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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.
| 1201. |
Power of a force acting on a block varies with time t as shown in figure. Then, angle between force acting on the block and its velocity is A. acute at = 1 sB. `90^(@)` at t = 3 sC. obtuse at t = 7 sD. change in kinetic energy from t = 0, to t = 10 s is 20 J |
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Answer» Correct Answer - A::C::D |
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| 1202. |
`F = 2x^(2)-3x-2`. Choose correct option.A. `x = - 1//2` is position of stable equilibriumB. x = 2 is position of stable equilibriumC. `x = - 1//2` is position of unstabl equilibriumD. x = 2 is position of neutral equilibrium |
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Answer» Correct Answer - A |
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| 1203. |
A force`F = (2hat(i)+5hat(j)+hat(k))N` is acting on a particle. The particle is first displacement from (0, 0, 0) to (2m, 2m, 0) along the path x = y and then from (2m, 2m, 0) to (2m, 2m, 2m) along the path x = 2m, y = 2 m. The total work done in the complete path isA. 12 JB. 8 JC. 16 JD. 10 J |
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Answer» Correct Answer - C |
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| 1204. |
The potential energy of the system is represented in the first figure. The force acting on the system will be represented by A. B. C. D. |
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Answer» Correct Answer - C |
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| 1205. |
A body of mass m dropped from a certain height strikes a light vertical fixed spring of stifness k. the height of its fall before touching the spring the if the maximum compression of the spring the equal to `(3 mg)/k` isA. `(3mg)/(2k)`B. `(2mg)/(k)`C. `(3mg)/(4k)`D. `(mg)/(4k)` |
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Answer» Correct Answer - A |
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| 1206. |
Force acting on a block moving along x-axis is given by `F = - ((4)/(x^(2)+2))N` The block is displaced from `x = - 2m` to `x = + 4m`, the work done will beA. positiveB. negativeC. zeroD. may be positive or negative |
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Answer» Correct Answer - B |
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| 1207. |
A block of mass 250 g is kept on a vertical spring of spring constant 100 N/m fixed from below. The spring is now compressed to have length 10 cm shorter than its natural length and the system is released from this position. How high does the block rise? take `g=10 m/s^2`.A. 20 cmB. 30 cmC. 40 cmD. 50 cm |
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Answer» Correct Answer - A |
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| 1208. |
From the fixed pulley, masses 2kg, 1kg and 3kg are suspended as shown in the figure. Find the extension in the spring if `k=100 N//m`. (Neglect oscillations due to spring) A. 0.1mB. 0.2mC. 0.3mD. 0 |
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Answer» Correct Answer - B |
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| 1209. |
Potential energy of a particle moving along x-axis under the action of only conservative force is given as `U=10+4cos(4pi x)`. Here, U is in Joule and x in metres. Total mechanial energy of the particle is 16 J. Choose the correct option.A. At x = 1.25 m, particle is at equilibrium positionB. Maximum kinetic energy of the particle is 20 JC. Both (a) and (b) are correctD. Both (a) and (b) and wrong |
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Answer» Correct Answer - A |
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| 1210. |
In a two dimensional space the potential energy function for a conservative force acting on a particle of mass m = 0.1 kg is given by U = 2 (x + y) joule (x and y are in m). The particle is being moved on a circular path at a constant speed of `V = 1 ms ^(-1)`. The equation of the circular path is `x^2 + y^2 = 42`. (a) Find the net external force (other than the conservative force) that must be acting on the particle when the particle is at (0, 4). (b) Calculate the work done by the external force in moving the particle from (4, 0) to (0, 4). |
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Answer» Correct Answer - (a) `(2hati+1.975 hatj)N` (b) zero |
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| 1211. |
A block of mass 50 kg is projected horizontal on a rough horizontal floor. The coefficient of friction between the block and the floor is 0.1. The block strikes a light spring of stiffness `k = 100 N//m` with a velocity `2 m//s`, the maximum compression of the spring is A. 1 mB. 2 mC. 3 mD. 4 m |
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Answer» Correct Answer - A |
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| 1212. |
A system consists of two cubes of mass `m_(1)`, and `m_(2)` respectively connected by a spring of force constant k. force (F) that should be applied to the upper cube for which the lower one just lifts after the force is removed, is A. mgB. `(m_(1)m_(2))/(m_(1)+m_(2))g`C. `(m_(1)+m_(2))g`D. `m_(2)g` |
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Answer» Correct Answer - C |
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| 1213. |
A block is suspended by an ideal spring of force constant force F and if maximum displacement of block from its initial mean position of rest is `x_(0)` thenA. increase in energy stored in spring is `kx_(0)^(2)`B. `x_(0)=(3F)/(2k)`C. `x_(0)=(2F)/(k)`D. work done by applied force F is `Fx_(0)` |
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Answer» Correct Answer - C::D |
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| 1214. |
A vertical spring supports a beaker containing some water in it. Water slowly evaporates and the compression in the spring decreases. Where does the elastic potential energy stored in the spring go? |
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Answer» Correct Answer - When a small amount of water evaporates, the spring relaxes a little bit. Water remaining in the beaker gains gravitational potential energy. Therefore, the spring energy gets converted into the gravitational potential energy of beaker + water. |
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| 1215. |
A particle of mass m = 1 kg is free to move along x axis under influence of a conservative force. The potential energy function for the particle is `U=a[(x/b)^(4)-5(x/b)^(2)]` joule Where b = 1.0 m and a = 1.0 J. If the total mechanical energy of the particle is zero, find the co-ordinates where we can expect to find the particle and also calculate the maximum speed of the particle. |
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Answer» Correct Answer - `-sqrt(5) le x le sqrt(5) ; v_(max) =5//sqrt(2) ms^(-1)` |
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| 1216. |
A block of mass m is pulled by a constant powert `P` placed on a rough horizontal plane. The friction coefficient the block and surface is `mu`. The maximum velocity of the block is.A. `(muP)/(mg)`B. `(mumg)/P`C. `mumgP`D. `P/(mumg)` |
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Answer» Correct Answer - D |
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| 1217. |
A block of mass `m_1` is lying on the edge of a rough table. The coefficient of friction between the block and the table is `mu`. Another block of mass`m_2` is lying on another horizontal smooth table. The two block are connected by a horizontal spring of force constant K. Block of mass `m_2` is pulled to the right with a constant horizontal force F. (a) Find the maximum value of F for which the block of mass m1 does not fall off the edge. (b) Calculate the maximum speed that `m_2` can acquire under condition that `m_1` does not fall. |
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Answer» Correct Answer - `(a) (mum_(1) g)/2 ` (b) `(mum_(1)g)/2 1/(sqrt(m_(2)k))` |
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| 1218. |
A block of mass `M_(1)` is attached with a spring constant `k`. The whole arrangement is placed on a vechile as shown in the figure. If the vehicle starts moving towards right with an acceleration a (there is no friction anywhere), then : .A. maximum elongation in the spring is `(Ma)/(k)`B. maximum elongation in the spring is `(2Ma)/(k)`C. maximum compression in the spring is `(2Ma)/(k)`D. maximum compression in the spring is zero |
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Answer» Correct Answer - B::D |
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| 1219. |
A block of mass m is attached to a frame by a light spring of force constant k. The frame and block are initially at rest with `x = x_(0)`, the natural length of the spring. If the frame is given a constant horizontal accelration `a_(0)` towards left, determine the maximum velocity of the block relative to the frame (block is free to move inside frame). Ignore any friction. A. `a_(0)sqrt((m)/(2k))`B. `a_(0)sqrt((2m)/(k))`C. `a_(0)sqrt((m)/(k))`D. `(1)/(2)a_(0)sqrt((m)/(k))` |
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Answer» Correct Answer - C |
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| 1220. |
The given graph represents the total force in x direction being applied on a particle of mass m = 2 kg that is constrained to move along x axis. What is the minimum possible speed of the particle when it was at x = 0? |
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Answer» Correct Answer - 0.5 m/s |
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| 1221. |
A block of mass m is pulled by a constant powert `P` placed on a rough horizontal plane. The friction coefficient the block and surface is `mu`. The maximum velocity of the block is.A. `(muP)/(mg)`B. `(mumg)/(P)`C. `mumgP`D. `(P)/(mumg)` |
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Answer» Correct Answer - D |
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| 1222. |
A block of mass M is placed on a horizontal smooth table. It is attached to an ideal spring of force constant k as shown. The free end of the spring is pulled at a constant speed u. Find the maximum extension (`x_0`) in the spring during the subsequent motion. |
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Answer» Correct Answer - `x_(0)=sqrt(M/k) u` |
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| 1223. |
A child of mass m is sitting on a swing suspended by a rope of length L. The swing and the rope have negligible mass and the dimension of child can be neglected. Mother of the child pulls the swing till the rope makes an angle of `theta_(0)=1` radian with the vertical. Now the mother pushes the swing along the arc of the circle with a force `F=(Mg)/2` and releases it when the string gets vertical. How high will the swing go? [Take cos(1 radian) `~= 0.5`] |
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Answer» Correct Answer - The swing gets horizontal |
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| 1224. |
A uniform chain of mass `m_0` and length l rests on a rough incline with its part hanging vertically as shown in the fig. The chain starts sliding up the incline (and hanging part moving down) provided the hanging part equals `eta` times the chain length (`eta` lt 1). What is the work performed by the friction force by the time chain slides completely off the incline. Neglect the dimension of pulley and assume it to be smooth. |
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Answer» Correct Answer - `-(l(1-eta)[eta-(1-eta)sin theta)/2 m_(0)g` |
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| 1225. |
A block of mass M slides along a horizontal table with speed `v_(0)`. At `x=0`, it hits a spring with spring constant k and begins to experience a friction force. The coefficient of friction is variable and is given by `mu=bx`, where b is a positive constant. Find the loss in mechanical energy when the block has first come momentarily to rest. .A. `(gbMv_(0)^(2))/(2k)`B. `(Mgbv_(0)^(2))/(2(k+gb))`C. `(gbMv_(0)^(2))/(k)`D. `(M^(2)gbv_(0)^(2))/(2(k+Mgb))` |
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Answer» Correct Answer - D |
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| 1226. |
Two small rings each of mass ‘m’ are connected to a block of same mass ‘m’ through inextensible light strings. Rings are constrained to move along a smooth horizontal rod. Initially system is held at rest (as shown in figure) with the strings just taut. Length of each string is ‘l’. The system is released from the position shown. Find the speed of the block (v) and speed of the rings (u) when the strings make an angle of theta=60^(@)` with vertical. (Take `g = 10 m//s^2`) |
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Answer» Correct Answer - `u=sqrt((gl)/5). V=sqrt((3gl)/5)` |
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| 1227. |
A block of mass 2 kg is connect to an ideal spring and the system is placed on a smooth horizontal surface. The spring is pulled to move the block and at an instant the speed of end A of the spring and speed of the block were measured to be 6 m/s and 3 m/s respectively. At this moment the potential energy stored in the spring is increasing at a rate of 15 J/s. Find the acceleration of the block at this instant. |
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Answer» Correct Answer - 2.5 `m//s^(2)` |
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| 1228. |
A uniform rope of linear mass density `lambda` (kg/m) passes over a smooth pulley of negligible dimension. At one end B of the rope there is a small particle having mass one fifty of the rope. Initially the system is held at rest with length L of the rope on one side and length `L/4` on the other side of the pulley (see fig). The external agent begins to pull the end A downward. Find the minimum work that the agent must perform so that the small particle will definitely reach the pulley. |
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Answer» Correct Answer - `(lambdaLg)/4` |
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| 1229. |
A ball is hanging vertically by a light inextensible string of length L from fixed point O. The ball of mass m is given a speed u at the lowest position such that it completes a vertical circle with centre at O as shown. Let AB be a diameter of circular path of ball making an angle `theta` with vertical as shown. (g is acceleration due to gravity) a) Let `T_A` and `T_B` be tension in string when ball is at A and B respectively, then find `T_A` – `T_B`. (b) Let `vec(a)_(A)` and `vec(a)_(B)` be acceleration of ball when it is at A and B respectively, then find the value of `|vec(a)_(A)+vec(a)_(B)|` |
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Answer» Correct Answer - `(a) 6mg cos theta ` `(b) gsqrt(4+12 cos^(2) theta)` |
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| 1230. |
Two block A and B are connected to a spring (force constant k = 480 N/m) and placed on a horizontal surface. Another block C is placed on B. The coefficient of friction between the floor and block A is `mu_1 = 0.5`, whereas there is no friction between B and the floor. Coefficient of friction between C and B is `mu_2 = 0.85`. Masses of the blocks are `M_A = 50 kg, M_B = 28` kg and `M_C = 2 kg`. The system is held at rest with spring compressed by `x_0 = 0.5 m`. After the system is released, find the maximum speed of block B during subsequent motion. |
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Answer» Correct Answer - 4 m/s |
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| 1231. |
A ideal spring of force constant k is connected to a small block of mass m using an inextensible light string (see fig). The pulley is mass less and friction coefficient between the block and the horizontal surface is `mu=1/(sqrt(3))`. The string between the pulley and the block is vertical and has length l. Find the minimum velocity u that must be given to the block in horizontal direction shown, so that subsequently it leaves contact with the horizontal surface. [Take `k=(2mg)/l`] |
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Answer» Correct Answer - `mu=2sqrt(gl)` |
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| 1232. |
A block of mass 2 kg is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force F = 40 N. The kinetic energy of the particle increases 40 J in a given interval of time. Then, `(g = 10 m//s^(2))`A. tension in the string is 40 NB. displacement of the block in the given interval of time is 2 mC. work done by gravity is `-20 J`D. work done by tension is 80 J |
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Answer» Correct Answer - A::B::D |
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| 1233. |
In the arrangement shown in the fig. string, springs and the pulley are mass less. Both the springs have a force constant of k and the mass of block B resting on the table is M. Ball A is released from rest when both the springs are in natural length and just taut. Find the minimum value of mass of A so that block B leaves contact with the table at some stage |
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Answer» Correct Answer - `m=M/2` |
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| 1234. |
The system is released from rest with both the springs in unstretched positions. Mass of each block is 1 kg and force constant of each spring is `10 N//m`. Extension of horizontal spring in equilibrium isA. 0.2 mB. 0.4 mC. 0.6 mD. 0.8 m |
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Answer» Correct Answer - B |
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| 1235. |
A rod of length L is pivoted at one end and is rotated with as uniform angular velocity in a horizontal plane. Let `T_1 and T_2` be the tensions at the points L//4 and 3L//4 away from the pivoted ends.A. `T_(1) gt T_(2)`B. `T_(2) gt T_(1)`C. `T_(1) = T_(2)`D. Then relation between `T_(1)` and `T_(2)` depends on wheter the rod rotates clockwise or anticlockwise. |
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Answer» Correct Answer - A |
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| 1236. |
A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall ?A. Kinetic energyB. Potential energyC. Total mechanical energyD. Total linear momentum |
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Answer» Correct Answer - C When a body is falling freely under the action of gravity alone in vacuum, its total mechanical energy remains constant during the fall |
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| 1237. |
Which of the diagrams in figure, correctly shows the change in kinetic energy of an iron sphere falling freely in a lake having sufficient depth to impart if a terminal velocity ?A. B. C. D. |
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Answer» Correct Answer - B When an iron sphere falls freely in a lake, its motion is accelerated dur to gravity and retarded due to viscous force. The overall efect is increase in velocity and hence increase in `K.E.` till the spher acquires terminal velocity, which is constant. hence `K.E.` of sphere beyond this depth of lake becomes constant. Choice (b) is most appropriat. |
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| 1238. |
The potential energy of a `1 kg` particle free to move along the x- axis is given by `V(x) = ((x^(4))/(4) - x^(2)/(2)) J` The total mechainical energy of the particle is `2 J` . Then , the maximum speed (in m//s) isA. `sqrt2`B. `(1)//(sqrt2)`C. `2`D. `(3)//(sqrt2)` |
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Answer» Correct Answer - D `V(x)=(x^4)/(4)-(x^2)/(2)` `F=-(dV_(x))/(dx)=-[x^3-x]=0impliesx(x^2-1)=0` `x=0`,`x=+-1` `(d^2V_(x))/(dx)=3x^2-1` At `x=+-1`,`(d^2V_(x))/(dx)=+ve`, i.e., at `x=+-1`, P.E. is minimum `V_(min)=-(1)/(4)` `E=K_(max)+V_(min)` `implies2=(1)/(2)xx1xxv_(max)^2-(1)/(4)impliesv_(max)=(3)/(sqrt2)(m)//(s)` |
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| 1239. |
The potential energy of a `1 kg` particle free to move along the x- axis is given by `V(x) = ((x^(4))/(4) - x^(2)/(2)) J` The total mechainical energy of the particle is `2 J` . Then , the maximum speed (in m//s) isA. `(3)/sqrt(2))`B. `sqrt(2)`C. `(1)/(sqrt(2))`D. `2` |
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Answer» Correct Answer - A (a) Velocity is muximum when K. E. is maximum for minimum P.E. ` (dV)/(dx) = 0 rArr x^(2) - x = 0 rArr x = +- 1` `rArr Min P.E. = (1)/(4) - (1)/(2) = - (1)/(4) J` `K. E _(max) + P.E_(min) = 2(Given)` `:. K.E _(max) = 2 + (1)/(4) = (9)/(4) = (1) /(2) m nu_(max) ^(2)` `rArr (1)/(2) xx 1 xx nu_(max)^(2) . = (9)/(4) rArr nu_(max) . = (3)/(sqrt(2))` |
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| 1240. |
Which of the diagrams in figure, correctly shows the change in kinetic energy of an iron sphere falling freely in a lake having sufficient depth to impart if a terminal velocity ?A. B. C. D. |
| Answer» (b) First velocity of the iron sphere increases and after sometime becomes constant called terminal velocity. Hence accordingly first KE increases and then becomes constant which is best represented by(b). | |
| 1241. |
The potential energy of a `1 kg` particle free to move along the x- axis is given by `V(x) = ((x^(4))/(4) - x^(2)/(2)) J` The total mechainical energy of the particle is `2 J` . Then , the maximum speed (in m//s) isA. `sqrt(2)`B. `1//sqrt(2)`C. `2`D. `3//sqrt(2)` |
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Answer» Correct Answer - C Max. `K.E.=(1)/(2)m upsilon_(max)^(2)0`total mechanical energy `:. (1)/(2)xx1xxupsilon_(max)^(2)=2` `upsilon_(max)=sqrt(2xx2)=2m//s` |
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| 1242. |
A particle move in a straight line with retardation proportional to its displacement its loss of kinectic energy for any displacement `x` is proportional to |
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Answer» Here, mass`=`m. Retardation `(-a)` of particle is proportional to displacement `(x)` `:. -apropx or a=-kx` ..(i) Now, `a=(dv)/(dt)=(dv)/(dx).(dx)/(dt)=(dv)/(dx)(v)` From (i), `v(dv)/(dx)=-kx` or `v dv=-kx dx ` …(ii) Suppose velocity of particle changes from u to v during its displacement x. `:.` Intefrating (ii), within proper limits, `int_(u)^(v)vdv=-kint_(0)^(x)xdx` `[(v^(2))/(2)]_(u)^(v)=-k[(x^(2))/(2)]_(0)^(x)` `(1)/(2)v^(2)-(1)/(2)u^(2)=-k(x^(2))/(2)` or `(1)/(2)u^(2)-(1)/(2)v^(2)=(1)/(2)kx^(2)` Loss in KE`=(1)/(2)m u^(2)-(1)/(2)mv^(2)=(1)/(2)mkx^(2)` |
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| 1243. |
A particle moves in a straight line with retardatino proportional it its displacement. Calculate the loss of `K.E.` for any displacement `x`. |
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Answer» Here, retardation `prop` displacement `-a prop x ` or `-a=Kx` where `K` is constant of proportionality. From (i) `as=-(dupsilon)/(dt)=Kx` or `(dupsilon)/(dt)(dx)/(dx)=-Kx` As `(dx)/(dt)=upsilon`, therefore, `upsilon dx=-Kx dx` `int_(0)^(upsilon)upsilon dx=-Kint_(0)^(x)x dx` or `[(upsilon^(2))/(2)]_(u)^(upsilon)=-K[(x^(2))/(2)]_(0)^(x)` or `(1)/(2)m upsilon^(2)-(1)/(2)m u^(2)=-(Kx^(2)xxm)/(2)` Thus loss in `K.E., DeltaK=-(Kmx^(2))/(2)` |
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| 1244. |
It requires 20 turns of the stem of a watch to wind the main spring and this stores sufficient energy to keep the watch running for `30 hrts.` If 10 turns are given to thestem, the watch will run forA. `20 h`B. `24 h `C. `7.5 h`D. `10 h` |
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Answer» Correct Answer - C The watch runs on account of potential energy stored by winding the spring . As `P.E. prop x^(2)` and `x` is halved (from 20 turns to 10 turns ) `:. P.E.` becomes `(1)/(4)th.` The watch will run for `(30)/(4)=7.5` hours |
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| 1245. |
A body is moving undirectionally under the influence of a source of constant power. It displacement in time t is proportional to (i) `t^(1//2)` (ii) t (iii) `t^(3//2)` (iv) `t^(2)`A. `t^(1//2)`B. `t`C. `t^(3//2)`D. `t^(2)` |
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Answer» Correct Answer - C `P=Fxxupsilno=(MLT^(2))(LT^(-1))` `=M^(1)L^(2)T^(-3)=` constant `:. L^(2) prop(1)/(T^(-3))` or `L^(2)propT^(3)` or `Lprop T^(3//2)` or `Lprop t^(3//2)` |
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| 1246. |
A bullet fired into a fixed target loses half of its velocity after penetrating 3cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?A. `1.0 cm`B. `1.5 cm`C. `2.0 cm`D. `3.0 cm` |
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Answer» Correct Answer - A Using law of conservation of total energy Case `(i)` , `(1)/(2)m u^(2)=(1)/(2)m ((u)/(2))^(2)+Fxx3` or `(3)/(4)((1)/(2)m u^(2))=Fxx3` …(i) Case `(ii)` `(1)/(2)m u^(2)=FxxS` ....(ii) or `(3)/(4)=(3)/(S)` or`S=4 cm` `:.` Further distance travelled `=4-3=1cm` |
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| 1247. |
The potential energy of a particle of mass 1 kg in a conservative field is given as `U=(3x^(2)y^(2)+6x)` J, where x and y are measured in meter. Initially particle is at (1,1) & at rest then:A. Initial acceleration of particle is `6sqrt(5)ms^(2)`B. Work done to slowly bring the particle to origin is 9 JC. Work done to slowly bring the particle to origin is `-9` JD. If particle is left free it moves in straight line |
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Answer» Correct Answer - A::C |
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| 1248. |
An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because,A. the two magnetic forces are equal and opposite, so they produce no net effectB. the magnetic forces do not work on each particleC. the magnetic forces do equal and opposite (butnon-zero) work on each particleD. the magnetic forces are necessarily negligible |
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Answer» Correct Answer - B (b) When electron and proton are moving under influence of their mutual forces, the magnetic forces will be perpendicular to their motion hence no work is doneby these forces. |
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| 1249. |
An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because,A. the two magnetic forces are equal and opposite, so they produce no net effectB. the magnetic forces do not work on each particleC. the magentic forces do equal and opposite (but non-zero) work on each particleD. the magnetic forces are necessarily negligible |
| Answer» (b) When electron and proton are moving under influence of their mutual forces, the magnetic forces will be perpendicular to their motion hence no work is done by these forces. | |
| 1250. |
A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is.A. constant and equal to mg in magnitude.B. constant and greater than mg in magnitudeC. variable but always greater than mg.D. at first greater than mg, and later becomes equal to mg. |
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Answer» Correct Answer - D In the process of getting straight up and stand from squatting position, the man exerts a variable force (F) on the ground to set his body in motion. This force is in addition to the force required to support his weight (mg). Once the man is in standing position, F becomes zero. |
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