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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 particle slides along a track with elevated ends and a flat central part as shown in Fig. The flat part has a length l= 3.0m. The curved portions of the track are fricionless. For the flat part the coefficient of kinetic friction is mu_(k)= 0.20, the particle is released at point A which is at height h= 1.5m above the flat part of the track. Where does the particle finally come to rest? |
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Answer» Solution :As INITIAL ME of the particle is mgh and final is zero, so loss in ME= mgh. This ME is lost its doing work against frictionn in the flat part so loss im ME= WD against friction `mgh= mu mgs`  i.e., `s= (h)/(mu)= (1.5)/(0.2)= 7.5m` After starting from B the particle will reach C and then will rise up TILL the remaining KE at C is converted into potential energy. It will then again descend and at C will have the same value as it had when ASCENDING, but now it will move from C to B, the same will be repeated and finally the particle will come to rest at E such that `BC + CB + BE = 7.5 = 3 + 3 + BE= 7.5` BE = 1.5 So the particle COMES to rest at the CENTRE of the flat part. |
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| 2. |
Which of the following functions does not indicates simple harmonic motion? |
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Answer» `y= a SIN OMEGA t` |
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| 3. |
In Column-I relaion and in Column-II corresponding equation is given. Match them properly : {:("Column-I","Column-II"),((1) Velocity rarr relation,(a)v=v_0+at),((2) Velocity rarr "displacement relation",(b)S=v_0t+1/2at^2),(,(c) v^2=v_(0)^2+2as):} |
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| 4. |
A table fan, rotating at a speed of 2400 rpm, is switched off and the resulting variation of the rpm with time is shown in the figure. The total number of revolutions of the fan before it comes to rest is |
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Answer» 420 Area = Total number of REVOLUTIONS `=(1)/(2)xx8xx((1800)/(60))+8xx(600)/(60)+(1)/(2)xx16xx((600)/(60))` `=120+80+80` `=280`. |
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| 5. |
In solids interatomic forces are: |
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Answer» TOTALLY repulsive |
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| 6. |
Two waves represented by the following equation are travelling in the same medium y_(1) = 5 sin 2pi (75 t - 0.25 x) and y_(2) = 10 sin 2 pi (150 - 0.25 x) The intensity ratio of the two waves is ………… |
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Answer» `1:2` |
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| 7. |
Explain that how does an open ended tube manometer measure pressure. |
Answer» Solution :An open tube MANOMETER is a simple device. By this device PRESSURE of gas contained in closed container can be measured. It cosists of a U- tube containing a suitable liquid. ONE end of the tube is open to the atmosphere and other end is connected to the SYSTEM whose pressure we WANT to measure. Consider pressure at point A is P. The pressure P at A is equalto pressure at point B. `P=P_(B)thereforeP=P_(A)=P_(B)P_(a)+hrhog` hence `P=P_(a)+hrhog` ...(2) where `P_(a)` is atmospheric pressure. h= Height of mercury column from B. `rho` = Density of liquid. |
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| 8. |
A glass bulb is filled in at 0^(@)C by 350 g of mercury. When a few steel balls are put in the bulb, it can then hold only 265 g of mercury. When the bulb is heated to 100^(@)C, with the steel ballsin mercury, 5 g of mercury flows out. Find the coefficient of linear expansion of steel. Given, the coefficient of real expansion of mercury =18 times 10^(-5@)C^(-1). Neglect expansion of glass. |
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Answer» Solution :Let `rho_(0) " and " rho_(100)` be the densities of mercury at `0^(@)C " and " 100^(@)C` respectively. `THEREFORE` Volume of mercury in the glass bulb at `0^(@)C=350/rho_(0)` Volume of mercury in the glass after the steel balls are put `=265/rho_(0)` `therefore` Volume occupied by the steel balls at `0^(@)C` ` ""=350/rho_(0)-265/rho_(0)=85/rho_(0)` Expansion of mercury at `100^(@)C` `=265/rho_(0) times 18 times 10^(-5) times 100=4.77/rho_(0)` Expansion of the steel balls at `100^(@)C` `""=85/rho_(0) times gamma_(s) times 100=(8500gamma_(s))/rho_(0)` `""[gamma_(s)=` coefficient of volume expansion of steel] Now, expansion of mercury + expansion of the steel balls =volume of mercury expelled i.e, `4.77/rho_(0)+(8500gamma_(s))/rho_(0)=5/rho_(100)` or, `4.77+8500gamma_(s)=(5rho_(0))/rho_(100)=(5rho_(100)(1+18 times 10^(-5) times 100))/rho_(100)` `""=5 times 1.018` `""=[therefore rho_(0)=rho_(100)(1+18 times 10^(-5) times 100)]` or, `8500gamma_(s)=0.32 " or, " gamma_(s)=0.32/8500=37.65 times 10^(-6@)C^(-1).` `therefore` Coefficient of LINEAR expansion `alpha_(s)=gamma_(s)/3=37.65/3 times 10^(-6)=12.55 times 10^(-6@)C^(-1).` |
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| 9. |
The following observations were made in an experiment with Poiseuille's apparatus: Diameter of tube = 0.1 cm Density of liquid=2.4 g/cm""^(3) Volume of liquid collected per minute = 10 cm^(3) Pressure head of liquid = 25 cm Length of tube = 20 cm Using the above data, calculate the viscosity of the liquid. |
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Answer» SOLUTION :The volume Q of liquid flowing out per second through the capillary tube is `Q=(pi Pr^(4))/(8 eta l)` where `P/l` is the pressure gradient r is the radius of the tube `eta ` is the viscosity Given, `Q =10 CM^(3)" per MINUTE "= 10/60 cm^(3)` per s `P=hpg =25 xx 2.4 xx 980` l=20 cm `r=(0.1 cm)/(2)=0.05 cm` `n=?` `eta=(pi Pr^(4))/(8 l Q)` `=(3.14 xx 25 xx 2.4 980 xx (0.05)^(4) xx 60)/(8 xx 20 xx 10)` =0.043 cgs unit |
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| 10. |
A small body of mass m is attached to a light thread which passes through a hole at the centre of a smooth table. The body is set into rotation in a circle of radius r_(1) with a speed v_(1) . The thread is pulled down so that the radius of the path is changedto r_(2) . What will be the new linear speed and the new angular speed of thebody ? |
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| 11. |
Angad and Nalin were observing a building having two different staircases. One slanting and other vertically spiral. Angad was of the opinion that a person using slanting staircase will be doing more work against gravity but Nalin though otherwise. They started quarrelling. Arif, their friend, explained and gave entirely different view and pacified them. (b) What appreciable values do you see in Arif ? |
| Answer» SOLUTION :Arif is HONEST and he has the ability to EXPLAIN the incident of nature SCIENTIFICALLY. | |
| 12. |
Angad and Nalin were observing a building having two different staircases. One slanting and other vertically spiral. Angad was of the opinion that a person using slanting staircase will be doing more work against gravity but Nalin though otherwise. They started quarrelling. Arif, their friend, explained and gave entirely different view and pacified them.(c ) Does it also follow with friction? why or why not ? |
| Answer» Solution :In case of slanting STAIRCASE, path TRAVERSED by a person in more than that of a vertically sprial staircase. For the longer path the work done against FRICTION will be more. So for friction TOTAL work done by a person USING slanting staircase will be more than that of spiral staircase. | |
| 13. |
A uniform disc of diameter R/2 is put over another uniform disc of diameter R of same thickness and density. The peripheries of the two disc touch each other. The centre of mass of the system from centre of big disc is |
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Answer» `R//10` |
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| 14. |
The product of 1.2, 2.54 and 3.257 is |
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Answer» 9.934 |
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| 15. |
A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to -a//r^(2), where a is a constant. What is the total energy of the particle ? |
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| 16. |
A cube of wood floating in water supports a 300 g mass at the centre of its top face. When the mass is removed, the cube rises by 3 cm. Determine the volume of the cube. |
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Answer» Solution :Let each side of the cube be l. The volume occuied by 3 cm depthof cube, `V = (3 cm) xx l^2 = 3 l^2 cm^3` According to the principle of floatation, we have, `V rho g = mg IMPLIES V rho = m` `rho` is density of water = `1000 kg m^(-3)` `implies (3 l^2 xx 10^(-2)m) xx (1000 kg m^(-3)) = 300 xx 10^(-3) kg` `l^2 = (300 xx 10^(-3))/(3 xx 10^(-2) xx 1000) m^2 implies l^2 = 100 xx 10^(-4) m^2` `l = 10 xx 10^(-2) m = 10 cm` THEREFORE, volumne of cube = `V = l^3 = 1000 cm^3` |
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| 17. |
What is the magnitude of torque for a body circulating with constant angular speed? Why? |
| Answer» Solution :Zero. Because `a_(T)=0 THEREFORE veca=a_(r)` and DIRECTION of `a_(r)` towards the CENTRE and hence `|vectau|=rFsintheta, theta=180^(@) therefore |vectau|=0` | |
| 18. |
An deal gas is expandedsuchthatpT^(2)=a constant findthe coefficientof volumeexpansion ofthe gas |
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Answer» SOLUTION :fromidealgas equation pV = nRT(or )`r= ( nRT)/(V )` Asgasexpandssuchthat `pT^(2)` = a constant =C so `((n RT)/(V ))T^(2)= C ` (or )` T^(3)alpha V` (or)`T^(3) = KV` differentiatingit w.r.t. T wehave `3T^(2) = k (dV)/(dT )` Dividingit by `T^(3)= kV ` `(3)/(T)= (1)/(V ). (dV)/(dT )` coefficientof volumeexpansion `r = (1)/(V ). (d V)/(dT )= (3)/(T )` |
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| 19. |
The acceleration will be positive in . |
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Answer» `(I) and (III)` In IV slope is positive and its magnitude is increasing with time, So velocity is increasing towards right, and so acceleration is positive. |
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| 20. |
A 2kg collar is attached to a spring of spring constant 800 Nm^(-1). It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10.0 cm and released. Calculate maximum acceleration of the collar. |
| Answer» SOLUTION :`a_("MAX") = 40 MS^(-2)` | |
| 21. |
A 2kg collar is attached to a spring of spring constant 800 Nm^(-1). It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10.0 cm and released. Calculate the maximum speed. |
| Answer» SOLUTION :`V_(m)= 2 MS^(-2)` | |
| 23. |
The strain stress curves of three wires of different material as shown in fig. A,B,C are the elastic limits of the wire. The figure show that |
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Answer» ELASTICITY of WIRE A is maximum |
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| 24. |
(A) : Stationary bomb explodes into two fragments. They move with different velocities in oppsite directions.(R ) : Explosion does not violate the low of conservation of linear momentum. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A) |
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| 25. |
In the question number 89, the force of friction acting on the cylinder is |
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Answer» 2/3 Mg SIN`theta` `=Mgsintheta-2/3Mgsintheta=1/3Mgsintheta` |
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| 26. |
(A) : All oscillatory motions are necessarily periodic motion in conservative fields but all periodic motion are not oscillatory. (R) : Simple pendulum is an example of oscillatory motion. |
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Answer» Both 'A' and 'R' are ture and 'R' is the correct EXPLANATION of 'A' |
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| 27. |
Arrive at an expression for power and velocity. Given some examples for the same. |
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Answer» Solution :(i) The work done by a force `vec(F)` for a displacement `d vec(r )` is `W = int vec(F).d vec(r )""`…(1) Left hand side of the EQUATION (1) can be written as `W=int dW = int(dW)/(dt)dt` (multiplied and DIVIDED by dt)` ""` .....(2) (ii) Since, velocity is `vec(v)=(d vec(r ))/(dt),d vec(r )= vec(v )dt`. Right hand side of the equation (1) can be written as `int vec(F).d vec(r )= int(vec(F).vec(v))dt[vec(v)=(d vec(r ))/(dt)]""` ....(3) Substituting equations (2) and (3) in equation (1), we get `int(dW)/(dt)dt = int (vec(F).vec(v))dt int((dW)/(dt)-vec(F).vec(v))dt = 0` (iii)This relation is true for any ARBITRARY value of dt. This implies that the term within the bracket MUST be equal to zero, i.e., `(dW)/(dt)= vec(F).vec(v)=0`(or)`(dW)/(dt)=vec(F).vec(v)` |
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| 28. |
The potential energy of a system increase if work is done ………. |
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Answer» UPON the system by non CONSERVATIVE FORCE |
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| 29. |
Plot P-V, V-T and p-T graph correspongidng to the P-T graph for an ideal gas shown in figure. |
Answer» SOLUTION :
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| 30. |
A particle is placed at rest inside a hollow hemisphere of radius R. The coefficient of friction between the particle and the hemisphere is mu=(1)/(sqrt(3)). The maximum height up to which the particle can remain stationary is |
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Answer» `(R )/(2)` |
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| 31. |
When an elastic material with Young's modulus Y is subjected to stretching stress S, elastic energy stored per unit volume of the material is : |
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Answer» `(YS)/(2)` `=1/2 xx ` STRESS `xx` STRAIN `=1/2xx` stress `xx (("stress")/("Young.s modulus"))` `=1//2 xx (("stress") ^(2))/("Young .s modulus") = (S ^(2))/( 2Y)` |
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| 32. |
Answer the following questions based on the P-T phase diagram of CO_(2): CO_(2)is heated to a temperature 70^(@) Cand compressed Isothermally. What changes in its properties do you expect to observe ? |
| Answer» SOLUTION : It will not exhibit any CLEAR transition to the liquid phase, but will depart more and more from IDEAL gas BEHAVIOUR as its PRESSURE increases. | |
| 33. |
A boat of mass 80 kg is floating on still water. A dog of mass 20 kg on the boat is at a distance of 10m from the shore. The dog moves on the boat by a distance of 2m towards the shore. The distance of the dog from the shore is |
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Answer» 11.6 m |
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| 34. |
A 2kg collar is attached to a spring of spring constant 800 Nm^(-1). It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10.0 cm and released. Calculate the period of the oscillation. |
| Answer» SOLUTION :`T= 0.314s` | |
| 35. |
Choose the correct statement(s) from the following |
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Answer» steel is more elastic than rubber |
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| 36. |
A simple pendulum of length L has been set up inside a railway wagon sliding down a frictionless inclined plane of inclination with the horizontal. Its period of oscillation as recorded by an observer inside the wagon is |
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Answer» `2PI sqrt((L)/(g COS THETA))` |
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| 37. |
The length of a spring is increased by 2.5 cm the elastic potential energy stored in the spring is U . If the length is increasedby 5 cmits elastic potential energy become …… |
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Answer» SOLUTION :The elastic potential energy STORED in the springis `U = 1/2 ky^(2)` ` :. U PROP y^(2)` ` :. (U_(2))/(U_(1)) = ((y_(2))/(y_(1)))^(2) = ((5.0)/(2.5))^(2) = (2)^(2) = 4 ` ` U_(2) = 4U "" [ :. U_(1) = U] ` |
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| 38. |
A hollow cylinder of height 100 cm and diameter 8 cm has its top end open. It contains water up to a height of 50 cm. Mass per square centimetre of the hollow cylinder is 9 g. Find the height of the centre of gravity of the water-cylinder system from the base of the cylinder. |
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Answer» Solution :Centre of gravity of the curved surface of the empty cylinder is 50 cm above its BASE at P Fig. Weight of the base of the empty cylinder `= PI(4)^(2)xx9xx "g dyn" ""` [g= acceleration due to gravity ] Weight of curved surface of the cylinder `=2pixx4xx100xx9xxg `dyn. Its centre of gravity is at `G_(1)` where `PG_(1)` = 50 cm. Weight of water in the cylinder `= pi(4)^(2)xx50xx1xxg` dyn and centre of gravity of this water is at `G_(2)` which is 25 cm above the base of the cylinder. Suppose the centre of mass of the cylinder filled with water is at G, which is h cm above its base. The centreof gravity COINCIDES with this centre of mass. Taking moments of all the weights about P, `2pixx4xx100xx9xxG_(1)P+pixx16xx50xx1xxG_(2)P` = `(pixx16xx9+2pixx4xx100xx9+pixx16xx50xx1)xxGP` or, `2pixx4xx100xx9xx50+pixx16xx50xx1xx25` =(`pixx16xx9+2pixx4xx100xx9+pixx16xx50xx1)xxh` or, h = 46.7 cm. 
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| 39. |
The upper half an inclined plane of inclination theta is perfectly smooth while the lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom. The coefficient of friction between the block and the lower half of the plane is given by |
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Answer» `mu =2 TAN THETA ` |
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| 40. |
A stone of 0.36 m^(2) cross -sectional area and 0.1 m thickness has its lower surface in contact with water at 100^(@)C and upper surface in contact with ice at 0^(@)C. If 4.8 kg ice melting in 1 hour, then what is the thermal conductivity of stone ? Latent heat of fusion of ice is 3.36xx10^(5)" J/kg". |
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Answer» `1.24" J/ms "^(@)C` `mL.=KA((T_(1)-T_(2))/(L))t` `:.K-(mL.xxL)/(A(T_(1)-T_(2))XXT)` `=(4.8xx3.36xx10^(5)xx0.1)/(0.36(100-0)xx3600)` `=1.24" J/ms "^(@)C` |
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| 41. |
the work done against force of friction is |
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Answer» 8.7 J `=0.1xx1 kg xx 10 m s^(-2) xx COS 30^@ xx 10 m` =8.7 J |
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| 42. |
Statement A : When an iron sphere on the ground is heated its centre of mass rises slightly Statement B : Centre of mass of system of particles depends on mass of the particles and their relative positions. |
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Answer» only A is correct |
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| 43. |
A double convex lens of focal length 30cm is made of glass. When it is immersed in a liquid of refractive index 1,4 the focal length is found to be 126cm. The critical angle between glass and the liquid is |
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Answer» `sin^-1(3/4)` |
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| 44. |
(A) : The position of centre of mass of a body does not depend upon shape and size of the body.(R ) : Centre of mass of a body always lies at the centre of the body. |
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Answer» Both (A) and (R ) are TRUE and (R ) is the CORRECT explanation of (A) |
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| 45. |
The motion represented by the eqution y(t)=Acos(Omega+phi) is called SHM.Which one of the following examples closely represents SHM?Substantiate your answer(i) The rotation of the earth about its axis(iil)Oscillation of spring |
| Answer» SOLUTION : Osilliction of a SPRING `(Falpha-x)`. | |
| 46. |
(A) It is possible to eliminate dispersion by combining two prism is different refracting angles and of different materials. (R ) The angular dispersion depends on refractive index of the material of the prism. |
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Answer» Both A and R are true and R is the correct EXPLANATION of A |
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| 47. |
A bodyof mass m= 3.513 kg is moving along the x - axis with a speed of 500ms^(-1). The magnitude of its momentum is recorded as |
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Answer» `17.6 "KGMS"^(-1)` |
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| 48. |
The pulley arrangements to Fig . (A) and (B) , are identical . The mass of the rope is negligible . In (A) the mass m is lifted up by attaching a mass 2m to the other end of the rope .In (B) m is lifted up by pulling the other end of the rope with a constant downward force F=2 mg . In which case the acceleration of .m. is more? |
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| 49. |
Bernoulli's equation is a consequence of law conservation of |
| Answer» Answer :A | |
| 50. |
The branch of physics that was developed to understandand improve the working heat engines is.............. |
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Answer» Optics |
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