InterviewSolution
Saved Bookmarks
InterviewSolution
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.
| 351. |
At what angle should a road be banked so that the vehicle may take a bend of radius 10 m travelling with a speed of `10m//s` `( g=10 m//s^(2))`A. `80^(@)`B. `45^(@)`C. `60^(@)`D. `90^(@)` |
|
Answer» Correct Answer - B ` tan theta =(v^(2))/(rg)=(10xx10)/(10xx10)=1` `theta =45^(@)`. |
|
| 352. |
A box is placed on the floor of a truck moving with an acceleration of `7 ms^(2)`.If the coeffecient of kenetic friction between the box and surface of the truck is `0.5`,find the acceleration of the box relative to the truckA. `1.7 ms^(-2)`B. `2.1 ms^(-2)`C. `3.5 ms^(-2)`D. `4.5 ms^(-2)` |
|
Answer» Correct Answer - B `f_(k)=mu_(k)N,N=mg,a^(1)=mu_(k)g,a_(rel)=a-a^(1)` |
|
| 353. |
A block is placed at a distance of `2m` from the rear on the floor of a truck `(g=10ms^(-2))`.When the truck moves with an acceleration of `8ms^(-2)` the block takes `2 sec` to fall off from the rear of the truck.The coefficient of sliding friction between truck and the block isA. `0.5`B. `0.1`C. `0.8`D. `0.7` |
|
Answer» Correct Answer - D `s=ut+1/2at^(2),a^(1)=mu_(k)g,a_(rel)=a-a^(1)` |
|
| 354. |
A car is travelling at 36 kmph on a road . The maximum turning radius of the car is 20 m ans `g=10m//s^(2)` .Then the coefficient of friction between its tyres and the road isA. `0.2`B. `0.4`C. `0.5`D. `0.8` |
|
Answer» Correct Answer - C `v=36xx(5)/(18)=10m//s, r=20m, g=10 m//s^(2), mu=?` `v=sqrt(mu rg)` `v^(2)=mu rg therefore mu=(v^(2))/(rg)` `=(10xx10)/(20xx10)=0.5.` |
|
| 355. |
The height of the centre of gravity of the truck above the ground is 1.5 m and the distance between the wheel is 1.5 m .If the maximum velocity at which a truck can safely tavel along the horizontal track withiout toppling on a curve of radius 250 m will beA. `30 m//s`B. `35 m//s`C. `40 m//s`D. `45 m//s` |
|
Answer» Correct Answer - B `v=sqrt((rgd)/(2h))` `=sqrt((250xx10xx1.5)/(2xx1.5))=sqrt(1250)=35 m//s` |
|
| 356. |
If a car moves with a velocity of 45 kmph, angular velocity of its wheel of diameter 50 cm isA. `50 rad //s`B. `25 rad//s`C. `100 rad//s`D. `5 rad//s` |
|
Answer» Correct Answer - C `v=45xx(5)/(18)=(25)/(2)` `omega=(v)/(r)=(25)/((2)/(0.5))=25 rad//s` |
|
| 357. |
A small spehre of mass `m` is connected by a stirng to a nail at `O` and moves in a circle of radius `r` on the smooth plane inclined at an angle ` theta` with the horizontal. If the sphere has a velocity `u` at the top position A. Mark the correct options. A. the tension in the string as the sphere passes the `90^(@)` position B is equal to `m((u^(2))/(l)-2gsin theta)`B. the tension in the string at the bottom most position C is equal to `m((u^(2))/(l)+5g sin theta)`C. the tension in the string as the sphere passes the `90^(@)` position B is equal to `m((u^(2))/(l)+2g sin theta)`D. the tension in the string at the bottom most position C is equal to `m((u^(2))/(l)-5g sin theta)` |
|
Answer» Correct Answer - B::C |
|
| 358. |
The centre of gravity of a car is at a height `h` and the distance between its wheel is 2a. The car moves on a level curve of radius r with speed `v`. Let `N_(1) and N_(2)` be the normal reactions on the inner and outer wheels of the car. ThenA. `N_(1)gtN_(2)`B. `N_(2)gtN_(1)`C. `N_(1)=N_(2)`D. the maximum value of v to avoid overturning is `sqrt((a r g)/(h))` |
|
Answer» Correct Answer - B::D |
|
| 359. |
A body of mass 40kg resting on a rough horizontal surface is subjected to a force `P` which is just enough to start the motion of the body. If `mu_(s)=0.5mu_(k)=0.4 , g=10ms^(-2)` an dthe force `P` is continuously applied on the body, then the accceleration of the body is.A. `0.98`B. `3.92`C. `4.90`D. Zero |
|
Answer» Correct Answer - A `F_(R)=f_(s)-f_(k),a=(mu_(s)-mu_(k))g` |
|
| 360. |
A person weighing `60kg` In a small boat of mass `140 kg` which is at rest, throws a `5 kg` stone in the horizontal direction with a velocity of `14m//s^(-1)`. The velocity of the boat immediately after the throw is (in `m//s`)A. `1.2`B. `0.5`C. `0.35`D. `0.65` |
|
Answer» Correct Answer - C `m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)` |
|
| 361. |
A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at `t=0`. At this instant of time, the horizontal component of its velocity is v. A bead Q of the same mass as P is ejected from A at `t=0` along the horizontal string AB, with the speed v. Friction between the bead and the string may be neglected. Let `t_P` and `t_Q` be the respective times taken by P and Q to reach the point B. Then: A. `t_(P)ltt_(Q)`B. `t_(P)=t_(Q)`C. `t_(P)gt t_(Q)`D. `(t_(P))/(t_(Q))=("length of arc ACB")/("length of chord AB")` |
|
Answer» Correct Answer - A P particle has `a_(1)=g sin theta` due to which its horizontal component of velocity gt v |
|
| 362. |
Assertion: A frame moving in a circle with constant speed can never be an inertial frame. Reason: It has a constant acceleration.A. If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. If both Assertion and Reason are true but Reason is not the correct explanation of Assertion.C. If Assertion is true, true but the Reason is false.D. If Assertion is false but the Reason is true. |
|
Answer» Correct Answer - C Direction of acceleration keeps on changing. So, it is variable acceleration.Further, it is acceleration so it is non-inertial. |
|
| 363. |
Two cars having masses `m_1 and m_2` miove in circles of radii `r_1 and r_2` respectively. If they complete the circle is equal time the ratio f their angular speedsd `omega_1/omega_2` isA. `m_1/m_2`B. `r_1/r_2`C. `m_1r_1/m_2r_2`D. 1 |
| Answer» Correct Answer - D | |
| 364. |
A particle o fmass m is observed from an inertial frame of reference and is found to miove in a circle of radius r with a uniform speed v. The centrifugal force on it isA. `(mv^2)/r` towards the centreB. `(mv^2)/r` away from the centreC. `(mv^2)/r` along the tangent through the particleD. zero |
| Answer» Correct Answer - D | |
| 365. |
A car of mass m moves in a horizontal circular path of radius r meter. At an instant its speed is `Vm//s` and is increasing at a rate of a `m//sec^(2)`. Then the acceleration of the car is:A. `(a^(2)+(v^(2))/(r))`B. `sqrt(a^(2)+((v^(2))/(r)))`C. `a^(2)+((v^(2))/(r))^(2)`D. `sqrt((a^(2))+((v^(2))/(r))^(2))` |
|
Answer» Correct Answer - D `a=sqrt(a^(2)+((v^(2))/(r)))^(2)` |
|
| 366. |
A car of maas M is moving on a horizontal circular path of radius r. At an instant its speed is v and is increasing at a rate a.A. The acceleration of the car is towards the centre of the pathB. The magnitude the frictional force on the car is greater than `(mv^2)/r`.C. The friction coefficient between the ground and the car is not less than `a/g`.D. The friction coefficient between the ground and the car is `mu=tan^-1 v^2/rg` |
| Answer» Correct Answer - B::C | |
| 367. |
Two moving particles `P` and `Q` are 10cm apart at any instant. Velocity of P is `8 m//s` at `30^(@)`, from line joining the P and Q and velocity of Q is `6m//s` at `30^(@)`. Calculate the angular velocity of P w.r.t. Q |
| Answer» `omega_(PQ)=(8sin30^(@)-(-6sin30^(@)))/(10)=0.7rad//s` | |
| 368. |
A particle is projected with velocity `20sqrt(2)m//s` at `45^(@)` with horizontal. After `1s` , find tangential and normal acceleration of the particle. Also, find radius of curveture of the trajectory at that point. (Take `g=10m//s^(2))` |
|
Answer» Correct Answer - A::B::D After `1s` , `v=u+at=20hat(i)+10hat(j)` , `v=sqrt(500)m//s=10sqrt(5)m//s` `a=-10hat(j)` `a_(t)=acostheta` `=(a.v)/(v)=(-100)/(10sqrt(5))` `=-2sqrt(5)m//s` `a_(n)=sqrt(a^(2)-a_(t)^(2))=sqrt((10)^(2)-(2sqrt(5)^(3)))` `=sqrt(80)m//s^(2)=4sqrt(5)m//s^(2)` `R=(v^(2))/(a_(n))=(10sqrt((5)^(2)))/(4sqrt(5))=25sqrt(5)m` . |
|
| 369. |
A ball suspended by a thread swing in a vertical plane that its acceleration values in the lowest possition and the extreme postition are equal . Find the thread deffection angle in the extreme possition.A. `30^(@)`B. `60^(@)`C. `53^(@)`D. `37^(@)` |
|
Answer» Correct Answer - C Acceleration at lowest position `a_(L)=(v^(2))/R` from energy conservation `mgR(1- cos theta)=(mv^(2))/2` `(v^(2))/R=2g(1-cos theta)` `a_(L)=2g(1-cos theta)` acceleration at highest position `a_(H)=g sin theta` according to problem `a_(L)=a_(H)` `2g(1-cos theta)=g sin theta` `2(1- cos theta)=sin theta` `rArr 2(1-1+2 sin^(2)theta//2)=2 sintheta//2 cos theta//2` `tan .(theta)/(2)=1/2` `tan theta=(2 tan .(theta)/(2))/(1-tan^(2).(theta)/(2))=(2(1/2))/(1-1/4)=4/3` `rArr theta=53^(@)` |
|
| 370. |
A block of mass `2kg` is placed on the surface of trolley of mass `20kg` which is on a smooth surface.The coefficient of friction between the block and the surface of the trolley is `0.25`.If a horizontal force of `2 N` acts on the block, the acceleration of the system in `ms^(-2)` is `(g=10ms^(-2))`A. `1.8`B. `1.0`C. `0.9`D. `0.09` |
|
Answer» Correct Answer - D `f=mu_(s)N=mu_(s)mg` Block does not move `a=F/(m+M)` |
|
| 371. |
In a two dimensional motion of a body, prove that tangentiol acceleration is nothing but component of acceleration along velocity. |
|
Answer» `v=v_(x)hat(i)+v_(y)hat(j)` Accleration `a=(dv_(x))/(dt)hat(i)+(dv_(y))/(dt)hat(j)` Component of a along `v` will be, `(a.v)/(|v|)=(v_(x)(dv_(x))/(dt)+v_(y).(dv_(y))/(dt))/(sqrt(v_(x)^(2)+v_(y)^(y))` Further, tangential acceleration of particle is rate of change of speed. or `a_(t)=(dv)/(dt)=(d)/(dt)(sqrt(v_(x)^(2)+v_(y)^(2)))` or `a_(t)=(1)/(2sqrt(v_(x)^(2)+v_(y^(2))))[2v_(x).(dv_(x))/(dt)+2v_(y)(dv_(y))/(dt)]` or `a_(t)=(v_(x).(dv_(x))/(dt)+v_(y).(dv_(y))/(dt))/(sqrt(v_(x)^(2)+v_(y)^(2)))` Form Eqs. (i) and (ii), we can see that or Tangential acceleration=component of acceleration along velocity. |
|
| 372. |
A point moves along a circle with a speed `v=kt` , where `K=0.5m//s^(2)` Find the total acceleration of the point the momenet when it has covered the `n^(th)` Fraction of the circle after the beginging of motion, where `n=(1)/(10)` . |
|
Answer» Correct Answer - B Let velocity of the particle be, `v=(ds)/(st)=kt` or `int_(0)^(s)ds=kint_(0)^(t)tdt` implies :. `s=(1)/(2)kt^(2)` For completion of `nth` fraction of circle, `s=2pirn=(1)/(2)kt^(2)` or `t^(2)=(4pinr)//k` Tangential acceleration `=a_(t)=(dv)/(dt)=k` Normal acceleration `=a_(n)=(v^(2))/(r)=(k^(2)t^(2))/(r)` Substituting the value of `t^(2)` from Eq. (i), we have or `a_(n)=4pirk` :. `a=sqrt((a_(t)^(2)+a_(n)^(2))=[k^(2)+16pi^(2)n^(2)k^(2)]^(1//2)` `=k[1+16pi^(2)n^(2)]^(1//2)` `=0.50[1+16xx(3.14)^(2)xx(0.10)^(2)]^(1//2)` `=0.8m//s^(2)` |
|
| 373. |
The minimum force required to move a body up on an inclined plane is three times the minimum force required to prevent it from sliding down the plane. If the coefficient of friction between the body and the inclined plane is `1/(2sqrt3)` the angle of the inclined plane isA. `60^(@)`B. `45^(@)`C. `30^(@)`D. `15^(@)` |
|
Answer» Correct Answer - C `F_(1)=mg(sin theta+mu_(k)cos theta)` `F_(2)=mg(sin theta-mu_(k)cos theta)` |
|
| 374. |
The speed of a particle moving in a circle of radius `r=2m` varies witht time `t` as `v=t^(2)`, where `t` is in second and `v` in `m//s`. Find the radial, tangential and net acceleration at `t=2s`. |
|
Answer» Given in the question, `v=t^(2)` `therefore` Linear speed of particle at `t=2s` is `v=(2)^(2)=4ms^(-1)` `therefore" Radial acceleration "a_(r)=(v^(2))/(r)=((4)^(2))/(2)=8ms^(-2)` The tangential acceleration is `a_(t)=(dv)/(dt)=(d)/(dt)(t^(2))=2t` `therefore` Tangential acceleration at t 2 s is `a_(t)=(2)(2)=4ms^(-2)` `therefore` Net acceleration of particle at t = 2 s is `a=sqrt((a_(r))^(2)+(a_(t))^(2))=sqrt((8)^(2)+(4)^(2))=sqrt(64+16)" or "a=sqrt(80)"ms"^(-2)` |
|
| 375. |
If the kinetic energy of a particle moving with a constant speed on a circular path of radius 4 m is 100 j .Then the centripetal force will beA. 30 NB. 50 NC. 40 ND. 60 N |
|
Answer» Correct Answer - B `K.E.=(1)/(2)mv^(2)=(1)/(2)((mv^(2))/(r))r=F_(r)/(2)` `therefore F=(2E)/(r)` |
|
| 376. |
If a car is taking turn along a circular track, then the necessary force reqyired to go round circular track will beA. gravitational forceB. restoring forceC. centripetal forceD. centrifugal force |
| Answer» Correct Answer - C | |
| 377. |
A car is travelling along a circular curve that has a radius of `50m` . If its speed is `16m//s` and is increasing uniformly at `8m//s^(2)` . Determine the magnitude of its acceleration at this instant. |
|
Answer» Given, tangential acceleration `a_(t)=8ms^(-2)` Radius R = 50 m, Speed v = 16`ms^(-1)` `therefore"Radial acceleration" a_(r)=(v^(2))/(R)=((16)^(2))/(50)=(256)/(50)` Magnitude of net acceleration of the car. `a=sqrt(a_(t)^(2)+a_(r)^(2))=sqrt((8)^(2)+((256)/(50))^(2))=9.5ms^(-2)` |
|
| 378. |
A car is travelling along a circular curve that has a radius of `50m` . If its is speed is `16m//s` and is increasing uniformly at `8m//s^(2)` . Determine the magnitude of its accleration at this instant. |
|
Answer» Correct Answer - B `a_(t)=8m//s^(2)` `a_(r)=(v^(2))/(R)=(16)^(2)/(50)=5.12m//s^(2)` `a=sqrt(a_(t)^(2)+a_(r)^(2))=9.5m//s^(2)` |
|
| 379. |
For a body travelling along a circle of radius 4 m and with a speed of ` m//s`, the force acting on the body towards the centre is 16 N.If the radius is 2 m and the speed is `3 m//s` the force towards the centre isA. 18 NB. 36 NC. 9 ND. 27 N |
| Answer» Correct Answer - A | |
| 380. |
A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force equal to `(-K//r^(2))`, where k is a constant. The total energy of the particle is -A. `-K//r`B. `-K//2r`C. `K//2r`D. `-2K//r` |
|
Answer» Correct Answer - B `F=(-K)/(r^(2))` `KE=(1)/(2)Fr=(1)/(2)(-K)/(r^(2))xxr=-(k)/(2r)` |
|
| 381. |
A body of mass 1 kg is suspended by a string 1 m long .The body is rotate in a vertical with a constant speed of `1m//s` . The tension in the steing when it is at horizontal position will beA. 1 NB. 2 NC. 3 ND. 4 N |
|
Answer» Correct Answer - C `T=(mv^(2))/(r)=(1xx1)/(1)=1N`. |
|
| 382. |
A boy is sitting on a horizontal platform of joy wheel at a distance of 5 m from its centre .The joy wheel begins to rotate and when the angular speed exceeds 10 revolutions per minute, the boy just slip , the cofficient of friction between (`g=10m//s^(2))`A. `pi ^(2)//6`B. `pi^(2)//18`C. `pi//6`D. `pi//2` |
| Answer» Correct Answer - B | |
| 383. |
The magnitude of displacement of a particle moving in a circle of radius a with constant angular speed `omega` varries with time t isA. `2 a sin omega t`B. `2a " sin " (omegat)/(2)`C. `2a cos omega t`D. `2a " cos " (omega t)/(2)` |
|
Answer» Correct Answer - B |
|
| 384. |
Consider the system as shown in the figure. The pulley and the string are light and all the surfaces are frictionless. The tension in the string is `(g=10m//s^(2))` . A. `0 N`B. `1 N`C. `2 N`D. `5 N` |
|
Answer» Correct Answer - D For the block on the surface `T=m_(1)a` For the hanging block `m_(2)g-T=m_(2)a` |
|
| 385. |
If a body is moving withuniform speed of `10 m//s` on a circular path of diameter 2 m , then the difference between the distance covered by it and displacement un half revolution will beA. 2 mB. 1.142 mC. 3.142 mD. 6.284 m |
|
Answer» Correct Answer - B `s=pir-2r` `=r(pi-2)` `=1(3.142-2)` `=1.142m` |
|
| 386. |
If a speed man is rotated at the end of a long beam of length 5m and the acceleration of 9g, then the number of revolution performed will be (`g=10 m//s^(2))`A. `(3)/(sqrt(2)pi) rps`B. `(3pi)/(sqrt2) rps`C. `(2sqrt5)/(3) rps`D. `(sqrt3)/(3 pi ) rps` |
|
Answer» Correct Answer - A Centripetal acceleration `(v^(2))/(r)= 9g ` `r omega ^(2)=9g` `omega^(2)=(9g)/(r)` `omega=sqrt((9g)/(r))` `2pi n = sqrt((9g)/(r))` `n=(1)/(2pi)sqrt((9g)/(r))`. |
|
| 387. |
A chain consisting of `5` links each of mass d `0.1 kg` is lifted vertically up with a constant acceleration of `2.5m//s^(2)`.The force of interaction between `1st` and `2nd` links as shown A. `6.15 N`B. `4.92 N`C. `9.84 N`D. `2.46 N` |
|
Answer» Correct Answer - B `T-mg=ma` |
|
| 388. |
A chain of 100 links is 1 m long and has a mass of 2 kg . With the ends fastened together it is set roation at 3000 rpm.then centripetal force on each link isA. 3.14 NB. 31.4 NC. 314 ND. 3140 N |
| Answer» Correct Answer - C | |
| 389. |
Two particles revolve concentrically in a horizontal plane in the same direction. The time required to complete one revolution for particle `A` is `3min` , while for particle `B` is `1min` . The time reqired for `A` to complete one revolution relative to `B` isA. `2` minB. `1` minC. `1.5` minD. `1.25` min |
|
Answer» Correct Answer - C `(omega_(1)-omega_(2))t=2pi` `t=(2pi)/(omega_(1)-omega_(2))=(2pi)/((2pi//T_(1))-(2pi//T_(2)))` `(T_(1)T_(2))/(T_(2)-T_(1))=(3xx1)/(3-1)` `=1.5min` |
|
| 390. |
A stone of mass 50 g is tied to the end of a string 2 m long and is set into rotation in a horizontal circle with a uniforn speed of `2m//s` .Then tension in the string isA. 0.1 NB. 0.2 NC. 0.4 ND. 0.8 N |
| Answer» Correct Answer - A | |
| 391. |
An object of mass M is tied to a string of l and revolve in a horizontal circle .If length is reduced by `l//2`, then period isA. `T//sqrt2`B. `2sqrt2 T`C. `T//4`D. `2 T` |
|
Answer» Correct Answer - C `omega =(2 pi)/(T)=(v)/(l)` Accounding to the law of conservation of angular momentum , `m_(1)v_(1)r_(1)=m_(2)v_(2)r_(2)` `therefore r_(1)^(2) omega _(1)=r_(2)^(2)omega _(2)implies (r_(1)^(2))/(T_(1))=r_(2)^(2)/(T_(2))` `therefore T_(2)=((r_(2)^(2))/r_(1)^(2))T_(1)((l^(2)//4)/(l^(2)))T=(T)/(4)`. |
|
| 392. |
A particle performing circular motion with its diameter d and velocity v. Then the angular displacement of the particle in time t isA. `(vt)/(d)`B. `(2vt)/(d)`C. `(vt)/(2d)`D. `(d)/(vt)` |
|
Answer» Correct Answer - B `theta= omegat` `=(V)/(r)t=(2V)/(2r)t` `theta=(2Vt)/(d)`. |
|
| 393. |
The speed of revolution of a particle around a circle is halved and its angular speed is doubled what happens to the radial acceleration ?A. remainsunchangedB. halvedC. doubledD. quadrupled |
|
Answer» Correct Answer - A `a =v xx omega` |
|
| 394. |
A particle is in motion on the x-axis. The variation ofits velocity with position is as shown. The graph is circle and its equation is `x^(2)+v^(2)=1`, where x is in m and v in m/s. The correct statement(s) is/are:- A. When x is positive, acceleration is negative.B. When x is negative, acceleration is positiveC. At Q, acceleration has magnitude `(1)/(sqrt2)m//s^(2)`D. At S, acceleration is infinite |
| Answer» Correct Answer - A::B::C | |
| 395. |
Three particles A, B, Care located at the comers of an equilateral triangle as shown in figure. Each of the particle is moving with velocity v. Then at the instant shown, the relative angular velocity of A. `A wrt. B "is"=(v cos 30^(@))/(a)" in z-direction"`B. `B wrt. C "is"=(v cos 30^(@))/(a)" in z-direction"`C. `A wrt. C "is"=(v cos 30^(@))/(a)" in z-direction"`D. `B wrt. A "is"=(v cos 30^(@))/(a)" in z-direction"` |
| Answer» Correct Answer - A::B::C | |
| 396. |
Centrifuges are speed to separate the particles ofA. light massesB. light and heavy massesC. heavy massesD. all the above are true |
| Answer» Correct Answer - B | |
| 397. |
Select the wrong statementA. centrifugal force has same magnitude as that of centripetal forceB. centrifugal force is along the radius, away from the centreC. centrifugal force exist in inertial frame of referenceD. centrifugal force is called pseudo force as its origin cannot be explained . |
| Answer» Correct Answer - C | |
| 398. |
A particle performing a U.C.M. has aA. radial velocityB. radial accelerationdirection towards the centreC. tangential accelerationD. radial acceleration , direction away from the centre |
| Answer» Correct Answer - B | |
| 399. |
If the centripetal force acting on a body performing U.C.G. is slowly decreased, then the body will moveA. along ellipitical pathB. towards centre along spiral pathC. towards centre along radiusD. outwards the centre along radius |
| Answer» Correct Answer - B | |
| 400. |
The centripetal force is real force which provides with the real interacting force ofA. mechanicalB. electricalC. magnetic or gravitationalD. all of these |
| Answer» Correct Answer - D | |