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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 steel drill is making 180 revolutions per minute, under a constant torque of 5 N-m. If it drills a hole in 7 s. in a steel block of mass 600 gm, rise in temperature of the block is ( s= 0.1 "cal/gm"//""^(@) C) |
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Answer» `2.6^(@) C` |
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| 2. |
Prove that the maximum horizontal range 'is four times the maximum height attained by the projectile, when fired at an inclination so as to have maximum horizontal |
| Answer» Solution :`R_max.=mu^2/g,` when `theta=45^@` Then the height ATTAINED `H=(mu^2(SIN45^@)^2)/(2G)=mu^2/(4g)=(R_max)/4` THUS `R_max=4H` | |
| 3. |
If a satellite is revolving around a planet of mass M in an elliptic orbit of semi-major axis a, show that the orbit al speed of the satellite when it is at a distance r from the focus will be given by v^(2) = GM((2)/(r)-(1)/(a)). |
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Answer» Solution :Total mechanical ENERGY of the system is `E = -(GMm)/(2A)` which remains conserved `rArr KE + PE = -(GMm)/(2a)` At a position .r. orbital speed of the satellite is V. Then `KE = (1)/(2) mv^(2), PE = -(GMm)/(r)` So, `(1)/(2) mv^(2) -(GMm)/(r) = -(GMm)/(2a)` (or) `v^(2) = GM((2)/(r) - (1)/(a))`. |
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| 4. |
What is sonometer ? Give its construction and working. Explain how to determine the frequency fo tuning fork using sonometer. |
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Answer» Solution :Sonometer is used for sound-related measurements. Using this device, the following quantities can be determined. (i) the frequency of the tuning fork or frequency of alternating current. (ii) The tension in the string. (iii) The unknow hanging mass.  Sonstruction : The sonometer is made up of a hollow BOX that is one meter long with a uniform metallic thin string attached to it. One end of the string is connected to a hook and the other end is connected to a weight hanger through a pulley as shown in Figure. Since only one string is used, it is also known as monochord. The weights are added to the free end of the wire to increases the tension of the wire. Two adjustable wooden knives are put over the board and theirpositions are adjusted to CHANGE the vibrating length of the stretched wire. Procedure : A transverse stationary or standing wave is produced. So, at the knife EDGES P adn Q, nodes are formed. In between the knife edges, anti-node sare formed. In between the knife edges, anti-nodes are formed. If the length of the vibrating element is l then `l=(lambda)/(2) rArr lambda=2l` Let f be the frequency of the vibrating element, T the TENTION of in the string and `mu` the mass per unit length of the string. then using the the EQUATION `v=sqrt((T)/(mu))` measured in `ms^(-1)`, we get `f=(v)/(lambda)=(1)/(2l) sqrt((T)/(mu))` in Hertz. Let `rho` be the density of the material of the string and d be the diameter of the string. then the mass per unit length `mu`. `mu="Area"xx"Density"= pir^(2)rho` `=(pirhod^(2))/(4)` Frequency `f=(v)/(lambda)` `=(l)/(2l)sqrt((T)/((pid^(2)rho)/(4)))` `:.f=(1)/(ld) sqrt((T)/(pirho))` |
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| 5. |
A particle moving along straight line has velocity v = mus^2where s is the displacement. If s = s_0then which of the following graph best represent s versus t |
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Answer»
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| 6. |
A stone is projected horzontally with a velocity 9.8 ms^(-1) from a tower of height 100m. Its velocity one second after projection is |
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Answer» `9.8 MS^(-1)` |
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| 7. |
The triple point temperature of water is _______ |
| Answer» Solution :100 K | |
| 8. |
A particle is moving with SHM in a straight line. When the distance of the particle from the equilibrium position has valuesx_(1) andx_(2), the corresponding value of velocities are u_(1) and x_(1). Show that the time period of oscillation is given by t = 2.... |
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Answer» SOLUTION :When `x= x^(1), V = U^(1) ` when `x = x^(2), v = u^(2) `As`v = omegasqrtA^(2)-x^(2) ` therefore u_(1)=omegasqrtA^(2)-x_(1)^(2) or u_(1)^(2)=OMEGA^(2)(A^(2)-x_(1)^(2)) and u_(2)=omegasqrtA^(2)-x_(2)^(2) or u_(2)^(2)=omega^(2)(A^(2)-x_(2)^(2)) SUBTRACTING (ii) from (i), we get |
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| 9. |
What will be distance covered byu vehicle after applying breakes with retardation 8 m//s ^(2)if its velocity is 72 km/h ? |
| Answer» SOLUTION :`d _(s) = 25 m` | |
| 10. |
Two blocks are kept on an inclined plane and tied to each other with a mass-less string. Coefficient of friction between m_(1) and inclined plane is mu_(1) & that betweeb m_(2) & the inclined is mu_(2). Then : |
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Answer» The tension in the string in ZERO if `mu_(1) gt mu_(2)` |
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| 11. |
If Kelvin temperature of an ideal black body is doubled, what will be the effect on energy radiated by it ? |
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Answer» Solution :`EalphaT^(4):.(E_(2))/(F_(1))=((T_(2))/T_(1))^(4)=((2T_(1))/(T_(1)))^(4)=16` `:. E_(2)=16E_(1)` |
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| 12. |
Which of the following represents (a) a progressive wave and (b) a stationary wave? (a) y=2 cos 5x sin 9t,""(b) y=2 sqrt(x-vt), (c) y=3 sin (5x-0.5 t)+4 cos (5x-0.5 t) (d) y= cos x sin t+cos 2x. Sin 2t. If progressive, find its velocity. |
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Answer» |
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| 13. |
When the number 0.046508 is reduced to 4 significant figures, then it becomes |
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Answer» `0.0465` |
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| 14. |
Bats can ascertain distances, directions, nature and size of the obstacle without any eye, explain how ? |
| Answer» Solution :Bats emit ultrasonic waves of very SMALL wavelength (high frequencise) and so high SPEED. The reflected waves from an obstacle in their path give them idea about the DISTANCE, DIRECTION,nature and size of the obstacle. | |
| 15. |
State total linear momentum is conserved in all collisions. |
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Answer» SOLUTION :(i) Linear momentum is conserved in all collision processes. When two bodies collide, the mutual implusive force acting between them during the collision time `(Delta t)` produces a change in their respective momenta. (ii) That is, the FIRST body exerts a force `VEC(F)_(12)` on the second body. From Newton's third law, the second body exerts a force `vec(F)_(21)` on the first body. This causes a change in momentum `Delta vec(p)_(1)` and `Delta vec(p)_(2)` of the first body and second body respectively. Now, the relations could be written as, `Delta vec(p)_(1)=vec(F)_(12)Delta t ""`.....(1) `Delta vec(p)_(2)=vec(F)_(21)Delta t ""`...(2) (III) Adding equation (1) and equation (2), we get `Delta vec(p)_(1)+Delta vec(p)_(2)=vec(F)_(12)Delta t + vec(F)_(21)Delta t=(vec(F)_(12)+vec(F)_(21))Delta t` According to Newton's third law `vec(F)_(12)=-vec(F)_(21)` `Delta vec(p)_(1)+Delta vec(p)_(2)=0 "" Delta (vec(p)_(1)+vec(p)_(2))=0` (iv) Dividing both sides by `Delta t` and taking linit `Delta t to 0` we get `lim_(Delta t to 0)(Delta (vec(p)_(1)+vec(p)_(2)))/(Delta t)=(d(vec(p)_(1)+vec(p)_(2)))/(dt)=0` `THEREFORE p_(1)+p_(2) =` constant. (v) The above expression implies that the total linear momentum is a conserved quantity. |
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| 16. |
The dimension of (1)/(2) epsi_(0)E^(2). Where epsi_(0) is permittivity of free space and E is electric field is |
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Answer» `[ML^(2)T^(-2)]` |
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| 17. |
A person swims at 135^(@) to current of river, to meet target on reaching opposite point. The ratio of person.s velocity to river water velocity is |
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Answer» `sqrt(3) :1` |
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| 18. |
The value of 'g' at a particle point on the surface of the earth is 9.8 ms^(-2). Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass, the value of g at the same point (assuming that the distance of the point from the centre of the earth does not shrink) will now be |
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Answer» `4.9 MS^(-2)` |
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| 19. |
A source and an observer are approaching one another with the relative velocity 40 ms^(-1). If the true source frequency is 1200 Hz, deduce the observed frequency under the following conditions: (i) All velocity is in the source alone. (ii) All velocity is in the observer alone. The source moves in air at 100 ms^(-1) towards the observer, but the observer also moves with the velocity v_(0) in the same direction. |
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Answer» SOLUTION :Here V=1200Hz, `v=340ms^(-1)` RELATIVE velocity `=40ms^(-1)` (i) Here source move TOWARDS the stationary observer, `v_(s)=+40ms^(-1),v_(0)=0`  `"Train"overset(S""+ve""v_(s))tooverset(*)"Observer "v_(0)=0` `v'=((v-v_(0)))/((v-v_(0)))xxv=(340-0)/(340-40)xx1200` `=(340xx1200)/(300)=1360Hz`. (ii)Here observer moves towards the stationary source, `v_(0)=-40ms^(-1),v_(s)=0` `S*v_(s)=0""overset(-ve""v_(0)""O)to` `v'=(v-v_(0))/(v-v_(s))xxv=(340+40)/(340-0)xx1200` `=(380)/(340)xx1200=1341Hz`.  (iii)Here observer and source move in the same direction, `v_(s)=100ms^(-1),v_(0)=100-40=60ms^(-1)` `overset(S""+ve""v_(s))to""overset(S""+ve""v_(s))to` `v'=(v-v_(0))/(v-v_(s))xxv=(340-60)/(340-100)xx1200` `=(280)/(240)xx1200=1400Hz`. 
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| 20. |
A mass of 2kg is suspended with thread AB Thread CD of the same type is attached to the other end of 2kg mass Lower thread is pulled gradually harder and harde in the downward direction so as to apply force on AB Which of the threads will break and why |
Answer» Solution :The THREAD `AB` will BREAK EARLIER than the thread `CD` This is becase force on `CD` = applied force `f` on `AB =f` plus force due to 2 kg weight as shown in . 
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| 21. |
A body is projected from the ground with some angle to the horizontal. What happens to the angular momentum about the initial position in tis motion? |
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Answer» decreases |
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| 22. |
Three point masses are at the corners of an equilateral triangle of side a. Their separations do not change when the system rotates about the centre of the triangle. For this, the time period of rotation must be proportional to a) a^(3//2)b) a c) m d ) m^(1//2) |
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Answer» only a & C are TRUE |
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| 23. |
Show that for a particle executing simple harmonic motion a. the average value of kinetic energy is equal to the average value of potential energy. b. average potential energy = average kinetic energy =1/2 (total energy) (Hint : average kinetic energy = < kinetic energy > =1/Tint_(0)^(T) (Kinetic energy) dt and average Potential energy = =1/Tint_(0)^(T) (potential energy) dt |
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Answer» <P> Solution :a). `(E_(k))"ave"=1/Tint_(0)^(T)E_(k)dt=1/Tint_(0)^(T)(1)/(2)mA^(2)omega^(2)cos^(2)omegatdt``=1/(4T)mA^(2)square^(2).T""...(1)` `(E_(p))"ave"=1/Tint_(0)^(T)E_(p)dt=1/Tint_(0)^(T)(1)/(2)mA^(2)omega^(2)SIN^(2)omegatdt` `=1/(2T)momega^(2)A^(2)int_(0)^(2)((1-cos2omegat))/(2).dt` `=1/(4T).momega^(2)A^(2)[t-(sin2omegat)/(2omega)]_(0)^(T)` `(E_(p))_("ave")=1/(4T)msquare^(2)A^(2)T""..(2)` `=msquare^(2)A^(2)//4` From EQUATIONS (1) and (2) it can show that the average K.E. is equal toaverage P.E. (Total Energy) `TE=1/2msquare^(2)A^(2)` `(E_(k))_("ave")=(E_(p))_("ave")=1/2.((1)/(2)momega^(2)A^(2))=1/2(TE)` b. average kinetic energy = `=1/Tint_(0)^(T)` (Kinetic energy) dt and average Potential energy = `=1/Tint_(0)^(T)` (Potentialenergy) dt |
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| 24. |
A small trolley of mass 2.0 kg resting on a horizontal turntable is connected by a light spring to the centre of the table. When the turntable is set into rotation at a speed of 300 rev/min the length of the stretched spring is 40 cm. If the original length of the spring is 35 cm, determine the force constant of the spring. |
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Answer» |
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| 25. |
An ideal gas has molar specific heat 5R/2 at constant pressure. If 300 J of heat is given to two moles of gas at constant pressure, the change in temperature is: |
| Answer» ANSWER :A | |
| 26. |
A block of mass m rigidly attached with a spring k is compressed through a distance A. If the block is released, the period of oscillation of the block for a complete cycle is equal to |
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Answer» `(4pi)/(3)sqrt((m)/(k))` `rArr t_(AO)=(T)/(4)=(1)/(4)[2pisqrt((m)/(k))]=(pi)/(2)sqrt((m)/(k))` Since the motion is simple harmonic `OB=OAsin"(2pi)/(T)t_(OB)`, where `t_(OB)` is the time of motion from `O` to `B`. `rArr t_(OB)=(T)/(2pi)sin^(-1)"(A//2)/(A)=(T)/(2pi)((pi)/(6))` `=(T)/(12)=(2pi)/(12)sqrt((m)/(k))=(pi)/(6)sqrt((m)/(k))` `:. ` The TOTAL time of motion for a complete CYCLE `=t=2(t_(AO)+t_(OB))` `rArr t=2[(pi)/(2)sqrt((m)/(k))+(pi)/(6)sqrt((m)/(k))]=(4pi)/(3)sqrt((m)/(k))`. |
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| 27. |
Calculate the force .F. needed to punch a 1.46 cm diameter hole in a steel plate 1.27 cm thick (as shown in fig). The ultimate shear strength of steel is 345 M N//m^(2) |
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Answer» Solution : As in punching, SHEAR elasticity is involved, the hole will be punched it `((F_(11))/(A)) gt` ultimate shears streass `: F_(11) gt` (shear STRESS) X Area `F_(11) min=(3.45xx10^(8)) (2PI RL)(A=2pi rL)` `=(3.45xx10^(8))(2xx3.14)xx0.73xx10^(-2)xx1.27xx10^(-2)~~200 KN` |
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| 28. |
The expression of mean free path, if we consider that only one molecule is moving, for whom we are considering mean free path |
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Answer» `(1)/(npid^(2))` |
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| 29. |
A drop of liquid pressed between two glass plate spreads to a circle of diameter 10cm. Thickness of liquid film is 0.5mm and coefficient of surface tension is 70xx10^(-3)Nm^(-1). The force required to pull them a part is |
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Answer» 4.4N |
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| 30. |
Find the impulse due to the force vecF=ahati+bthatj, where a=2N and b=4Ns^(-1) if this force acts from t_(i)" to "t_(f)=0.3s |
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Answer» SOLUTION :`J=int_(t_(i))^(t_(F))vecF.dt=int_(0)^(0.3)(ahati+bthatj)dt` `J=aint_(0)^(0.3)dt hati+bint_(0)^(0.3)tdthatj=a[t]_(0)^(0.3)hati+b[(t^(2))/(2)]_(0)^(0.3)HATJ` `=2xx0.3xxhati+4xx((0.3)^(2))/(2)xxhatj=0.6hati+0.18hatj" NSec"` |
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| 31. |
State the rules for finding the significant figures in the addition and subtraction of two numbers, with example. |
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Answer» SOLUTION :Addition and sub TRACTION: The final result should RETAIN as many DECIMAL places as there are in the number with the smallest number with the smallest number of decimal places in addition and subtraction. Eg: (i) `4.1+1.780+3.046=8.926` Here the least number of significant digits after the decimal is one. Hence the result WIL be 8.9. (ii) `14.637 - 4.52 =10.117` Here the least number of significant digits after the decimal is two. Hence the result will be 10.12 |
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| 32. |
Why do we feel dizzy on a merry-go-round rotating in vertically upward direction ? |
| Answer» Solution :`implies ` When merry - GO - ROUND COMES down our weight SLIGHTLY decreases and it go up our weight slightly INCREASES and due to this it happens. | |
| 33. |
. Statement A: Energy is the cause for doing work and work is the effect Statement B: Energy is a scalar but power is a vector quantity. |
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Answer» A & B is CORRECT |
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| 34. |
The length of a rubber cord is l_1 meters when the tension is 4N and l_2 meters when thetension is 5N. The length in meters when the tension is 9 N is |
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Answer» `5l_(1) - 4l_(2)` |
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| 35. |
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 running in the direction of motion of the belt is |
| Answer» Answer :D | |
| 36. |
A body covers 100 cm in first 2 seconds and 128 cm in the next four seconds moving with constant acceleration. Find the velocity of the body at the end of 8 sec ? |
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Answer» SOLUTION :In first 2 sec. `100=2u+1/2, a.4…….(1)` In first 6 seconds : `228=6u+1/2a. 36 to (2)` `(1) xx 3-(2) " GIVES "72=-12a` `a=-6 cm//s^(2) ……(3)` `2u=112u=56cm//s…….(4)` Now velocity of the body at the end of 8 sec is `V=u+at=56-6xx8` V=8cm/s |
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| 37. |
The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe 2 m long. The length of the open pipe is |
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Answer» (A) 8m 3V)(2L) = ` frequency of second OVERTONE For closed pipe, `3f._(1) = (3v)/(4L.)=` frequency of first overtone Here, ` (3v)/(2L)= (3v)/(2L) = (3v)/(4L.)` `therefore L = 2L. = 2 xx 2 =4m` |
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| 38. |
Answer the following questions based on the P-T phase diagram of carbon dioxide: Is CO_(2) solid, liquid or gas at (a) - 70^(@) Cunder 1 atm, (b) - 60^(@) C under 10 atm, (c)15 ^(@) C under 56 atm? |
| Answer» SOLUTION :(a) vapour(B) SOLID (c ) liquid | |
| 39. |
Fig shows the Maxwell's velocity distribution curves 'p' and 'q' for two ideal gases contained in separate containers. The ordinates corresponding to curve q are double that of p. If n_(p), n_(q) : E_(p) , E_(q) and v_(p) , v_(q) represent the number of moles, internal energy and root mean square speeds corresponding to the curves p and q, then |
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Answer» `n_(p) = n_(Q) , E_(p) = E_(q) , v_(p) = v_(q)` |
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| 40. |
Which of the following has the highest specific heat capacity? |
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Answer» Water |
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| 41. |
In the classification of sound waves which one of the following statements is correct. The sound waves having frequencies between 20 Hz to 20 KHz are : |
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Answer» ULTRASONIC WAVES |
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| 42. |
In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at150^(@) C is dropped in a copper calorimeter (of water equivalent 0.025 kg containing 150 cm^(3)of water at27^(@)C. The final temperature is40^(@) C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible. Is your answer greater or smaller than the actual value for specife heat of the metal ? |
| Answer» Solution :`0.43 J g^(-1) K^(-1)` , SMALLER | |
| 43. |
The direction of the angular velocity vector is along |
| Answer» SOLUTION :PERPENDICULAR to the XY PLANE. | |
| 44. |
Consider an attractive force which is central but is inversely proportional to the first power of distance. If such a particle is in circular orbit under such a force, which of following statements are correct |
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Answer» `V PROP r ` |
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| 45. |
A bomb of mass 6 kg initially at rest explodes in to three identical fragments. One of the fragments moves with a velocity of10sqrt(3)hatim//s , another fragment moves with a velocity of 10hatjm//s, then the third fragment moves with a velocity of magnitude. |
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Answer» Solution :`P_(3)=-(P_(1)+P_(2)), P_(3)=SQRT(P_(1)^(2)+P_(2)^(2))`, `m_(3)v_(3)=sqrt((m_(1)v_(1))^(2)+(m_(2)v^(2))^(2))` |
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| 46. |
"Energy of a body is defined as its capacity of doing work". The energy possessed by a body by virtue of motion is know as ……….. |
| Answer» SOLUTION :KINETIC ENERGY | |
| 47. |
A wooden plank of length 1m and uniform cross-section is hinged at one end of the bottom of tank as shown in figure. The tank is filled with water upto a height of 0.5 m. Find the angle theta that the plank makes with the vertical in the equilibrium position. [Exclude the case thea = 0] |
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Answer» |
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| 48. |
If the length of the simple pendulum is equal to the radius of the earth, and it oscillates just above the surface of the earth then its time period is (a) 2pi sqrt((R)/(2g))(b) 2pi sqrt((R)/(g)) (c) nearly 59.5 minutes (d) 84.6 minutes |
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Answer» a is only correct |
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| 49. |
A uniformly tapering conical wire is made from a material of Youngs's gamma and has a normal, unextended length L. The radii, at the upper and lower ends of this conical wire, have values R and 3R, respectively. The upper end of the wire is fixed to a rigid support and a mass Mis suspended from its lower end. The equilibrium extended length, of this wire would equal : |
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Answer» ` L ( 1+ (2)/( 9) ( MG)/( pi gamma R^(2) ) )` |
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| 50. |
One mole of an ideal gas undergoes a cyclic change ABCD. Calculate from the diagram. (P is in atm, and V in litre) (i) Work done along AB, BC, CD and DA. (ii) What is the net work done in the process ? (iii). What is the efficiency of the process ? (iv) What is the net change in internal energy of the gas ? Given 1 atm = 1. 01 xx 10 ^(5) Nm ^(-2) |
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Answer» Solution :(i) Work done ALONG AB= Pressure `xx` Change in volume = PdV `= 5 xx 1.01 xx 10 ^(5) xx 2 xx 10 ^(-3)` `= 1010` Joule Work done along `BC = 0` because DV =0 Word done along `CD = PdV` `= 2 xx 1.01 xx 10 ^(5) xx (1-3) xx 10 ^(-)` `=- 404 J ` Work done along `DA = PdV` =0, Because `dV =0` (ii) Net word done `=1010 - 404 = 6060 J` (iii) Efficiency `= ("Useful work")/("Total work") xx 100` `= ( 606)/(1010) xx 100 = 60%` (IV) Net change in interna energy of the gas =0, because the process is a cyclic process. |
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