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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 swimmer wishes to cross a river 500m width flowing at a rate u. his speed w.r.t. still water is v. for this he makes an angle theta with the vertical as shown in the given figure then: |
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Answer» to CROSS the river in minimum time `theta=0^(@)` |
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| 2. |
Molecular forces exist between the molecules of a liquid in a container. The molecules on the surface have unequal force leading to a tension on the surface. If this is not compensated by a force, the equilibrium of the liquid will be difficult task. This leads to an excess pressure on the surface. The nature of the meniscus can inform us of the direction of the excess pressure. The angle of contact of the liquid decided by the forces between the molecules, air and container can make the angle of contact.The direction of the excess pressure in the meniscus of a liquid of angle of contact ( 2pi )/(3)is |
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Answer» UPWARD |
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| 3. |
If y=x-1/2x^(2) is the equation of a trajectory, find the time of flight. |
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Answer» Solution :We have `y=X-1/2x^(2)-x(1-x//2)` If y=0, then either x=0 or x=2 HENCE the arange of the motion is 2. For HALF the range x=1 then `y=1//2` Hence maximum height attained `=1//2` Time to reach maximum height `t=sqrt((2Y)/g)=sqrt(1/g)` Time of FLIGHT `T=2t=2sqrt(1/g)` i.e. `T=2/(sqrt(g))` |
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| 4. |
A stationary bomb explodes into three fragments. If the linear momentum of two A pieces are 4hati and 3hatJ unit respectively, then what is the value of the third fragment ? |
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Answer» <P>7 UNIT  `:. , p= sqrt(p_(1)^(2) + p_(2)^(2)) = sqrt(4hat(i)^(2)+ 3HAT(j)^(2))` `:. , p=sqrt(16+9)= sqrt(25) = 5 units` |
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| 5. |
A body is projected at angle 30^(@) to horizontal with a velocity 50ms^(-1). Its time of flight is (g=10m//s^(2)) |
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Answer» 4s |
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| 6. |
If a source emitting waves of frequency f moves towards an observer with a velocity v//3 and the observer moves away from the source with a velocity v//4, the apparent frequency as heard by the observer will be (v= velocity of sound) |
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Answer» `9f//8`  `n.=n((V-V_(0))/(V-V_(S)))=f((V-(V)/(4))/(V-(V)/(3)))=f(((3V)/(4))/((2V)/(3)))` |
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| 7. |
A ball hangs from a string inside a rail road car moving along a straight track. The string is observed to be inclined towards the rear of the car making a constant small angle with the vertical. It shows that the car is |
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Answer» MOVING with a uniform acceleration |
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| 8. |
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27^(@)C. What is the change in the diameter of the hole when the sheet is heated to 227^(@)C ? (Coefficent of linear expansion of copper = 1.70 xx 10^(-5) ""^(@)C^(-1)) |
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Answer» `1.44 xx 10^(-2)cm` `S_(1) = (pi D_(1)^(2))/(4) = (pi)/(4) xx (4.24)^(2) cm^(2)` If `D_(2)` cm is the DIAMETER of the hole at `227^(@)C`, then area of the hole at `227^(@)C`, `S_(2) = (pi D_(2)^(2))/(4) cm^(2)` Coefficient of superficial expansion of copper is `beta= 2 ALPHA = 2 xx 1.70 xx 10^(-5) = 3.4 xx 10^(-5) ""^(@)C^(-1)` Increase in area `=S_(2) - S_(1) = beta S_(1) Delta T` or `S_(2) = S_(1) + beta S_(1) Delta T = S_(1) (1 + beta Delta T)` or `(pi D_(2)^(2))/(4) = (pi)/(4) (4.24)^(2) [1 + 3.4 xx 10^(-5) xx (227-27)]` or `D_(2)^(2) = (4.24)^(2) xx 1.0068 or D_(2) = 4.2544 cm` Change in diameter `=D_(2)-D_(1)= 4.2544-4.24` `=0.0144cm = 1.44 xx 10^(-2)cm` |
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| 9. |
These questions consists of two statements each printed as Assertion and Reason. While answering these question you are required to choose any one of the following five reponses (a) If both Assertion and Reason arecorrect and Reason is the correct explanation of Asserrtion. (b)If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion. (c ) If Assertion is true but Reason is false. (d) If Assertion is false but Reason is true. AssertionAtstable equilibrium position of a body, kinetic energy cannot be zero. Because it is maximum. ReasonDuring oscillations of a body, potential energy is minimum at stable equilibrium position. |
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Answer» |
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| 10. |
The frequency of a particle performing SHM is 12 Hz. Its amplitude is 4 cm. Its initial displacement is 2 cm towards positive extreme position. Its equation for displacement is |
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Answer» `x= 0.04 cos(24pit+ (PI)/(6))` |
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| 11. |
The orbiting time .T. of a surface satellite is given by the relation |
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Answer» `2pisqrt(R/G)` |
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| 12. |
The diagram shows a concavo- convex lens. What is the condition on the refractive indices so that the lens is diverging? The refractive index of the lens is mu_2 |
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Answer» Solution :`mu_3/v_f+mu_1/u_0=(mu_2-mu_1)/(-2R)+(mu_3-mu_2)/(-R)` For DIVERGENCE `V_f` is NEGATIVE so, `(mu_1-mu_2)/(2R) lt (mu_3-mu_2)/ R` `IMPLIES mu_1-mu_2lt 2mu_3 -2mu_2 implies mu_1+mu_2 lt 2mu_3` |
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| 13. |
On a long horizontally moving belt, a child runs to and fro with a speed 9 km h""^(-1) (with respect to the belt) between his father and mother located 50 m apart. The belt moves with speed of 4 km h""^(-1). For an observer on a stationary platform, the speed of the child runing in the direction of motion of the belt is |
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Answer» `4km H^(-1)` |
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| 14. |
(A): Workdone by an ideal gas in a cyclicprocess must be zero (R): During a cyclic process, a gas will cometo same state of its parameters. |
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Answer» If both (A) and (R) are TRUE and (R) is THECORRECT EXPLANATION of (A) |
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| 15. |
A wire length L and cross sectional area A is made of material of Young's modulus Y . It is strectched by an amount X . The work done is |
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Answer» '(YXA/2L)' |
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| 16. |
The displacement y of a wave travelling in the x- direaction is given by y = 10^(-4) sin (600t -2x + pi//3) Where x is expreseend in metres and t is seconds. What is the speed of the wave motion (in ms^(-1)) ? |
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Answer» Solution :`y = 10^(-4) sin(600t - 2x + pi/3) and y = a sin (omegat -kx +pi/3)` `:.omega = 600 RED s^(-1),K = 2 red^(-1)m` `v = omega/k= 600/2 = 300ms^(-1)` |
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| 17. |
(A): Power developed in circular motion is always zero. (R ): Workdone in case of circular motion is zero. |
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Answer» Both 'A' and 'R' are TRUE and 'R' is the correct EXPLANATION of 'A's |
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| 18. |
Ray optics is valid, when the dimensions of objects interacting with light |
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Answer» of the same ORDER as the WAVELENGTH of light |
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| 19. |
Obtain an expression of acceleration produced by gravity of earth. OR Obtain general equation of gravitation force at distance r from the centre of earth and derive the equation of acceleration due to gravity on the surface of earth. |
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Answer» SOLUTION :`implies` Imagined a earth to be a sphere made of a large number of concentric spherical shells with the smallest one at the centre and the LARGEST one at its surface. `implies` A point outside the earth is outside all the shells. `implies` Thus, all the shells exert a gravitational force at the point outside just as if their masses are concentrated at their common centre. `implies` For a point inside the earth, the situation is different. This is illustrated in figure. ` (##KPK_AIO_PHY_XI_P1_C08_E01_012_Q01.png" width="80%"> `implies` Consider the earth to be made up of concentric shells. `implies`A point mass m is situated at a distance r `(r ltR_E)` `implies`from the centre. The point P LIES outside the sphere of radius r. For the shells of radius greater than r, the point P lies inside. Hence, no gravitational force exert on mass m kept at P `implies`If mass of particle is m at P, mass of sphere m, with radius r, then the force on the mass m at P has a magnitude of force `implies F =(GmM_(r))/r^2 ""...(1)` `implies`We assume that the entire earth is of uniform density. Hence its mass is `M_E = (4/3 piR_E^3)rho` `:. 4/3 pirho = M_E/ R_E^3 ""...(2)` `implies` But `M_r = 4/3 PIR^(3) rho ""...(3)` `implies`Putting value of equation (3) in equation (1) `implies :. F = (Gm)/(r^2)((4)/(3) pir^3rho)` `implies F = Gm (4/3pi rho)r "" ... (4)` `implies`Value of equation (2) in equation (4), Equation indicates for F r, r `:. F = (GM_Em)/(R_E^2) ""....(6)` `implies ` The acceleration experienced by the mass m is denoted by g. According to Newton.s second law, `F = mg ""... (7)` `implies`Equating equation (6) and (7), `:. mg = (GM_(E)m)/(R_E^2)` `:. g = (GM_E)/(R_E^2)""....(8)` `implies` Acceleration g is readily measurable. `G = 6.61 x 10^(-11)"Nm"^(2) //kg^(2), M_E = 6 xx 10^(24) kg` `R_E = 6400 km,`then value g can be obtained. `g = 9.77050 m//s^(2) = 9.8 m//s^(2)`. `implies`The measurement of G is combined with knowledge of g and `R_E`enables one to estimate `M_E`. This is the reason why there is a popular statement regarding Cavendish : "Cavendishweighed the earth". |
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| 20. |
A jar is filled with two non mixing liquide 1 and 2 having densities rho_(1) and rho_(2) respectively A solid ball, made of a material of density rho_(3), is dropped in the jar. It comes to equilibrium in the position shown in the figure. Which of the following is true for rho_(1),rho_(2) and rho_(3) |
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Answer» `rho_(1)gtrho_(3)gtrho_(2)` |
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| 21. |
Calculate the ratio of electric and gravita-tional forces between two electrons. (charge of the electron (e) = 1.6 xx 10^(-19) c, mass of the electron (m) = 9.1 xx 10^(-31) kg, (1)/(4pi epsilon_(0)) = 9 xx 10^(19) Nm^(2)C^(-2), universal gravitational constant G = 6.67 xx 10^(-11) Nm^(2) kg^(-2)) |
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Answer» Solution :The electric force `Fe = (E^(2))/(4pi epsilon_(0) r^(2))` The gravitational force `Fg = (Gm^(2))/(r^(2))` `therefore (Fe)/(Fg) = (1)/(4 pi epsilon_(0)) xx (e^(2))/(m^(2)G)` `= 9 xx 10^(9) xx ((1.6 xx 10^(-19))^(2))/((9.1 xx 10^(-31))^(2) xx 6.67 xx 10^(-11)) = 4.17 xx 10^(42)` |
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| 22. |
If ecternalforce actingon a systemis constant . It meansthat the velocityof the CM. |
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Answer» is CONSTANT |
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| 23. |
In question. Let us take position of mass when the spring is unstretched as x=0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass it at the moment we start the stoop watch (t=0) , the mass is (a) at the mean position (b) at the maximum streteched position, and (c) at the maximum compressed position. In what way do these functions for S.H.M. differ from each other, in frequency, in amplitude or initial phase ? |
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Answer» Solution :Here, `a=2.0cm, omega=sqrt((K)/(m))=sqrt((1200)/(3))=20s^(-1)` (a) As time is noted from the mean position, HENCE using `x=asinomegat`, we have `x=2sin20t` (b) At maximum stretched position, the body is at the extreme RIGHT position, with and intial phase of `pi//2` rad. Then, `x=1sin(omegat+(pi)/(2))=acosomegat=2cos20t` (c) At maximum compressed position, the body is at the extreme left position, with an intial phase of `3ph//2` rad. Then, `x=asin(omegat+(3pi)/(2))=-acosomegat=-2cos20t` The functions neither differ in AMPLITUDE nor in frequency. They differ in initial phase. |
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| 24. |
A ball is dropped from the top of a tower of height 78.4 m Another ball is thrown down with a certain velocity 2 sec later. If both the balls reach the ground simultaneously, the velocity of the second ball is |
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Answer» `29.4m//s` |
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| 25. |
A vehicle is travelling along unbanked curved path. If the friction between the road and tyres suddenly disappears then the vehicle |
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Answer» Moves along TANGENTIAL DIRECTION |
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| 26. |
A boxer becomes more tired boxing in the air than against the opponet. Why? |
| Answer» SOLUTION :The boxer does not gain any momentum fighting in the AIR and more likely loses energy QUICKLY. | |
| 27. |
A body is thrown vertically up reaches a maximum height of 78.4 m. After what time it will reach the ground from the maximum height reached? |
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Answer» 4 sec |
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| 28. |
What happens to the change in internal energy of a gas during (a) isothermal expansion |
| Answer» SOLUTION :During ISOTHERMAL expansion, temperature remains constant and `Delta=0`. | |
| 29. |
What is the moment of inertia of a rod of mass M, length l about an axis perpendicular to it through one end? |
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Answer» Solution :For the rod of mass M and length L, `I = Ml^(2)//12`. Using the parallel axes theorem, `I ′ = I + Ma^(2)` with`a = l//2`we GET, `I.=M(l^(2))/(12)+M((l)/(2))^(2)=(Ml^(2))/(3)` We can check this INDEPENDENTLY since I is HALF the moment of INERTIA of a rod of mass 2M and length 2l about its midpoint, `I.=2M.(4l^(2))/(12)xx(1)/(2)=(Ml^(2))/(3)` |
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| 30. |
Which parts of the curve in the figure shown represent compression and rarefaction for a longitudianal wave? |
| Answer» SOLUTION :ABC represent compression (NEGATIVE slope region) and CDE represent rarefaction (positive slope region). | |
| 31. |
Define thrust and pressure and give their unit and dimensional formula. |
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Answer» Solution :(i) The force applied on a SURFACE in a direction perpendicular to the surface is called thrust. Due to thrust LIQUID flow out from the hole of container . Thrustis a force . ItsSI unit is Newton (N).Its dimensional formula is `[M^(1)L^(1)T^(-2)]`. (ii)Pressure : Force ACTING on a surface per unit area , in a direction perpendicular to it is called pressure on the surface. If F acting perpendicular to the surface of area A then pressure `P=(F)/(A)` SI unit of pressure `(N)/(m^(2))`Pascal (Pa) CGS unit of pressure is `("dyne")/(cm^(2))`. Dimensional formula of pressure is `[M^(1)L^(-1)T^(-2)]`. |
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| 32. |
When a tap is open so that water comes out in a stream, one would observe the cross-section of water decreases as water descends? Why? |
| Answer» SOLUTION :There will be loss in gravitational POTENTIAL energy and so KINETIC energy or fluid VELOCITY INCREASES from equation of continuity cross section must decrease. | |
| 33. |
A yo-yo is resting on a perfectly rough horizontal table. Forces F_(1), F_(2) and F_(3) are applied separately as shown. The correct statement is . |
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Answer» When `F_(3)` is applied the centre of mass will MOVE to the right. |
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| 34. |
Force F on a particle moving in a straight line varies with distance d as shown in figure. The work done on the particle during its displacement of 12 m is |
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Answer» 18 J = Area under (F - d) graph = Area of rectangule ABCD + Area of triangle DCE `= 2 xx (7 - 3) + 1/2 xx 2 xx (12 - 7)` `= 8 + 5 = 13 J`. 
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| 35. |
A flywheel has a mass 200 kg and radius of gyration 0.6 m. It is given an angular speed of 150 rpm in 90 rotations starting from rest. Determine the torque assuming it to be constant that acted on the flywheel. |
| Answer» SOLUTION :`a=omega^(2)//2THETA = (2pi xx 2.5)^(2)//2 xx 2pi xx 90 =0.22 "rads^(-2), tau=Mk^(2)ALPHA = 200 xx 0.62 xx 0.2^(2)=15.8 Nm` | |
| 36. |
The tyre of a motor car contains air at 15^@C. If the temperature increases to 35^@C , the approximate percentage increase in pressure is (ignore expansion of tyre) |
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Answer» 7 |
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| 37. |
A body of mass 60 kg is pushed up with just enough force to start it moving on a rough surface with mu_(s)=0.5 and mu_(k)=0.4 and the force continues to act afterwards. What is the acceleration of the body ? |
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Answer» Solution :`m=60 kg , mu_(s)=0.5, mu_(K)=0.4` Force `f =ma=m(mu_(s)-mu_(k))G` `THEREFORE a=(mu_(s)-mu_(k))g` `= (0.5-0.4)9.8=0.98 m//s^(2)` |
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| 38. |
In the system shown in the figure all surfaces are smooth and both the pulleys are mass less. Block on the incline surface of wedge A has mass m. Mass of A and B are M = 4 m and M_(0) = 2 m respectively. Find the acceleration of wedge A when the system is released from rest. |
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Answer» |
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| 39. |
Find the weight of the atmosphere around the earth (take the radius of earth as 6400km.) |
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Answer» 1600 km `:.phi/g = -r ` `= - (5.4 xx10^7)/(6.0) =-r` `:. 9 xx 10^(7) =r` `:. 9 xx 10^(7) = Re +h` `:. H = 9000 xx 10^3 - Re` `:.h = 9000 km - 6400 km ` `= 2600 km ` |
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| 40. |
A partcle is projected with a velocity of 10sqrt(2)m//s at an angle of 45^(@) with the horizontal. Find the interval between the moments when speed is sqrt(125)m//s. |
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Answer» Solution :`(G = 10m//s^(2)), u_(n) = 10, u_(y)=0` `v^(2)=v_(x)^(2)+v_(y)^(2)` `125 = 100 + v_(x)^(2)`  `v_(y) = 5""Deltat = (2v_(y))/(g)=(2xx5)/(10)` = 1 |
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| 41. |
Answer the following: (a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)? (b) There were two fixed points in the originalCelsuis scales as mentioned above which were assigned the numver 0^(@)C and 100^(@)C respectively. on the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16K. what is the other fixed point on this (Kelvin) scale? (c ) The absolute temperature (Kelvin scale) T is related to the temperature t_(c) on the Celsius scale by t_(c) = T-273.15 Whay do we have 273.15 in this relation, and not 273.16? (d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale? |
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Answer» Solution :(a) This is on ac count of the fact that the triple point of water has a unique value i.e., `273.16K` at a unique point, where exists unique of pressure and volume. On the other hand, the melting point of ice and boiling point of water do not have unique set of values as they cnange with the change in pressure and volume. (b) The other fixed point on the Kelvin absolute scale is the absolute zero itself. (c) On Celsius scale `0^(@)C` corresponds to melting point of ice at normal pressure. The corresponding value of abslute temperature is `273.15K`. the temperature `273.16K` corresponds to the triple point of water. form the given relation, the corresponding value of triple point of water on Celsius scale =`273.16-273.15 = 0.01^(@)C`. (d) We know that Fahrenheight scale and absolute scale are related as `(T_(F)-32)/(180) = (T_(K)-273.15)/(100)` ....(i) For ANOTHER set of TEMERATURE `T_(F)^(')` and `T_(K)^(')` `(T_(F)^(')-32)/(180) = (T_(K)^(')-273.15)/(100)` ....(ii) Subtracting (i) from (ii) , we have ` (T_(F)^(')-T_(F))/(180) = (T_(K)^(')-T_(K))/(100)` or `T_(F)^(') - T_(F) = 180/100 (T_(K)^(")-T_(K))` if `T_(K)^(')-T_(K) =1K` , then, `T_(F)^(') - T_(F) = 180/100xx1=9/5` For a temperature of triple point i.e., `273.16K`, the temperature on the NEW scale is `=273.16xx9/5 = 491.69`. |
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| 42. |
Find the frequency of note emitted by a string of length 10sqrt(10)cm, under tension of 3.14kg.Radius of string is 0.5 mm and density =9.8g//cc. |
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Answer» Solution :Here, `l=10sqrt(10)cm=0.1m,` `T=3.14kg=3.14xx9.8N` `D=2r=2xx0.5mm=1mm=0.001m,` `rho=9.8g//c c=9.8xx10^(3)kg//m^(3)` As, `v=(1)/(LD)sqrt((T)/(pirho))` `:.v=(1)/(0.1sqrt(10)xx0.001)xxsqrt((3.14xx9.8)/(3.14xx9.8xxx10^(3)))` |
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| 43. |
When a particle moves in a circle with auniform speed …………. |
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Answer» its VELOCITY and ACCELERATION are both CONSTANT |
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| 44. |
A simple pendulum has a length L and a bob of mass m. The bob is oscillating with amplitude A. Show that the maximum tension (T) in the string is (for small angular displacement). T_("max") =mg [1+(A/L)^(2)]. |
Answer» Solution :Here, Mass of bob =m,Amplitude of oscillation = A, Length of PENDULUM = L, `/_BAC = THETA`  Tension `T= mg cos theta +(mv^2)/(L)""[(mv^(2))/(L) = " centripetal force "]` If `cos theta =1` `v= v_("max") " and " T= T_("max")` `T_("max") = mg +(mv_("max")^(2))/(L)"""..........."(1)` `=mg +(mA^(2)omega^(2))/(L)" but "omega^(2) = (g)/(L)` `T_("max") = mg+ (mA^(2) g)/(L xx L)""".........."(2)` `THEREFORE T_("max") = mg +(mgA^(2))/(L^2)` `therefore T_("max") = mg[1+(A/L)^(2)]`. |
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| 45. |
A force produces an acceleration of a_(1) in a body and the same force produces an acceleration of a_(2) in another body. If the two bodies are combined and the same force is applied on the combination, the acceleration produced in it is |
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Answer» `a_(1)+a_(2)` |
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| 46. |
If .A. is the A real velocity of a planet of mass M, its angular momentum is |
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Answer» M/A |
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| 47. |
A satellite orbitsthe earth at a heightof R/6 , its orbital speed ? |
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Answer» `sqrt((2GM)/R)` |
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| 48. |
What is parallax ? |
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Answer» Solution :When you hold a pencil in front of you against some specific point on the background (like WALL) and look at the pencil first through your left eye (A) (closing the RIGHT eye) and then look at the pencil through right eye (B) (closing the left eye.) You WOULD notice that position of the pencil seems to CHANGE with respect to point on wall. This is called parallax. Angle formed by two eye at position of pencil is called parallax angle The distance between the two points of observation is called basis. |
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| 49. |
A family uses 8 kW of power. Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? |
| Answer» SOLUTION :`= 200 m^(2)` | |
| 50. |
(A) : When a body is projected from the surface of the earth with a velocity greater than the escape velocity, its total mechanical energy is negative. (R) : When the body is not under the influence of the earth, its P.E is negative |
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Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
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