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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.
| 15351. |
A small steel ball falls through a syrup at a constant speed of 1.0m/s. If the steel ball is pulled upwards with a force equal to twice its effective weight, how fast will it move upward? |
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Answer» SOLUTION :`W_(e)=` effective WEIGHT In equilibrium `kv=w_(e)` Again `2w_(e)-w_(e)=kv^(1)` From eqs Iand II `v^(1)=v=1.0m//s` 
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| 15352. |
An elevator can carry a maximum load of 1800kg (elevator + pasengers) is moving up with a constant speed of 2ms^(-1). The frictional force opposing the motion is 4000N. Determine the minimum power delivered by the motor to the elevator in watts as well as in horse power. |
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Answer» SOLUTION :The downword force on the elevator is `F= mg + F_(f)= (1800 XX 10) + 4000= 22000N` The motor MUST supply enough power of balance this force. Hence, `P= Fv = 22000 xx 2 = 44000` W= 59 hp |
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| 15353. |
A small metal ball falls in liquid with a terminal velocity of V. If a ball of radius twice of first ball but same mass falls through a same medium, calculate the terminal velocity with which it falls. |
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Answer» Solution :Given`v = (2R^(2)rhog)/(9eta)` mass `= 3/4 pi r^(3) RHO = 4/3 pi (2r)^(3)rho_(1)`or`rho_(1) = (rho)/(8)` TERMINAL velocity of second ball is `v_(1) = (2(2r)^(2)(rho/8)G)/(9eta) = v/2` `v_(1) = v/2` |
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| 15354. |
A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one receded with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v_(0)= 1400Hz and the velocity of sound in air is 350m/s. The speed of each tuning fork is close to |
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Answer» `(1)/(4)`  `f_(0) ((c )/(c-v)) - f_(0) ((c )/(c+ v)) =2` [Velocity of sound C= 350 m/s] `1400 ((c^(2) + cv- c^(2) + cv)/(c^(2)- v^(2))) =2` `700 xx 2cv = c^(2) -v^(2)` `v^(2) + 1400cv - c^(2) = 00` By TAKING C= 350 `v^(2) + 1400 xx 350 v- [350]^(2)= 0` `:. v^(2) + 490000 v- 122500= 0` Which is QUADRATIC equation `:. a=1, b= 490000c, = -122500` `:. sqrtDelta = sqrt(b^(2) - 4ac)` `= sqrt((490000)^(2)- 4 xx 1 xx (-122500))` `=sqrt(2401 xx 10^(8) + 490000)` `= sqrt(24010049 xx 10^(4))` `= 4900.00499 xx 10^(2)` =490000.5 Now `v= (-b +- sqrtDelta)/(2a)` `=(-490000 +- 490000.5)/(2 xx 1)` `= (-490000 + 490000.5)/(2) or (-490000-490000.5)/(2)` `= (0.5)/(2) = (1)/(4)`= This is impossible Because `(v LT c)` velocity v should be less than c. |
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| 15355. |
Under the action of force P, the constant acceleration of block B is 3 ms^(-2) to the right. At the instant when the velocity of B is 2 ms^(-1) to the right, determine the velocity of B relative to A, the acceleration of B relative to A and the absolute velocity of point C of the cable. |
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Answer» 2 `(V_(B)+V_(C))/2=V_(A),V_(C)=2V_(A)-V_(B),V_(C)=3-2=1` |
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| 15356. |
When the angle of inclination of on inclined plane is theta, an object slides down with uniform velocity. If the same object is pushed up with an initial velocity u on the same inclined plane, it goes up the plane and stops at certain distance on the plane. There after the body. |
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Answer» Slides down the inclined plane and reaches the GROUND. With VELOCITY ''U'' |
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| 15357. |
The radiation emitted by a star A is 10,000 times that of the Sun. If the surface temperature of the Sun and the star are 6000 K and 2000 K respectively, the ratio of the radius of star A and the Sun is. . . .. . . . |
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Answer» `300:1` `W=esigmaAT^(4)=esigma(4piR^(2))T^(4)(becauseA=4piR^(2))` `:.WpropR^(2)T^(4)` `:.(W_(1))/(W_(2))=((R_(1))/(R_(2)))^(2)((T_(1))/(T_(2)))^(4)` `W_(1)=10000W_(2),T_(1)=2000K,T_(2)=6000K` `(100)^(2)=((R_(1))/(R_(2)))^(2)((2000)/(6000))^(4)` `:.100=(R_(1))/(R_(2))XX(1/3)^(2)` `:.(R_(1))/(R_(2))=(900)/(1)` `:.(R_(1))/(R_(2))=900:1` |
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| 15358. |
If the mean radius of the Earth is R, its angular velocity is omega and the acceleration due to gravity at the surface of the Earth is g, then what will be the cube of the radius of the orbit of a geostationary satellite. |
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Answer» Solution :Let R the radiusof thegeostationaryorbit . Angularvelocity of revolution of a geostationarysatellite is same as the angularvelocity of ROTATION of the EARTH . `mromega^(2) = (GMm)/(r^(2))` ` r^(3) = (GMm)/(momega^(2)) xx (R^(2))/(R^(2)) = (GM)/(R^(2)) ((R^(2))/(omega^(2))) "" [ g = (GM)/(R^(2))] ` ` r^(3) = (gR^(2))/(omega^(2))` |
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| 15359. |
Select the incorrect alternatives(s): |
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Answer» has a FIXED value |
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| 15360. |
Which has greater angular velocity (i) rotation of the hour hand of watch (ii) rotation of earth around its axis ? |
| Answer» Solution :The HOUR hand, because its PERIOD T= 12 hours WHEREAS that of earth is `T = 2 xx 12` hours. | |
| 15361. |
A is a fixed point at a height H above a perfectly inelastic smooth horizontal plane. A light inextensible string of length L (gtH) has one end attached to A and the other to a heavy particle. The particle is held at the level of A with string just taut and released from rest. Find the height of the particle above the plane when it is next instantaneously at rest. |
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Answer» |
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| 15362. |
A uniform thin rod of mass 'm' and length L is held horizontally by two vertical strings attached to the two ends. One of the string is cut. Find the angular acceleration of the rod soon after it is cut : |
| Answer» Answer :C | |
| 15363. |
What is meant by right-handed Cartesian coordinate system ? |
Answer» Solution :If the X , y and zaxes are DRAWN in anticlockwise direction then the corrdinate system is called as "right-handed CARTESIAN coordinate system"
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| 15364. |
If vec(A) + vec(B) = vec(R ) and 2 vec(A) + vec(B) is perpendicular to vec(B) then |
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Answer» A=R |
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| 15365. |
When a body is hung from a string. It does not snap. But when the same mass is set into rotation along a horizontal path at high speed holding the other end of the string. The string snaps. What is the reason? |
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Answer» Solution :If the mass of the body is m, then in the first CASE , TENSION in the string, T.= mg…(1) Let the length of the thread be l, linear velocity of the body be v, and tension in the thread beT, when the body is rotated in the horizontal PLANE. Required centripetal force for revolution, ` (mv^(2))/(l sin theta) = T sin theta""`...(2) Also,mg = T COS`theta""` ...(3) From EQUATIONS (1) and (3) we get, T. = T cos `theta` `therefore "" 0 le theta lt (pi)/(2)` `therefore "" 0 lt cos theta le 1 "" therefore T. lt T` So, tension in the thread is greater in the second case than that in the first case, and hence, in the first case though the thread does not snap, in the second case the thread my snap. 
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| 15366. |
The molar heat capacity of a gas at constant volume is C_(v) . If n moles of the gas undergo DeltaT change is temperature, its internal energy will change by nC_(v)DeltaT |
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Answer» Only if the change of TEMPERATURE occurs at CONSTANT volume |
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| 15367. |
A disc is given an angular velocity omega_(0) and a linear velocity v_(0) as shown in the figure. It is released on a rough horizontal surface of friction coefficient mu. Mark the correct statement (omega_(0)=3v_(0)//R) |
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Answer» The frictional force will be `mu`mg during the entire motion. |
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| 15368. |
What is the significance of moment of a force about a point? |
| Answer» Solution :Moment of a force on a body about a POINT is an external influence that can ROTATE the body about that point. Higher TORQUE would produce more ROTATION if the body REMAINS the same. | |
| 15369. |
The pressure P and volume V of an ideal gas both increase in a process |
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Answer» It is not possible to have such a PROCESS |
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| 15370. |
Consider the motion of the tip of the minute hand of a clock. In one hour (a) the displacement is zero (b) the distance covered is zero (c ) the average speed is zero (d) the average velocity is zero |
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Answer» a & B are CORRECT |
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| 15371. |
Let .A. be the area swept by the line joining the earth and the sun during feb 2008. The area swept by the same line during the first week of that month is |
| Answer» ANSWER :B | |
| 15372. |
A point sized sphere of mass .m. is suspended from a point using a string of length .r.. It is then pulled to a side till the string is horizontal and released. As the mass passes through the portion where the string is vertical, magnitude of its angular momentum is |
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Answer» `mlsqrt(GL)` |
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| 15373. |
Derive the work energy theorem for a variable force exerted on a body in one dimension . |
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Answer» Solution :If a body of mass m and speed v moving in X - direction in one DIMENSIONS then its kinetic energy . `K = 1/2 mv^(2)` Intergating on both the SIDE , `(dK)/(dt) =d/(dt) (1/2 mv^(2))` ` =1/2 mxx2v . (dv)/(dt)` ` :. (dK)/(dt) =m . (dv)/(dt) xxv` ` :. (dK)/(dt) = mav [ :. (dv)/(dt) =a] ` ` :. (dK)/(dt) = F (DX)/(dt) [ :. ma = F and V = (dx)/(dt)] ` ` :. dK = F dx ` Intergating on both side from intial position `x_(i)` to final position `x_(f)` `int_(K_(i))^(K_(f)) dK = int_(x_(i))^(x_(f)) F dx` `K_(f) - K_(i) = F(x_(f)-x_(i))` ` = FDeltac ` where `Deltax = x_(f) -x_(i)` which is a work energy theorem for a variable force . Work energy theorem does notincorporate the COMPLETE dynamical information of Newton.s second law. Work energy theorem involves an integral over an interval of time , the timeinformation contained in the Newton.s second law is integrated over and is not available explicity . Newton.s second law for two or three dimensions is in vector form whereas the work energy theorem is in scalar form . |
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| 15374. |
Find the efficiency of a Carnot's engine working between 127^(@)C and 27^(@)C. |
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| 15375. |
The escape velocity on the surface of earth is 11.2 km/s. What will be its value on a planet having double the radius and eight times the mass of earth? |
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Answer» SOLUTION :`implies v_p/v_e=SQRT((2GM_e)/R_p)XX sqrt((R_e)/(GM_e))=sqrt(M_p/M_exxR_e/R_p)` `v_p/v_e=sqrt(8xx1/2)=sqrt4 =2` `:. v_p=2v_e=2 (11.2km//s)= 22.4 (KM)/s` |
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| 15376. |
When two water drops merge to form a large drop |
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Answer» ENERGY is LIBERATED |
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| 15377. |
The fraction of ice that melts by mixing equal masses of ice at -10°C and water at 60°C is |
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Answer» `(6)/(11)` |
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| 15378. |
The centreof massof a body |
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Answer» depends on the choiceof co - ordinatesystem |
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| 15379. |
Figure shows two hydrogen atoms. Show on as separate diagram all the electric forces acting on different particle of the system. |
Answer» Solution :Each PARTICLE EXERTS electric forces on the REMAINING three particles. Thus there EXIST `4xx3=12` forcers in all. ure SHOWS them 
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| 15380. |
A block of mass 5kg is lying on a rough horizontal surface. The coefficient of static and kinetic friction are 0.3 and 0.1 and 10m//s^(-2). If a horizontal force of 50N is applied on the block, the frictional force is |
| Answer» Solution :`f_(s)=mu_(S)N, f_(K)=mu_(k)N, N=mg` | |
| 15381. |
If the length and tiome period of an oscillating pendulum have errors of 1% and 3% respectgively then the error in measurement of acceleratinon due to gravity is |
| Answer» ANSWER :D | |
| 15382. |
A block of mass 0.50 kg is moving with a speed of 2.00 m s^(-1) on a smooth surface. It strikes another mass of 1.00 kg which is at rest and then they move together as a single body. The energy which is lost during the collision is |
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Answer» Solution :Here, `m_1 = 0.50 kg, u_1 = 2.00 MS^(-1)` `m_2 = 1.00 kg, u_2 = 0` If v is the velocity of the combination after collision , then according to the PRINCIPLE of conservation of linear momentum `m_1 u_1 + m_2u_2 = (m_1 + m_2) v` or `v = (m_1 u_1)/(m_1 + m_2) = (0.50 XX 2.00)/(0.50 + 1.00) = (1.00)/(1.50) = 2/3 ms^(-1) ( :. u_2= 0)` Energy loss = Initial energy - Final energy `=1/2 m_1 u_1^2 - 1/2 (m_1 + m_2)v^2` `= 1/2 xx 0.50 xx (2.00)^2 - 1/2 xx (0.50 + 1.00) xx (2/3)^2` `1.00 - (1.50)/2 xx 4/9` = 0.67 J`. |
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| 15383. |
A bomb in the steady state explodes into three fragments. Two fragments of equal masses move with velocity 15 ms^(-1) in mutually perpendicular directions. The mass of the third fragment is equal to three times the mass of each of these two fragments. The magnitude of velocity of third fragment is ms^(-1) . |
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Answer» 20 `0=mv_(1) HAT( I ) =mv^(2) hat(J ) + 2mv_(3)` `:. ,vec( V )_(3)= (-5)hat(I ) + (-5)hat(j )` `|vec(v)_(3)| = sqrt(-5)^(2) + (-5)^(2))` `=sqrt(25 xx 2) = 5 squr(2) m//s` |
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| 15384. |
A body is in pure rotation. The linear speed v of the particle, the distance r of the particle from the axis and the angular velocity w of the body are related as omega=(v)/(r). Thus |
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Answer» `OMEGA alpha(1)/(r)` |
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| 15385. |
A cyclist is moveing on a smooth horizontal curved surface with a speed of 5ms^(-1). If angle of leaning is 30^(@) radius of curvature of Road should be |
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Answer» `5sqrt(3)` |
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| 15386. |
An object of mass m is ejected from a satellite revolving around earth at a distance r with constant speed v. If the object excapes from gravitational pull of earth, then the kinetic energy with which the object is thrown is |
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Answer» `-0.5 mv^(2)` `T.E. = P.E. + K.E.` `T.E. = -(GMm)/(2a)` For object to be ejected with velocity v in orbit r T.E. = - K.E. `rarr T.E. = -(1)/(2) mv^(2)` …(i) If we want that the object escapes earth.s GRAVITATIONAL pull, it should be supplied ENERGY that is EQUAL to the NEGATIVE of the total energy. Therefore, energy required to escape `K.E.. = - (T.E.) = (1)/(2) mv^(2)` |
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| 15387. |
If for a gas ( R)/(C_v) =0.67, the gas is made of molecules which are |
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Answer» DIATOMIC |
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| 15388. |
A load 'm' is altached to a spring of force constant 'k' and stretched it through 'x' and released, it makes oscillation in a vertical plane with a time period T'. It is further pulled down through another 'x' and released. Now the time period of vertical oscillation will be |
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Answer» `T//2` |
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| 15389. |
A siren emitting sound of frequency 800 Hz is going away from a static listener with a speed of 30 m/s. Frequency of the sound to be heard by the listener is (Take velocity of sound as 330 m/s) |
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Answer» `733.3 Hz `  We have `(F _(L))/( v + v _(L)) = (f _(S))/( v + v _(S ))` `THEREFORE (f _(L))/( 330 + 0) = (800 )/( 330 + 30)` `therefore f _(L) = 800 xx (330)/(360) = 733.3 Hz` |
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| 15390. |
Compare the periods of revolution of two planets with their semi major axes. |
| Answer» SOLUTION :`(T_(1))/(T_(2))=((a_(1))/(a_(2)))^(1//2)` | |
| 15391. |
The specific heat of a substance varies as (3t^(2) + t) xx 10^(-3) "cal"//g- ^(0)C. What is the amount of heat required to raise the temperature of 1 kg of substance from 10^(@)C to 20^(@)C? |
| Answer» Solution :For small change in temperature dt, heat required , dQ =med t < br> `therefore Q = int_(t_(1))^(t_(2))` mcdt Given , m= 1000 g, `therefore Q= int_(10)^(20)1000(3t^(2) + t xx 10^(-3) dt =|t^(3)+t^(2)/2|_(10)^(20) -(20^(3) + 20^(2)/2) -(10^(3)+10^(2)/2)= 8200-1050=7150 "CAL".` | |
| 15392. |
Which one of the following is not a force? |
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Answer» Impulse |
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| 15393. |
A 10 W electric heater is used to heat a container filled with 0.5 kg of water. It is found that the temperature of water and container rises by 3^(@)K in 15 min. The container is then emptied, dired and filled with 2kg of oil. The same heater now raises the temperature of container oil system by 2K in 20 min. Assume there is no heat loss in the process and the specific heat of water is 4200Jkgi^(-1)K%^(-1), the specific heat of oil in the same limit is equal to |
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Answer» <P>`1.50 xx 10^(3)` i.e., `m_(1)c_(1)Deltat+m_(2)c_(2)Deltat`= work done So, `m_(1)c_(1)Deltat+m_(2)c_(2)Deltat=P_(1)t_(1)` where, `m_(1) = 0.5` kg, specific heat `c_(1)` = 4200 J`kg^(-1)K^(-1)`, `Deltat = Deltat_(1) = Deltat_(2)=3K` `P_(1)=P_(2)`=10W `t_(1)` = 15 xx60=900 s `c_(2)`= specific heat capacity of container. So, `0.5 xx4200xx(3-0) + m_(2)c_(2)xx(3-0) = 10 xx15 xx60` `2100 xx3 +m_(2)c_(2)xx3=9000` `m_(2)c_(2)=(900-6300)/3 = 900` Similarly, in the cae of oil, `m_(1)c_(0)Deltat + m_(2)c_(2)Deltat=P_(2)t_(2)` Where, `c_(0)` = specific heat capacity of oil `P_(1)=P_(2)`=10W `2 xxc_(0)xx2+900 xx2=10 xx20 xx60` `4c_(0)` + 1800 = 12000 `c_(0)` = 2550 = `2.55 xx 10^(3)`J`kg^(-1)K^(-1)` |
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| 15394. |
A man stands on a rotating table stretching his arms. He is rotating with a definite angular velocity. Now the man draws his arms closer. His moment of inertia is reduced to 75% of its initial value. The angular kinetic energy of the man |
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Answer» will INCREASE by 33.3 % |
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| 15395. |
A solid sphere of mass 2 kg is rolling on a frictional horizontal surface with velocity 6 m/s. It collides on the free end of an ideal spring whose other end in fixed. The maximum compression produced in the spring will be (Force constant of the spring is 36 N/m). |
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Answer» `sqrt(14) `m `= (1)/(2) mv^(2) + (1)/(2) I OMEGA^(2) = (1)/(2) mv^(2) + (1)/(2) xx (2)/(5)mr^(2) omega^(2) = (1)/(2) mv^(2) + (1)/(5) mv^(2)` `= (7)/(10) mv^(2) = (7)/(10) xx 2 xx (6)^(2) = (7)/(10) xx 2 xx 36 = (252)/(5)` The potential energy of the spring of MAXIMUM compression X , `= (1)/(2) kx^(2) or (1)/(2) kx^(2) = (252)/(5) , kx^(2) = (252 xx 2)/(5)` `x^(2) = (252 xx 2)/(5 xx 36) = 2.8 or x = sqrt(2.8)` m |
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| 15396. |
A mass attached to a spring is free to oscillate, with angular velocity omega, in a horizontal plane without friction or damping. It is pulled to a distance x_(0) and pushed towards the centre with a velocity v_(0) at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters omega,x_(0)andv_(0). Hint : Start with the equation x=acos(omegat+theta) and note that the initial velocity is negative. |
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Answer» |
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| 15397. |
Two identical glass equiconvex lenses of focal length f each are kept in contact. The space between the two lenses is filled with water . The focal length of the combination is |
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Answer» `F//3` |
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| 15398. |
State Newton's law of cooling. |
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Answer» Solution :Newton.s law of cooling states that the RATE of loss heat of a body is directly proportional to the mean EXCESS of temperature of the body over surroundings. This law is valid only up to `30^@ C` temperature difference. Also, the loss of heat by RADIATION depends upon the nature of the surface of the body and the AREA of the exposed surface. Rate of cooling`(delta theta)/(dt)prop (theta - theta_(0))` |
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| 15399. |
Two wires have masses in the ratio 3:4. They are made of the same material. Under the same load what is ratio of their elongations ? Explain why? |
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Answer» |
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| 15400. |
coefficient of real expansion of mercury is 0.18xx10^(-3)//""^(@)C. If the density of mercury at 0^(@)C is 13.6 gm/c.c., its density at 573 K will be |
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Answer» 12.90 gm/c.c |
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