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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. |
Consider two points, one above the surface of the earth and the other inside the surface of the earth at the same distance of 1 km. Where is .g. greater ? |
| Answer» SOLUTION :Below the SURFACE of the EARTH. | |
| 2. |
In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10^(-10) m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus. |
| Answer» Solution :LOCATED on the LINE joining H and C1 NUCLEI at a distance of `1.24 Å` from the H end. | |
| 3. |
A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18.The trolley accelerates from rest with 0.5 m s^(-2) for 20 s and then moves with uniform velocity.Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley. |
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Answer» Solution : (a) MAXIMUM frictional force possible for opposing impending relative motion between the block and the TROLLEY `= 150 xx 0.18 = 27 N`,which is more than the frictional force of `15 xx 0.5 = 7.5 N ` needed to accelerate the BOX with the trolley. When the trolley MOVES with uniform velocity, there is no force of friction acting on the block. (b) For the accelerated (non-inertial) observer, frictional force is opposed by the pseudo- force of the same magnitude, keeping the box at rest relative to the observer.When the trolley moves with uniform velocity there is no pseudo-force for the moving (inertial) observer and no friction. |
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| 4. |
Two identical cars A and B are moving at 36kmph. A goes on a bridge, convex upward and 'B' on concave upward. If the radius of curvature of bridge is 20m, the ratio of normal forces exerted on the cars when they are at the middle of bridges (g= 10 ms^(-2)) |
| Answer» ANSWER :A | |
| 5. |
A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K, If radius of the ball be R, then the fraction of total energy associated with its rotational energy will be |
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Answer» `(K^(2))/(K^(2)+R^(2))` `:. K_(R)=((K^(2))/(K^(2)+R^(2)))K_("Total") or (K_(R))/(K_("Total"))=(K^(2))/(K^(2)+R^(2))`. |
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| 6. |
If density of Earth increased 4 times and its radius become half of then our weight will be........ |
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Answer» FOUR TIMES its present value |
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| 7. |
One person moves towards a stationary source of sound with a speed (1)/(5) times the speed of sound. Sound emitted from the source has wavelength lamdaand frequency f. Frequency and wavelength of sound heard by this person are respectively ...... |
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Answer» `0.8f,0.8 LAMDA`  We have, `(f _(L))/( v + v _(L)) = (f _(S))/( v + v _(S))` `therefore (f _(L))/( v + (v)/(5)) = (f _(S))/( jv + 0)` `therefore (f_(L))/( (6v)/( 5)) = (f)/(v)` `therefore f _(L) = (6)/(5) f = 1.2 f` Here, velocity of sound wave w.r.t. listener is, `v = v _(L) = v + (v)/(5) = (6v)/(5) = 1.2v` Frequency of sound wave heard by listener is, `f _(L)1.2f` Now, if wavelength of sound wave heard by the person is `lamda _(L)` then, `v = v _(L) = f _(L) lamda _(L)` `thereforef 1.2 v = 1.2 f xx lamda _(L)` `therefore lamda _(L) = (v)/(f)` `therefore lamda _(L) = lamda ""(because v = f lamda)` |
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| 8. |
Find the gravitational field and potential due to these two concentric thin spherical shells at distance r( gt R_2)from the centre. |
| Answer» SOLUTION :`-G ((m_1+m_2))/(R ), (G(m_1+m_2))/(r^2) ` | |
| 9. |
A uniform disc of radius r and mass m is free to rotate on a frictionless pivot through point P on its rim as shown in Fig. The disc is at rest and is released from the position (1). (i) What is the speed of the centre of mass when the disc reaches the position (2). (ii) Also rind the speed of the lowest point on the disc in the dashed position. |
| Answer» SOLUTION :`mgh = (1//2) IOMEGA^(2), I = ml^(2), I = ml^(2)//3, l = 1//2, omega = sqrt(6)g, alpha = tau//l = MG XX (1//4) (m//12) = 3 g` | |
| 10. |
The volume of one mole of an ideal gas with the adiabatic exponent y is changed according to relation V = a/T , where a is a constant. Find the amount of heat absorbed by the gas in the process, if the temperature is increased by DeltaT |
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Answer» |
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| 11. |
A physical quantity P is given by P=(A^(3)B^((1)/(2)))/(C^(-4)D^((3)/(2))). Due to which physical quantity produced the maximum percentage error in P ? |
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Answer» |
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| 12. |
Two forces are such that the sum of their magnitudes is 18 N, the resultant is sqrt(228)when they are at 120°. Then the magnitude of the forces are |
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Answer» 12N, 6N |
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| 13. |
Two wedges each of mass 600 g are placed next to each other on a rough horizontal surface. The coefficient of static friction between the wedges and the surface is 0.4. A cube of mass 'M' is balanced on the wedges as shown in the figure. If there is no friction between the cube and wedges, the largest mass 'M' of the cube that can be balanced without motion of the wedges in __________ kg. |
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Answer» `0.8` |
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| 14. |
The initial and final temperatures are recorded as (40.6 pm 0.3)^@C and (50.7pm0.2)^@C . The rise in temperature is |
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Answer» `10.1^@C` |
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| 15. |
Newton's second and third laws of motion lead to the conservation of |
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Answer» LINEAR momentum |
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| 16. |
In case of stationary wave, amplitudes of particles in any one loop are same. |
| Answer» SOLUTION :False. (ACTUALLY their PHASE are same in any given loop). | |
| 17. |
A boat of mass 40 kg is floating on water. A boy of mass 10 kg on the boat moves by 8m towards the shore. The distance by which the boat moves away from the shore is |
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Answer» 4 m |
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| 18. |
Displacement of an object is proportional to t^3. Show that its acceleration is increasing with time |
| Answer» SOLUTION :GIVE `xproptot^3,x=kt^3` where K is a constant.v=dx/dt=3kt^2a=(DV)/dt=6kt` i.e,ACCELERATION `aproptot` | |
| 19. |
A particle of mass m moving with velocity v_(0) strikes a smooth wedge of mass M. After collision the ball satrts up moving an inclined plane and raises to height 'h'. Then : |
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Answer» Particle is at REST w.r.t. GROUND |
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| 20. |
A black body is at a temperature of 2880 K. The energy of radiation emitted by this object witth wavelength between 499nm and 500 nm is U_(1), between 999 nm and 1000 nm is U_(2) and between 1499 nm and 1500 nm is U_(3). The Wien constant =2.88xx10^(6) nm K. |
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Answer» `U_(1) =0` |
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| 21. |
Blood flows through a main artery of area of cross section 2cm^2 with a speed of 0.25 ms^(-1). If it branches into several capillaries of total area of cross section 0.4 m^(2), the average speed of blood through each capillary is |
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Answer» `0.25 MS^(-1)` |
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| 22. |
An insect crawls up a hemispherical surface as shown the coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemisperical surface to the insect makes an angle theta with the vertical, the maximum possible value of theta is given by |
Answer» Solution : `N=mg cos theta` …….(1) `mu N = mg sin theta` …….(2) From (1) and (2) we get : `COT theta =1//mu=3 RARR theta = cot^(-1)(3)` |
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| 23. |
Two blocks of masses 10kg and 30kg are placed on x-axis. The first mass is moved on the axis by a distance of 2cm right. By what distance should the second mass be moved to keep the position of centre of mass unchanged. |
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Answer» Solution :mass of the first block, `m_(1)=10kg` mass of the second block, `m_(2)=30kg` Let `x_(1)` and `x_(1).` are POSITIONS of `m_(1)` `x_(2)` and `x_(1).` are positions of `m_(2)` initially and later respectively  In this case if `x_(cm)` is the position of centre of mass then `x_(cm)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))` then the new position of CM when blocks are shifted `x_(cm).=(m_(1)x_(1).+m_(2)x_(2))/(m_(1)+m_(2))` subtracting the above equations `x_(cm).-x_(cm)=(m_(1)(x_(1).-x_(1))+m_(2)(x_(2).-x_(2)))/(m_(1)+m_(2))` `Deltax_(cm)=(m_(1)Deltax_(1)+m_(2)Deltax_(2))/(m_(1)+m_(2))` `0=(10xx2+30Deltax_(2))/(40)` `:.Deltax_(2)=-(2)/(3)` Therefore the second block should be moved LEFT through a distance `(2)/(3)cm` to keep the position of centre of mass UNCHANGED. |
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| 24. |
The density of earth increased 4 times and its radius becomes half then our weight will .... |
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Answer» be four times its present VALUE `= ((G)/(R_e^2) xx4/3piR_e^3rho)` `W = m (4/3 PI R_e) rho ""...(1)` New weight if the density of EARTH increased four TIME and radius becomes half is `W. = m (4/3 piR_e/2G) 4rho` `= 2 [m(4/3 piR_e/2G) 4rho` `= 2 [ m(4/3 pi GR_e) rho] "" ...(2)` `:. W. = 2W` |
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| 26. |
When the object is at rest, the resultant of gravitational force and upward normal force is |
| Answer» SOLUTION :zero | |
| 27. |
A steel wire of length 1.5 m and area of cross section 1.5mm^2is stretched by 1.5cm. Then the work done per unit volume. |
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Answer» `1xx10^(7)J//m^(3)` |
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| 28. |
A compensated pendulum shown in figure is in the form of an isosceles triangle of base length l_(1)=10cm and coefficient of linear expansion alpha_(1)=18xx10^(-6) and side length l_(2) and coefficient of linear expansion alpha_(2)=12xx10^(-6). Find l_(2) so that the distance of centreof mass of the bob-from suspension centre O may remain the same at all the temperatures. |
Answer» SOLUTION : In this case the distance of center of mass from suspensionpoint O is `h=sqrt(l_(2)^(2) -(l_(1)^(2))/(4))` At some higher temperature `Deltat`, new value of h will become `h.=sqrt(l_(2)^(2)(1+alpha_(2) Deltat)^(2) -(l_(1)^(2))/(4)(1+alpha_(1)Deltat)^(2)) ` `l_(2)^(2)-(l_(1)^(2))/(4)=l_(2)^(2)(1+2alpha_(2) Deltat)-(l_(1)^(2))/(4)(1+2alpha_(1)Deltat)` (or) `alpha_(2)l_(2)^(2) =(l_(1)^(2))/(4) alpha_(1) or l_(2) =(l_(1))/(2) sqrt((alpha_(1))/(alpha_(2))) =(10)/(2) sqrt((18xx10^(-6))/(12xx10^(-6)))=6.12cm` |
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| 29. |
A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 N m^(-1). The block is pulled to a distance x = 10 cm from its equilibrium position at x = 0 on a frictionless surface from rest at t = 0. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position. |
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Answer» Solution :The block EXECUTES SHM, its angular frequency, as given by Eq. (14.14b), is `omega=sqrt(k/m)` `=sqrt((50Nm^(-1))/(1kg))` `=7.07rads^(-1)` Its displacement at any time t is then given by, x(t) = 0.1 cos (7.07t) Therefore, when the particle is 5 cm away from the MEAN position, we have 0.05 = 0.1 cos (7.07t) Or cos (7.07t) = 0.5 and hence `sin(7.07t)=(sqrt3)/2=0.866` Then, the velocity of the block at x = 5 cm is `=0.1xx7.07xx0.866ms^(-1)` `=0.61ms^(-1)` Hence the K.E. of the block, `=1/2mv^(2)` `=1/2[1kgxx(0.6123ms^(-1))^(2)]` `=0.19J` The P.E. of the block, `=1/2kx^(2)` `=1/2(50Nm^(-1)xx0.05mxx0.05m)` `=0.0625J` The TOTAL energy of the block at x = 5 cm, = K.E. + P.E. = 0.25 J we also know that at maximum displacement, K.E. is zero and hence the total energy of the system is equal to the P.E. Therefore, the total energy of the system, `=1/2(50Nm^(-1)xx0.1mxx0.1m)` =0.25J which is same as the sum of the two energies at a displacement of 5 cm. This is in conformity with the principle of CONSERVATION of energy. |
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| 30. |
Prove that there is a loss of KE during one dimensional inelastic collision. |
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Answer» Solution :In a perfectly inelastic collision, the loss in kinetic energy during collision is transformed to ANOTHER form of energy like sound, thermal, heat, light etc. Let `KE_(i)` be the total kinetic enery before collision and `KE_(f)` be the total kinetic energy after collision. Total kinetic energy before collision, `KE_(i)=(1)/(2)m_(1)u_(1)^(2)+(1)/(2)m_(2)u_(2)^(2)""...(1)` Total kinetic energy after collision, `KE_(f)=(1)/(2)(m_(1)+m_(2))v^(2)""...(2)` Then the loss of kinetic energy is Loss of `KE, DeltaQ=KE_(f)-KE_(i)` `=(1)/(2)(m_(1)+m_(2))v^(2)-(1)/(2)m_(1)u_(1)^(2)-(1)/(2)m_(2)u_(2)^(2)""...(3)` SUBSTITUTING equation `v=(m_(1)u_(1)+m_(2)u_(2))/((m_(1)+m_(2)))` in equation (3), and on simplifying expand v by using the ALGEBRA `(a+B)^(2)=a^(2)+b^(2)+2ab`, we get Loss of KE, `DeltaQ=(1)/(2)((m_(1)m_(2))/(m_(1)+m_(2)))(u_(1)-u_(2))^(2)""...(4)` |
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| 31. |
A body is projected with a velocity 60ms^(-1) vertically upwards the distance travelled in last second of its motionis [g=10ms^(-1)] |
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Answer» 35 m |
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| 32. |
Two bodies of masses m_(1)" and "m_(2) are initially at rest at infinite distance aprat. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is |
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Answer» `[2G ((m_(1)-m_(2)))/(R )]^(1/2)` |
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| 33. |
Select the correct statement from the following . |
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Answer» The angular momentum iof a comet revolving around a star remains conserved. |
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| 34. |
Four round objects namely a ring, a disc, a hollow sphere and a solid sphere with same radius R and made of same material start to roll down an inclined plane at the same time. The object that will reach the bottom third is |
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Answer» Solution :For all the FOUR objects namely the ring, disc, hollow sphere and SOLID sphere, the radii of gyration K are `R, sqrt((1)/(2))R, sqrt((2)/(3))R, sqrt((2)/(5))R`. With numerical values the RADIUS of gyration K are IR, 0.707R, 0.816R 0.632R respectively. The expression for time taken for rolling has the radius of gyration K in the numerator as per equation `Vsqrt((2gh)/((1+(K^(2))/(R^(2)))))` `t= sqrt((2h(1+(K^(2))/(R^(2))))/(g sin^(2) THETA))` The one with least value of radius of gyration K will take the shortest time to reach the bottom of the inclined plane. The ORDER of objects reaching the bottom is first, solid sphere, second, dise: third, hollow sphere and last, rin |
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| 35. |
In the given figure, an ideal gas changes its state from state A to state C by two paths ABC and AC. Find the path along which work done is the least. |
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Answer» Solution :As work done `W= int PdV`= area under P-V curve, so `W_(ABC)=W_(AB) +W_(BC )i.e.,W_(ABC) = 0+15 XX 4 =60andW_(AC )=1/2[5+15]xx (6-2) =40 J` So work done along AC is LEAST |
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| 36. |
The physical quantities related to linear motion and the physical quantities related to angular motioin are given in column I and column II respectively. Match the physical quantities in both column. |
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Answer» 1-I,2-v,3-iv,4-ii |
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| 37. |
Energy required to break one bond in DNA is approximately |
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Answer» `-1EV` |
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| 38. |
A bullet of mass m moving with velocity v strikes and becomes embedded at the edge of a cylinder of mass M and radius R_(0). As shown in Fig. The cylinder, initially at rest, begins to rotate about its symmetry axis, which remains fixed in position. Assuming no frictional torque, what is the angular velocity of cylinder after this collision (mv)/((1/xM+m)R_(0)) where 'x' is |
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Answer» |
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| 39. |
A body is moving up an inclined plane of angle q with an initial kinetic energy E. The coefficient of friction between the plane and body is m. The work done against friction before the body comes to rest is |
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Answer» `(mu COS theta)/(E cos theta + sin theta)` |
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| 40. |
A person, with outstretched arms, is spinning on a rotating stool. He suddenly brings his arms down to his sides. Which of the following is true about his kinetic energy K and angular momentum L? |
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Answer» Both K and L increase |
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| 41. |
(A): The displacement of a particle starting from rest is directly proportional to the cube of the time of travel then the particle is moving with non uniform acceleration. (R): Acceleration overset(-)(a)=(d^(2) vecs)/(dt^(2)) |
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Answer» Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A) |
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| 42. |
How many radians are there in one minute? |
| Answer» Solution :`180^(@)" degrees " = PI` RADIANS or `1^(@)=pi"/"180"rad"= pi"/"10,800` rad. | |
| 43. |
Two bodies of masses m_(1) and m_(2),respectively, are thrown vertically upwards withequal initial linear momenta. Then the ratio of maximum heights attained by the first to that by the second body is |
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Answer» `m_(2):m_(1)` |
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| 44. |
How can you lower a 100 kg body from the roof of a house using a cord with a breaking strength of 80 kg weight without breaking the rope ? (1 kg weight = g N) |
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Answer» `1.96 m//s^(2)` DOWNWARDS |
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| 45. |
A ball of radius 'a' is made to oscillate in a bowl of radius 'b' then time period of vibration of the ball is |
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Answer» `2PI SQRT(a/G)` |
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| 46. |
A diatomic ideal gas is heated at constant volume until its pressure is trebled (three times). It is again heated at constant pressure until its volume is doubled. Find the molar capacity for the whole process. |
| Answer» SOLUTION :`C=(31)/( 10)R ` | |
| 47. |
How can general body be obtained? |
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Answer» SOLUTION :General body is considered as a system of particle INSTEAD of considering only a one particle. By considering a general body as a system whole mass of it concentrated on its centre of mass and all external forces acting on the centre of mass. Such type of assumption LEADS to DETERMINE the MOTION of centre of mass of system. |
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| 48. |
Find the point at which the gravitational force acting on any mass is zero due to the earth and the moon system. (such a point is called neutral point). The mass of the earth is 81 times the mass of the moon and the distance between the earth and the moon is 3,85,000km. |
Answer» Solution : Let `m_1 and m_2`be the masses of the earth and the MOON separated by a distance d. Consider an OBJECT of mass mat a point P, which is at a distance X from `m_1` . The FORCE due to the mass `m_1` onthe mass m is `F_(1) = (G"mm"_1)/X^2`towards `m_1`and The force due to the mass m, on the mass m is `F_(2) = ("Gmm")/((d-x)^2)` towards `m_2` If the resultant force on the mass .m. is to be zero, `F_1`must be equal to `F_2`in magnitude and they are oppositely directed `("Gmm"_1)/x^2= ("Gmm"_2)/((d-x)^2)implies`x = 38,500 KM from moon, here `m_1`= M mass of the moon, `m_2 = 81 M` mass of the earth |
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