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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 12251. |
A box is placed on the floor of a truck moving with an acceleration of 7 ms^(-2). If the coefficient of kinetic friction between the box and surface of the truck is 0.5, find the acceleration of the box relative to the truck |
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Answer» `1.7 MS ^(-2)` |
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| 12252. |
A wire length l is bent into the shape of an n sided regular polygon of mass m, then |
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Answer» Its moment of inertia about it natural axis `(ml^(2))/(4N^(3))(1/3+cot^(2)(pi/n))` |
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| 12253. |
A bomb in the steady state explodes into three fragments. Two fragm ents of equal masses move w ith velocity 30 m /s in m utually perpendicular direction. The mass of the third fragment is equal to five times the mass of each of these two fragm ents. Find the magnitude and direction of the velocity of this third fragment. |
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Answer» `6sqrt(2) m//s, THETA=45^(@)` with -x and -y direction `M vec( v) = mvec( v )_(1)= mvec(v ) _(2)+ 5 m vec( v)_(3)` `O=m (30hat(i) )+ m(30hat(J))+ 5 m vec_(v ) ^(3)` `=sqrt(36+36)` `=sqrt(72)` `=6sqrt(2) ms^(-1)` `tan theta= (-6)/( -6) =1` `theta=45^(@)` `6sqrt(2) m//s theta =45^(@)` with- X and `-Y direction |
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| 12254. |
A ball at rest is dropped from a height of 12 m. It losses 25% of its kinetic energy on striking the ground and bounces back to a height 'h'. then value of 'h' is |
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Answer» 3 m |
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| 12255. |
A man walks on a straight road from his home to a market 2.5 km away with speed of 5 \frac { k m } { h \r }. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 \frac { k m } { h \r }. The average speed of the man over the interval of time 0 to 40 min is equal to |
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Answer» Solution :Time taken by woman to go from her home to market `t_(1) =("Distance") /("Speed") = (2.5)/(5) = (1)/(2)` h Time taken by woman to go from market to her home, `t_(2) = (2.5)/(7 . 5) = (1)/(3) h` `:.`Total time`= t_(1) + t_(2) = (1)/(2) +(1)/(3) = (5)/(6)`h =50 MIN (a)Average velocity `vec(v) _(av) = ("Displacement")/("Time")` (b)Average sped `v_(av) = ("Distance")/("Time") ` Between 0 to 30 min `vec(v) _(av) = (2.5)/((1)/(2))` = 5 KM /h TOWARDS market |
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| 12256. |
An woman walks on a straight road from her home to a market 2.5 km away with a speed of 5 km/h . Finding the market closed .he instantly turns and walks back with a speed of 7.5 km/hr. What is the(a)magnitude of an average velocity ,(b)average speed of the man,over the interval of time .0 to 50 min and |
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Answer» SOLUTION :TIME taken by woman to go from her home to market `t_(1) =("Distance") /("Speed") = (2.5)/(5) = (1)/(2)` h Time taken by woman to go from market to her home, `t_(2) = (2.5)/(7 . 5) = (1)/(3) h` `:.`TOTAL time`= t_(1) + t_(2) = (1)/(2) +(1)/(3) = (5)/(6)`h =50 min (a)Average velocity `vec(v) _(av) = ("Displacement")/("Time")` (b)Average sped `v_(av) = ("Distance")/("Time")` Between 0 to 50 min ` {:("Total distance"),("travelled"):}}= 2 . 5 + 2 . 5 = 5 km ` `{:("Total"),("displacement"):}}= zero` `vec(v)_(av) = 0` `v_(av) = (5)/((5)/(6))= 6 `km /hr |
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| 12257. |
In an experiment related to interference , Quinck's tube was employed to determine the speed of sound in air. A tunning fork of frequency 1328 Hz was usedas the sounding source . Initially , the apparatus , yielded a maximum sound intensity . Later , when the slidable tube was drawn by a distance of 12.5 cm, the intensity was found to be maximum for the first time . Determine the speed of sound in air . |
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Answer» Solution :A distance of `12.5 cm` drawn aside implies a DOUBLE path difference of `25 cm`. Since the shortest path difference , which enables the intensity to BECOME MAXIMUM once again is , obviously , `LAMDA`. so ` LAMBDA = 25 cm` `v = 1328 s^(-1) xx 25 cm = 332 m//s` Therefore , the speed of sound in air is `332 m//s`. |
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| 12258. |
In the indicator diagram shown the work done along path AB is: |
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Answer» ZERO |
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| 12259. |
A body executing SHM has a maximum velocity 1m/sec and maximum acceleration of 4m//sec^(2). Its amplitude in meters is |
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Answer» 1 |
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| 12260. |
Define simple harmonic motion (S.H.M) |
| Answer» Solution :A particle is SAID to EXECUTE simple harmonic motion if it MOVES to and fro about a MEAN position under the action of a restoring force which is directly proportional to its displacement from the mean position and is always directed towards the mean position. | |
| 12261. |
A boat is moving with a velocity (3hati-4hatj) with respect to ground. The water in river is flowing with a velocity (-3hati-4hatj) with respect to ground. What is the relative velocity ofboat with respect to river? |
| Answer» SOLUTION :`vecV_(BW)=vecV_(B)` | |
| 12262. |
A rigid rod of mass m slides along a fixed circular track followed by a flat track. At the given instant, velocity of end B is v along horizontal plane. Then at the given instant: |
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Answer» angular speed of rod is `(V)/( r)` |
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| 12263. |
A steel wire of length 80 cm and of area of cross section 1.5 mm^(2)is stretched. Find the work done in increasing its length from 80.2 cm to 80.3 cm.(Y=2.0xx10^(11)Nm^(-2)) |
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Answer» 0.9375 J |
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| 12264. |
The figure given below depicts two circular motions. The radius of the circle, the period of revolution, the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the x-projection of the radius vector of the rotating particle P in each case. |
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Answer» Solution :(a) At t = 0, OP makes an angle of `45^(@) = pi//4` rad with the (positive direction of ) x-axis. After time t, it covers an angle `(2pi)/Tt` in the anticlockwise sense, and makes an angle of `(2pi)/Tt+pi/4` with the x-axis. The projection of OP on the x-axis at time t is given by, `x(t)=Acos((2pi)/Tt+pi/4)` For `T=4s`. `x(t)=Acos((2pi)/4t+pi/4)` which is a SHM of amplitude A, period 4 s, and an INITIAL phase*=`pi/4`. (b) In this case at t = 0, OP makes an angle of `90^(@)=pi/2` with the x-axis. After a time t, it covers an angle of `(2pi)/Tt` in the CLOCKWISE sense and makes an angle of `(pi/2-(2pi)/Tt)` with the x-axis. The projection of OP on the x-axis at time t is given by `x(t)=Bcos(pi/2-(2pi)/Tt)` `=Bsin((2pi)/Tt)` For T = 30 s, `x(t)=Bsin(pi/15t)` Writing this as x (t ) = B cos`(pi/15t-pi/2)`, and comparing with Eq. (14.4). We find that this REPRESENTS a SHM of amplitude B, period 30 s, and an initial phase of `-pi/2`. |
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| 12265. |
What should be the lengths of a brass and an aluminium rod at 0^(@). if the difference between their lengths is to be maintained equal to 5 cm at any temperature ? (For brass alpha=18xx10^(-6)""^(@)C^(-1) and for aluminium alpha=24xx10^(-6)" "^(@)C^(-1) ) |
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Answer» Solution :Suppose `l_(1)` and `l_(2)` are the lengths of aluminium and brass rods at `0^(@)C` respectively. At any TEMPERATURE, DIFFERNCE of their lengths remains the same. Hence, increase in their lengths with increase in temperature MUST also be the same. `:.Deltal_(1)=Deltal_(2)` `:.alpha_(1)l_(1)DeltaT=alpha_(2)l_(2)DeltaT` `:.(l_(1))/(l_(2))=(alpha_(2))/(alpha_(2))=(24xx10^(-6))/(18xx10^(-6))=(4)/(2)`. . . (1) Now according to the given condition, `l_(1)-l_(2)=5` cm. .. . . (2) From equations (1) and (2). `(l_(1))/(l_(1)-5)=(4)/(3)` `:.3l_(1)=4l_(1)-20` `:.l_(1)=20` cm and `l_(2)=15` cm Thus, at `0^(@)C` lengths of brass and aluminium rods should be 20 cm and 15 cm respectively. |
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| 12266. |
The work done in stretching a wire by 0.1mm is 4J. Find the work done in streching another wire of same material, but with double the radius and half the length by 0.1 mm in joules. |
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Answer» |
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| 12267. |
Hooke's law states that stress prop strain.What is the necessary condition for the above law to be valid? |
| Answer» SOLUTION :Hooke.s LAW is VALID only for SMALL DEFORMATIONS. | |
| 12268. |
A pendulum clock is taken to the bottom of a deep mine. Will it gain or lose time? How should its length be altered to correct the time? |
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Answer» looses TIME, length to be increased |
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| 12269. |
Give examples of shim in which elastic constants are not involved. |
| Answer» SOLUTION :The motion can be represented using simple SINE or COSINE functions which are harmonic functions and the motion is of constant AMPLITUDE and SINGLE frequency. | |
| 12270. |
The angular velocity of a particle 5 cm away from the axis of rotation is 10 rad/s. What will its linear velocity at 10 cm away from axis? |
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Answer» SOLUTION :`100cm//s` The velocity of all PARTICLES of rigid BODY is same and from linear velocity `v=romega`. |
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| 12271. |
Two infinitely long straight current carrying wires are placed parallel to y-axis in xy plane. Each wire carries current I in opposite directions as shown. A charged particle of charge +q and mass m is project from position P(0,0,0a) with initial velocity v=i. Then |
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Answer» The magnitude of initial ACCELERATION of the particle `(mu_(0)lqv)/(2piam)` |
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| 12272. |
The speed of a whirlwind in a tornado is alarmingly high . why ? |
| Answer» Solution :Air from the nearby regions GETS CONCENTRATED in a small SPACE in a whiriwind in a TORNADO. | |
| 12273. |
Deduce the dimensional formula for the following physical quantities : (i) Young's modulus (ii) Co-efficient of viscosity (iii) Surface tension (iv) Bulk modulus (v) Force constant. |
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Answer» Solution :(i) Young.s `=("Longitudinal stress")/("Longitudinal strain")` `=(F"/"A)/(trianglel"/"l)= (F)/(A)*(l)/(trianglel)` `[Y]= ([MLT^(-2)][L])/([L^(2)]*[L])= [ML^(-1)T^(-2)]` (ii) Coefficient of viscosity, `eta= ("Force")/("AREA"xx"Velocity gradient")` `=("Force")/("Area")xx("Distance")/("Velocity")` `[eta]=([MLT^(-2)]*[L])/([L^(2)]*[LT^(-1)])= [ML^(-1)T^(-1)]` (iii) Surface tension, `SIGMA= ("Force")/("Length")` `[sigma]=([MLT^(-2)])/([L])= [ML^(0)T^(-2)]` (iv) BULK modulus `=("Pressure applied")/("Relative decrease in volume")` `(P)/(triangleV"/"V)=(P)/(triangleV)*V` `=("Force")/("Area")*(V)/(triangleV)=([MLT^(-2)]*[l^(3)])/([L^(2)]*[L^(3)])` `[ML^(-1)T^(-2)]` (v) Force CONSTANT `=("Force")/("Displacement")` `=([MLT^(-2)])/([L])= [ML^(0)T^(-2)]`. |
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| 12274. |
Power is given by .......... . |
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Answer» <P>`P = (vec F)/(vec V)` |
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| 12275. |
A blackened metal container of negligible heat capacity is filled with water. The container has sides of length l=10cm. It is placed in an evacuated enclosure whosewalls are kept at 27^(@)C. How long will it take for the temperature of th water to change from 30^(@)C to 29^(@)C. (sigma=5.67xx10^(-8)Wm^(-2)K^(-4)) |
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Answer» |
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| 12276. |
(a) In the system shown in figure 1, the uniform rod of length L and mass m is free to rotatein vertical plane about point O. The string an pulley are mass less. The block has mass equal to that of the rod. Find the acceleration of the block immediately after the system is released with rod in horizontal position. (b) System shown in figure 2 is similar to that in figure 1 apart from the fact that rod is mass less and a block of mass m is attached to the centre of the rod with the help of a thread. Find the acceleration of both the blocks immediately after the system is released with rod in horizontal position. |
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Answer» (b) `(2G)/(5) " and " (g)/(5)` |
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| 12277. |
What is therange of nuclear forces ? |
| Answer» Solution :The RANGE of nuclear FORCES is `~~ 10` FERMI. | |
| 12278. |
A person can swim in still water at 5m/s. He moves in a river of velocity 3m/s. Fist down the stream next same distance up the steam the ratio of times taken are |
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Answer» `1:1` |
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| 12279. |
Two blocks of m asses 6 kg and 2 kg are placed in contact on a horizontal frictionless surface. If a horizontal force of 2N is applied to m ass 6 kg to move them together, what will be the acceleration of 2 kg block ? What will be the force on this block ? |
Answer» Solution :  ACCELERATIONOF the systemof twoblock `a= ("totalforce")/("total mass ") = (F)/( m_(1) +m_(2)) = (2 ) /(6+2)= (2)/(8)= 0.25 MS^(-2)` Forceon 2 kgblockF= `m_(2)a = 2xx0.025=0.50 N` |
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| 12280. |
Statement I : When a wave goes from one medium to other, then average power transmitted by the wave may change .Statement II : Dueto change in medium, amplitude, speed,wavelenght and frequency of wave may change . |
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Answer» STATEMENT I is true, statement IIis true, statement II is a correct EXPLANATIONFOR statement I . |
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| 12281. |
The length of a sonometer wire is 90cm and the stationary wave setup in the wire is representedby an equation y=6sin((pix)/(30))cos(250t) where x , y are in cm and t is in second. The number of loops is |
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Answer» Solution :By comparing with `y=2Asin(KX)cos(250t)` `K=(pi)/(30)implieslambda=(2pi)/(k)`, `L=90cm`, `l=(plambda)/(2)` |
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| 12282. |
A golfer standing on the ground hits a ball with a velocity of 52 m/s at an angle theta above the horizontal if tan theta = (5)/(12) find the time for which the ball is at least 15m above the ground ? (g = m//s^(2)) |
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Answer» SOLUTION :`v_(y) = SQRT(u_(y)^(2) - 2gy)` `= sqrt(52 XX 52 xx (5 xx 5)/(13 xx 13) - 2 xx 10 xx 15)` `= sqrt(16 xx 25 - 300) = 10` `Delta t = (2u_(y))/(10) = (2 xx 10)/(10) = 2s` 
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| 12283. |
State the first law of thermodynamics. Explain its mathematical notation. |
| Answer» SOLUTION :FIRST LAW of thermodynamics : The heat energy (DQ) supplied to a system is equal to the sum of the INCREASES in the internal energy (DU) of the system and external work DONE (dW) by it i.e., dQ = dŲ +dW | |
| 12284. |
The excess pressure inside a spherical soap bubble of radius lcm is balanced by a column of oil (Sp.gr =0.8),2mm high. The surface tension of the bubble is : |
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Answer» 3.92 N/m |
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| 12285. |
What was mistake in Aristotle’s idea regarding motion ? |
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Answer» SOLUTION :GREEKPHILOSOPHER Aristotledue tohis EXPERIENCEIN DAYTO daylifeassumed thatto continueuniformmotionof the bodysomeextrnalagencyis REQUIRED. Thiscalledaristotle law . Accordin tohiman arrowreleasedfrombowcontinueits motionbecauseairbehindthearrowpushesitforward. Aristotlesismistakewas thathe consideredday todaylifeexperienceas fundamentallawof nature. |
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| 12286. |
For a system to be in equilibrium the torques acting on it must balance. This is true only if the torques are taken about |
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Answer» the CENTRE of the system |
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| 12287. |
An object is allowed to fall freely towards the earth from a distance r ( gt R_e)from the centre of the earth. Find the speed of the object when it strikes the surface of the earth. |
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Answer» Solution : Initial speed of BODY is zero, HENCE total energy, `E=(GM_em)/R "" .......(1)` Suppose mass of body is ` m_1`mass of earth `M_(e)`radius of earth `R_eand v`is the speed of body when it hit the earth. Total energy of body when body hit the earth, `E = 1/2 mv^(2) - (GM_(e)m)/R_e ""..........(2)` Neglecting air resistance, according to conservation of MECHANICAL energy, `1/2mv^2-(GM_em)/R_e=(GM_em)/r` `:.v^2=2GM_e(1/R_e-1/r)` but `g = (GM_e)/R_e^2` `:. v^2=2R_e^2g(1/R_e-1/r)` or `:. v =R_e[2g(1/R_e-1/r)]^(1/2)` |
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| 12288. |
A man carefully brings down a glass sheet from a height 2 m to the ground. The work done by him is |
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Answer» negative |
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| 12289. |
If a body moves on the circumference of a circle with a speed equal to that which it would acquire by falling freely through half the radius of the circle, prove that its centripetal acceleration is equal to the acceleration of the free fall. |
| Answer» Solution :`v^(2) =0 + 2g xx (r//2) = GR, a = v^(2)//r = rg//r = G` | |
| 12290. |
At what depth in water can an air bubble remain stationary? Density of air under normal pressure and temperature is 0.001293 g.cm^-3 and the atmospheric pressure is equal to 76 cmHg. |
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| 12291. |
A child's pogo stick(figure) stores energy in a spring with a force constant of 2.5xx10^4Nm. At position (A) (x_A=0.10m), the spring compression is a maximum and the child is momentarily at rest. At position (B) (x_B=0), the spring is relaxed and the child is moving upward. At position (C), the child is again momentarily at rest at the top of the jump. The combined mass of child and pogo stick is 25kg. (a) Calculate the total energy of the child-stick-earth system if both gravitational and elastic potential energies are zero for x=0. (b) Determine x_C. (c) Calculate the speed of the child at x=0. (d) Determine the value of x for which the kinetic energy of the system is a maximum. (e) Calculate the child's maximum upward speed. |
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Answer» Solution :`k=2.5xx10^4Nm^-1`, `m=25kg`, `X_A=-0.10m`, `U_g:|_(x=0)=U_s:|_(x=0)=0` a. `E_(mech)=K_A+U_(gA)+U_(sA)=0+mgx_A+1/2kx_A^2` `=25xx10xx(-0.100)+1/2xx2.5xx10^4xx(-0.100)^2` `=-25J+125J=100J` b. Since only conservation forces are involved, the TOTAL energy of the child-pogo stick-earth system at point C is the same as that at point A. `K_C+U_(gC)+U_(sC)=K_A+U_(gA)+U_(sA)` `0=25xx10xxx_C+0=0-25J+125J` `x_C=0.40m` c. `K_B+U_(gBB)+U_(sB)=K_A+U_(gA)+U_(sA)` `1/2xx25xxv_B^2+0+0=0+(-25)+125` `v_B=2sqrt2ms^-1` d. K and V are at a maximum when `a=sum F//m=0` (i.e., when the MAGNITUDE of the upward spring force equals the magnitude of the downward gravitational force). This occurs at `xlt0`, where `k|x|=mg`. `|x|=(25xx10)/(2.5xx10^4)=10xx10^-3m=10mm` Thus, `K=K_(max)` at `x=-10mm` e. `1/2xx25xxv_(max)^2=25xx10xx[(-0.100)-(-0.01)]+1/2(25xx10^4)[(-0.100)^2-(-0.01)^2]` `impliesv_(max)=2.85ms^-1` |
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| 12292. |
Explain longitudinal stress. |
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Answer» Solution :In FIGURE, a cylinder is stretched by two equal forces applied normal to its cross-sectional area is SHOWN. The restoring force per unit area is called TENSILE STRESS. If the cylinder is compressed under the action of applied forces the restoring force per unit area is known as compressive stress. Tensile or compressive stress `(sigma)` is known as LONGITUDINAL stress. |
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| 12293. |
A small sphere, dropped from a height of 10 cm, on striking the floor, reboundes and rises to a height of 15 cms Is this possible ? Explain your answer. |
| Answer» Solution :Here` e = sqrt (( h _(2))/( h _(1))) = sqrt ((15)/(10)) GT 1.`Since the maximum value of e is 1, it is not possible. This ALSO violates energy CONSERVATION law. | |
| 12294. |
What is the shortest time interval measured indirectly so far ? |
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Answer» SOLUTION :The shortest interval is the TIME taken by light to CROSS a distance of NUCLEAR SIZE `(~~ 10^(-22)s).` |
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| 12295. |
Statement : I The angular velocity of a rigid body in motion is defined for the whole body. Statement II : All points on a rigid body performing pure rotational motion are having same angular velocity. |
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Answer» STATEMENT I is TRUE, statement II is true , statement II is a correct EXPLANATION for statement I. |
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| 12296. |
For a projectile the ratio of maximum height reached to the square of flight time is (g = 10ms^(-2)) |
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Answer» `5:4` |
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| 12297. |
A body of weight W and density rho floats on liquid . What will its apparent weight ? |
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Answer» `W(RHO-rho_(1))` `=W-m_(1)g` `=rhoV_(g)-rho_(1)Vg"" (becauseW=rhoVgandm_(1)=rho_(1)V)` `=rhoVg(1-(rho_(1))/(rho))` `=W(1-(rho_(1))/(rho))` |
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| 12298. |
A communication satellite appears stationary from a place on the equator. Find the height of the satellite from the surface of the earth. |
| Answer» SOLUTION :35930 KM | |
| 12299. |
The bimetallic strip of above question is cooled. What shape does it acquire? |
| Answer» SOLUTION :It BENDS with BRASS on CONCAVE SIDE. | |
| 12300. |
If the total mechanical energy of a particle of mass m is E, its speed can be given as v = sqrt((2(E-U))/(m)). where U= potential energy function of the particle. Using the above expression find the speed of the particle at earth.s surface when the particle falls freely in gravity from rest from a height =R, where R = radius of earth. |
Answer» Solution : The gravitational POTENTIAL ENERGY of the particle m is given as `U = - (GMm)/(r ) = E = U_0 + K_0 = -(GMm)/(r_0) + 0 =- (GMm)/(r_0), r_0 = 2R` Then, SUBSTITUTING E and U in the EXPRESSION ` v = sqrt((2(E - U))/(m))` we have` v= sqrt((2{(-(GMm)/(r_0)) - (-(GMm)/(r )) })/(m) )` , this gives , ` v = sqrt((2GM(r_0-r))/(r r_0))` where `r_0 - r = h = R , because r = R ` , then , `v = sqrt((GM)/(R )) = sqrt(gR) `, where `g = (GM)/(R^2)` |
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