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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. |
In a system of particles `8 kg` mass is subjected to a force of `16 N` along + ve `x`-axis and another `8 kg` mass is subjected to a force of `8 N` along + ve ` y`-axis. The magnitude of acceleration of centre of mass and the angle made by it with `x`-axis are given, respectively, byA. `(sqrt(5))/2ms^(2),theta=45^(@)`B. `3sqrt(5)ms^(2),theta=tan^(-1)(2/3)`C. `(sqrt(5))/2ms^(2),theta=tan^(-1)(1/2)`D. `1ms^(2),theta=tan^(-1)sqrt(3)` |
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Answer» Correct Answer - C `veca_(CM)=vecF/(m_(1)+m_(2))=(16hati+8hatj)/(8+8)=hati+1/2hatj` `|veca(CM)|=sqrt(1^(2)(1/2)^(2))=sqrt5/2m//s^(2)` `tantheta=(a_(y))/(a_(x))=(1/2)/1=1/2impliestheta=tan^-(1)(1/2)` |
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| 152. |
A uniform metallic spherical shell is suspended form celling. It has two holes A and B at top and bottom respectively. Which of the following is/are true :- A. If B is closed and sand is poured from A, centre of mass first rises and then fallsB. If shell is completely filled with sand and B is opened then centre of mass falls initiallyC. If shell is slightly filled with sand and B is opened, then centre of mass falls.D. None of these |
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Answer» Correct Answer - B Initinally when the shell is empty the C.M. lies at its geometric centre. Also when the shell is filled with sand CM lies at its geometric centre. |
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| 153. |
A ball of mass `m` is released from rest relative to elevator at a height `h`, above the floor of the elevator. After making collision with the floor of the elevator it rebounces to height `h_2`. The coefficient of restitution for Collision is `e`. For this situation, mark the correct statement(s). A. If elavator is moving done with constant velocity `v_(0), then h_(2)=e^(2)h_(1)`B. If elevator is moving down with constasnt velocity `v_(0)`, then `h_(2)=e^(2)h_(1)-v_(0)^(2)/(2g)`C. if elevator is moving down with constant velocity `v_(0)`, then impulse imparted by floor of the elevator of the ball is `m(sqrt(2gh_(2))+sqrt(2gh_(1))+2v_(0))` is the upward direction.D. If elevator is moving with constant acceleration of `g//4` in upward direction, then it is not possible to determine a relation between `h_(1)` and `h_(2)`from the given information. |
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Answer» Correct Answer - A As coefficient of restitution equation is valid in all fames of reference. So in the framd of lift `e=("velocity of separation")/("velocity of approach") v_(f)/v_(i)=(sqrt(2gh_(2)))/(sqrt(2gh_(1)))`………i `h_(2)=e^(2)h_(1)` Impulse delivered by the elavator to ball is `J=vecv_(mv)_(f)-vecv_(mv)_(i)=msqrt(2gh_(2))+sqrt(2gh_(1))` If the elevator is accelarating then replacing g with `g+g/4` we get `h_(2)=e^(2)h_(1)` |
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| 154. |
Two block A and B of masses m and `2m` respectively are connected by a spring of spring cosntant k. The masses are moving to the right with a uniform velocity `v_(0)` each, the heavier mass leading the lighter one. The spring is of natural length during this motion. Block B collides head on with a thrid block C of mass `2m`. at rest, the collision being completely inelastic. The maximum compression of the spring after collision is -A. `sqrt((mv_(0)^(2))/(12k))`B. `sqrt((mv_(0)^(2))/(5k))`C. `sqrt((mv_(0)^(2))/(10k))`D. None of these |
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Answer» Correct Answer - (B) At maximum compression, velocity of all block are same & equal to velocity of centre of mass. `(1)/(2)kx_(m)^(2) = [(1)/(2)mv_(0)^(2) + (1)/(2)(4m) ((v_(0))/(2))^(2)] - (1)/(2)(5m)((3v_(0))/(5))^(2) rArr (1)/(2)kx_(m)^(2) = (1)/(10)mv_(0)^(2) rArr x_(m) = sqrt((mv_(0)^(2))/(5k))` |
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| 155. |
Two block A and B of masses m and `2m` respectively are connected by a spring of spring cosntant k. The masses are moving to the right with a uniform velocity `v_(0)` each, the heavier mass leading the lighter one. The spring is of natural length during this motion. Block B collides head on with a thrid block C of mass `2m`. at rest, the collision being completely inelastic. The velocity of block B just after collision is -A. `v_(0)`B. `(v_(0))/(2)`C. `(3v_(0))/(5)`D. `(2v_(0))/(5)` |
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Answer» Correct Answer - (B) by applying conservation of linear momentum `2mv_(0) = (2m + 2m)v rArr v = (v_(0))/(2)` |
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| 156. |
Two block A and B of masses m and `2m` respectively are connected by a spring of spring cosntant k. The masses are moving to the right with a uniform velocity `v_(0)` each, the heavier mass leading the lighter one. The spring is of natural length during this motion. Block B collides head on with a thrid block C of mass `2m`. at rest, the collision being completely inelastic. The velocity of centre of mass of system of block A, B, & C is -A. `v_(0)`B. `(3v_(0))/(5)`C. `(2v_(0))/(5)`D. `(v_(0))/(2)` |
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Answer» Correct Answer - (B) `v_(cm) = (mv_(0) + 2mv_(0))/(m + 2m + 2m) = (3v_(0))/(5)` |
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| 157. |
A circular disc rolls on a horizontal floor without slipping and the center of the disc moves with a uniform velocity v. Which of the following values of the velocity at a point on the rim of the disc can have?A. vB. `-v`C. 3D. Zero |
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Answer» Correct Answer - D In this case (pure rolling), the velocity of point of contact is zero. |
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| 158. |
A body of mass `0.25` kg is projected with muzzle velocity `100ms^(-1)` from a tank of mass 100 kg. What is the recoil velocity of the tankA. `5 ms^(-1)`B. `25 ms^(-1)`C. `0.5 ms^(-1)`D. `0.25 ms^(-1)` |
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Answer» Correct Answer - D `Using law of conservation of momentum, we get `100 x v = 0.25 xx 100` `rArr v=0.25 ms^(-1)` |
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| 159. |
The velcoity of centre of mass of the system as shown in the figure :- A. `((2 - 2sqrt3)/(3)) hat(i) - (1)/(3)hat(j)`B. `((2 + 2sqrt3)/(3)) hat(i) - (2)/(3) hat(j)`C. `4 hat(i)`D. None of these |
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Answer» Correct Answer - B `vec(v)_(cm) = (m_(1)vec(v)_(1) + m_(2)vec(v)_(2))/(m_(1) + m_(2)) = (1 xx 2hat(i) + 2 xx (2 cos30hat(i) - 2sin 30hat(j)))/(3)` `= ((2 + 2sqrt3)/(3))hat(i) - (2)/(3)hat(j)` |
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| 160. |
Find the center of mass(x,y,z) of the following structure of four identical cubes if the length of each side of a cube is 1 unit. A. (1/2,1/2,1/2B. (1/3,1/3,1/3)C. (3/4,3/4,3/4)D. (1/2,3/4,1/2) |
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Answer» Correct Answer - (C ) First we find the centre of mass of each cube. It is located by symmetry: `(0.5, 0.5, 0.5), (1.5, 0.5, 0.5), (0.5, 1.5, 0.5), (0.5, 0.5, 1.5)`. Now we find the centre of mass by treating the COM of each cube as a point particle: `x_(COM) = (0.5 + 1.5 + 0.5 + 0.5)/(4) = 0.75 , y_(COM) = (0.5 + 0.5 + 1.5 + 0.5)/(4) = 0.75` `z_(COM) = (0.5 + 0.5 + 0.5 + 1.5)/(4) = 0.75` |
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| 161. |
Assertion: Two identical spherical spheres are half filled with two liquids of densities `rho_(1)` and `rho_(2) (gt rho_(1))`. The center of mass of both the spheres lie at same level. Reason: The center of mass will lie at center of the sphere.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - C We know that center of mass of half disc depends only on radius and only the density of the material of disc similarly in this case center of mass of half filled sphere will depends only on radius and not on density of liquid inside. Since both sphere are of same radius so both have CM at the same level. |
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| 162. |
The centre of mass of a non uniform rod of length L, whose mass per unit length varies as `rho=(k.x^2)/(L)` where k is a constant and x is the distance of any point from one end is (from the same end)A. `3L//4`B. `L//4`C. `2L//3`D. `L//3` |
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Answer» Correct Answer - A `bar(x) = (intxdm)/(intdm) = (int_(0)^(L) x(kx^(2))/(L)dx)/(int(kx^(2))/(L)dx) = (3L)/(4)` |
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| 163. |
A set of n identical cubical blocks lies at rest parallel to each other along a line on a smooth horizontal surface. The separation between the near surfaces of any two adjacent blocks is L . The block at one end is given a speed v towards the next one at time 0 t . All collisions are completely inelastic, thenA. The last block starts moving at `t = n (n - 1) (L)/(2v)`B. The last block starts moving `t = (n - 1) (L)/(v)`C. The centre of mass of the system will have a final speed `(v)/(n)`D. The centre of mass of the system will have a final speed `v`. |
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Answer» Correct Answer - A::C `t_(1) = (L)/(v)` (time for `1st` collision) `t_(2) = (2L)/(v)` (time for IInd collision ) `t_(3) = (3L)/(v)` (time for 3rd collision) `t_(n - 1) = (L)/(v) (n - 1)` (time for `(n^(th))` collision) `underset(1 - 1)overset(h)sum t_(1) = (n(n - 1))/(2) (L)/(v)` |
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| 164. |
In a two block system in figure match the following. `{:("Column1","Column2"),("A Velocity of center of mass","P Keep on changing all the time"),("B Momentum of center of mass","Q First decreases then become zero"),("C Momentum of 1kg block","R Zero"),("D Kinetic energy of 2kg block","S Constant"):}` |
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Answer» Correct Answer - A `rightarrow`r,s, B`rightarrow`r s, C`rightarrow`q, D`rightarrow`q `v_(CM)` = `(m_(1)v-(1) + m_(2)v-(2))/(m_(1)+m-(2)` = `((1)(10) + (2)(-5))/(3)` = 0` Similarly, `p_(CM)` =0` Net force on the system is zero, hence `V_(CM)` and `p_(CM)` will remain constant. Velocity of 1 kg and 2kg blocks keep on changing initially and finally both of them stop as `V_(CM)` was zero. |
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| 165. |
Four point masses are placede at four corners of a square of side 4m as shown. Match the following columns. `{:("Column1","Column2"),("A x-coordinate of center of mass of 4kg and 2kg","P (7//2)m"),("B x-coordinates of center of mass of 4kg 2kg and 3kg","Q (4//3)m"),("C y-coordinate of center of mass of 1kg 4kg and 3kg","R 3m"),("y-coordinate of center of mass of 1kg and 3kg","(20/9)m"):}` |
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Answer» Correct Answer - (A `rightarrow`q, B`rightarrow`s, C `rightarrow`p, D `rightarrow` r Applying `X_(CM) = (m_)(1)x_(1) + ........+ (m_(n)x_(n))` And `Y_(CM) = (m_(1)y_(1) + .......+m_(n)y_(n))/(m_(1) + ..........+m_(n))` |
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| 166. |
The friction coefficient between the horizontal surface and blocks A and B are `(1)/(15)` and `(2)/(15)` respectively. The collision between the blocks is perfectly elastic. Find the separation (in meters) between the two blcoks when they come to rest. |
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Answer» Correct Answer - `5 m` Velocity of block A just before collision `v_(A) = sqrt(u_(A)^(2) - 2mugx) = sqrt(16 - 2 (1)/(15) (10) (2)) = sqrt((40)/(3))` Velocity of Block B just after collision `v_(B) = v_(A) = sqrt((40)/(3))` Velocity of Block A just after collision `= 0` Total dsitance travelled by block B `= (v_(B)^(2))/(2mug) = (40//3)/(2(2)/(15) (10)) = 5m` |
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| 167. |
If the linear momentum is increased by 50%, then KE will be increased by :A. 0.5B. 1C. 1.25D. 0.25 |
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Answer» Correct Answer - C Let `p_(1) = p, p_(2)=p_(1)+50%` of `p_(1)=p_(1) + p_(t)/2 = (3p_(1)/(2)` `E alpha p^(2)` `rArr E_(2)/E_(1) = (p_(2)/p_(1))^(2) = ((3p_(1)//2)/p_(1))^(2) = 9/4` `rArr E_(2) = 2.25E_(1) = E_(1) + 1.25E_(1)` `therefore E_(2) = E_(1) + 125% of E_(1)` i.e., Kinetic energy will increase by 125%. |
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| 168. |
A body of mass a moving with a velocity b strikes a body of mass c and gets embered into it. The velocity of the systems after collision isA. `(a+c)/(ab)`B. `(ab)/(a+c)`C. `(a)/(b+c)`D. `(a)/(a+b)` |
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Answer» Correct Answer - B `m_(1)v_(2) + m_(2)v_(2) = (m_(1) + m_(2))v` `therefore a.b+c.0 = (a+c)v` `rArr v=(ab)/(a+c)` |
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| 169. |
In Fig., a hollow tube of mass `M` is free in horizontal direction. The system is released from rest. There is no friction present. The tube and blocks are taken as system. i. Momentum of the system is conserved in x-direction. ii. Speed of `A` w.r.t. `M =` speed of `B` w.r.t. `M`. iii. Trajectory of centre of mass is `X`-constant. iv. Centre of mass has finite acceleration. Evaluate the above statements and choose the correct option from the following: A. Statements i, ii are true and iii, iv are false.B. Statements i, ii are false and iii, iv are true.C. All statements are true.D. All statements are false. |
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Answer» Correct Answer - C Momentum of the suystem in `d`-direction is conserved because all the forces on the system in `x`-direction are balanced. But the fore in `y`-direction is not balanced so, trajector of motion of centre of the system is `X-constant. |
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| 170. |
A machine gun can fire bullets of `50` grams at a Speed of `2000 m s^(-1)`, the man holding the gun can exert an average force of `200 N` against the gun. Calculate the maximum number of bullets which he can fire per minute. |
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Answer» We have force `"mn"u` ltbr where `w=m `of bullet, `u =` speed ofbullet, `=` number of bullets per unit time. `:. F=mnu` or `200=50/1000xx2000xxn` Hence, number of bullets fired per minute `=60xx2=120` |
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| 171. |
Three identical balls, ball I, ball II and ball III are placed on a smooth floor on a straight line at the separation of `10m` between balls as shown in figure. Initially balls are stationary. But I is given velocity of `10m//s` towards ball II, collision between balls I and II is inelastic with coefficient of restitution `0.5` but collision between balls II and III is perfectly elastic. What is the time interval between two consecutive collisions between ball I and II? |
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Answer» Let velocity of Ist ball and Iind ball after collision be `v_(1)` and `v_(2)` `v_(2) -v_(1) = 0.5 xx 10` ...............(i) `mv_(2) + m_(1)v_(1) = m xx 10` .......................(ii) `rArr v_(2) + v_(1) = 10` Solving Eqs. (i) and (ii) `v_(1) = 2.5 ms^(-1)`, `v_(2) = 7.5 ms^(-1)` Ball II after moving 10 m collides with ball III elastically and stops. But ball I moves towards ball II. Time taken between two consecutive collisions. `(10/7.5) + (10 - 10 xx (2.5)/(7.5))/(2.5) = 4s` |
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| 172. |
Two identical balls `A` and `B` lie on a smooth horizontal surface, which gradually merges into a curve to a height `3.2 m`. Ball `A` is given a velocity of `10 m//s`, to collide head-on with ball `B`, which then takes up the curved path. What is the minimum coefficient of restitution, `e`, for the collision between `A` and `B` in order that ball `B` reaches the highest point `C` of the curve. |
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Answer» Velocity of `B` after collision `v_(2)((1+e)m_(1)u_(1))/((m_(1)+m_(2)))=((1+e)mxx10)/(m+m)impliesv_(2)=5(1+e)` Height raised by` B=v_(2)^(2)/(2g)` Now `v_(2)^(2)/(2g)ge3.2implies((5(1+e))^(2))/(2xx10)ge3.2impliesege3/5` |
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| 173. |
Simple Atwood Machine as System of Particles The system shown in the figure is known as simple Atwood machine. Initially the masses are held at rest and then let free. Assuming mass `m_(2)` more than the mass `m_(1)`, find acceleration of mass center and tension in the string supporting the pulley. |
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Answer» We know that accelarations `a_(1)` and `a_(2)` are given by the following equations. `a_(2) = (m_(2) - m_(1))/(m_(2) + m_(1))g downarrow` and `a_(1)=(m_(2) - m_(1))/(m_(2) + m_(1))g uparrow` Making use of eq. , we can find acceleration `a_(c )` of the mass center. We denote upward direction positive and downward direction negative signs respectively. `Mvec(a)_(c ) = Sigma(m_(i)bar(a)_(i))rarr" "(m_(1) + m_(2))a_(c) = m_(1)a_(1) - m_(2)a_(2)` Substituting values of accelerations `a_(1)` and `a_(2)`, we obtain `a_(C ) = 2m_(1)m_ (2) - (m_(1)^(2) + m_(2)^(2))/(m_(1)+m_(2)^(2))` To find tension T in the string supporting the pulley, we again use eq.(9) `Sigma vec(F)_(i) = Mvec(a)_(c )rarr " " T - m_(1) g - m_(2) g = (m_(1) + m_(2))a_(c)` Substituting expression obtained for `a_(c )`, we have `T = (4m_(1) m_(2))/(m_(1) + m_(2)) g` |
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| 174. |
A body of mass M moves in outer space with vleocity v. It is desired to break the body into two parts so that the mass of one part is one-tenth of the total mass. After the explosion, the heavier part comes to rest whole the lighter part continues to move in the original direction of motion. The velocity of the small part will beA. `v`B. `((v)/(2))`C. `((v)/(10))`D. `10 v` |
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Answer» Correct Answer - D `Mv = 0 + (M)/(10)v_(2) = 10v` |
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| 175. |
In the figure shown, when the persons A and B exhange their position, then (i) the distance moved by the centre of mass of the system is ____. (ii) the plank moves towards_____. (iii) the distance moved by the plank is _________. (iv) the distacne moved by A with respect to ground is _____. (v) the distance moved by B with respect to ground is _____. |
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Answer» Correct Answer - (i)zero (ii) right (iii) `0.2 m` (iv) `2.2 m` (v) `1.8 m` (i) CM does not shift (ii) Plank moves towards right. (iii) `Deltax_(cm) = (m_(1)Deltax_(1) + m_(2)Deltax_(2) + M Deltax)/(m_(1) + m_(2) + M) = 0` `rArr 0 = (50(x + 2) + 70(x - 2) + 80x)/(50 + 70 + 80)` `rArr x = 0.2m` (right) (iv) `Deltax_(m_(1)) = x + 2 = 2.2 m` (right) (v) `Deltax_(m_(2)) = x - 2 = -1.8m` (rleft) |
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| 176. |
Assertion: Two spherical bodies of mass ratio `1:2` travel towards each other (starting from rest) under the action of their mutual gravitational attraction. Then, the ratio of their kinetic energies at any instant is `2:1` Reason: At any instant their momenta are same.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - D `K = p^(2)/(2m)` As momentum is equal and opposite `therefore K alpha 1/m` |
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| 177. |
Assertion: A projectile gets exploded at its highest point. All the piece get only horizontal velocities. Reason: The weight of the projectile is the external force for projectile.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - D It will fall at the same point |
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| 178. |
A body falling vertically downwards under gravity breaks in two parts of unequal masses. The centre of mass of the two parts taken together shifts horizontally towardsA. havier pieceB. lighter pieceC. does not shift horizontallyD. depends on the vertical velocity at the time of breaking |
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Answer» Correct Answer - C Center of mass does not change its path during explosion. Therefore, it will keep on falling vertically and will not shift horizontally, as `F_(x)=0` |
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| 179. |
An object is moving so that its kinetic energy is 150 J and the magnitude of its momentum is 30.0 kg-m/s. With what velocity is it travelling? |
| Answer» `K=(p^2)/(2m)=1/2pvrarr " " v=2xx150/30.0=10.0m//s` | |
| 180. |
An inverted `T`-shaped object is placed on a horizontal floor as shown in Fig. A force `F` is applied on the system as shown in Fig. The value of `x` so that the system performs pure translational motion is A. `L/4`B. `(3L)/4`C. `L/2`D. `(3L)/2` |
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Answer» Correct Answer - A The force has to be applied at the centre of mass of the system for pure translation motion. |
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| 181. |
a. A rail road car of mass M is moving without friction on a straight horizontal track with a velocity ct. A man of mass m lands on it normally from a helicopter. What will be the new velocity of the car? ltbgt b. If now the man begins to run on it with speed um with respect to car in a direction opposite to motion of the car, what will be the new velocity of the car? |
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Answer» Let `v_(1)` be the new velocity of the car. Then according to the principle of conservatioin of linear momentum `Mu=(M+m)v_(1)` `implies v_(1)=m/(M+m)u`…………..i b. Let `v_(2)` be the new velocity of the car relative to the earth, then velocity of the man relative of the earth `=(v_(2)-v_(rel))` `:. v_(2)=v_(1)+(mv_(rel))/(M+m)` |
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| 182. |
In a two block system an initial velocity `v_(0)` with respect to ground is given to block A A. the momentum of block A is not conservedB. the momentum of system of blocks A and B is conservedC. the increase in momentum of B is equal to the decrease in momentum of block AD. All of the above |
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Answer» Correct Answer - D Force of friction on A is backward and force of friction on B is forward. Net external force on the system is zero. Hence, momentum of system will remain conserved. As the momentum of system is conserved, increase in momentum of B is equal to decrease in momentum of A. |
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| 183. |
Assertion: The centre mass of an electron and proton, when released moves faster towards proton. Reason: Proton is heavier than electron.A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.C. If Assertion is true but Reason is false.D. If Assertion is false but Reason is true. |
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Answer» Correct Answer - D Center of mass remains stationary. |
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| 184. |
Two trains A and B are running in the same direction on the parallel rails such that A is faster than B. Packets of equal weight are transfcrred from A to B. What will happen due to this :A. A will be accelerated but B will be retarded.B. B will be accelerated but A will be retardedC. there will be no change in A but B will accelerated.D. there will be no change in B, but A will be accelerated |
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Answer» Correct Answer - B Packet from A falls with greater momentum on B. Therefore, B is slightly accelerated. |
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| 185. |
Figure shows the system is at rest initially with `x = 0`, A man and a woman both are initially at the extreme carrier of the platform. The man and the woman start to move towards each other. Obtain an expression for the displacement `s` of the platform when the two meet in terms of the displacement `x_1` of the man relative to the platform. ` |
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Answer» As no external force is acting on the system the diplacemnte of the centre of mass should be zero. Let the displacement of the platform toward right be `s`. `/_vecx_(1)=/_vecx_(1,p)+/_vecv_(p)=(-x_(1)+s)` `/_vecx_(2)=/_vecx_(2,p)+/_vecx_(p)=(l-x_(1)+s)` Hence, `0=(m_(1)(-x_(1)+s)+m_(2)(l-x_(1)+s)+m_(0)s)/(m_(1)+m_(2)+m_(3))` `s(m_(1)+m_(2)+m_(0))-m_(1)x_(1)+m_(2)(l-x_(1))=0` `s=(m_(1)x_(1)-m_(2)(l-x_(1)))/(m_(1)+m_(2)+m_(0))` |
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| 186. |
A body from height h is dropped, if the coefficient of restitution is e, then calcualte the height achieved after one bounce.A. `h_(1) = e^(4)h`B. `h=eh_(1)`C. `h_(1)=he^(2)`D. `h = h_(1)//e` |
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Answer» Correct Answer - C When a body falls from height h, it strikes the ground with a velocity `v = sqrt(2gh)`. Let it rebounced with a velocity v and rise to a height `h_(1)`. `v = sqrt(2gh_(1) rArr e=v/u = (sqrt(h_(1)/h)` Clearly, `h_(1) = e^(2)h` |
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| 187. |
A ball released from a height ho above a horizontal surface rebounds to a height `h_(1)`, after one bounce. The graph that relates `h_(0)` to `h_(1)` is shown Fig. If the ball (of the mass `m`) was dropped from an initial height `h` and made three bounces, the kinetic energy of the ball immediately after the third impact with the surface was A. `(0.8)^(3)mgh`B. `(0.8)^(2)mgh`C. `0.8mg(h//3)`D. `[1-(0.8)^(3)]mgh` |
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Answer» Correct Answer - A `h_(1)/h_(0)=e^(2)implies 80/100=e^(2)impliese=sqrt0.8` velocity before the first impact `u=sqrt(2gh)` velocity after the third impact `v_(3)=e^(3)sqrt(2gh)` `KE=1/2mv_(3)^(2)=e^(6)mgh=(0.8)^(3)mgh` |
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| 188. |
A particle moves in the X-Y plane under the influence of a force such that its linear momentum is `oversetrarrp(t)=A[haticos(kt)-hatjsin(kt)]`, where A and k are constants. The angle between the force and the momentum isA. `0^(@)`B. `30^(@)`C. `45^(@)`D. `90^(@)` |
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Answer» Correct Answer - D `vec(P)(t) = A[hat(i) cos (kt) - hat(j) sin (kt)]` `vec(F) = (dvec(p))/(dt) = Ak [- hat(i) sin (kt) - hat(j) cos (kt)]` `vec(F).vec(p) = Fp cos theta` But`vec(F).vec(p) = 0 rArr cos theta = 0 rArr theta = 90^(@)` |
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| 189. |
In an elastic collision between two particlesA. the total kinetic energy of the system is always conserved.B. the kinetic energy of the system before collision is equal to the kinetic energy of the system after collision.C. the linear momentum of the system is conserved.D. the mechanical energy of the system before collision is equal to the mechanical energy of the system after collision. |
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Answer» Correct Answer - B::C::D In a elastic collision, linear momentum of the system ws always conserved i.e., the in all the three stages of collision. But the kinetic energy of the system is not always conserved. It is conserved only before and after the collision. During collision it gets converted into potential energy. |
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| 190. |
A uniform rope of linear mass density `lambda` and length `l` is coiled on a smooth horizontal surface. One end is pulled up with constant velocity v. Then the average power applied by the external agent in pulling the entire rope just off the ground is :- A. `(1)/(2)lambdalv^(2) + (lambdal^(2)g)/(2)`B. `lambdalgv`C. `(1)/(2)lambdav^(3) + (lambdalvg)/(2)`D. `lambdalgv + (1)/(2)lambdav^(3)` |
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Answer» Correct Answer - C Average power `(DeltaW)/(Deltat) = ((DeltaK + DeltaU)/(Deltat)) = (1)/(2) ((lambdaDeltax)v^(2))/(Deltat) + (lambdaDeltaxg(Deltax//2))/(Deltat)` ltPgt `= (1)/(2)lambdav^(3) + (lambdal)/(2)vg` |
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| 191. |
A ball of mass mis released from the top of an inclined plane of inclination `theta` as shown. It strikes a rigid surface at at a distance `(3l)/(4)` from top elastically. Impulse imparted to ball by the rigid surface is A. `msqrt(3/2)gh`B. `msqrt(3gh)`C. `2msqrt(3gh)`D. `msqrt(6gh)` |
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Answer» Correct Answer - D Loss in PE = gain in KE `mgh_(1) = 1/2mv^(2)` `mg xx 3/4h = 1/2mv^(2) rArr v=sqrt(3gh)/2` Now, impulse imparted, `J = 2mv = 2msqrt(3gh)/2` = `msqrt(6gh)` |
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| 192. |
A particle with a mass of `1 kg` a velocity of is having `10 m//s` in `+ve x`-direction at `t = 0`. Forces `vecF_(1)` and `vecF_(2)`, act on the particle whose magnitudes are changing with time according to the variation shown in Fig. The magnitude of the velocity of the particle at `t = 3 s` (neglect gravity effect) is found to be `root(n)(5)` Find the value of `n` |
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Answer» Correct Answer - 2 `F_(1)=4N=-4hati 0lttle1s` `=2N=-2hati, 1letle3ss` ` F_(2)1N=-hatj, 0lttle2s` `=2t-3N-(2t-3)hatj 2letle3s` Initial velocity of the particle is `vecu=10hati` From impulse momentum theorem, `int vec(dp)=int vecF dt` `vec("mv")-vec("mu")=int^(3)(vecF_(1)+vecF_(2))dt` (where `vecv` is the required velocity) `1xvecv-1xx10hati=-8hati-4hatj` or `vecv=2hati-4hatj` or `v=sqrt(2^(2)+4^(2))=2sqrt(5) m//s` |
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| 193. |
In which o the following cases the centre of mass of a rod is certainly not at its centre?A. the density continuously increases from left to rightB. the density continuously decreases from left to rightC. the density decreases from let to right upto the centre and then increasesD. the density increases from left to right upto the centre and then decreases. |
| Answer» Correct Answer - A::B | |
| 194. |
Two identical balls, each of mass `m`, are tied with a string and kept on a frictionless surface. Initially, the string is slack. They are given velocities `2u` and `u`. in the same direction. Collision between the balls is perfectly elastic. After the first collision, what is the total loss in kinetic energy of the balls?A. `2"mu"^(2)`B. `"mu"^(2)`C. `3"mu"^(2)`D. zero |
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Answer» Correct Answer - D Since the collision is elastic there is no loss in `KE` |
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| 195. |
Two blocks of msses 10 kg and 30 kg are placed along a vertical line. The first block is raised through a height of 7 cm. By what distance should the second mass be moved to raise the centre of mass by 1 cm?A. 2 cm upwardB. 1 cm upwardC. 2cm downwardD. 1 cm downward |
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Answer» Correct Answer - D `y_(CM) = (m_(1)y_(1) + m_(2)y_(2))/(m_(1) + m_(2))` or `+1= ((10)(7) + (30)y_(2))/(10 + 30)` `therefore y_(2) = -1cm` |
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| 196. |
Two identical balls, each of mass `m`, are tied with a string and kept on a frictionless surface. Initially, the string is slack. They are given velocities `2u` and `u`. in the same direction. Collision between the balls is perfectly elastic. What is the impulse generated in the string during the second collision?A. `"mu"//2`B. `"mu"//4`C. `(2"mu")//3`D. none of these |
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Answer» Correct Answer - A For left ball: `I=mv-"mu"=m(3u)/2-"mu"="mu"//2` |
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| 197. |
Two persons, A of mass `60 kg` and `B` of mass `40 kg`, are standing on a horizontal platform of mass `50 kg`. The platform is supported on wheels on a horizontal frictionless surface and is initially at rest. Consider the following situations. (i) Both `A` and `B` jump from the platform simultaneously and in the same horizontal direction. (ii) A jumps first in a horizontal direction and after a few seconds `B` also jumps in the same direction. In both the situations above, just after the jump, the person (`A` or `B`) moves away from the platform with a speed of `3 m//s` relative to the platform and along the horizontal. In situation (i), just after both `A` and `B` jump from the platform, velocity of centre of mass of the system (`A, B` and the platform) isA. `2m//s`B. `6m//s`C. `5m//s`D. none of these |
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Answer» Correct Answer - D Centre of mass will remain at rest because there isno external force on the system of `A,B` and platform |
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| 198. |
Two persons of masses `55 kg` and `65 kg` respectively are at the opposite ends of a boat. The length of the boat is `3.0 m` and weights `100 kg`. The `55 kg` man walks up to the `65 kg` man and sits with him. If the boat is in still water the centre of mass of the system shifts by.A. 3.0 mB. 2.3 mC. zeroD. 0.75 m |
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Answer» Correct Answer - C Here on the entire system net external force on the system is zero, hence center of mass remains unchanged. |
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| 199. |
Two persons A and B of weight `80 kg` and `50 kg` respectively are standing at opposite ends of a boat of mass `70 kg` and length `2 m` at rest. When they interchange their positions then displacement of the centre of mass of the boat will be :- A. `60 cm` towards leftB. `30 cm` towards rightC. `30 cm` towards leftD. stationary |
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Answer» `Deltabar(x) = (m_(1)Deltax_(1) + m_(2)Deltax_(2) + m_(3)Deltax_(3))/(m_(1) + m_(2) + m_(3))` `Deltabar(x) = ((80 xx 2) - (50 xx 2) + (70 xx 0))/(80 + 50 + 70)` `= 3 cm` towards right |
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| 200. |
A particle of mass `2m` is projected at an angle of `45^@` with horizontal with a velocity of `20sqrt2m//s`. After `1s` explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. Find the maximum height attained by the other part. Take `g=10m//s^2`.A. `50m`B. `25m`C. `40m`D. `35m` |
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Answer» Correct Answer - D `u_(x)=20sqrt(2)cos45^(@)=20m//s` `u_(y)=20sqrt(2)sin45^(@)=20m//s` After `1 s`, horizontal component remains unchanged while vertical component becomes `v_(y)=u_(u)-gt=20-10=10m//s` Due to explosion one part comes to rest. Hence, from conservatuion of linear mometum vertical component of second pard will become `v_(y1)=20m//s`. therefore, maximum height attained by the second part will be `H=h_(1)+h_(2)` Here `h_(1)=`height attained in `1 s` `=(20)(1)-1/2xx10xx1^(2)=5m` and `h_(2)`= height attaine dafter `1 s` `v_(y1)^(2)=(2g)=((20)^2)/(2x10)=20m` `:. H=15+20=35m` |
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