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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.
| 151. |
The radius of gyration of a body about an axis at a distance of 4 cm from the centre of gravity is 5 cm. Its radius of gyration about a parallel axis through centre of gravity isA. `sqrt(31)cm`B. 1 cmC. 3 cmD. none |
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Answer» Correct Answer - C `K_(0)^(2)=K_(c )^(2)+h^(2)` `K_(c )^(2)= K_(0)^(2)-h^(2)=25-16` `K_(c )^(2)=9` `K_(c ) = 3 cm`. |
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| 152. |
In q. 30. acceleration of the point A isA. `agt-0`B. gt0C. lt0D. 0 |
| Answer» Correct Answer - d) | |
| 153. |
An Indian bread ‘‘Roti’’ is a uniform disc of mass M and radius R. Before eating a person usually folds it about its diameter (say about x axis). After folding it a sector of angle `60^(@)` is removed from it. Find the moment of inertia of the remaining ‘‘Roti’’ about Z-axis |
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Answer» Correct Answer - `(MR^(2))/(3)` |
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| 154. |
Consider a uniform square plate shown in the figure. `I_(1), I_(2), I_(3)` and `I_(4)` are moment of inertia of the plate about the axes 1, 2, 3 and 4 respectively. Axes 1 and 2 are diagonals and 3 and 4 are lines passing through centre parallel to sides of the square. The moment of inertia of the plate about an axis passing through centre and perpendicular to the plane of the figure is equal to which of the followings. (a) `I_(3) + I_(4)` (b) `I_(1) + I_(3)` (c) `I_(2) + I_(3)` (d) `(1)/(2) (I_(1) + I_(2) + I_(3) + I_(4))` |
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Answer» Correct Answer - a, b, c, d |
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| 155. |
A uniform rectangular plate has side length `l` and `2 l`. The plate is in `x - y` plane with its centre at origin and sides parallel to x and y axes. The moment of inertia of the plate about an axis passing through a vertex (say A) perpendicular to the plane of the figure is `I_(0)`. Now the axis is shifted parallel to itself so that moment of inertia about it still remains `I_(0)`. Write the locus of point of intersection of the axis with `xy` plane |
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Answer» Correct Answer - `x^(2) + y^(2) = (5l^(2))/(4)` |
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| 156. |
A flywheel rotates with a uniform angular acceleration. Its angular speed increases from `2pirad//s` to `10pirad//s` in `4 s`. Find the number of revolutions in this period. |
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Answer» `omega_(0)=2pi rad//s, omega=10 pi rad//s` `omega=omega_(0)+alpha t` ` 10 pi = 2 pi+alphaxx4implies alpha=2pi rad//s^(2)` `theta = omega_(0) t + (1)/(2)alpha t^(2)` `= 2pixx4+(1)/(2)xx2pixx(4^(2))` `= 8pi+16pi=24 pi rad` Number of revolutions `N=(theta)/(2pi)=(24pi)/(2pi)=12` |
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| 157. |
In the figure shown a uniform rod of length l and mass m is kept at rest in horizontal position on an elevated edge. The value of x is such that the rod will have maximum angular acceleration `alpha`, as soon as it is set free.A. x is equal to `(l)/(2 sqrt(3))`B. `alpha` is equal to `(g sqrt(3))/(2l)`C. `alpha` is equal to `(g sqrt(3))/(l)`D. x is equal to `(l)/( sqrt(3))` |
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Answer» Correct Answer - A::C |
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| 158. |
There are two point masses A and B, situated at origin and point (5m,0m) respectively . At a certain time `v_(A)` and `v_(B)` are respectively the velocities of point masses A and B . Match the situations under Table-1 with their correct option under Table-2 |
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Answer» Correct Answer - (A)P,(B)S,(C)Q,(D)R |
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| 159. |
Two identical uniform thin rods have been connected at right angles to form a ‘T’ shape. One end of a rod is connected to the centre of the other rod. Length of each rod is L. The upside down ‘T’ can swing like a pendulum about a horizontal axis passing through the top end (see fig.). Axis is perpendicular to plane of the fig. The speed of the meeting point of the two rods is `u = 2 sqrt(gL)` when it is at its lowest position. Calculate the angular acceleration of the ‘T’ shaped object when it is at extreme position of its oscillation. |
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Answer» Correct Answer - `(sqrt935)/(34) (g)/(L)` |
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| 160. |
Four masses m, 2 m, 3 m and 4 m are connected by a rod of negligible mass. The moment of inertia of the system about the axis AB is A. `52 ma^(2)`B. `9 ma^(2)`C. `50 ma^(2)`D. `20 ma^(2)` |
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Answer» Correct Answer - C `l=m_(1)r_(1)^(2)+m_(2)r_(2)^(2)+m_(3)r_(3)^(2)` `= 2m(a^(2))+3m(4a^(2))+4m(9a^(2))` `= 2ma^(2)+12ma^(2)+36ma^(2)` `50 ma^(2)`. |
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| 161. |
Two particles `A` and `B` are situated at a distance `d = 2m` apart. Particle `A` has a velocity of `10 m//s` at an angle of `60^@` and particle `B` has a velocity `v` at an angle `30^@` as shown in figure. The distance `d` between `A` and `B` is constant. the angular velocity of `B` with respect to `A` is : .A. `5sqrt3 rad//s`B. `(5)/(sqrt3) rad//s`C. `10sqrt3rad//s`D. `(10)/(sqrt3)rad//s` |
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Answer» Correct Answer - B |
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| 162. |
Two circular iron discs are of the same thickness. The diameter of A is twice of B . The moment of inertia of A as compared to that of B isA. twice as largeB. four as largeC. 8 time as largeD. 16 times as large |
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Answer» Correct Answer - D M.I. of disc `=(1)/(2)mR^(2)=(1)/(2)(pi R^(2)t)rho R^(2)=(1)/(2)pi R^(4)t rho` [`rho` = density, t = thickness] If discs are made of same material and same thickness then `I prop R^(4) prop ("Diameter")^(4)` `therefore (I_(1))/(I_(2))=((D_(A))/(D_(B)))^(4)=((2)/(1))^(4)-(16)/(1)`. |
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| 163. |
Two particles of masses 2 kg and 3 kg are separated by 5 m. Then moment of inertia of the system about an axis passing through the centre of mass of the system and perpendicular to the line joining them isA. `10 kg m^(2)`B. `20 kg m^(2)`C. `30 kg m^(2)`D. `40 kg m^(2)` |
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Answer» Correct Answer - C `r_(2)=((m_(1))/(m_(1)+m_(2)))r=(6)/(5)xx25=30 kg. m^(2)`. |
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| 164. |
Two particles of masses 5 gm and 10 gm are 12 cm apart. What will be the location of the centre of mass of the system of these two particles from the lighter particle is ?A. 8 cmB. 6 cmC. 4 cmD. 9 cm |
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Answer» Correct Answer - A `r_(1)=((m_(2))/(m_(1)+m_(2)))r=(10)/(15)xx12=8 cm` |
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| 165. |
A uniform sphere of radius R has a spherical cavity of radius `(R)/(2)` (see figure). Mass of the sphere with cavity is M. The sphere is rolling without sliding on a rough horizontal floor [the line joining the centre of sphere to the centre of the cavity remains in vertical plane]. When the centre of the cavity is at lowest position, the centre of the sphere has horizontal velocity V. Find: (a) The kinetic energy of the sphere at this moment. (b) The velocity of the centre of mass at this moment. (c) The maximum permissible value of V ( in the position shown ) which allows the sphere to roll without bouncing |
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Answer» Correct Answer - (a) `(31)/(40) MV^(2)` (b) `(15V)/(14)` (c) `V le sqrt(14 gR)` |
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| 166. |
Particles of masses 3 kg, 4 kg, 5 kg nd 2 kg are placed at points `A(3,2), B(-2,2), C(-2,-3)` and `D(1,-2)` respectively of a co-ordinate system. The centre of mass of the system is atA. `(-(1)/(2),-(1)/(2))`B. `(-(1)/(2),-(5)/(14))`C. `(0,-(1)/(2))`D. (0,0) |
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Answer» Correct Answer - B (i) `x = (m_(1)n_(1)+m_(2)x_(2)+m_(2)x_(3)+m_(4)x_(4))/(m_(1)+m_(2)+m_(3)+m_(4))=(-7)/(14)=(-1)/(2)` (ii) `y=(m_(1)y_(1)+m_(2)y_(2)+m_(3)y_(3)+m_(4)y_(4))/(m_(1)+m_(2)+m_(3)+m_(4))=(-5)/(14)` (x,y) of centre of mass `=((-1)/(2), (-5)/(14))`. |
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| 167. |
Two hydrogen atoms are located at the distances `r_(1)` and `r_(2)` from origin. Their centre of mass is atA. `(r_(1)-r_(2))//2`B. `r_(1)-r_(2)`C. `(r_(1)+r_(2))//2`D. `r_(1)+r_(2)` |
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Answer» Correct Answer - C `x_(cm)=(m(r_(1)+r_(2)))/(2m)` `r_(cm)=(r_(1)+r_(2))/(2)` |
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| 168. |
Four identical bodies each of mass 1 kg are placed touching to each other with centres on a straight line. If their centres are marked A, B, C and D respectively, then the distance of centre of mass of the system from B will be A. `(AB+BC+BC+CD)/(4)`B. `(AB+BC+BD)/(4)`C. `AB+(AC)/(4)+AD`D. `(AB+BC+CD)/(4)` |
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Answer» Correct Answer - B `R_(g)=(m_(1)x_(1)+m_(2)x_(2)+m_(3)x_(3)+m_(4)x_(4))/(m_(1)+m_(2)+m_(3)+m_(4))` `m_(1)=m_(2)=m_(3)=m_(4)=m` `R=((x_(1)+x_(2)+x_(3)+x_(4)))/(4)` `= (AB+BC+BD)/(4)` |
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| 169. |
A hollow sphere and solid sphere of the same radius rotate about their diameters with the same angular velocity and angular momentum. Then the ratio of their masses isA. `3:8`B. `5:8`C. `3:5`D. `5:3` |
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Answer» Correct Answer - C `(I_(1))/(I_(2))=((2)/(3)m_(1)R_(1)^(2))/((2)/(5)m_(2)R_(2)^(2))=(5m_(1))/(3m_(2)) " " therefore (m_(1))/(m_(2))=(3)/(5)` |
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| 170. |
A trough has two identical inclined segments and a horizontal segment. A ball is released on the top of one inclined part and it oscillates inside the trough. Friction is large enough to prevent slipping of the ball. Time period of oscillation is T. Now the liner dimension of each part of the trough is enlarged four times. Find the new time period of oscillation of the ball. |
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Answer» Correct Answer - 2T |
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| 171. |
The M.I. of two spheres of equal masses about their diameters are equal. If one of them is solid and other is hollow, the ratio of their radius isA. `sqrt(3):sqrt(5)`B. `3:5`C. `sqrt(5):sqrt(3)`D. `5:3` |
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Answer» Correct Answer - C `I_(s)=(2)/(5)MR_(s)^(2)` and `I_(h)=(2)/(3)MR_(h)^(2)` As `I_(s)=I_(h) therefore (2)/(5)MR_(s)^(2) = (2)/(3)MR_(h)^(2)` `therefore (R_(s))/(R_(h))=(sqrt(5))/(sqrt(3))` |
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| 172. |
If the diameter of fly wheel is increases by 1%, then increase in its M.I. about an axis passing through centre and perpendicular to the plane will beA. `1 %`B. `0.5 %`C. `2 %`D. `4 %` |
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Answer» Correct Answer - C `I=MR^(2)` Differentiating we get, `(dI)/(I)= 0+2 (dR)/(R )` `therefore (dI)/(I)xx 100 = 2((dR)/(R ))xx100` `= 2xx1% xx 100 = 2%` |
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| 173. |
A uniform rod of mass M is moving in a plane and has a kinetic energy of `(4)/(3) MV^(2)` where V is speed of its centre of mass. Find the maximum and minimum possible speed of the end point of the rod. |
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Answer» Correct Answer - `V_("max") = (sqrt5 + 1) V; V_("min") = (sqrt5 - 1)V` |
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| 174. |
If a body is lying in the Y-Z plane, then according to theorem of perpendiculr axes the correct expression will beA. `I_(z)=I_(x)+I_(y)`B. `I_(y)=I_(x)+I_(z)`C. `I_(x)=I_(y)+I_(z)`D. `I_(y)=I_(z)+Md^(2)` |
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Answer» Correct Answer - C According to perpendicular axis theorem, `I_(x)=I_(y)+I_(z)`. |
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| 175. |
An asteroid in the shape of a uniform sphere encounters cosmic dust. A thin uniform layer of dust gets deposited on it and its mass increases by `2%`. Find percentage change in its moment of inertia about diameter |
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Answer» Correct Answer - `(10)/(3) %` |
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| 176. |
A uniform rod of length `b` can be balanced as shown in figure. The lower end of the rod is resting against a vertical wall. The coefficient of friction between the rod and the wall and that between the rod and the support at A is `mu`. Distance of support from the wall is `a`. (a) Find the ratio `(a)/(b)` if the maximum value of `theta` is `theta_(1)`. (b) Find the ratio `(a)/(b)` if the minimum value of `theta` is `theta_(2)`. |
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Answer» Correct Answer - (a) `(a)/(b) = (1)/(2) sin^(2) theta_(1) [(1 - mu^(2)) sin theta_(1) - 2 mu cos theta_(1)]` (b) `(a)/(b) = (1)/(2) sin^(2) theta_(2) [(1 - mu^(2)) sin theta_(2) + 2 mu cos theta_(2)]` |
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| 177. |
The position vector `vacr` of the z-component of `vecL` is `55ma^2omega`. Following equation `vacr(t)= alphat^3hati+beta^2hatj,` where `alpha = 10//3ms^(-3), beta = 5 ms^(-2)` and m = 0.1kg. At t=1s, which of the following statement (s) is (are) true about the particle?A. The velocity `vacv` is given by `vacv = (10hati+10hatj) ms^(-1)`B. The angular momentum `vacL` with respect to the origin is given by `vacL = -5//3) hatk N ms.`C. The force `vacF` is given by `vacF = (hati+2hatj)ND. The torque `vactau` with respect to the origin is given by `vactau = -(20//3) hatk Nm` |
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Answer» Correct Answer - A::B (a,b) `vecr = alpha t^3 + betat^2 hatj` `vecr = (10)/(3) t^3 hati + 5t^2 hatjm` `vecv = (dvecr)/(dt) = 10t^2 hati + 10thatj ms^(-1)` `vecalpha = (dvecv)/(dt) = 20t hatj + 10hatjms^(-2)` At t = 1s `vecr_(t=1) = (10)/(3)hat I + 5hatjm` `vecv_(t =1) = 10hati + 10hatj ms^(-1)` `hatP_(t =1) = hati + hatj kgms^(-1)` `alpha_(t =1) = 20hati + 10 hatj ms^(-2)` `vecL = vecrxxvecP = |(hat i,hat j,hat k ),(10/3,5,0),(1,1,0)| = hatk[(10)/(3) -5]=-5/3 hatk kgms^(-1)` `vecF = mveca = (2hati + hatj)N` `vectau = vecrxxvecF=|(hat i,hat j,hat k ),(10/3,5,0),(2,1,0)|=hatk[(10)/(3) -10]=(-20)/(3) hatkN-m.` |
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| 178. |
The speed of a motor increases from 600 rpm to 1200 rpm in 20 s. What is the angular acceleration and how many revolution does it make during this time ? |
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Answer» Correct Answer - `pi rad s^(-2), 300` |
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| 179. |
A uniform rod of length l is pivoted at point A. It struk n=by an horizontal force which delivers an ikmpulse J at a distance c from point A as shown in figure. Impulse delivered by pivot is zero, if x is equal to A. `(l)/(2)`B. `(l)/(3)`C. `(2l)/(3)`D. `(3l)/(4)` |
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Answer» Correct Answer - C |
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| 180. |
The density of as rod graduallly decreases from one end to the other. It is pivoted at an end so that ilt can move about a vertical axius through the pivot. A horizontal force F is applied on the free end n a direction perpendicular to the rod. The quantities, that do not depend on which end of the rod is pivoted areA. angular accelerationB. angular velocity when the rod completes one rotationC. angular momentum when the rod completes one rotationD. torque of the applied force |
| Answer» `M.I.` depends on the disctribution of mass. | |
| 181. |
A ring of mass M and radius R is held at rest on a rough horizontal surface. A rod of mass M and length `L = 2 sqrt3 R` is pivoted at its end A on the horizontal surface and is supported by the ring. There is no friction between the ring and the rod. The ring is released from this position. Find the acceleration of the ring immediately after the release if `theta = 60^(@)`. Assume that friction between the ring on the horizontal surface is large enough to prevent slipping of the ring. |
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Answer» Correct Answer - `a = (g)/(4 sqrt3)` |
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| 182. |
If the moment of inertia of a ring about transverse axis passing through its centre is 6 kg `m^(2)`, then the M.I. about a tangent in its plane will beA. `3kg m^(2)`B. `9kg m^(2)`C. `6kg m^(2)`D. `12 kg m^(2)` |
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Answer» Correct Answer - B `MR^(2)=6 kg m^(2)` `I_(t)=(3)/(2)MR^(2)` |
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| 183. |
A L shaped uniform rod has both its sides of length `l`. Mass of each side is m. The rod is placed on a smooth horizontal surface with its side AB horizontal and side BC vertical. It tumbles down from this unstable position and falls on the surface. Find the speed with which end C of the rod hits the surface. |
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Answer» Correct Answer - `sqrt(3gl)` |
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| 184. |
A particle moves in a circle with constant angular velocity `omega` about a point P on its circumference. The angular velocity of the particle about the centre C of the circle isA. `2omega`B. `(omega)/(2)`C. `omega`D. Not constant |
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Answer» Correct Answer - A |
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| 185. |
Two equal and opposite forces act on a rigid body at a certain distance. ThenA. the body is in equilibriumB. the body will rotate about its centre of massC. the body may rotate about any point other than its centre of massD. the body cannot rotate about its centre of mass |
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Answer» Correct Answer - B |
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| 186. |
Find the `M.I.` |
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Answer» `Mass//area = (m)/(pi(r_(1)^(2)-r_(2)^(2)))` Taking a ring of radius `r` and thickness `dr` `dA=2pidr` `dm=(m)/(pi(r_(1)^(2)-r_(2)^(2)))xx2pirdr=(2mrdr)/((r_(1)^(2)-r_(2)^(2)))` `I= intr^(2)dm=(2m)/((r_(1)^(2)-r_(2)^(2))) int_(r_(2))^(r_(1))r^(3) dr` ` =(2m)/((r_(1)^(2)-r_(2)^(2)))|(r_(4))/(4)|_(r_(2))^(r_(1))` `= (1)/(2)(m(r_(1)^(4)-r_(2)^(4)))/((r_(1)^(2)-r_(2)^(2)))=(1)/(2)(m(r_(1)^(2)+r_(2)^(2))(r_(1)^(2)-r_(2)^(2)))/((r_(1)^(2)-r_(2)^(2)))` `=(1)/(2)m(r_(1)^(2)+r_(2)^(2))` |
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| 187. |
A ball of mass 100 g is suspended by a string `40 cm` long. Keeping the string taut, the ball describes a horizontal circle of radius 10 cm. Find the angular speed. |
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Answer» Correct Answer - `1.59 rad s^(-1)` |
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| 188. |
A body whose moment of inertia is `3 "kgm"^(2)`, is at rest. It is rotated for 20 s with a moment of force 6 Nm. Find the angular displacement of the body. Also calculate the work done.A. 60 JB. 2400 JC. 120 JD. 4800 J |
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Answer» Correct Answer - B `tau = I alpha` and `theta = omega_(1) t+(1)/(2) alpha t^(2)` Work done `= tau theta` |
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| 189. |
A disc of radius `R` rolls without slipping at speed `v` along positive `x`-axis. Velocity of point `P` at the instant shown in Fig. is A. `v_(p)=(v+(vrsintheta)/(R))hati+(vrcostheta)/(R)hatj`B. `v_(p)=(v+(vrsintheta)/(R))hati-(vrcostheta)/(R)hatj`C. `v_(p)=v+(vrsintheta)/(R)hati+(vrcostheta)/(R)hatj`D. `v_(p)=v+(vrsintheta)/(R)hati-(vrcostheta)/(R)hatj` |
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Answer» Correct Answer - B |
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| 190. |
The M.I. of a circular ring of radius R and mass M about a tangent in its plane isA. `MR^(2)//2`B. `3//2 MR^(2)`C. `MR^(2)`D. `2MR^(2)` |
| Answer» Correct Answer - C | |
| 191. |
A uniform rod AB of mass m and length 2a is falling freely without rotationb under gravity with AB horizontal. Suddenly the end A is fixed when the speed of the rod is v. The angular speed which the rod begains to rotate isA. `(v)/(2a)`B. `(4v)/(3a)`C. `(v)/(3a)`D. `(3v)/(4a)` |
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Answer» Correct Answer - D |
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| 192. |
A solid sphere and a hollow sphere of equal mass and radius are placed over a rough horizontal surface after rotating it about its mass centre with same angular velocity `omega_(0)`. Once the pure rolling starts let `v_(1)` and `v_(2)` be the linear speeds of their centres of mass. ThenA. `v_(1)=v_(2)`B. `v_(1)gtv_(2)`C. `v_(1)ltv_(2)`D. data is insufficient |
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Answer» Correct Answer - C |
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| 193. |
In the above question No.326, the reason for the bodies to have different times of descent isA. they have same massB. they have same radiusC. they have different radii of gyrationD. all |
| Answer» Correct Answer - C | |
| 194. |
If the kinetic energy of a rotating body about an axis is decreased by 36%, its angular momentum about that axis isA. increases by 72%B. decreases by 72 %C. increases by 20%D. decreases by 20% |
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Answer» Correct Answer - D `(L_(2)-L_(1))/(L_(1))=(1)/(2)(Delta KE)/(KE_(1))` |
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| 195. |
Solid sphere, hollow sphere, solid cylinder and hollow cylinder of same mass and same radii are simultaneously start rolling down from the top of an inclined plane. The body that takes longest time to reach the bottom isA. solid sphere, hollow sphereB. solid sphere, discC. ring, solid sphereD. solid sphere, ring |
| Answer» Correct Answer - D | |
| 196. |
When a body starts to roo on an inclined plane, its potential energy is converted intoA. translational kinetic energy onlyB. rotational kinetic energy onlyC. both translational and rotational kinetic energiesD. neither translational nor rotational kinetic energies |
| Answer» Correct Answer - C | |
| 197. |
In rotatory motion, linear velocities of all the particles of a bodyA. are sameB. are differentC. can not be predictedD. are zero |
| Answer» Correct Answer - B | |
| 198. |
Consider the moment of inertia I of the rigid homogerneous disc of mass M as shown in the figure about an axis through its centre (different shadings only differentiate the two parts of the disc each with equal mass `M//2`). Which one of the following statements concerning I is correct? A. The inner and outer parts of the disc, each with mass `M//2`, contribute equal amounts to /.B. The inner part of the disc contributes less to// than outer part.C. The inner part of the disc contributes less to//than the outer part.D. The inner part of the disc may contribute more or less to//depending on the actul numerical value to the mass M of the disc. |
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Answer» Correct Answer - C |
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| 199. |
The point at which total mass of a body is suppose to be concentrated is known asA. deep centreB. centre of gravityC. centre of massD. geometric contre |
| Answer» Correct Answer - C | |
| 200. |
The center of mass of a system of two particles divides the distance between them.A. inverse ratio of square of masses of particleB. direct ratio of square of masses of particleC. inverse ratio of masses of particleD. direct ratio of masses of particle |
| Answer» Correct Answer - C | |