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. |
Abody of mass 1kg falls from the top of a cliff 50 m high and buries itself one metre deep in mud.Find the average resistance offered by a mud and the time of penetration. |
| Answer» SOLUTION :500N,00065 S | |
| 2. |
A freely falling body takes 4s to reach the ground. One second after release, the percentage of its initial potential energy, that is still retained is |
|
Answer» 0.0625 |
|
| 3. |
A thin circular disc of mass 'm' and radius R is rotating bout its axis with a constant angular velocity omega. It has two masses each of mass 'm' (same mass as that of disc) at the opposite ends of a diametre of the disc. Now if the masses are gently removed, the final angular velocity of disc would become |
| Answer» Answer :A | |
| 4. |
A ball of mass 10 gm dropped from a height of 5 m hits the floor and reboundsd to a height of 1.25 m. If the ball is in contact with the ground for 0.1 s, the force exerted by the ground on the ball is (g=10 m//s^(2)) |
|
Answer» 0.5 N |
|
| 5. |
Obtain tau=Ialpha from angular momentum of rigid body. |
|
Answer» Solution :The total angular momentum of a motion of rigid body about a fixed axis `vecL=VEC(L_(z))+vec(L_(_|_))` Differentiating w.r.t. time `(dvecL)/(DT)=(DVEC(L_(z)))/(dt)+(dvec(L_(_|_)))/(dt)` but `(dvec(L_(z)))/(dt)=tauhatk and (dvec(L_(_|_)))/(dt)=0` `therefore (dvec(L))/(dt)=tauhatk` but `vec(L_(Z))=Iomegahatk` `therefore (d(Iomega))/(dt)=tauhatk` `I(domega)/(dt)=tau` `therefore Ivecalpha=vectauimpliesvectau=Ivecalpha` Scalar form `tau=Ialpha` is similar to `F=ma` in linear motion. |
|
| 6. |
A stationary particle explodes into twoparticle of masses m1 and m2 and move in opposite directions with velocities v1 and v2.The ratio of their K.E is |
|
Answer» m2/m1 |
|
| 7. |
The length and breadth of a rectangular object are 25.2 cm and 16.8 cm respectively and have been measured to an accuracy of 0.1 cm. Find the relative error and percentage error in the area of the object. |
|
Answer» Solution :Area `A = 1 xx B`, `b= 16.8 cm and Deltab = 0.1 cm` `THEREFORE` Relative ERROR in area `((DeltaA)/(A) ) = ( ( Delta l)/( l)) + ((Deltab)/(b)) = ((0.1)/(25.2) ) + ((0.1)/( 16.8))` `= 0.004+ 0.006 = 0.01` `((DeltaA)/( A)) 100 = 0.01 xx 100` Percentage error `= 1%` |
|
| 8. |
A stationary shell breaks into three fragments. The momentum of two of the fragments is P each and move at 60^(@) to each other. The momentum of the third fragment is |
|
Answer» <P>P |
|
| 9. |
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 100ms^(-1). Find when and where the two stones meet. |
|
Answer» `2 s, 200.9 m` |
|
| 10. |
Select the incorrect statement of the following statements. |
|
Answer» Reynold's number is used to FIND the nature of flow of liquid. |
|
| 11. |
Two ends of a plank are placed on two wedges. The plank gets depressed by 1mm when a load of 2 kgis hung from its mid-point. What will be the depression of the plank if a load of 5 kg is hung from its mid-point. |
|
Answer» |
|
| 12. |
A black body radiates energy at the rate of 1.452 xx 10^(10) erg s^(-1) m^(-2). Calculate the value of Stefan's constant. |
|
Answer» |
|
| 13. |
Three particles are projected in air with the minimum possible speeds, such that the first goes from Ato B, the second goes from 5 to C and the third goes from C to A. Points A and Care at the same vertical level. The two inclines make the same angle a with the horizontal as shown. Thenthe relation among the projection speeds of the three particles is : |
|
Answer» `u_(3)=u_(1)+u_(2)` |
|
| 14. |
A uniform sheet of steel is cut into the shape as shown. Compute the X and Y co-ordinated of the centre of mass of the piece. (Width of each plate is 10cm ) |
Answer» SOLUTION : `(X_(CM),Y_(cm))=(22.5,17.5)` |
|
| 15. |
In the constant flow moethof of callendar and Barnes it was found that when the rate of flow water was 11g per minute the heating current was 2 amperes and the difference in potential between the ends of the heating wire was 1 volt, the rise in temperature was 2.5^(@)C. On increasing the rate of flow to 25.4g per minute, the heating current to 3 amperes and the p.d.to 1.51 volts, the rise of temperature of water was still 2.5^(@)C.Calculate the value of J. |
|
Answer» |
|
| 16. |
What is the difference between isothermal and adiabatic process? |
| Answer» SOLUTION :In an osothermal process, the system is in good THERMAL contact with the surroundings and the TEMPERATURE remains constant. In an adiabatc process the system is insulated from its surroundings so that no haet EXCHANGE takes place between them. | |
| 17. |
You can shled a charge from electrical forces by putting it inside a hollow conductor. Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ? |
|
Answer» SOLUTION :(a)No. A body cannot be shielded from the gravitational influence of nearby matter. Gravitation forces are independent of medium. (b) Yes. If the size of the spaceship is EXTREMELY large then the gravitational effect of the spaceship may become measurable. The variation in g can also be detected. (c) Tidal effect depends inversely on the cube of the distance, unlike force which depends inversely on the square of the distance. Since distance of moon from the ocean water is very small as COMPARED to the distance of SUN from the ocean water on earth. Therefore, the tidal effect of moon.s pull is greater than the tidal effect of sun. These effect can be shown below for moon and sun. `:. T_m/T_s=((d_s)/(d_m))^3=((1.5xx10^11)/(3.8 xx10^8))^3` `:. T_m/T_s=6.15xx10^6` |
|
| 18. |
A particle is vibrating in SHM with an amplitude of 4cm. At what displacement from the equilibrium position it has mechanical energy which is half potential and half kinetic |
|
Answer» 1 CM |
|
| 19. |
An amount of heat passing through a metallic rod in time t is given byQ=(KA(T_(1)-T_(2))t)/(I) where k= thermal conductivity A= Cross sectional area, T_(1) and T_(2) are temeprature of hot and cold ends respectively and L= length. So the dimesional formula for k=..... |
|
Answer» Solution :`Q=(KA(T_(1)-T_(2))t)/(L)` is given `:.k=(QL)/(A(T_(1)-T_(2))t)....(i)` Where, HEAT energy `[Q]=M^(1)L^(2)T^(-2)` Length `[L]=L^(1)` area `[A]=L^(2)` differerence of temperature. `(T_(1)-T_(2))=[DeltaT]=K^(1)` time `[t]=T^(1)` Note that here we have included K (for temperature) along with M, L and T. Substituting these dimensional FORMULA in equation (1), we GET `[k]=(M^(1)L^(2)T^(-2)L^(1))/(L^(2)L^(1)T^(1)=M^(1)L^(1)T^(-3)K^(-1)` |
|
| 20. |
During the motion of a projectile , the speed of the projectile is minimum at the top of its path. Is this statement true ? Give reason . |
| Answer» SOLUTION :The statement is ture. The vertical COMPONENT of the velocity will be zero at the top. Hence the RESULTANT velocity of the PROJECTILE will be minimum at the top . | |
| 21. |
In a diagram, the weight of an extended body is shown to act from the centre of mass. Does it mean that other panicles are not attracted by the earth ? What is the relation between the weight of the body and that of all other particles ? |
| Answer» SOLUTION :No. All the PARTICLES are atuacted by the earth. The weight of the BODY is EQUAL to sum of its particles. | |
| 22. |
The difference in the number of wavelengths, when yellow light propagates through air and vacuum columns of the same thickness, is one. The thickness of the air column is: (Refractive index of air mu_(a)= 1.0003,Wavelength of yellow light in vacuum = 6000 overset(circ)(A)) |
|
Answer» 1.8nm |
|
| 23. |
Initial angular speed of wheel is 20 rad s^(-1). Its angular displacement during 10 s is 100 rad then how many revolution it makes from initially to remain rest. What is its angular acceleration? |
| Answer» SOLUTION :`alpha=-2" rads"^(-2),N=(50)/(PI)` | |
| 24. |
Does the nature of forces differ in solids and liquids ? |
| Answer» Solution :the NATURE of forces DIFFER in solids and liquids . In solids there is a STRONG FORCE of repulsive between the ATOMS whereasin case of liquids there is force of attaraction between them . | |
| 25. |
According to Kepler's planet move in |
|
Answer» Circular orbits AROUND the SUN |
|
| 26. |
A body moving on a surface is brought to rest suddenly. What happens to the friction between the two ? |
| Answer» Solution :The FORCE of FRICTION disappears, because there is no RELATIVE MOTION. | |
| 27. |
(A): Two gases with the same average translational kinetic energy have same temperature even if one has greater rotational kinetic energy than the other. (R): Only average translational kinetic energy of a gas contributes to its temperature. |
|
Answer» Both (A) and ( R) are TRUE and (R ) is the CORRECT explanation of (A) |
|
| 28. |
The difference in the value of 'g' at pole and at a latitude is (3)/(4)R omega^(2) then latitude angle is |
|
Answer» `60^(@)` |
|
| 29. |
If a soap bubble of radius 3 cm coalesce in vacuum under isothermal conditions with another soap bubble of radius 4 cm find the radius of the big bubble so formed. |
|
Answer» SOLUTION :`r_(1)=3cm,r_(2)=4cm` Let R be the radius of the big drop surface energy of big drop = Surface enegy of FIRST drop + surface energy of second drop `:.8piR^(2)S=(8pir_(1)^(2)+8pir_(2)^(2))S` `R=sqrt(r_(1)^(2)+r_(2)^(2))=sqrt(3^(2)+4^(3))=5cm` |
|
| 30. |
A brass wire of diameter 1mm and of length 2m is stretched by applying a force of 20N. If the increase in length is 0.51mm, find i) the stress, ii) the strain and iii) the young's modulus of the wire. |
|
Answer» `2.546xx10^(7)N//m^(2)` |
|
| 31. |
If L is the inductance, C is the capacitance and R is the resistance, then Rsqrt(C/L). has the dimension |
|
Answer» `MLT^(-2) I^(-2)` |
|
| 32. |
The dimensions of resistivity in terms of M, L, T and Q, where Q stands for the dimensions of charge is |
|
Answer» `ML^(3)T^(-1)Q^(-2)` |
|
| 33. |
A particle is moving in a potential region defined by U=K(x^2 + y^2 + z^2). Calculate the force acting on the particle. |
|
Answer» Solution :`vecF = -hati (DEL U)/(del X) - HATJ (del U)/(del y) - HATK (del U)/(del z) ` or `vecF = -K [hati xx 2x + hatj xx 2y + hatk xx 2z]` or `vecF = -2K[x hati + y hatj + zhatk]`. |
|
| 34. |
Sttopping distance is proportional to initial velocity. |
|
Answer» |
|
| 35. |
In the above question, the time taken by the ball to come again to the water surface is |
|
Answer» 2 s `t=2((V)/(a)) =2((19.6)/(9.8)) =4S` |
|
| 36. |
The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate by using the formula P=(F)/(l^(2)). If the maximum errors in the measurement of force and the length are 4% and 2% respectively and the maximum error in the measurement of pressure is x %, then the value of x is. |
|
Answer» |
|
| 37. |
The area under a 'force-displacement' curve gives: |
|
Answer» impulse |
|
| 38. |
A pendulum bob of mass 'm' is held out in the horizontal position and then released from rest. If the string is of length l(1) what is the velocity of the bob and tension in the string when the bob reaches the lowest position (2) If the breaking strength of the string is 'nmg'. Find the maximum angle 'theta' with respect to the vertical at which the bob is to be released to avoid breaking of the string. |
|
Answer» Solution :(1) From law of consevation of mechanical energy. <BR> `0 + MG l = (1)/(2) mv^(2) + 0 " " v= sqrt(2gl)` The tension in the string at the lowest position br> `=(mv^(2))/(r ) + mg = (m(2gl))/(l) + mg ( because r=l) = 3mg` (2) `mgl (1- COS theta) = (1)/(2) mv^(2) or v^(2)= 2gl (1- cos theta)` Tension in the string at the lowest position `T= (mv^(2))/(l) + mg` `therefore nmg= (m)/(l) [2gl(1- cos theta)] + mg` on simplification `cos theta= (3-n)/(2)` |
|
| 39. |
A very small hole in an electric furnance is used for heating metals . The hole nearly acts as a black body . The area of the hole is 200 mm^(2). To keep a metal at 727^@C , heat energy flowing through this hole per sec, in joules is (sigma= 5.67 xx 10^(-8)Wm^(-2)K^(-4)) |
|
Answer» 22.68 |
|
| 40. |
In cold countries, water does not freeze at the bottom of a lake in winter, on account of |
|
Answer» anomalous expansion of water between `4^(@) and 0^(@)` C |
|
| 41. |
We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63s, 2.56s, 2.42s , 2.71s and 2.80s. Calculate the absolute errors, relative error or percentage error. |
|
Answer» Solution :The mean PERIOD of oscillation of the pendulum `T = ((2.63+2.56+2.42+2.71+2.80)s)/(5)` `=(13.12)/(5)`s =2.624s =2.62s As the periods are measured to a resolution of 0.01s, all times are to the second decimal, it is proper to put this mean period also to the second decimal. The errors in the measurements are `2.63s=2.62s = 0.01s` 256s - 2.62s = -0.06s 2.42s-2.62s = -0.20s 2.71s-2.62s = 0.09s 2.80s-2.62s = 0.18s Note that the errors have the same units as the quantity to be measured. The arithmetic mean of all the absolute errors (for arithmetic mean, we take only the magnitudes ) is `Delta T_("mean") =[(0.01 + 0.06 + 0.20 + 0.09 + 0.18)s]//5` `=0.54 s//5` `=0.11s` That means , the period of oscillation of the simple pendulum is `(2.62 pm 0.11) ` s i.e. . It lies between (2.62 + 0.11) s and (2.62 - 0.11) s or between 2.73 s and 2.51 s . As the arithmetic mean of all the absolute errors is 0.11 s, there is already an error in the tenth of a second . Hence there is no point in giving the period to a hundredth. A more correct way will be to WRITE . `T = 2.6 pm 0.1s` Note that the LAST numeral 6 is unreliable, SINCE it may be anything between 5 and 7 . We indicate this by saying that the measurement has TWO significant figures are 2, which is reliable and 6, which has an error associated with it. You will learn more about the significant figures in section 2.7 . For this example, the relative error or the percentage error is `deltaa =(0.1)(2.6) xx 100 = 4%` |
|
| 42. |
Two Electric bulbs have filaments of lengths 1 and 21, diameters 2d and d. The emissivities of the bulbs are 3e, 4e with temperatures in the ratio 2 : 3. Their radiant powers will be in the ratio |
|
Answer» `4:27` |
|
| 43. |
A body of mass 12 kg is at the bottom of a lake of depth 5 m. If the work done in raising the body to the water surface is 350 J, the specifice gravity of the body is |
|
Answer» 1.2 |
|
| 44. |
A spherical ball of volume 1000 cm^3is subjected to a pressure of 10 atmosphere. The change in volume is 10^(-2)cm^3 . If the ball is made of iron find its bulk modulus, (Atmospheric pressure =1xx10^5 Nm^(-2)) |
|
Answer» `1XX10^(11)N//m^(2)` |
|
| 45. |
Copper of fixed volume 'V' is drawn into wire of length 'l'. When this wires is subjected to a constant force 'F', the extenstion produced in the wire is ' Delta l'which of the following graphs is a straight line? |
|
Answer» `DELTA L` VERSUS `l^2` |
|
| 46. |
Absolute error of measuring instrument is equal to least count of measuring device. |
|
Answer» |
|
| 47. |
Describe the formation of beats. |
|
Answer» Solution :Formation of beats : When two more waves SUPERIMPOSE each other with SLIGHTLY different frequencies , then a sound of periodically varying amplitude at a point is observed . This phenomenon is known as beats . The number of amplitude maxima per second is called beat frequency . If we have two sources , then their DIFFERENCE in frequency GIVES the beat frequency . Number of beats per second `n = |f_(1) - f_(2)|` per second . |
|
| 48. |
Find the torque about the point 'A' in the following diagram |
Answer» Solution : (1) Torque DUE to 20N : about A=O (2) Torque due to 10N`=10xx0.8=8N` (C wise) Torque due to 25N `=(25sin25^(@))` (AC wise) `=25xx0.8sin25^(@)` `=25xx0.8xx0.4225=845nm` Net torque `=0-8N`(CW)+8.45(ACW) =0.45 anticlockwise |
|
| 49. |
A lead piece of mass 25 g gives out 1200 calories of heat when it is cooled from 90^(@)C to 10^(@)C. What is its i) Specific heat ii) thermal capacity and iii) Water equivalent. |
|
Answer» Solution :MASS of lead piece (m) = 25 G = 0.025 kg AMOUNT of heat GIVEN out = 1200 calories Amount of heat energy given out (dQ) = `1200xx4.2J` The fall in temperature (dT) = `(90-10)^(@)C=80^(@)C=80K` (i) Specific heat S = `1/m(dQ)/(dT)=1/0.025xx(1200xx4.2)/80` = `5040/2=2520Jkg^(-1)K^(-1)` (ii) Thermal capacity = MS = `0.025xx2520=63JK^(-1)` (iii) Water equivalent = W = `63/4200kg` |
|