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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. |
Calculate the area of the triangle for which two of its sides are given by the vectorsvec(A) = 5 hat(i) - 3 hat(j) , vec(B) = 3 hat (i) + 5 hat(j) |
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Answer» Solution :Given :`vec(A) = 5 hat(i) - 3 hat(J)` `vec(B) = 3 hat(i) + 5 hat(j)` Formula: Area of the TRIANGLE`= (1)/(2) |vec(A) xx vec(B)|` `= (1)/(2) |{:(5,3),(-3,5):}|` `= (1)/(2) (25 - (-9)) = (34)/(2)` = 17 SQUARE units . |
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| 2. |
The total energy of a satellite having mass m orbiting around the earth has mass M in a circular orbit with velocity v is |
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Answer» `(1)/(2) mv^(2)` `P.E. = (-GMm)/(r ) = - mv^(2)` `T.E. = K.E. + P.E.` `= (1)/(2) mv^(2) - mv^(2) = -(1)/(2) mv^(2)` |
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| 3. |
A body of mass 10 kg revolves in a circle of diameter 0.40 m, making 1000 revolutions per minute. Calculate its linear velocity and centripetal acceleration. |
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Answer» |
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| 4. |
A person can jump safely from a height of 2m on the earth. On a planet where acceleration due to gravity is 2.45 ms^(-2) , the maximum height from which he can jump safely |
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Answer» 8m |
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| 5. |
Two heavy spheres each of mass 100kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational field and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium ? If so , is the equilibrium stable or unstable . |
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Answer» Solution :GRAVITATIONAL field at the mid - point of the line joining the centre of the TWO spheres `(GM)/((R//2)^(2))(-hatr)+(GM)/(r//2)^(2)hatr=0` Gravitational potential at the mid point of the line joining the centres of the two spheres is `V=(-GM)/(r//2) +((-GM)/(r//2))=` `(-4GM)/r = (-4xx6.67 xx10^(-11) xx100)/(0.1) = -2.7 xx10^(8) J//kg ` As the effective force on the body placed at mid - point is zero, so the body is in equilibrium . If the body is displaced a little TOWARDS either mass body from its equilibrium position , it will not return backto its initial position of equilibrium. Hence, the body is in unstable equilibrium. |
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| 6. |
The reading of a spring balance when a block is suspended fro it in air is 60 newton. This reading is changed to 40 when the block is submerged in water. The specific gravity (relative density) of the block is ____ |
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Answer» |
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| 7. |
The approximate depth of an ocean is 2700m. The compressibility of water is 45.4xx10^(-11) Pa and density of water is 10^(3) kg. What fractional compression of water will be obtained the bottom of the ocean? |
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Answer» <P> SOLUTION :Bulk modulus `B = (Delta P)/(-((DeltaV)/(V)))`Compressibility `K =1/B = (-((DeltaV)/(V)))/(DELTAP)` `(DeltaV)/(V) = - K DeltaP = - K h rho g =- 45.4 xx 10^(-11) xx 2700 xx 10^(3) xx 10` `(DeltaV)/(V) = -1.226 xx 10^(-2)`. |
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| 8. |
Two bodies of masses 1 kg and 2 kg are connected by a very light string passed over a clamped light smooth pulley. If the system is released from rest, find the acceleration of the two masses and the tension in the string |
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Answer» `3.27 m//s^(2), 13 N` |
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| 9. |
A particle executes the motion described by x(f) =x_(0) (1- e ^(lamda t) ) , t ge 0, x_(0) gt 0. (a) Where does the particle start and with what velocity ? (b) Find maximum and minimum values of x(t), v(t), a(t). Show that X(t) and alt) increase with time and v(t) decreases with time. |
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Answer» SOLUTION :Given, `x (t) = x _(0) (1- e ^( - gammat ))` `v (t) = (dx(t))/( dt ) = x _(0) gamma e ^(- gammat )` `a (t) = (DV(t))/(dt) =- x _(0) gamma ^(2) e ^(- gamma t)` (a) When `t =0, x (t) = x _(0) (1- e ^(-0)) =x_(0) (1-1) =0` `v (t) = x _(0) gamma e ^(-0) = x _(0) gamma (1) = gamma x _(0)` (b) `x (t)` is maximum when `t = oo` `[x (t) ] _(max) = x _(0) (1-0) = x _(0)` `x(t)` is minimum when `t =0` `[x (t) ]_(min) =0` v(t) is maximum when `t =0, v (0) = x _(0) gamma` v (t) is maximum when `t = oo, v (oo) =0` ` a (t)` is maximum when `t =oo, a (oo) =0` a (t) is minimum when `t =0, a (0) =- x_(0) gamma ^(2)` Note: We should he careful about NATURE of variation of the curve and maximum and minimum value will be DECIDED accordingly. |
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| 10. |
A fielder in a cricket match throws ball from the boudnary line to the wicket keeper. The ball describes at parabolic path. Which of the following quantities remain constant during the motion in air? (Neglecting air resistance) |
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Answer» Kinetic ENERGY |
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| 11. |
________ can also be treated as the momentary rotation about the point of contact. |
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Answer» SLIDING |
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| 12. |
The moon is acted upon by the gravitational force of earth. Why does not moon fall on to earth? |
| Answer» Solution :`implies` The gravitational PULL of the earth PROVIDES the necessary centripetal force to the moon for its orbital motion around the earth. Also, this force is perpendicular to the linear VELOCITY of force of moon. THEREFORE, the moon revolves around the earth in a STABLE orbit, so didn.t fall. | |
| 13. |
Whatis the shape, when- wettingliquidis placedin a capillarytube ? |
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Answer» convexupward |
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| 14. |
Assume that a tunnel ils dug along a chord of the earth, at a perpendicular distanceR/2 from the earth's centre where R is the radius of the earth. The wall of the tunnel is frictionless. a. Find the gravitational force exerted by the earth on a particle of mass m placed in the tunnel at a distance x from the centre of the tunnel. b. Find the component of this fore along the tunnel and perpendicular to the tunnel. c. Find the normal force exerted by the wall on the particle. d. Find the resultant force on the particle. e.Show that the motion of the particle in the tunnel is simple harmonic and find the time period. |
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Answer» `M^1=4/3pix_1^3rho` `M^1=(M/R^3)pi_1^3` a. F=GravitatioN/A force Exerted by the earth on the particle of mass m is `F=(GM_1m)/x_1^2=(GMm)/R^3x_1^3/x_1^2` `=(GMm)/R^3 x_1=((GMm)/R^3)/([x^2+(R^2/4)])` `F_y=Fcostheta` `=(GMmx_1)/R^3 x/x_1=(GMmx)/R^2` `F_x=Fsintheta` `=(GMmx_10/R^3 R/(2x_1)=(GMm)/(2R^2)` `F-x=(GMm)/R^2` [since NORMAL force exerted by the wall N=Fx] RESULTANT force =`(GMmx)/R^3` Acceleration=Driving force/mass `=(GMmx)/(R^3m)=(GMx)/R^3` So a ALPHAX (the body makes SHM). `:.a/x=w^2=(GM)/R^3` `RARR v=sqrt((GM)/R^3)` `rarr T=2pi(R^3/(GM))` |
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| 15. |
Springs of spring constants K, 2K, 4K, 8k, .... 2048 K are connected in series. A mass 'm' is attached to one end the system is allowed to oscillation. The time period is approximately |
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Answer» `2pisqrt((m)/(2K))` |
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| 16. |
Using a Vernier Caliper, the length of a cylinder in different measurements is found to be 2.36 cm, 2.27 cm, 2.26 cm, 2.28 cm, 2.31 cm, 2.28 cm and 2.29 cm. Find the mean value, absolute error the relative error and the percentage error of the cylinder. |
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Answer» Solution :The given readings are 2.36 cm, 2.27 cm, 2.26 cm, 2.28 cm, 2.31 cm, 2.28 cm, and 2.29 cm The mass value vec(l) = (2.36 + 2.27 + 2.26 + 2.28 + 2.31 + 2.28 + 2.29)/(7) = (16.05)/(7) = 2.29` cm Abolute ERRORS in the measurement are: `|Deltal_(1)| = |2.29 - 2.36| = 0.07` `|Deltal_(2)| = |2.29 - 2.27| = 0.02` `|Deltal_(3)| = |2.29 - 2.26| = 0.03` `|Deltal_(4)| = |2.29 - 2.28| = 0.01` `|Deltal_(5)| = |2.29 - 2.31| = 0.02` `|Deltal_(6)| = |2.29 - 2.28| = 0.01` `|Deltal_(7)| = |2.29 - 2.29| = 0.00` Mean A bsolute error `Deltal_("mean") = (0.07+0.02+0.03+0.01+0.02+0.01+0.00)/(7) = (0.16)/(7) = 0.02` RELATIVE error = `(Deltal_("mean"))/(overset(-)(l)) = pm (0.02)/(2.29) = pm 8.7 XX 10^(-3)` Percentage error `= pm 8.7 xx 10^(-3) xx 100 = 0.87% xx 100 = pm (8.7 xx 10^(-1)) = 0.9%` |
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| 17. |
If the kinetic energy of a body increases by 125 %, the percentage increase in its momentum is, |
| Answer» ANSWER :A | |
| 18. |
According to triangle law of vector addition, the direction of the resultant is given by : |
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Answer» `tan ALPHA = (P COS THETA)/(P +Q sin theta)` |
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| 19. |
A standard metre is equal to k wavelengths in vacuum, of the radiation from Krypton 86, where k is |
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Answer» 165076.37 |
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| 20. |
One mole of an ideal gas with gamma= 1.4 is adiabatically compressed so that its temperature rises from 27°C and 35°C. The change in the internal energy of the gas is |
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Answer» `-166` J |
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| 21. |
For meaurements of very high temperature say around 5000^(@)C(of sun), one can use: |
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Answer» Gas THERMOMETER |
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| 22. |
Force of 4 N is applied on a body of mass 20 kg. The work done in 3^(rd) second is |
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Answer» Solution :`a = F/M = 4/20 = 1/5 m//s^2` `S = 1/2 at^2 = 1/2 xx 1/5 xx 3 xx 3` work done in 3 SECOND = ` 4 xx 1/2 xx 1/5 xx 3 xx 3 = 18/5 ` Joule Work done in 2 second = `4 xx 1/2 xx 1/5 xx 2 xx 2 = 8/5 `Joule Work done in `3^(RD)` second = `18/5 - 8/5= 10/5 = 2` Joule. |
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| 23. |
When a force acts on a body of mass 200 gram the change in its velocity is15 cm s^(-1)per second. Themagnitude of this force is ......Newton. |
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Answer» 0.3 |
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| 24. |
The displacement y (in metre) of a body varies with time t (in seconds) as y = -(2)/(3)t^(2)16t+2.The body comes to rest in atime |
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Answer» 8s Compare it with STANDARD equation: We have : `u=16ms^(-1), (a)/(2)=-(2)/(3)` or `a=-(4)/(3)ms^(-2)` for a body to come to rest , V = 0 Using above values in V = u at we get `0=16+(-(4)/(3))t` or t =12SEC |
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| 25. |
A bird flying in the sky is an example of |
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Answer» ONE DIMENSIONAL motion |
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| 26. |
The value of accelerationdue to gravity on the surface of earth is x. At altitude of .h. from the surface of earth, its value is y.If .R. is the radius of earth, then the value of h is |
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Answer» `(SQRT(x/y)-1)R` |
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| 27. |
Consider a source of sound S and an observer P.Thesound source is of frequencyn_(0). The frequency observed by P is found to ben_(1) if P approaches S at speed v and S is stationary,n_(2) if S approaches P at a speed v and P is stationary andn_(3)if each of P and S has speedv/2towerds one another. Which of the following conclusion are correct ? |
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Answer» ` n_(1) = n_(2) = n_(3)` |
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| 28. |
In the question number 97, the speed of the child running opposite to the direction of motion of the belt is |
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Answer» `4 KM h^(-1)` `therefore` For an observer on a stationary platform, speed of child running opposite to the DIRECTION of motion of the belt is `v_(CG) = v_(CB) + v_(BG) = -9 km h^(-1) + 4 km h^(-1) = -5 km h^(_1)` -ve sign SHOWS that the child appears to be MOVING from right to left with a speed of `5 km h^(-1)`. |
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| 29. |
Kepler's third law states that square of period of revolution (T) of a planet around the sun is proportional to third power of average distance r between sun and planet. That means T^2 = Kr^3here K is constant. If the masses of sun and planet are M and m respectively then as per Newton's law of gravitation force of attraction between them is F = (GMm)/r^2, here G is gravitational constant the relation between G and K is described as |
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Answer» `GK = 4pi^2` `:. T = (2pir)/sqrt(GM) xxsqrtr "" [ :. v = sqrt((GM)/r)]` `:. T = (2pi) / sqrt(GM) .r^(3/2)` `:. T^(2)= (4pi^2)/(GM) .r^3` COMPARING this equaotion with equation `T^2 =Kr^3` `:. K = (4pi^2)/(GM)` `:. GMK = 4pi^2)` |
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| 30. |
Mention two factors on which the degrees of freedom of a gas depends. |
| Answer» SOLUTION :TEMPERATURE and ATOMICITY. | |
| 31. |
When sound waves travel from air to water, which of the following remains constant? |
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Answer» VELOCITY |
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| 32. |
Under the source of constant power a body start to move . Which one of following graph truly represents the change of displacement (s) with time (t) ? |
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Answer»
`= ma s/t "" [ :. F = ma ,v = s/t]` ` = m XX s/(t^(2)) xxs/t "" [ :. a = s/(t^(2))] ` = In `(MS^(2))/(t^(3))` P,m are constant ` :.s^(2) propt^(3) "" :. "GRAPH of " sto t` ` :.S prop t(3/2) "" `same as in (D) |
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| 33. |
Distinguish between kinetic equation for linear and angular motion . |
| Answer» Solution :`{:("LINEAR motion","Angular motion"),(V = u + at,omega= omega_(0)+alpha t ),(s = UT + (1)/(2)at^(2),theta = omega_(0)t + (1)/(2) alpha t^(2)),(s = ((v + u)t)/(2),theta = ((omega_(0) + omega)t)/(2)),(v^(2)= u^(2)+ 2AS,omega^(2)= omega _(0) .^(2)2 alpha theta):}` | |
| 34. |
A force of 50 N acts on a body of mass 10 kg at rest. Its velocity after 5 seconds if the same force acts, is |
| Answer» ANSWER :C | |
| 35. |
If E - V lt 0,then this position is possible? |
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Answer» Solution :No,because `E = K +V ` ` :. K = E - V ` and kinetic energy NEVER be ZERO .So `E - V ` cannot be negative . |
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| 36. |
A string of length l and mass m is fixed at both ends with tension t and excited in its fundamental mode.The maximum displacementat the mid point is A Find (a) the maximum kinetic energy of the string ,9b0 the mean kinetic energy of the srting averaged over one oscillation period hint Take sine distribution of amplitudes |
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| 37. |
In a Kundt's tube experiment , with a wooden rod170 cm along , clamped at the middle , the lycopodium powder gets heated up at regular intervals of 13.4 cm , the experiment being performed with air . If the frequency of vibrations be 1270 Hz , find the velocity of sound in air and in the wooden rod. |
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Answer» Solution :The wavelengths of sound waves in air and in rod will be respectively , ` lambda_(a) = 2 XX 13.4 cm = 26.8 cm` , and `lambda_(r) = 2 xx 170 cm = 340 cm` Since VELOCITY ` v= LAMBDA F` `:. "Velocity od sound in air" v_(a) = l_(a) f` `26.8 cm xx 1270 s^(-1) = 340 m//s` `:. "Velocity of sound in wooden rod"v_(r) = lambda_(r) f` ` = 340 cm xx 1270 s^(-1)` `= 4318 m//s` |
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| 38. |
In a two stage heat engine a quantity of heat Q_(1) is absorbed at a temperature T work W_(1) is done and a quantity of heat Q_(2) expelled at a lower temperature T_(2) by the first state. The second state absorbs the heat expelld by the first, does work W_(2) and expels a quanity of heat Q_(3) at a lower temperature T_(2) Calculate the effiency of the combination engine. |
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Answer» |
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| 39. |
Find the thickness of a transparent plastic plate which will produce a change in optical path equal to the wavelength lamda of the light passing through it normally, The refractive index of the plastic plate is mu. |
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Answer» Solution :When light travel a distance x in a medium of REFRACTIVE index `m`, its optical path `=MU x` change in optical path `=mu x-x =(mu-1) x`. this is to be EQUAL to `LAMDA`. But `(mu-1) x= lamda` The THICKNESS of the plate `x= lamda/(mu -1)` |
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| 40. |
While measuring the thermal conductivity of a liquid we keep the upper part hot and the lower part cool so that, |
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Answer» Convection MAY be stopped |
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| 41. |
A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading : 0mm Circular scale reading: 52 divisions Given that 1 mm on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is |
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Answer» 0.52cm |
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| 42. |
A body of mass Mkg is on the point of a smooth hemisphere of radius 5m. It is released to slide dewon the surface of hemisphere. It leaves the surface when its velocity is 5 ms^(-1). At this instant the angle made by the radius vector of the body with the vertical is (g= 10 ms^(-2)) |
| Answer» Answer :C | |
| 43. |
In steel, the Young's modulus and the strain at the breaking point are 2 xx 10^(11) Nm^(-2) and 0.15 respectively. The stress at the breaking point for steel is therefore……………. . |
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Answer» `2 xx 10^(8) NM^(-2)` `= 0.15 xx 2 xx 10^(11)= 3 xx 10^(10) Nm^(-2)` |
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| 44. |
A boy of mass 40 kg climbs up a rope with an acceleration of 2 ms^(-2). What is the tension in the rope ? |
| Answer» ANSWER :A | |
| 45. |
A body weights w_(1) in air, w_(2) in water and w_(3) in a liquid. The density of the liquid is given by |
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Answer» `(W_(1))/(W_(2)-W_(3))` |
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| 46. |
In planetary motion the areal velocity of position vector of a planet depends on angular velocity (m) and the distance of the planet from sup (r). If so the correct relation for areal velocity is |
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Answer» `(DA)/(DT) prop omegar` |
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| 47. |
A light string of length 20cm is carrying a bob of mass 100gm. It is pulled through an angle 60^(@) and released. Its maximum velocity is, |
| Answer» Answer :C | |
| 48. |
Assertion: Cohesive and adhesive forces obey inverse square law. Reason: While writing, graphite sticks to the paper on account of cohesive forces. |
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Answer» If both assertion and REASON are correct and remad is a correct EXPLANATION of the assertion. |
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| 49. |
A water proofing agent changes the angle of contact |
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Answer» From an obtuse to acute value |
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| 50. |
The thermal emissivities of two bodies A and B are 0.02 and 0.09, respectively. lamda_A and lamda_Bare the wavelengths corresponding to maximum intensity of radiations and lamda_Ais shifted from lamda_Bbymu m . If the surface area and the total power radiated by the bodies are same, then calculate the temperature of B, given temperature of A is 5000 K. |
| Answer» SOLUTION :`6.35 MU K` | |
