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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. |
An ideal gas mixture filled inside a balloon expands according to the relation PV^(2//3)= constant. What is the temperature inside the balloon |
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Answer» Solution :`PV^(2//3)="CONSTANT" RARR (PV^(2//3))/(PV)=("constant")/(RT)` or `(1)/(V^(1//3)) = ("constant")/(RT) rArr V prop T^(3)` Temperatureincreases with increase in volume. |
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| 2. |
(A) : Work done by or against force of friction in moving a body through any round trip is zero.(R ) : Friction is a conservative force. |
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Answer» Both (A) and (R ) are true and (R ) is the CORRECT EXPLANATION of (A) |
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| 3. |
Assertion : If a pendulum is suspended in a lift and liftis fallingfreely , then its time period becomesinfinite. Reason : Free fallingbody has acceleration equal to acceleration due to gravity . |
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Answer» Both assertion and reason are true and reason explains assertion correctly. `thereforeT=2pisqrt((l)/(0))="infinite"` Reason:For a FREE falling BODY ACCELERATION is equal to `sqrt(g)`. |
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| 4. |
Two satellites of masses 100kg and 200 kg are revolving the earth at altitude 6400 km and 44800 km. The ratio of orbital velocity of the satellite is |
| Answer» ANSWER :B | |
| 5. |
Explain why friction is necessary to make the disc roll (refer to Q.28) in the direction indicated. (a) give the direction of frictional force at B, and the sense of frictional torque, before perfect rolling begins. (b) What is the force of friction after perfect rolling begins ? |
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Answer» Solution :To ROLL a disc, we require a torque, which can be PROVIDED only by a tangential FORCE. As force of friction is the only tangential force in this case, it is necessary. (a) As frictional force at `B` OPPOSES the velocity of point `B`, which is to the left, the frictional force MUST be to the right. The sense of frictional torque will be perpendicular to the plane of the disc and outwards. (b) As frictional force at `B` decreaese the velovity of the point of contact `B` with the surface, the perfect rolling begins only when velocity of point `B` becomes zero. Also, force of friction would become zero at this stage. |
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| 6. |
The work done by the external forces on a system equals the change in |
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Answer» totalenergy |
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| 7. |
Four thin rods, each of mass M and length d=1m are rigidly connected in the form of a positive sign. The entire assembly rotates in a horizontal plane around a vertical axle at the centre with initial (clockwise) angular velocity omega_(i)="2rads"^(-1). A mud ball with mass m and initial speed v_(i)="12 ms"^(-1) is thrown at and sticks to the end of one rod. Let M=3m. What is the final angular velocity omega_(f) of the ''plus sign + mudd ball'' system, if the initial path of the balss is each of the four paths shown in figure |
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Answer» Path 1 CONTACT is made when the velocity of the ball is perpendicular to the rod `omega_(f)=-4"rads"^(-1)` |
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| 8. |
A simple pendulum is suspended fromthe roof of a school bus which moves in a horizontal direction with an acceleration a , then the time |
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Answer» `Talpha(1)/(G^(2)+a^(2))` |
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| 9. |
Two particles are projected from the same point on level ground simu ltaneously with the same velocity u but at the angles (alpha+beta)and (alpha-beta) of projection betalt45^(@) respectively. Which of the following statements will be true? |
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Answer» They will have the same horizontal RANGE if `alpha=45^(@)` Time for REACHING the maximum heights `t_(1)=(U sin (45^(@)+theta))/(g)` and `t_(2)=(u sin(45^(@)-theta))/(g)` or `Deltat=(u)/(g)[sin(alpha+theta)-sin(alpha-theta)]` Similarly, `R_(1)=(u^(2))/(g)sin 2(alpha+beta)` and `R_(2)=(u^(2))/(g)sin 2(alpha-beta)` |
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| 10. |
A projectile is fired horizontally with a velocity u. obtainthe expression for resultant velocity of the projectile at any instant . |
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Answer» Solution :At any instant t, at the VELOCITY components of the projectile along both x-axis and y-axis. The resultant of these two components gives the velocity of the projectile at that instant t, as shown in the figure . The velocity COMPONENT at any instant of the t along horizontal (x-axis) is`v_(x) = u_(x) + a_(x)t` Since , `u_(x) = u, a_(x) = 0 ` , we get `v_(x) = u` the component of velocity along vertical DIRECTION (y-axis) is given by`v_(y) = u_(y) + a_(y) t` `v_(y) = g t` `:.`The velocity of the particle at any instant is `vec(v) = u hat(i) + g t hat(j)` The speed of the particle at any instant t is given by `:. v = sqrt(v_(x)^(2)+ v_(y)^(2))` `v = sqrt(u^(2) + g^(2) t^(2))` 
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| 11. |
If the sum of two unit vectors is also a vector of unit magnitude, the magnitude of the difference of the two unit vectors is |
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Answer» 1 UNITS |
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| 12. |
In above problem ratio of distance traveled in first second of upward motion to first second of downward motion is |
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Answer» `1:7` |
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| 13. |
If the coefficient of friction is (1)/(sqrt(3)) , the height up to which a particle can rise and stay inside a hollow sphere of radius r is ( inner surface of the sphere is rough ) |
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Answer» 0.5 R |
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| 14. |
Weightlessness in satellite is due to: |
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Answer» ZERO GRAVITATIONAL acceleration W = mg, When g = 0, W = 0. |
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| 15. |
If energy 'E' momentum 'P' and force 'F' are choosen as fundamental units.Dimension of accelerations in new system is |
| Answer» Answer :A | |
| 16. |
As shown in figure a spherical air lens of radii R_1=R_2=10cm is cut in a glass cylinder. Determine the focal length and nature of air lens. IF a liquid of refractive index 2 is filled in the lens, what will happens to its focal length and nature? |
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Answer» Solution :ACCORDING to lens MAKER formula, `1/f=(mu-1) [1/R_1-1/R_2]` with `mu=mu_1/mu_2` Intially `mu=mu_a/mu_u=1/((3//2))=2/3, R=+10 cm and R_2=-10cm` So `1/f=[2/3-1][1/(+10)-1/(-10)]=-2/30 i.e., f=-15 cm` i.e. the air lens in glass behaves as divergent lens of focal LENGTH 15 cm When the liquid of `mu=2` is filled in the air CAVITY. `mu=mu_1/mu_u=2/1.5=4/3` So that now `1/f.=[4/3-1][1/10-1/10]=2/30` f.=15CM i.e., the liquid lens in glass will behaves as a convergent lens of focal length 15cm. |
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| 17. |
Three uniform spheres each having mass 'm' and radius 'r are kept in such a way that each touches the other. The magnitudeof the gravitational force on any sphere due to the other two is |
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Answer» Solution :Let `F_(1) = F_(2) = F` here `F = (Gm.m.)/((2R)^(2)) = (Gm^(2))/(4R^(2))` `therefore` The resultant gravitational FORCE `F_(R) = 2F COS theta//2` `F_(R)= 2F cos 30^(@) = sqrt(3)F = sqrt(3)(Gm^(2))/((2R)^(2)) = sqrt(3)(Gm^(2))/(4R^(2))` 
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| 18. |
In the above problem, find the height to which the lift takes the passenger. |
| Answer» SOLUTION :S = Area of the GRAPH `= (1)/(2)(8+12)xx3.6 = 36 m`. | |
| 19. |
A satellite lanching station should be |
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Answer» NEAR the EQUATOR region |
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| 20. |
Two wires of the same material and length but diameters in the ratio 1:2 are stretched by the same force. The elastic potential energy per unit volume for the wires, when stretched by the same force will be in the ratio |
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Answer» `16:1` |
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| 21. |
A solid sphere rolls down without slipping from rest on a 30^(@) incline. Its linear accelearation is |
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Answer» 5g/7 |
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| 22. |
Bernoulli's theorem is applicable for processes which are |
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Answer» isothermal |
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| 23. |
The mean distance of the moon from the earth is 3.84 xx 10^5 km and it rotates around the earth in 27.3 days. What is the angular momentum of the moon around the earth ? Mass of the moon = 7.3 xx 10^(22) kg |
| Answer» Solution :`L = mr^(2)OMEGA =7.3 xx 10^(22) xx (3.84 xx 10^(8))^(2) xx 2 xx 3.14//27.3 xx 24 xx 60 xx 60 = 2.8 xx 10^(24) JS` | |
| 24. |
An earth satellite is revolving in a circular orbit of radius 7000km. If its period of revolution is 5500 sec, its orbital velocity is de |
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Answer» `1 KMS^(-1)` |
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| 25. |
What is the vector sum of n coplanar forces, each of magnitude F, if each force makes an angle (2pi)/(n) with the preceding force? |
| Answer» SOLUTION :RESULTANT FORCE is ZERO. | |
| 26. |
A man with normal near point (25cm) reads a book with small print using a magnifying glass a thin convex lens of focal length 5cm. a) What is the closest and farthest distance at which he can read the book when viewing through the magnifying glass? b) What is the maximum and minimum magnifying power possible using the above simple microscope? |
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Answer» Solution :A for normal eye FAR and near points are `infty` and 25CM respectively so for magnifier `v_(max)=- infty and v_(mi N)=-25cm` However for a lens as `1/v-1/u=1/f` i.e., `u=f/((f//v)-1)` So u will be minimum when v=min =-25cm i.e., `(u)_(mi n)=5/(-(5//25)-1)=-25/6=-4.17 CM` and u will be maximum when v=max= `infty` i.e., `(mu)_(max)=5/((5/infty)-1)=-5cm` So the closest and farthest distances of the book from the magnifier (or eye) for clear viewing are 4.17cm and 5cm respectively b) As in case of simple magnifier `MP=(D//u)`. So MP will be minimum when u=max=5cm i.e, `(MP)_(mi n)=(-25)/(-5) =5 [=D/F]` And MP will be maximum when u=min=`(25//6)cm` `(MP)_(max)=(-25)/(-(25//6))=6[1+D/f]` |
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| 27. |
A sphericalbodymovingwith a speed of 10 m/s strikesanother identicalbody at rest such that after collision , the directionof motionof eachball makesan angleof15^(@) with theoriginaldirection of motion . Calculate the speedof eachbody aftercollision. |
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Answer» Solution :Massof eachbody = m The givencollisionprocess is shown below:  Here `u_(1)`and `u_(2)`areintialspeedsof body moving with 10 m/sand THATOF body , RESPECTIVELY. `v_(1) and v_(2)`are speedsof the bodyafter COLLISION Usinglawof conservation of MOMENTUM alongtheline of motion , `"mu"_(1) + "mu"_(2) = mv_(1) COS 15^(@) +mv_(2) cos15^(@)` `mxx 10 + 0 = mxxv_(1) xx 0.966 +mv_(2) xx0.966` ` 10 m = m xx 0.966 (v_(1) +v_(2))` ` rArr"" v_(1) +v_(2) = 10.35.....(i)` Now usinglawofconservationof momentumalong thedirectionperpendicularto the original line of motion , ` m xx 0 + m xx 0 = mv_(1) sin 15^(@) -mv_(2) sin 15^(@)` ` rArr "" v_(1) sin 15^(@) = v_(2) sin 15^(@)` or `v_(1) =v_(2) ""...(ii)` Using (i) and (ii) , we get ` rArr v_(1) = 5.175 ` m/s ` rArr "" v_(2) = 5.175 ` m/s |
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| 28. |
A monoatomic gas is compressed adiabatically to 1//4^(th) of its original volume, the final pressure of gas in terms of initial pressure P is |
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Answer» 7.08 P |
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| 29. |
(A): A spring has potential energy, both when it is compressed or stretched. (R) : In compressing or stretching, work is done on the spring against the restoring force. |
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Answer» Both .A. and .R. are true and .R. is the CORRECT EXPLANATION of .A. |
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| 30. |
How many astronomical units make 1 parsec? |
| Answer» SOLUTION :1 PARSEC `=3.08xx10^16m,1AU=1.496xx10^11mtherefore(1 parsec)/(1AU)=(3.08xx10^16)/(1.496xx10^11)=2.06xx10^5` 1 parsec `=2.06xx10^5AU` | |
| 31. |
The length of a rod is measured with a meter rod having least count 0.1cm. Its diameter is measured with vernier calipers of least count 0.01cm. If length is 5.0cm and radius 2.00 cm, then the percentage error in volume is |
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Answer» 0.04 |
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| 32. |
A tunnel is dug through the centre of the earth. Show that a body of mass m when dropped from rest from one end of the tunnel will execute simple harmonic motion. |
Answer» SOLUTION :SUPPOSE a body of MASS m reaches at point P with depth d from the surface of earth and P point at distance r from the centre of earth. So there is no force exerted on body of mass m from EXTERNAL part at the distance r from earth but force is exerted only by a mass of earth of radius r.  ACCELERATION due to gravity at depth d from the surface of earth `g. = g(1-(d)/(R ))= g((R-d)/(R ))` Let `R- d = y`, `g. = (gy)/(R )` Force on a body of mass m at point P, `F= -mg.""` (force is considered negative toward the centre) `F= -(mg)/(R )*y"""........."(1)` `therefore F propto -y` Now, `ma= -(mg)/(R )y""` (From equation (1)) `therefore a= -(g)/(R )y"""........."(2)` comparing this equation `a= -omega^(2)y` with (2)(acceleration of SHM oscillator) `omega^(2) = (g)/(R )` `therefore omega = sqrt((g)/(R ))` `therefore (2pi)/(T)= sqrt((g)/(R ))` `therefore T= 2pi sqrt((R )/(g))`. |
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| 33. |
Two completely identical samples of the same ideal gas are in equal volume containers with the same pressure and temperature in containers labeled A and B. The gas in containers A performs non-zero positive work W on the surroundings during an isobaric process before the pressure is reduced isochorically to 1/2 of its initial amount. The gas in container B has its pressure reduced isochorically to 1/2 of its initial amount. The gas in container B has its pressure reduced isochorically to 1/2 of its initial value and then the gas performs same non-zero positive work W on the surroundings during an isobaric process. After the processes are performed on the gases in containers A and B, which is at the higher temperature? |
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Answer» The gas in contaienr A |
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| 34. |
A long, thin bar ( alpha, L_0) is clamped rigidly at its ends at temperature theta_0. When the temperature ( theta )is increased, the expanding bar will bow out as shown. If the bowing is not too large, a fair first approximation to the shape of the bar is two equal straight segments in the form of a wide V. The arc of the bow (elevation) y_0 as a function of theta is expressed as y_0= ( L_0)/( 2) sqrt( K alpha ( theta - theta_0) ) where K=______ |
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Answer» |
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| 35. |
A man with a wrist watch falls from a tall building. Will the watch give correct time? |
| Answer» SOLUTION :YES, INDEPENDENT of G | |
| 36. |
A hollow aluminium sphere of mass 150 kg floats on water. It is observed that an additional mass of 30 kg is required to just submerge it at a temperature of 15°C. If gamma_("water") = 150 xx 10^(-6) //""^(@) C, alpha_(At) = 23xx10^(-6) //""^(@) C How much less mass is required to submerge the sphere, if the temperature is at + 35^(@) . |
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Answer» 30 kg |
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| 37. |
When air resistance is taken into account while dealing with the motion of the projectile which of the following properties of the projectile, shows an increases? |
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Answer» range |
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| 38. |
In a Quinck's experiment , the sound intensity being detected at an appropriate point , changes from minimum for the second time , when the slidable tube is drawn apart by 9.0 cm. If the speed of sound in air be 336 m//s, then what is the frequency of this sounding source ? |
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Answer» Solution :As the SLIDING tube is pulled out by a distance of `9.0 cm`, the path DIFFERNCE EFFECTED is ` 2 xx 9.0 cm = 18.0 cm`. INITIALLY , the position of the tube corresponds to the minimum intensity . When maximum intensity is heard for the first time , tjhe path differnce increased by `lambda//2` and when heard for the second time , there is an additional path differnce of `lambda`. THEREFORE , the total path difference ` = ( lambda)/(2) + lambda = ( 3 lambda)/(2)` `( 3 lambda)/(2) = 18.0 cm` `lambda = 12.0 cm` The frequency of sound emitted by the source will be ` f = (v)/( lambda) = ( 336 m//s) /( 12.0 xx 10^(-2) m) = 2.8 kHz` |
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| 39. |
An aeroplane requires for take off a speed of 108 kmph the run on the ground being 100m. Mass of the plane is 10^(4)kg and the coefficient of friction between the plane and the ground is 0.2. Assuming the plane accelerates uniformly the minimum force required is (g = 10ms^(-2)) |
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Answer» `2XX10^(4)N` |
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| 40. |
A cylindrical drum, open the top, contains 30 liters of water. It drains out through a small opening at the bottom. 10 liters of wates comes out in time t_(1) the next 10 liters in further time t_(2) and the last 1 liters in futher time t_(3). Then |
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Answer» `t_(1)=t_(2)=t_(3)` |
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| 41. |
Torque of a force overline(F)-2hat(i)+3hat(j) acting at apoint (1m, 0,3m) about an axis in xy plane whose equation is y=3x is : |
| Answer» Answer :A | |
| 42. |
A Satellite orbits the earth at a height of 400 km above the surface . How much energy must be expended to rocketthe satellite out of the gravititionalinfluence of earth? Mass of the satellite is 200 kg , mass of the earth = 6.0xx10^24 kg , radius of the earth =6.4xx10^6 m , G=6.67xx10^(-11) Nm^2 kg^(-2) |
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Answer» Solution :Total ENERGY of orbiting satellite at a HEIGHT `h=(-GMm)/(2(R+h))` Energy expended to ROCKET the satellite out of the earth.s gravititional field = - (total energy of the satellite) `=(GMm)/(2(R+h))=((6.67xx10^(-11))XX(6xx10^24)xx200)/(2(6.4xx10^6+4xx10^5))=5.9xx10^9` J |
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| 43. |
Four spheres A, B, C and D of different metals but of same radius and same surface finish are kept at same temperature. The ratio of their unit area at a mean distance 'b' from the sun, densities and specific heats are 2: 3: 5: 1 and solar constant Sand is given by 3 : 6: 2 : 4. Which sphere will show the fastest rate of cooling (initially) (1) A(2) B(3) C(4) D |
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Answer» Solution :Since, radius or VOLUME of all the four spheres are equal, the ratio of their masses will be 2 : 3 : 5 : 1. Heat capacity= (mass) (specific heat) `therefore`Ratio of heat capacities will be 6: 18 : 10 : 4. The sphere having the minimum heat capacity will show the fas test rate of cooling. `therefore ` CORRECT option is (4) |
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| 44. |
("he")/( 4pim) has the same dimensions as (h= Planck.s constant, e = charge, m = mass,) |
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Answer» MAGNETIC moment |
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| 45. |
Which of the following terms represent the same with reference to radiation ? |
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Answer» Emissivity |
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| 46. |
For a Hydrogen gas a) C_(P)=5R//2 b) C_(V)=5R//2 c) C_(P)//C_(V) is always greater than 1 |
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Answer» both a and B are TRUE |
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| 47. |
Graphs below show the variation of the frictional force against the force on a block applied parallel to the surface of contact of the block with a rough horizontal plane. Out of A, B, C,D which one is thecorrect graph? |
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Answer»
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| 48. |
Which of the following sets have different dimension ? |
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Answer» Pressure, Young's MODULUS, Stress |
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| 49. |
Explain why water does not run out of a dropper unless it's rubber bulb is pressed. |
| Answer» Solution :Because the TIME till RUBBER is not pressed the pressure inside DROPPER is equal to pressure outside the dropper .But when rubber is pressed, pressure increases and the WATER GOES outside the dropper. | |
| 50. |
A person throws a bottle into a dustbin at the same height as he is 2m away at an angle of 45^(@). The velocity of thrown is |
| Answer» ANSWER :D | |
