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.
| 101. |
A disc is rolling without slipping with angular velocity `omega`. P and Q are two points equidistant from the centre C. the order of magnitude fo velocity is A. `v_Q gt v_C gt v_P`B. `v_P gt v_C gt v_Q`C. `v_P = v_C, v_Q = v_C/2`D. `v_P lt v_C gt v_Q` |
|
Answer» Correct Answer - B (b) Note: in pure rolling, the point of contact is the instantaneous centre of rotation of all the particles fo the disc. On applying `v = romega` We find `omega` is same for all the particles then `v prop r.` Farther the particcles from O, higher is its velocity. |
|
| 102. |
Moment of inertia of a straight wire about an axis perpendicular to the wire passing through one of its end is I. This wire is now framed into a circle (a ring) of single turn. The moment of inertia of this ring about an axis passing through centre and perpendicular to its plane would beA. `((3)/(pi^(2)))l`B. `((3)/(4pi^(2)))l`C. `((pi^(2))/(3))l`D. `((4pi^(2))/(3))l` |
|
Answer» Correct Answer - B |
|
| 103. |
The triangular plate described in the last question has angle `lt A = theta`. Now find its moment of inertia about an axis through A perpendicular to the plane of the plate. |
|
Answer» Correct Answer - `(1)/(2) Ma^(2) (1 - (2)/(3) sin^(2) (theta)/(2))` |
|
| 104. |
ABC is an isosceles triangle right angled at A. Mass of the triangular plate is M and its equal sides are of length `a`. Find the moment of inertia of this plate about an axis through A perpendicular to the plane of the plate. Use the expression of moment of inertia for a square plate that you might have studied. |
|
Answer» Correct Answer - `(Ma^(2))/(3)` |
|
| 105. |
A uniform rod of mass 2M is bent into four adjacent semicircles each of radius r, all lying in the same plane. The moment of inertia of the bent rod about an axis through one end A and perpendicular to plane of rod is A. `22Mr^(2)`B. `88Mr^(2)`C. `44Mr^(2)`D. `66Mr^(2)` |
|
Answer» Correct Answer - C |
|
| 106. |
A uniform rectangular plate has been bent as shown in the figure. The two angled parts of the plate are of identical size. The moment of inertia of the bent plate about axis xx is I. Find its moment of inertia about an axis parallel to `x x` and passing through the centre of mass of the plate. |
|
Answer» Correct Answer - `(7I)/(16)` |
|
| 107. |
A billiard ball of mass m and radius r, when hit in a horizontal direction by a cue at a height h above its centre, acquired a linear velocity `v_(0).` The angular velocity `omega_(0)` acquired by the ball isA. `(2v_(0)h)/(5r^(2))`B. `(5v_(0)h)/(2r^(2))`C. `(5v_(0)r^(2))/(5h)`D. `(5v_(0)r^(2))/(2h)` |
|
Answer» Correct Answer - B |
|
| 108. |
A rod of weight w is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is.. And on B is......A. `N_(A)=2w(1-x//d),N_(B)=wx//d`B. `N_(A)=w(1-x//d),N_(B)=wx//d`C. `N_(A)=2w(1-x//d),N_(B)=2wx//d`D. `N_(A)=wx//d,N_(B)=w(1-(x)/(d))` |
|
Answer» Correct Answer - B |
|
| 109. |
A gridstone has a constant acceleration of `4 rad s^(-1)`. Starting from rest, calculate the angular speed of the grindstone 2.5 s later. |
|
Answer» Correct Answer - `10 rad s^(-1)` |
|
| 110. |
A disc rolls on ground without slipping . Velocity of centre of mass is v. There is a point P on circumference of disc at angle `theta` . Suppose `v_(p)` is the speed of this point. Then, match the following the following table. |
|
Answer» Correct Answer - (A)Q,(B)P,(C)S,(D)R |
|
| 111. |
A thin metal hoop of radius `0.25 m` and mass `2 kg` stars from rest and rolls down an inclined plane. If its linear velocity on reaching the foot of the plane is `2 m//s`, what is its rotational `KE` at that instant ? |
|
Answer» Correct Answer - 4 J |
|
| 112. |
A hoop rolls on a horizontal ground without slipping with linear speed `v`. Speed of a particle `P` on the circumference of the hoop at angle `theta` is : A. `2vsin((theta)/(2))`B. `v sin theta`C. `2vcos((theta)/(2))`D. `v cos theta` |
|
Answer» Correct Answer - A |
|
| 113. |
The angle turned by a body undergoing circular motion depends on time as `theta=theta_(0)+theta_(1) t+theta_(2)r^(2)`. Then the angular acceleration of the body isA. `theta_(1)`B. `theta_(2)`C. `2theta_(1)`D. `2theta_(2)` |
|
Answer» `theta=theta_(0)+theta_(1) t+theta_(2) t^(2), omega=(d theta)/(dt)=theta_(1)+2theta_(2)t` `alpha=(d omega)/(dt)=2theta_(2)` |
|
| 114. |
A solid body rotates with deceleration about a stationary axis with an angular deceleration `betapropsqrt(omega)`, where `omega` is its angular velocity. Find the mean angular velocity of the body averaged over the whole time of rotation if at the initial moment of time its angular velocity was equal to `omega_0`. |
|
Answer» `betapropsqrtomega implies beta=komega^(1//2)` (`k`: constant) `(domega)/(dt)= -komega^(1//2)` `int_(0)^(omega)omega^(-1//2)domega=-kint_(0)^(t)tdt` `(|omega^(1//2)|_(omega_(0))^(omega))/(1//2)=-kt` `sqrtomega-sqrtomega_(0)=(-kt)/(2)` `omega=(sqrtomega_(0)-(kt)/(2))^(2)` where `omega` is an angular velocity at time `t`. The body will stop, when `omega=0`. `sqrtomega_(0)-(kt)/(2)=0 implies t=(2sqrtomega_(0))/(k)` Again `beta=k omega(1//2)` ` implies omega(domega)/(d theta)=-komega^(1//2)` `int_(omega_(0))^(omega)omega^(1//2)domega=-kint_(o)^(theta) d theta implies (|omega^((3)/(2))|_(omega_(0))^(omega))/((3)/(2))=-ktheta` `omega^(3//2)-omega_(0)^(3//2)=-(3)/(2)ktheta` when body stops, `omega=0` ` implies theta= (2omega_(0)^(3//2))/(3k)` Average angular velocity `undersetomega(-)=(theta)/(t)=((2omega_(0)^(3//2))/(3k))/((2sqrtomega_(0))/(k))=(omega_(0))/(3)` OR `undersetomega(-)=(int_(0)^(t)omegadt)/(int_(0)^(t)dt)` ` = (int_(0)^(t)(sqrtomega_(0)-(kt)/(2))^(2)dt)/(t)=(|(sqrtomega_(0)-(kt)/(2))^(3)|_(0)^(t))/((-k)/(2)txx3)` ` =((sqrtomega_(0)-(k)/(2)(2sqrtomega_(0))/(k))^(3)-(sqrtomega_(0))^(3))/((-k)/(2)xx(2sqrtomega_(0))/(k)xx3)=(omega_(0))/(3)` |
|
| 115. |
A uniform stick of length l and mass m lies on a smooth table. It rotates with angular velocity `omega` about an axis perpendicular to the table and through one end of the stick. The angular momentum of the stick about the end isA. `ml^(2)omega`B. `(ml^(2)omega)/(3)`C. `(ml^(2)omega)/(12)`D. `(ml^(2)omega)/(6)` |
|
Answer» Correct Answer - B |
|
| 116. |
A wheel has mass of the rim 1 kg, having 50 spokes each of mass 5g. The radius of the wheel is 40 cm. The moment of inertia isA. 0.273 kg `m^(2)`B. 1.73 kg `m^(2)`C. 0173 kg `m^(2)`D. 2.73 kg `m^(2)` |
|
Answer» c) l =` mr^(2) + 50 (ml^(2))/3` `= 1 xx (0.4)^(2) + (50(5 xx 10^(-3))(0.4)^(2))/3` `0.16(1.083) =0.173kgm^(2)` |
|
| 117. |
A solid sphere rolls down two different inclined planes of the same height but of different inclinationsA. the speed and time of descend will be sameB. the speed will be same but time of descend will be differentC. the speed will be different but time of descend will be sameD. the speed and time of descend will be different |
|
Answer» Correct Answer - B |
|
| 118. |
An inclined plane makes an angle of `60^(@)` with horizontal. A disc rolling down this inclined plane without slipping has a linear acceleration equal toA. `(g)/(3)`B. `(3)/(4)g`C. `(g)/sqrt3`D. `(g)/(2)` |
|
Answer» Correct Answer - C |
|
| 119. |
Two identical thin rods are moving on a smooth table, as shown. Both of them are rotating with angular speed `omega`, in clockwise sense about their centres. Their centres have velocity V in opposite directions. The rods collide at their edge and stick together. Length of each rod is L. (a) For what value of `(V)/(omega L)` there will be no motion after collision ? (b) If the ratio `(V)/(omega L)` is half the value found in (a) above, what fraction of kinetic energy is lost in the collision? |
|
Answer» Correct Answer - (a) `(1)/(6)` (b) `(49)/(52)` |
|
| 120. |
A thin circular ring of mass M and radius R is rotating about its axis with a constant angular velocity omega. Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will beA. `((M+4m) omega)/(M)`B. `((M-4m)omega)/(M+4m)`C. `(M omega)/(4m)`D. `(M omega)/(M+4m)` |
|
Answer» Correct Answer - d |
|
| 121. |
A mass m in attached to a mass less string and swings in a horizontal circle, forming a conical pendulum, as shown in the figure. The other end of the string passes through a hole in the table and is dragged slowly so as to reduce the length `l`. The string is slowly drawn up so that the depth `h` shown in the figure becomes half. By what factor does the radius `(r)` of the circular path of the mass `m` change? |
|
Answer» Correct Answer - `((1)/(2))^(4)` |
|
| 122. |
A particle of mass m is doing horizontal circular motion with the help of a string (conical pendulu) as shown in the figure. If speed of the particle is constant then, A. the angular momentum of the particle about O is changingB. magnitude of angular momentum about O remains constantC. z component of the angular momentum remains conservedD. z component of torque is always zero. |
|
Answer» Correct Answer - A::B::C::D |
|
| 123. |
A cord is wound round the circumference of wheel of radius `r`. The axis of the wheel is horizontal and fixed and moment of inertia about it is `I`. A weight `mg` is attached to the end of the cord and falls from rest. After falling through a distance `h`, the angular velocity of the wheel will be.A. `[mgh]^(1//2)`B. `[(2mgh)/(I+2mr^(2))]^(1//2)`C. `[(2mgh)/(I+mr^(2))]^(1//2)`D. `[(mgh)/(I+mr^(2))]^(1//2)` |
|
Answer» Correct Answer - C PE = Rolling KE `mgh = (1)/(2) mv^(2) [1+k^(2)//r^(2)]` `v^(2)=(2mgh)/(m[1+k^(2)//r^(2)])` `therefore v = sqrt((2mgh)/(m[1+k^(2)//r^(2)]))` `omega = sqrt((2mgh)/(mr^(2)+mk^(2)))=sqrt((2mgh)/(I+mr^(2)))` |
|
| 124. |
Angular momentum of body changes by 80 kg `m^(2)//s`, when its angular velocity changes from 20 rad/s to rad/s. Then the change in its kinetic energy isA. 1200 JB. 1800 JC. 1600 JD. 2400 J |
|
Answer» Correct Answer - D `L_(2)-L_(1)=I(omega_(2)-omega_(1))` `KE_(2)-KE_(1)=(1)/(2)I(omega_(2)^(2)-omega_(1)^(2))` |
|
| 125. |
The rotational kinetic energy of a body is E. In the absence of external torque, the mass of the body is halved and its radius of gyration is doubled. Its rotational kinetic energy isA. 2 EB. E/2C. ED. E/4 |
|
Answer» Correct Answer - B `I_(1)omega_(1)=I_(2)omega_(2)` if no ecternal torqe acts on the system total angular momentum remains constant. `I_(1)=m_(1)K_(1)^(2) " " I_(2)=m_(2)K_(2)^(2)` `(I_(1))/(I_(2))=(m_(1))/(m_(2))(K_(1)^(2))/(K_(2)^(2))=2xx(1)/(4)=(1)/(2)` `therefore I_(2)=2I_(1)` `therefore omega_(2)=(I_(1)omega_(1))/(I_(2))=(I_(1)omega_(1))/(2I_(2))=(omega_(1))/(2)` `(KE_(1))/(KE_(2))=(I_(1))/(I_(2))xx((omega_(2))/(omega_(1)))^(2)` `=(1)/(2)xx4=2` |
|
| 126. |
Calculate the moment of inertia of a rod of mass 2 kg and length 0.5 m in each of the following cases, as shown in Fig. 8.52. |
|
Answer» Correct Answer - (i) `0.042 kg m^(2)` (ii) `0.166 kg m^(2)` |
|
| 127. |
A disc is free to rotate about an axis passing through its centre and perpendicular to its plane. The moment of inertia of the disc about its rotation axis is I. A light ribbon is tightly wrapped over it in multiple layers. The end of the ribbon is pulled out at a constant speed of u. Let the radius of the ribboned disc be R at any time and thickness of the ribbon be `d (lt lt R)`. Find the force (F) required to pull the ribbon as a function of radius R. |
|
Answer» Correct Answer - `F = (Idu^(2))/(2piR^(4))` |
|
| 128. |
When the mass is rotated in a plane about an fixed point, then angualr momentum is directed along theA. radiusB. tangent to orbitC. axis of rotationD. line at an angle of `45^(@)` to plane of rotation |
| Answer» c) The angular momentum will be directed along the axis of rotation. | |
| 129. |
A roller in a printing press turns through an angle `theta=3t^(2)-t^(3) rad`. (a) Calculate the angular velocity and angular acceleration as a function of time `t`. (b) What is the maximum positive angular velocity and at what time `t` does it occur? |
|
Answer» (a) `theta=3t^(2)-t^(3)` `omega=(d theta)/(dt)=6t-3t^(2)` `alpha=(domega)/(dt)=6-6t` (b) `omega` is maximum when `alpha=(domega)/(dt)=0` `6-6t = 0 implies t = 1 s` `omega_(max)=6(1)-3(1)^(2)= 3 rad//s` |
|
| 130. |
A body A of mass M while falling wertically downwards under gravity brakes into two parts, a body B of mass `(1)/(3)` M and a body C of mass `(2)/(3)` M. The center of mass of bodies B and C taken together shifts compared to that of body A towardsA. does not shiftB. depends on height of breakingC. body BD. body C |
|
Answer» Correct Answer - A (a) Does not shift as no external force acts. The centre of mass of the system continues its original path. It is only the internal forces which comes inot play while breaking |
|
| 131. |
A force of `-Fhatk` acts on O, the origin of the coodinate system. The torque about the point (1,-1) is A. `-F(hati+hatj)`B. `F(hati+hatj)`C. `-F(hati-hatj)`D. `F(hati-hatj)` |
|
Answer» `vecr=hati-hatj`, `vecF= -Fhatk` `vectau=vecrxxvecF=(hati-hatj)xx(-Fhatk)` `= -F[hatixxhatk-hatjxxhatk]` `= -F[-hatj-hati]=F[hati+hatj]` |
|
| 132. |
The moment of inertia of a uniform semicircular disc of mass disc through the centre isA. `(2)/(5)Mr^2`B. `(1)/(4)Mr^2`C. `(1)/(2)Mr^2`D. `Mr^2` |
|
Answer» Correct Answer - C (c ) The disc may be assumed as combination of two semi circular parts. Let I be the moment of inertia of the uniform semicircular disc `rArr 2I = (2Mr^2)/(2) rArr I=(Mr^2)/(2)` |
|
| 133. |
A force of `-Fhatk` actgs on O, the origin of the coodinate system. The torque about the point (1,-1) is A. `F(hati-hatj)`B. `-F(hati+hatj)`C. `F(hati+hatj)`D. `-F(hati-hatj)` |
|
Answer» Correct Answer - B (b) Torque `vectau = vecrxxvecF = (hatj-hati)xx(Fhatk)=-F(hati+hatj)` |
|
| 134. |
An ice skater spins with arms outstreched at 1.9 rps. Her moment of inertis at this instant is `1,33 kg m^(2)`. She pulls in her arms to increase her rate of spin. If the moment of inertia is `0.48 kg m^(2)` after she pulls in her arms, what is her new rate of rotation? |
|
Answer» Correct Answer - 5.26 rps |
|
| 135. |
Two rod 1 and 2 are released from rest as shown in figure . Given, `l_(1)=4l,m_(1)=2m,l_(2)=2l "and "m_(2)=m.` There is no friction between the two rods . If `alpha` be the angular acceleration of rod 1 just after the rods are released . Then, What is the horizontal force on rod 1 by hinge. A at this instant ? A. `((32-12sqrt(3))/(3sqrt(3))m//alpha`B. `((16-2sqrt(3))/(sqrt(3))m//alpha`C. `(14+2sqrt(3))m//alpha`D. `sqrt(3)m//alpha` |
|
Answer» Correct Answer - A |
|
| 136. |
Two rod 1 and 2 are released from rest as shown in figure . Given, `l_(1)=4l,m_(1)=2m,l_(2)=2l "and "m_(2)=m.` There is no friction between the two rods . If `alpha` be the angular acceleration of rod 1 just after the rods are released . Then, What is the normal reaction between the two rods at this instant ? A. `16sqrt(3)m//alpha`B. `(4m//alpha)/(sqrt(3))`C. `(32m//alpha)/(3sqrt(3))`D. `12sqrt(3)) m//alpha` |
|
Answer» Correct Answer - C |
|
| 137. |
`(L^(2))/(2I)` representsA. rotational kinetic energy of a particleB. potential energy of a particleC. torque on a particleD. power |
| Answer» Correct Answer - A | |
| 138. |
`(L^(2))/(2I)` representsA. rotational P.E.B. total energyC. rotational K.E.D. translational K.E. |
| Answer» Correct Answer - C | |
| 139. |
The kinetic energy of a rotating body depends uponA. distribution of mass onlyB. angular speed onlyC. distribution of mass and angular speedD. angular acceleration only |
| Answer» Correct Answer - C | |
| 140. |
A disc of radius `r = 0.1 m` is rolled from a point A on a track as shown in the figure. The part AB of the track is a semi-circle of radius R in a vertical plane. The disc rolls without sliding and leaves contact with the track at its highest point B. Flying through the air it strikes the ground at point C. The velocity of the center of mass of the disc makes an angle of `30^(@)` below the horizontal at the time of striking the ground. At the same instant, velocity of the topmost point P of the disc is found to be `6 m//s` (Take `g = 10 m//s^(2)` ). (a) Find the value of R. (b) Find the velocity of the center of mass of the disc when it strikes the ground. (c) Find distance AC. |
|
Answer» Correct Answer - (a) 1m (b) `12 m//s` (c) `3.6 sqrt3 m` |
|
| 141. |
A particle describes circular motion with its angular displacement `theta` at anytime `t` is given by `theta=5t-3` radian. If the radius of circular path is `2 m`, find the total acceleration of particle. |
|
Answer» `theta=5t-3` `omega=(d theta)/(dt)=5` `alpha=(domega)/(dt)=0` `a_(c)=omega^(2)R=(5)^(2)xx2=50 m//s^(2)` `a_(1)=Ralpha=0` Total acceleration `a=a_(c)=50 m//s^(2)` |
|
| 142. |
A flywheel of moment of inertia 0.4 kg `m^(2)` and radius 0.2 m is free to rotate about a central axis. If a string is wrapped around it and it is pulled with a force of 10N. Then its angular velocity after 4s will beA. `10 rad s^(-1)`B. `5 rad s^(-1)`C. `20 rad s^(-1)`D. None |
|
Answer» c) As , `rho= F xx r = lalpha` `=l(omega_(2)-omega(1))/t` `therefore` `omega_(2)-omega_(1) = (Fxxrxxt)/l` `=(10 xx 0.2 xx 4)` `=20rads^(-1)` |
|
| 143. |
A uniform block of mass M and dimensions as shown in the figure is placed on a rough horizontal surface and given a velocity `V_(0)` to the right. A is a point on the surface to the left of the block. (a) Write the angular momentum of the block about point A just after it begins to move (b) Due to friction the block stops. What happened to its angular momentum about point A? Which torque is responsible for change in angular momentum of the block? |
|
Answer» Correct Answer - (a) `(MV_(0)b)/(2)` (b) Normal reaction produces a torque |
|
| 144. |
If the radius of the earth decreases by 20% such that mass does not change, then the length of the new day would beA. `(24xx16)/(25)h`B. `(24xx25)/(16)h`C. `(25xx16)/(24)h`D. `(24)/(16)h` |
|
Answer» Correct Answer - A `(T_(2))/(T_(1))=((R_(2))/(R_(1)))^(2)` |
|
| 145. |
Mass remaining constant, if the earth suddenly contracts to one third of its present radius, the length of the day would be shorted byA. 8/3hB. 12hC. 8hD. 64/3 h |
|
Answer» Correct Answer - D `(R_(1))/(R_(2))=sqrt((T_(1))/(T_(2))) therefore T_(2)=(24)/(9)=(8)/(3)h` `therefore Delta T=T_(1)-T_(2)=24-(8)/(3)=(64)/(3)`. |
|
| 146. |
A disc of mass 16 kg and radius 25 cm is rotated about its axis. What torque will increase its angular velocity from 0 to `8 pi` rad/s in 8 s ?A. `pi Nm`B. `pi//2 Nm`C. `pi//4 Nm`D. `2pi Nm` |
|
Answer» Correct Answer - B `tau = (MR^(2))/(2)xx((omega_(2)-omega_(1)))/(t)` |
|
| 147. |
A rod of length 2 m is kept vertical inside a smooth spherical shell of radius 2m. The rod starts slipping inside the shell . Mass of the rod is 4 kg. Velocity of centre of the rod (in m/s) at the instant is approximately A. 5.5B. 6.2C. 3.2D. 10.2 |
|
Answer» Correct Answer - A |
|
| 148. |
If a spherical ball rolls on a table without slipping, the fraction of its total energy associated with rotation isA. `2/5`B. `2/7`C. `3/5`D. `3/7` |
|
Answer» Correct Answer - b |
|
| 149. |
A cube of mass m and side a is moving along a plane with constant speed `v_(o)` as shown in figure. The magnitude of angular momentum of the cube about z -axis would be. A. `(mv_(0)b)/(2)`B. `(sqrt3mv_(0)b)/(2)`C. `mv_(0)(b-(a)/(2))`D. none of these |
|
Answer» Correct Answer - D |
|
| 150. |
A uniform cube of mass M and side length a is placed at rest at the edge of a table. With half of the cube overhanging from the table, the cube begins to roll off the edge. There is sufficient friction at the edge so that the cube does not slip at the edge of the table. Find - (a) the angle `theta_(0)` through which the cube rotates before it leaves contact with the table. (b) the speed of the centre of the cube at the instant it breaks off the table. (c) the rotational kinetic energy of the cube at the instant its face AB becomes horizontal. |
|
Answer» Correct Answer - (a) `cos^(-1) ((6)/(11))` (b) `sqrt((3ga)/(11))` (c) `(mga)/(11)` |
|