Saved Bookmarks
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.
| 38451. |
A missile is launched with a velocity less than the escape velocity. The sum of its kinetic and potential |
|
Answer» POSITIVE |
|
| 38452. |
(a) A boy on a skateboard is sliding down on a smooth incline having inclination angle q. He throws a ball such that he catches it back after time T. With what velocity was the ball thrown by the boy relative to himself ? (b)Barrel of an anti aircraft gun is rotating in vertical plane (it is rotating up from the horizontal position towards vertical orientation in the plane of the fig). The length of the barrel is L = sqrt(2) m and barrel is rotating with angular velocity omega = 2 rad//s.At the instant angle theta is 45° a shell is fired with a velocity 2 sqrt(2) m//s with respect to the exit point of the barrel. The tank recoils with speed 4 m//s. What is the launch speed of the shell as seen from the ground? |
|
Answer» |
|
| 38453. |
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 bothing point of water as standard fixed points as was originally done in the Celsius scale) ? |
| Answer» Solution :Triple - POINT has a UNIQUE temperature, fusion point and BOILING point temperature DEPEND on pressure . | |
| 38454. |
A polished metal with rough black spot on it is heated about 1400K and quickly taken to a dark room. Which of the following statements will be true ? |
|
Answer» The spot will appear brighter than the PLATE |
|
| 38455. |
A mass m is released from the top of a vertical circular track of radius r with a horizontal speed v_(theta). Calculate the angle theta with respect to the vertical where it leaves contact with the track. |
|
Answer» Solution :The FORCES acting on the body are its weight mg and REACTION N as shown in Fig. So for circular motion of the body at any position `theta`  `(mv^(2))/(r)=mgcostheta-NorN=mgcostheta-(mv^(2))/(r)` The body will have contact where N = 0 i.e., mg `cos theta = (mv^(2))/(r) = 0` i.e, `cos theta = (v^(2))/(rg) "" .............(2)` Now applying conservationof mechanical energy between H and P, we get `(1)/(2)mv^(2)=(1)/(2)mv_(0)^(2)+MGR(1-costheta)` where .v. is the velocity at .P. `or v^(2)=v_(0)^(2)+2gr(1-costheta).........(3)` substituting the value of `v^(2)` in Equation (2) `cos theta=(v_(0)^(2))/(rg)+2(1-costheta)` `i.e., cos theta=[(v_(0)^(2))/(3rg)+(2)/(3)]ortheta=cos^(-1)[(v_(0)^(2))/(3rg)+(2)/(3)]` |
|
| 38456. |
A bob executes shm of period 20 s. Its velocity is found to be 0.05 ms^(-1)after 2 s when it has passed through its mean position. Find the amplitude of the bob. |
|
Answer» SOLUTION :`T = 20 s, v=0.05 MS^(-1), t=2 s` `omega = (2PI)/T =2/20 pi = (pi)/10 rad//s` `v = aomega COS omegat` `=0.05 = (api)/10 cos""(pi)/10 xx2 = (pia)/10 cos""pi/5` `0.05 = (pia)/10 xx0.809` `a= (0.05xx10)/(3.14 xx0.809)=0.197m` |
|
| 38457. |
A railway engine of mass 50 tons is pulling a wagon of mass 40 tons with a force of 4500N. The resistance force acting is 1N per ton. The tension in the coupling between the engine and the wagon is |
|
Answer» Solution :`F=ma, a=(F_("net"))/(m_(1)+m_(2))` For engine `F-f-T=m_(1)a` |
|
| 38458. |
The conceptof CMis applicable |
|
Answer» onlyfor rigidbodies |
|
| 38459. |
Using constraint equations find the relation between a_(1)"and" a_(2) |
|
Answer» Solution :Points 1,2,3 and 4 are movable . Let their dis-placements from a fixed line be `x_(1),x_(2),x_(3) "and" x_(4)` `x_(1)+x_(3)=l_(1)` `(x_(1)-x_(3))+(x_(4)-x_(3))=l_(2)` `(x_(1)-x_(4))+(x_(2)-x_(4))=l_(3)` On double differentiating with respect to time , we will get following three constraint RELATIONS :`a_(1)+a_(3)=0`....(i) `a_(1)+a_(4)-2a_(3)=0`....(ii) `a_(1)+a_(2)-2a_(4)=0`....(iii) Solving Eqs. (i), (ii), and(iii) we get `a_(2)=-7a_(1)` Which is desired RELATION between `a_(1) "and" a_(2)` 
|
|
| 38460. |
Force F is given by F=at+bt^(2). Where ''t'' is time. What are the dimensions of ''a'' and ''cb''? |
|
Answer» `MLT^(-3)` and `MLT^(-4)` According to principle of homogeneity , the FORCE becomes, `MLT^2 = MLT^(-2) + MLT^(-2)` Comparing (1) and (2) , we get a `= MLT^(-3) , b= MLT^(-4)` |
|
| 38461. |
The total energy of a particle executing simple harmonic motion is: |
|
Answer» `PROP X` |
|
| 38462. |
A particle of mass m performs SHM along a stright line with frequency f and amplitude A. |
|
Answer» The average kenetic energy of the particle is zero. |
|
| 38463. |
A new unit of length is chosen such that the speed of sound at room temperature is taken as unit. The sound of an explosion reaches a listener in 10s. Find the distance between the listener and the place where explosion took place. |
|
Answer» |
|
| 38464. |
A spring balance reads W_(1) when a ball is suspended from it. A weighing machine reads W_(2) when a tank of liquid is kept on it. When the ball is immersed in the liquid, the spring balance readsd W_(3) and the weighting machine reads W_(4). |
| Answer» Answer :A::C | |
| 38465. |
Due to higher specific heat of water compared to other solids and liquids |
|
Answer» WATER WARMS up QUICKLY but COOLS down slowly |
|
| 38466. |
What is the work done in increasing the angular frequency of a circular ring of mass 2 kg and radius 25 cm from 10 rpm to 20 rpm about is axis ? |
|
Answer» Solution :WORK done = Increase in ROTATIONAL kinetic ENERGY `=(1)/(2)I(omega_(f)^(2)-omega_(i)^(2))=(1)/(2)MR^(2)(omega_(f)^(2)-omega_(i)^(2))` `= (1)/(2)xx2xx(0.25)^(2)(((2PI)/(3))^(2)-((pi)/(3))^(2))=0.2058 J` |
|
| 38467. |
You are given total mass M. How do you divide it into two parts so that the gravitational force between them at a given distance is maximum |
| Answer» Answer :D | |
| 38468. |
The escape velocity of a body on the earth.s surface is v_(e ). A body is thrown up with a speed of kv_(e) where k gt 1 . Assuming that the sun and planets do not influence the motion of the body, then the velocity of the body at the infinite distance in |
|
Answer» `(v_(e))/(sqrt(K^(2)-1))` |
|
| 38469. |
The velocity of the image of a moving object in situation shown in figure is (given velocity w.r.t ground) |
|
Answer» SOLUTION :We know for the perpendicular component `(vecV_(1//M))_(bot)=-(vecV_(o//M))_(bot)` `(V_(IG))_(bot)-(V_(MG))_(bot)=-[(V_(OG))_(bot)-(V_(MG))_(bot)]` `(V_(MG))_(bot)= ((V_(OG))_(bot)+(V_(IG))_(bot))/2 IMPLIES -30=(10+(V_(IG))_(bot))/2 implies` `(V_(IG))_(bot)=-70` THe parallel component of velocity of image is same as that of OBJECT w.r.t ground . It does not depend on the velocity of mirror. So `(V_(IG))_(II)=5m//s` |
|
| 38470. |
At what distance 'd' from the surface of asolid sphere of radius 'R', (a)potential is sme as at a distance (R)/(2) from the centre ? (b) field strengt is ame as at a distance (R)/(4) fromcentre. |
|
Answer» No Such POINT will EXIST. Because potential at CENTRE is `(-1.5GM)/(R)`. Potential at surface is `(-GM)/(R)` and potential at infinity is zero . From centre to surface potential between `(-1.5GM)/(R)` and `(-GM)/(R)`. From surface to infinity potential varies between `(-GM)/(R)` and zero . Given, `E_("inside") = E_("outside")` `:. (GM)/(R^(3))(r_1) = (GM)/(r_(2)^(2))` Here, `r_(1)` and `r_(2)` are the distances from centre. `:. (GM)/(R^(3)) ((R)/(4)) = (GM)/((R+d)^(2))` Solving this EQUATION , WE get `d=R`. |
|
| 38471. |
Which of the following is an example of completely inelastic collision |
|
Answer» The collision between NUCLEI and FUNDAMENTAL particles |
|
| 38472. |
You lift a suitcase from the floor and keep it on a table. The work done by you on the suitcase does not depend on (a) the path taken by the suitcase (b) the time taken by you in doing so ( c) the weight of the suitcase (d) your weight |
|
Answer» a, B & d are correct |
|
| 38473. |
The displacement vector of a mass m is given by r (t) = hati A cos omegat + hatj B sin omegat (a) Show that the trajectory is an ellipse (b) Show that F = - m omega^(2) r . |
|
Answer» Solution :Here (a) ` vec(r) (t) = HATI A cos omegat + hatj B sin omegat, :. x = A cos omegat, y = B sin omegat ` ` x^(2)/(A^(2)) + y^(2)/(B^(2)) cos^(2) omegat + sin^(2) omegat = 1` which is the equation of an ELLIPSE `:.` The trajectory of the particle is elliptical (b) Now ` vec(upsilon) =vec(dr)/(dt) = -hatiomega A sin omegat + hatjomega Bcos omegat ` ` vec(a) = vec (d upsilon)/(dt) = hati omega^(2) A cos omegat - hatj omega^(2) B sin omegat = - omega^(2) [hati A cos omegat + hatjBsin omegat] = - omega^(2) vec(r) ` ` vec(F) = m vec (a) = - m omega^(2) vec(r) ` . |
|
| 38474. |
A body is projected with an initial Velocity 20 m/s at 60° to the horizontal. Its initial velocity vector is- (g=10 m//s^2) |
|
Answer» `10I - 20 J` |
|
| 38475. |
stateprevosttheoryof heatexchnage. |
| Answer» Solution :Prevost theory STATES that all BODIES emit thermal radiation at all TEMPERATURES above absolute zero irrespective of the NATURE of the SURROUNDINGS. | |
| 38476. |
What are the conditions in which force can not produce torque ? |
|
Answer» Solution :The torque is zero with `vecr` and `vecF` are parallel or antiparallel. If parallel then `theta=0^(@)` and `SIN 0^(@)=0`. If antiparallel, then `theta=180^(@)` and `sin 180^(@)=0`. Hence `tau=0` The torque is zero if the force ACTS at the reference point. i.e. as `vecr=0, tau=0`. |
|
| 38477. |
An electron starting from rest has a velocity that increases linearly with the time that is v=kt,where k=2 m//sec^(2).The distance travelled in the first 3 seconds will be |
|
Answer» 9m |
|
| 38478. |
The Young's modulus for a perfect rigid body is: |
|
Answer» 0 |
|
| 38479. |
The two gases are seperated by a fixed diathermic wall, with initial states as (P_(A), V_(A)) and (P_(b), V_(B)) |
|
Answer» The thermodynamics states of A and B will not change with time |
|
| 38480. |
Masses 8, 2, 4, 2 kg are placed at the corners A, B, C, D respectively of a square ABCD of diagonal 80 cm. The distance of centre of mass from A is |
| Answer» Answer :B | |
| 38481. |
If wavelengths of maximum intensity of radiations emitted by the sun and the moon are 05.xx 10^(-6) m and 10^(-4) m respectively, the ratio of their temperatures is |
|
Answer» `1//100` |
|
| 38482. |
What are heat rays ? |
| Answer» SOLUTION :Infrared rays are CALLED heat rays, because these rays GIVE heat. | |
| 38483. |
At what common temperature would a block of wood and a block of metal feel equally cold or equally hot when touched with? |
| Answer» SOLUTION :This will HAPPEN when both the blocks are at the temperature of the body in case when either of blocks is touched , there will be no TRANSFER of heat from the block to the body (human) or vice VERSA .So ,on touching they will be filled equally HOT aur equally cold. | |
| 38484. |
Write down the equation of a freely falling body under gravity. |
|
Answer» Solution :(i) Distanced traveled by an object falling for time .t., u = 0 , s = d , a = g `s = ut + 1/2 at^2` ` therefore = 1/2 "GT"^2` Time .t. TAKEN for an object to FAIL distance .d.. ` t = sqrt((2d)/(g))` Instantaneous VELOCITY `v_i`of a falling object after elapsed time .t. ` v_i = sqrt(2gd)` Instantaneous velocity `v_i`of a falling object that has traveled distance .d. ` v_i = sqrt(2gd)` (ii) ` u = - 19.6 ms^(-1)` `u = 9.8 ms^(-2)` t = 6 sec height of the building ` s = ut + 1/2 at^2` ` = (-19.6 XX 6) + 1/2 (9.8)(6)^2 = - 117.6 + 176.4` s = 58.8 m |
|
| 38485. |
A wire whose cross-sectional area is 4 mm^2is stretched by 0.1 mm by a certain load. If a similar wire of double the area of cross section is under the same load, then the elongation would be |
| Answer» ANSWER :2 | |
| 38486. |
In which case, weight of body will be maximum (i) in air (ii) vacuum (iii) in water |
| Answer» SOLUTION :A BODY will WEIGH MAXIMUM in VACUUM. | |
| 38487. |
A stone is dropped from a height of 10cm above the top of a window 80 cm high.the time taken by the stone to cross the window is (g=9.8 ms^(-2)) |
|
Answer» `1/7s` |
|
| 38488. |
An infinite number of particles each of mass m are placed on the positive x-axis at 1m, 2m, 4m, 8m,…. from the origin. The magnitude of the resultant gravitational force on mass 'm' kept at the origin is |
Answer» Solution : The resultant gravitational FORCE `F = (Gm^(2))/(1) + (Gm^(2))/(4) + (Gm^(2))/(16)+…..` `= Gm^(2)((1)/(1)+(1)/(4)+(1)/(16)+….) = Gm^(2)((1)/(1-(1)/(4)))` (`because` sum of the INFINIT TRMS in G.P is `S_(prop) = (a)/(1-r)`) `= (4)/(3)Gm^(2)` |
|
| 38489. |
Two parallel rail tracks run north-south. Train A moves north with a speed of 54 km h^(-1), and train B moves south with a speed of 90 km h^(-1). What is the (a) velocity of B with respect to A ?, (b) velocity of ground with respect to B ?, and (c) velocity of a monkey running on the roof of the train A against its motion (with a velocity of 18 km h^(-1) with respect to the train A) as observed by a man standing on the ground ? |
|
Answer» Solution :Choose the positive direction of x-axis to be from SOUTH to north. Then, `upsilon_(A)=+54 km h^(-1) = 15 ms^(-1)` `upsilon_(B)=-90 km h^(-1)=-25 ms^(-1)` Relative velocity of B with respect to `A=upsilon_(B)-upsilon_(A)=-40 ms^(-1)`, i.e. the TRAIN B appears to A to move with a speed of 40 m s–1 from north to south. Relative velocity of ground with respect to `B=0-upsilon_(B)=25 ms^(-1)`. In (c), let the velocity of the monkey with respect to ground be `upsilon_(M)`. Relative velocity of the monkey with respect to A, `upsilon_(MA)=upsilon_(M)-upsilon_(A)=-18 km h^(-1)=-5 ms^(-1)`. Therefore, `upsilon_(M)=(15-5)ms^(-1)=10 ms^(-1)`. |
|
| 38490. |
A uniform cylinder of height h and radius r is placed with its circular face on a rough inclined plane and the inclination of the plane to the horizontal gradually in creased. If mu is the coefficeint of friction, then under what conditions the cylinder will (a) slide before topping (b) topple before sliding |
Answer» Solution : (a) The Cylinder will slide if `Mg SIN theta gt mu Mg cos theta rArr tan theta gt mu` …..(1) the cylinder will topple if `(Mg sin theta) (h)/(2)gt (M G cos theta) r rArr tan theta gt (2r)/(h)` ...(2) Thus, the condition of sliding is `tan theta gt mu` and condition of toppling `tan theta gt (2r)/(h)`. Hence, the cylinder will slide before toppling if `mu LT (2r)/(h)` (B) The cylinder will topple before sliding if `mu gt (2r)/(h)` |
|
| 38491. |
Two balls of equal masses are thrown at the same time in vacuum. While they are in vacuum, the acceleration of their centre of mass |
|
Answer» depends on MASSES of the balls |
|
| 38492. |
A particle is fastened at the end of a string and is whirled in a vertical circle with the other end of the string being fixed. The motion of the particle is |
|
Answer» Periodic |
|
| 38493. |
A particle moves along a circle with uniform acceleration. Workdone by the resultant force acting on that particle |
|
Answer» is ALWAYS zero |
|
| 38494. |
Weighing the earth : Youare given the following data : g =9.81 ms^(-2).R_(E)= 6.37 xx10^(6) m, the distance to the moon R = 3.84xx10^(8)m and the time period of the moons revolution is 27 .3 days . Obtain the mass of the earth M_E in two different ways. |
|
Answer» Solution :We know that `M_E=(gR_E^2)/G=(9.81 xx(6.37xx10^(6))^2)/(6.67 xx10^(-11))=5.97xx10^(24)kg` The moon is a SATELLITE of the EARTH. According to Kepler.s Third law `T^(2) = (4PI^2R^3)/(GM_E) M_(E) =(4pi^(2)R^3)/(GT^2)` `= (4xx3.14 xx3.14 xx(3.84)^3xx10^(24))/(6.67xx10^(-11) xx(27.3xx24xx60xx60)^(2))=6.02xx10^(24)` kg Both methods yields almost the same answer , the difference between them being less than 1% . |
|
| 38495. |
Should the centre of mass of a body necessarily lie inside the body ? |
| Answer» Solution :No, example the centre of mass of a RING LIES in its HOLLOW PORTION. | |
| 38496. |
(A): The path followed by one projectile as observed by another projectile is a straight line in uniform gravitation field. (R) : The relative velocity between two projectiles at a given place doesn't change with time. Because their relative acceleration is zero. |
|
Answer» Both (A) and (R) are true and (R) is the CORRECT EXPLANATION of (A) |
|
| 38497. |
The greatest height to which a man can throw a stone is h. What will be the greatest distance upto which he can throw the stone? |
|
Answer» SOLUTION :Maximum height: `H=(U^(2)sin^(2)THETA)/(g)impliesH_(max)=(u^(2))/(2G)=h(at""theta=90^(@))` Maximum range `R_(max)=(u^(2))/(g)=2H` |
|
| 38498. |
A ball is projected vertically upward with speed v. another ball of the same mass is projected at an angle of 60^(@) with the vertical with the same speed. The ratio of their potential energy at the highest point is: |
|
Answer» |
|
| 38499. |
Efficiency of a heat engine whose sink is at temperature of 300 K is 40% . To increase the efficiency to 60 % keeping the sink temperature same, the source temperature must be increased by |
|
Answer» Solution :`(T_(2))/(T_(1))=1-eta=1-(40)/(100)=(3)/(5)` `RARR T_(1)=(5)/(3)T_(2) rArr T_(1)=(5)/(3)xx300=500K` New efficiency `eta=60%` `(T_(2))/(T_(1))=1-eta=1-(60)/(100)=(2)/(5)""T_(1)=(5)/(2)xx300=750K`, `DeltaT=750-500=250K` |
|
| 38500. |
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N ? |
|
Answer» `5 m//s^(2)` |
|