(1)=100 cm` For longer arm, Initial volume of air `=A times 100 cm^(3)` Let `l_(0)` be the length of mercury, then volume of air `=A times(100-l_(0))` Let `P_(2)` be the pressure of the arm in the longer air, then by Pascal.s law `P_(A)=P_(B)` `l_(0)+P_(2)=10+P_(1)` `P_(2)=(10+P_(1))-l_(0)=(10+100)-l_(0)=(110-l_(0))` cm Now by PV = constant, we have `(110-l_(0)) times A(100-l_(0))=80 times (A times 100)" (or) "(110-l_(0))(100-l_(0))=8000` `l_(0)^(2)=210l_(0)+3000=0, l_(0)=15.5cm` 
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| 19. |
The instantaneous angular position of a point on a rotating wheel is given by the equation theta_((t))=2t^(3)-6t^(2). The torque on the wheel becomes zero at ………. s |
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Answer» 1 Now `theta_((t))=2t^(3)-6t^(2)` `therefore` Angular velocity `omega=(d theta_((t)))/(dt^(2))=6t^(2)-12t` and angular accelertion `alpha=(domega)/(dt)=(d)/(dt)(6t^(2)-12t)` `alpha=12t-12` putting `t=1 s, alpha=12-2=0` `therefore t=1 sec` |
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| 20. |
Ther vernier constant of a travelling microscope is 0.001 cm. If 49 main scale divisions coincide with 50 vermier scale divisions, then the value of IMSD is |
| Answer» Answer :B | |
| 21. |
The rms velocity of a gas at a given temperature is 300 m/s .What will be the rms velocity of a gas having twice the molecular weight and half the temperature in J |
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Answer» 300 m/s |
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| 22. |
When 'n' number of particles of masses m,2m,3m,…..nm are at distances x_(1)=1,x_(2)=2,x_(3)=3,…..x_(n)=n units respectively from origin on the x-axis, then find the distance of centre of mass of the system from origin. |
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Answer» SOLUTION :`x_(CM)=(m(1)+2m(2)+3M(3)+……+(NM)N)/(m+2m+3m+….+nm)` `x_(cm)=(m(1^(2)+2^(2)+3^(2)+…..+n^(2)))/(m(1+2+3+….+n))` `X_(cm)=(((n(n+1)(2n+1))/(6)))/(((n(n+1))/(2)))=(2n+1)/(3)` |
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| 23. |
Which one of the following graphs represents the behaviour of an ideal gas. |
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Answer»
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| 24. |
Water drops fall at regular intervals form a hole at the bottom of a vessel placed at a high level. The ninth drop is about to fall when the first drop just falls on the floor after being in the air for 2 second. The distance between the 3rd and the 5 th drop at the instant is : (Taking g = 10 m//s^(2)) |
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Answer» `(40)/(9) m` `H = 20 m` Time interval between two drops `= (2)/(8) = (1)/(2) sec`.  Let the distance TRAVELLED by 3 rd drops is `s_(3) = (1)/(2) xx g xx (1.5)^(2)` Distance travelled by 5th drops `s_(5) = (1)/(2) xx g (1)^(2)` `Delta s = s_(3) - s_(5)` |
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| 25. |
Neglecting end corrections, determine the lengths of the three shortest closed pipes and the shortest open pipes, which would resound to a note of frequency 512 Hz when air in the pipes is at a temperature of 298 K.[Velocityof sound at 273 K = 330 m * s^(-1) ] . |
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Answer» |
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| 26. |
A Black metal foil receives radiation of power P from a hot sphere at absolute temperature T, kept at a distance d. If the temperature is doubled and distance is halved, then Power will be |
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Answer» 64P |
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| 27. |
A bodystarting from rest has an acceleration of 20 ms^(-2)the distance travelled by it in the sixth second is ... |
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Answer» Solution :Distance TRAVELLED in nth second ,u = 0 `S_a = u + 1/2 a (2 n- 1) ` `S_6= 0+ 1/2 xx 20 xx (2 xx 6) , S_6= 110 m ` |
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| 28. |
A light rod of length 2 m is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of Steel the wires is made of steel and is of cross-section 10^(-3)m^(2) and the other wire is of brass of cross-section 2xx 10^(-3)m^(2). Find out the position along the rod at which a weight may be hung to produce: (i) equal stress in both wires (ii) equal strains in both wires Young.s modulus of brass = 1 xx 10^(11) N//m^(2) Young.s modulus of steel =2xx10^(11)N//m^(2) |
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Answer» Solution :SUPPOSE a, and a, are the cross-sectional areas, and `Y_(1) and Y_(2)` are the Young.s moduli of steel and brass wire respectively. Let `T_(1) and T_(2)` are tensions in the steel and brass wires respectively. Let x is distance of the position of the hanging WEIGHT from the steel wire. (i) First case : For equal stress in both wires, we have `(T_(1))/(a_(1))=(T_(2))/(a_(2) or (T(1))/(10^(-3))=(T_(2))/(2xx10^(-3)) (or) T_(2)=2T_(1)...(i)` As the whole system is in equilibrium, so `sum bar tau=0`. TAKING moment of all the forces ACTING on the rod about C, we have `T_(1)x-T_(2)(2-x)=0 "...(ii)` Solving equation(i) and (ii) we get `x=(4)/(3)m` (ii) Second case : For equal strains in both the wires`e_(1)=e_(2)` `(T_(1)l)/(a_(1)y_(1))=(T_(2)l)/a_(2)Y_(2)) or (T_(1))/(10^(-3)xx2xx10^(12))=(T_(2))/(2xx10^(-3)xx10^(11) (or) T_(1)=T_(2).... (ii)` From equation (ii) and (III) , we get x=1m |
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| 29. |
The minimum force required to start pushing a body up a rough (frictional coefficient mu) inclined plane is F_(1) while the minimum force needed to prevent it from sliding down is F_(2). If the inclined plane makes an angle theta with the horizontal such that tan theta = 2mu, then the ratio (F_(1))/(F_(2)) is |
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Answer» 4 `F_("down")=mg(sin theta-mu_(k)cos theta)`. |
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| 30. |
Three identical solid spheres moves down three incline A, B and C-all of the same dimensions . A is without friction, the friction between B and sphere is sufficient to cause rolling without slipping, the friction between C and sphere causes rolling with slipping. The kinetic energies A,B,C at the bottom of the inclines are E_(A), E_(B), E_(C ) |
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Answer» `E_(A)=E_(B)-E_(C )` |
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| 31. |
The period of revolution of a surface satellite around a planet of radius R is T. The period of another satellite around the same planet in an orbit of radius 3R is |
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Answer» T |
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| 32. |
A physical quantity x is given by x = (a^(2)b^(3))/(csqrtd) . If the percentage errors of measurement in a, b, c and d are 4%, 2%, 3% and 1% respectively, then calulate the percentage error in the calculation of x. |
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Answer» Solution :Given x = `(a^(2)b^(3))/(csqrtd)` The PERCENTAGE ERROR in x is given by `(Deltax)/(x) xx 100 = 2 (Deltaa)/(a) xx 100 + 3(Deltab)/(b) xx 100 + (DELTAC)/(c) xx 100 + 1/2 (Deltad)/(d) xx 100` = `(2xx4%)+(3xx2%)+(1xx3%)+(1//2xx1%)` = `8% + 6% + 3% + 0.5%` The percentage error is x = `17.5%` |
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| 33. |
An astronaut of mass m is working on a satellite orbiting the earth at a distance h from the earth.s surface . The radius of the earth is R, while its mass is M. The gravitational pull F_G on the astronaut is : |
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Answer» Zero since astronaut feels weightless |
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| 34. |
The bon of a simple pendulum of mass 100 g is oscillating with a time period of 1.42 s. If the bob is replaced by another bob of mass 150 g but of same radius what will be the newtime period of osicillation? |
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Answer» 2.24 sec |
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| 35. |
A batsman deflects a ball by an angle of 60° without changing its initial speed of 20 ms^(-1). what is the impulse imparted to the ball if its mass is 0.15 kg? |
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Answer» 9Js |
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| 36. |
Represents two simple harmonic motions. The parameter which has different values in the two motions is |
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Answer» AMPLITUDE |
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| 37. |
The diameter of a wire is measured as 0.25 cm using a screw gauge of least count 0.005 cm. The percentage error is |
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Answer» 0.01 |
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| 38. |
A particle traveling horizontally with speed u collides and sticks with a particle of equal mass hanging at rest at the end of a light inextensible string of length 2l. If the string rotates through an angle of 60^(@) before its velocity becomes zero, then u is |
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Answer» `SQRT(4GL)` |
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| 39. |
Define gravitational potential energy. |
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Answer» Solution :Potential energy of a body at a point in a gravitational field is the work done by an external agent in moving the body from infinity to that point. ` VEC(E)_("total")= vec(E)_(1)+vec(E)_(2)+vec(E)_(n)= - (Gm_(1))/(r_(1)^(2))hat(R)_(1)-....(Gm_(n))/(r_(n)^(2))hat(r)_(n)` `vec(E)_("total")=sum_(i=1)^(n) (Gm_(i))/(r_(1)^(2)) hat(r)_(i)` |
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| 40. |
A vector perpendicular to the vector (3hat(i) + 5hat(j)) is |
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Answer» `5 HAT(i) - 3HAT(K)` |
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| 41. |
List the six point difference between translational and rotational quantities. |
Answer» SOLUTION :
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| 42. |
A circular racetrack of radius 300 m is banked at an angle of 15^@ . If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the (a) optimum speed of the race- car to avoid wear and tear on its tyres, and (b) maximum permissible speed to avoid slipping ? |
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Answer» Solution :On a BANKED road, the horizontal component of the normal force and the frictional force contribute to provide centripetal force to keep the car moving on a circular turn without slipping.At the optimum speed, the normal reaction’s component is enough to provide the needed centripetal force, and the frictional force is not needed.The optimum speed `v_0` is given by Eq. `v_0 = (Rg tan THETA)^(1//2)` Here `R = 300 m , theta = 15^@ , g =9.8 MS^(-2) ` , we have `v_0 = 28.1 ms^(-1)` The MAXIMUM permissible speed `v_(MAX) ` is given by eq. `v_(max) = (Rg (mu_s + tan theta)/(1 -mu_s tan theta))^(1//2) = 38.1 ms^(-1)` |
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| 43. |
An aeroplane is flying with the velocity of V_(1) = 800 kmph relative to the air towards south. A wind with velocity of V_(2) = 15 ms^(-1) is blowing from West to East. What is the velocity of the aeroplane with respect to the earth. |
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Answer» `222.7 MS^(-1)` |
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| 44. |
What do you mean by the term weightlessness ? Explain the state of weightlessness of a freely falling body. |
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Answer» Solution :When a man is standing in the elevator, there are two FORCES acting on him. 1. Gravitational force which acts downward. If we take the vertical DIRECTION as positive y direction, the gravitational force acting on the man is `vecF_G = -mghatj` 2. The normal force EXERTED by floor on the man which acts vertically UPWARD `vecN = N hatj` Weightlessness of freely falling bodies: Freely falling objects experience only gravitational force. As they fall freely, they are not in contact with any surface (by neglecting air friction). The normal force acting on the object is zero. The downward acceleration is equal to the acceleration due to the gravity of the Earth. i.e., (a = g) Newton.s 2nd law acting on the man `N = m (g -a)` `a= g thereforeN = m (g - g) = 0` |
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| 45. |
Advanced sunset and delayed sunset is due to |
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Answer» ATMOSPHERIC reflection |
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| 46. |
Of which conservation law is the Kepler's second law of planetory motion, a consequence> |
| Answer» SOLUTION :LAW of CONSERVATION of ANGULAR MOMENTUM. | |
| 47. |
One seond is defined as |
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Answer» 1650763.73 PERODS of krypton clock |
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| 48. |
What will be the force of contact if the same force is applied on the mass m_2 in the above problem. |
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Answer» |
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| 49. |
What is the maximum number of components into which a vector can be resolved ? |
| Answer» SOLUTION :INFINITE. | |
| 50. |
A small sleeve of mass m=1//10kg can move along the diameter of a horizontal disc, which slides without friction along a guide rode. The sleeve is tied to the end of the rod with the aid of a massless spring whose force constant k=10 N//m. When the spring is not under tension, the sleeve is at the oscillations of the sleeve when the disc rotates about its axis at the angular speed omega equal to a) 6 rad/s, b) 15 rad/s. |
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Answer» |
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