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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.
| 351. |
A spring is compressed between two toy carts to masses `m_1` and `m_2`. When the toy carts are released, the spring exerts on each toy cart equal and opposite forces for the same small time t. If the coefficients of friction `mu` between the ground and the toy carts are equal, then the magnitude of displacements of the toy carts are in the ratioA. (a) `S_1/S_2=m_2/m_1`B. (b) `S_1/S_2=m_1/m_2`C. (c) `S_1/S_2=(m_2/m_1)^2`D. (d) `S_1/S_2=(m_1/m_2)^2` |
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Answer» Correct Answer - C Minimum stopping distance=s Work done against the friction`=W=mumgs` Initial momentum gained by both toy carts will be same because same force acts for same time. Initial kinetic energy of the toy cart `=((p^2)/(2m))` Therefore, `mumgs=(p^2)/(2m)` or `s=((p^2)/(2mugm^2))` For the two toy carts, momentum is numerically the same. Further `mu` and g are the same for the toy carts. So, `s_1/s_2=(m_2/m_1)^2` |
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| 352. |
A block of mass m is released from rest at point A. The compression in spring (force constant k) when the speed of block is maximum is found to be `(nmg cos theta)/(4k)`. What should be the value of n? |
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Answer» Correct Answer - `(4)` Speed of block will be maximum when `mg cos theta=kximpliesx=(mg cos theta)/(K)` Where x is compression. So `n=4` |
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| 353. |
When two bodies collide elastic, then:A. `KE` of the system along is conservedB. only momentum is conservedC. both `KE` and momentum are conservedD. neither `KE` nor momentum is conserved |
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Answer» Correct Answer - C In case of collision of two bodies (whether elastic or inelastic), as the impulsive force acting during collision is interval, hence the total momentum of the system always remain conserved . Further , if in a collision , kinetic energy after collision is equal to kinetic energy before collision. The collision is said to be elastic: otherwise inelastic. |
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| 354. |
An inelestc ball of mass `m` moves at speed a toward mother ineclasits ball of mass `m` at rest They collide and stick togather both moving at speed is nothing else is known about the condition under which the collision take plane which of the fpllowing statement is the most correct?A. Nether total kinetic energy not total linear momenr=tum can be conservedB. This is at clastic collision in which both total kinetic energy and total linear momentum are conserved the fixed speed is `v = u//2`C. This is an inelistic collision and in such collision , total linear momentum is always conserved the final speed is `v = u//4`D. This is an inelistic collision and in which total linear momentum is conserved provided no external force can deliver to the system (of two ball) di=uring the collision |
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Answer» Correct Answer - D Both the balls collide and stick together, so it is perfectly inelastic collision. If no internal impulasive forces are present , then by momentum of conservation `m u + 0 = (m + m) v` `v = (u)/(2)` |
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| 355. |
A block `(B)` is attached to two unstriched sprig `S_(1)` and S_(2) with spring constant `K` and `4K` , respectively (see fig 1) The other ends are atteched in identical support `M_(1)` and `M_(2)` not attached in the walls . The springs and supports have negligible mass . There is no friction anywhere . The block `B` is displaced toword wall `1` by a small distance `z` (figure (ii)) and released . THe block return and moves a maximum displacements `x` and `y` are musured with reoact to the equalibrum of the block `B` and the ratio `y//x` is A. (a) `4`B. (b) `2`C. (c) `1/2`D. (d) `1/4` |
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Answer» Correct Answer - C From energy conservation, `1/2kx^2=1/2(4k)y^2` `:. y/x=1/2` |
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| 356. |
A block of mass m strikes a light pan fitted with a vertical spring after falling through a distance h. If the stiffness of the spring is k, find the maximum compression of the spring. |
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Answer» Let the maximum compression in the spring be x. The reference level for potential energy is assumed at the position of maximum compression. Applying mechanical energy conservation, `DeltaK+DeltaU=0` As block is released from rest and finally comes to rest. Hence, net change in kinetic energy, `DeltaK=0`. Net change in potential energy, `DeltaU=DeltaU_(gr)+DeltaU_(sp)=[-mg(x+h)]+(1/2kx^2)` `0+[-mg(h+x)]+(1/2kx^2-0)=0` `x^2-2((mg)/(k))x-2((mg)/(k))h=0` After solving, we get `x=(mg)/(k)[1+sqrt(1+(2kh)/(mg))]` |
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| 357. |
Two identical balls `A and B` having velocity of `0.5 m//s and -0.3 m//s` respectively collide elastically in one dimension. The velocities of `B and A` after the collision respectively will beA. `-0.3 m//s and 0.5 m//s`B. `0.3 m//s and 0.5 m//s`C. `-0.5 m//s and 0.3 m//s`D. `0.5 m//s and -0.3 m//s` |
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Answer» Since both bodies are identical and collision is elastic. Therefore velocity will be interchanged after collision `v_(A) = - 0.3 m//s and v_(B) = 0.5 m//s` |
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| 358. |
A block of mass m is attached with a massless spring of force constant K. the block is placed over a rough inclined surface for which the coefficient of friction is `mu =3/4` find the minimum value of M required to move the block up the place. (Neglect mass of string and pulley. Ignore friction in pulley). .A. (a) `3/5m`B. (b) `4/5m`C. (c) `6/5m`D. (d) `3/2m` |
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Answer» Correct Answer - C As long as the block of mass m remains staionary, the block of mass M released from rest let it comes down by distance x, applying conservation of mechanical energy `DeltaK+DeltaU=0implies0+(DeltaU_(gravity)+DeltaU_(spri ng))=0` `(-Mgx+1/2kx^2)=0impliesx=(2Mg)/(k)` Thus at the maximum extension in spring is `kx=2Mg` ...(1) For block of mass m to just move up the incline `kx=mg sin theta+mu mg cos theta` ...(2) `2Mg=mgxx3/5+3/4mgxx4/5` or `M=3/5m` |
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| 359. |
A massless platform is kept on a light elastic spring as shown in figure. When a small stone of mass 0.1 kg is dropped on the pan from a height of 0.24 m, the spring compresses by 0.01m. From what height should the stone be droppped to cause a compression of 0.04m in the spring ? A. (a) `0.96m`B. (b) `2.96m`C. (c) `3.96m`D. (d) `0.48m` |
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Answer» Correct Answer - C Let the particle be dropped from a height h and the spring be compressed by y. According to the conservation of mechanical energy, loss in PE of the particle=gain in elastic potential energy of the spring. `mg(h+y)=1/2ky^2` Now, as the particle and spring are same for second case, `(h_1+y_1)/(h_2+y_2)=(y_1/y_2)^2` or `((0.24+0.01)/(h_2+0.04))=((0.01)/(0.04))^2` Solving, we get `h_2=3.96m` |
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| 360. |
A bob of mass M is suspended by a massless string of length L. The horizonta velocity v at position A is just sufficient to make it reach the point B. The angle `theta` at which the speed of the bob is half of that at A, satisfies A. (a) `theta=pi/4`B. (b) `pi/4ltthetaltpi/4`C. (c) `pi/2ltthetalt(3pi)/(4)`D. (d) `(3pi)/(4)ltthetaltpi` |
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Answer» Correct Answer - D `v=sqrt(5gL)` `(v/2)^2=v^2-2ghimpliesh=L(1-costheta)` Solving the three equations, we get `cos theta=7/8` or `theta=cos^-1(-7/8)=151^@` |
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| 361. |
Two blocks `A` and `H`. each of mass `m`, are connected by a massless spring of natural length `I`. and spring constant `K`. The blocks are initially resting in a smooth horizontal floor with the spring at its natural length, as shown in Fig. A third identical block `C`, also of mass `m`, moves on the floor with a speed `v` along the line joining `A` and `B`. and collides elastically with `A`. Then A. (a) The KE of the AB system at maximum compression of the spring is zero.B. (b) The KE of the AB system at maximum compression of the spring is `(1//4)mv^2`.C. (c) The maximum compression of the spring is `vsqrt(m/k)`.D. (d) The maximum compression of the spring is `vsqrt((m)/(2k))` |
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Answer» Correct Answer - B::D Initially, there will be collision between C and A which is elastic. So by conservation of momentum, we have `mv=mv_A+mv_C` `v=v_A+v_C` (i) And as in elastic collision, KE after collision is same as before collision, hence `1/2mv^2=1/2mv_A^2+1/2mv_C^2` i.e., `v^2=v_A^2+v_C^2` Subtracting Eq. (ii) from the square of Eq. (i), we have `2v_Av_C=0` So, either `v_A=0` or `v_C=0` `v_A=0` corresponds to no interaction between A and C, so the only physically possible solution is `v_C=0`, which is the light of Eq. (i) gives `v_A=v`, i.e., after collision C stops and A starts moving with velocity v. Now A will move and compress the spring which in turn accelerates B and retards A and finally both A and B will move with same velocity (say V). In this situation, compression of the spring will be maximum. As external force is zero, momentum of the system `(A+B+spri ng)` is conserved, i.e., `mv=(m+m)VimpliesV=v/2` By conservation of mechanical energy, `1/2mv^2=1/2(m+m)V^2+1/2kx_0^(2_2)` or `mv^2-2m(v_0/2)^2=kx^2` i.e., `k=(mv^2)/(2x_0^2)` or `x_0=vsqrt((m)/(2k))` |
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| 362. |
When two blocks connected by a spring move towards each other under mutual interaction,A. (a) Their velocities are equal and oppositeB. (b) Their accelerations are equal and oppositeC. (c) The forces acting on them are equal and oppositeD. (d) Their momenta are equal and opposite |
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Answer» Correct Answer - C::D When two blocks connected by a spring move towards each other, the law of conservation of momentum leads to the following results: I `vecp_2=-vecp_1` i.e., at any instant two blocks will have equal and opposite momenta. ii. `vecv_2=-(m_1/m_2)vecv_1`, i.e., two blocks move in opposite directions with lighter block moving faster. iii. It is quite clear that forces acting on them will be equal and opposite. Acceleration may be different if masses are different. So options (c) and (d) are correct. |
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| 363. |
A ball of mass `M_(1)` collides elastically and head on with another ball of mass `M_(2)` initially at rest . In which the following cases the transfer of momentum will be maximum?A. `M_(1) = M_(2)`B. `M_(1) gt M_(2)`C. `M_(1) lt M_(2)`D. Data is not sufficient to predict it |
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Answer» Correct Answer - B In this case `M_(1)` stop and its whole momentum is transferred to `M_(2)` |
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| 364. |
Two identical blocks, each having mass M, are placed as shown in figure. These two blocks A and B are smoothly conjugated, so that when another block C of mass m passes from A to B there is no jerk. All the surfaces are frictionless, and all three blocks are free to move. Block C is released from rest, then A. (a) Block C will move for a very small duration.B. (b) Block A will move for a very small duration.C. (c) Block B will acquire maximum speed when C is at the lowest point on B and moving towards leftD. (d) Block B will acquire maximum speed when C is at the topmost point of B |
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Answer» Correct Answer - C When block C is released from rest, it slides down on A, pushing it against the wall (because of a component of normal force between A and C), as on left of A a wall is there, it does not move. The kinetic energy of block C increases and it becomes `mgh` when it is just going to pass on B. When block C comes in contact with B, due to horizontal component of normal force between C and B, B starts moving towards right and kinetic energy of C gets converted into kinetic energy of B and potential energy of C, so block C is not able to reach the topmost point of B. Then block C starts sliding down B, but velocity of B will still be increasing and becomes maximum when C reaches its bottom-most point and is moving left. |
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| 365. |
Two identical blocks A and B are placed on two inclined planes as shown in figure. Neglect resistance and other friction. Read the following statements and choose options. Statement I: The kinetic energy of A on sliding to J will be greater than the kinetic of B on sliding to O. Statement II: The acceleration of A will be greater than acceleration of B when both are released on the inclined plane. Statement III: The work done by external agent to move the block slowly from position B to O is negative.A. (a) Only statement I is trueB. (b) Only statement II is trueC. (c) Only I and III are trueD. (d) Only II and III are true |
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Answer» Correct Answer - D Statement I: Work done by gravity is same for motion from A to J and B to M for equal mass. So KE will be equal. Statement II: Acceleration `=gsin theta` `sin theta_Agt sin theta_B=h/lgt(h)/(2l)` Statement III: `W_g+W_(ext)=0` (Because to block moves slowly) `W_(ext)=-W_g` From B to O: `W_g` is positive, so `W_(ext)lt0`. |
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| 366. |
A small mass slides down an inclined plane of inclination `theta` with the horizontal. The co-efficient of friction is `mu=mu_(0)` x where x is the distance through which the mass slides down and `mu_(0)` a constant. Then, the distance covered by the mass before it stops isA. `(2)/(mu_(0)) tan theta`B. `(4)/(mu_(0)) tan theta`C. `(1)/(2mu_(0))tan theta`D. `(1)/(mu_(0)) tan theta` |
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Answer» Correct Answer - A |
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| 367. |
A block of mass `m` is moving with a speed `v` on a horizontal rought surface and collides with a horizontal monted spring of spring constant `k` as shown in the figure .The coefficient of friction between the block and the floor is `mu` The maximum cobnpression of the spring is A. `- (mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))`B. ` (mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))`C. `- (mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) - mkv^(2))`D. `(mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))` |
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Answer» Correct Answer - A In pressent of friction both the spring force and the frictional act so as to oppose the compression of the spring Work done by the net force `W = -(1)/(2) kx_(m)^(2) - mu mhs_(m)` where `x_(m)` is the maximum compression of the spring change in kinetic energy `Delta K = K_(1) - K_(1) = 0 - (1)/(2) mv^(2)` Accerding to worik energy therem `W = Delta K` `- (1)/(2) kx_(m)^(2) - mu mgx_(m) = -(1)/(2)mv^(2)` ` kx_(m)^(2) + mu mgx_(m) = (1)/(2)mv^(2)` ` kx_(m)^(2) + 2mu mgx_(m) = -mv^(2)= 0` ` x_(m)^(2) +(2 mu mgx_(m))/(k) - (mv^(2))/(k) = 0` it is a quadratic equation in `x_(m)` Solving this equation for `x_(m)` is positive we get `x_(m) = -(mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))` |
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| 368. |
There are two inclined planes with the same vertical height, but with different angles of inclination. Two bodies of mass m each are placed at the bottom of the inclined plane. Two external forces of equal magnitude do work in taking the bodies to the top of the inclined plane. Which force does more work? |
| Answer» The force that does more work is the force that acts on the body placed on the bottom of the inclined plane with lesser angle of inclination. As the angle of inclination is less, the body will travel more distance. The frictional force will be more as the `cos theta` component will be high. Work done is force multiplied by the displacement, and in this case, the distance as well as the frictional force is hight. The force acting on the body, which is at the bottom of the inclined plane with lesser angle of inclination, does more work. | |
| 369. |
A stationary partical explodes into two partical of a masses `m_(1) and m_(2)` which move in opposite direction with velocities `v_(1) and v_(2)`. The ratio of their kinetic energies `E_(1)//E_(2)` isA. 1B. `(m_(1)v_(1))/(m_(2)v_(2))`C. `(m_(2))/(m_(1))`D. `(m_(1))/(m_(2))` |
| Answer» Correct Answer - c | |
| 370. |
A partical is realeased from the top of two inclined rought surface of height `h` each. The angle of inclination of the two planes are `30^(@)` and `60^(@)` respectively. All other factors (e.g. coefficient of friction , mass of the block etc) are same in both the cases. Let `K_(1) and K_(2)` be the kinetic energy of the partical at the bottom of the plane in two cases. ThenA. `K_(1)= K_(2)`B. `K_(1) gt K_(2)`C. `K_(1) lt K_(2)`D. Data insufficient |
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Answer» Correct Answer - C |
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| 371. |
A partical is realeased from the top of two inclined rought surface of height `h` each. The angle of inclination of the two planes are `30^(@)` and `60^(@)` respectively. All other factors (e.g. coefficient of friction , mass of the block etc) are same in both the cases. Let `K_(1) and K_(2)` be the kinetic energy of the partical at the bottom of the plane in two cases. ThenA. (a) `K_1=K_2`B. (b) `K_1gtK_2`C. (c) `K_1ltK_2`D. (d) Data insufficient |
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Answer» Correct Answer - C Work done by friction: `W=(mumgcostheta)S=(mumgcostheta)(h)/(sin theta)=mumghcottheta` Now `cot theta_1=cot30^@=sqrt3` `cot theta_2=cot 60^@=1/sqrt3` i.e., kinetic energy `(KE=mgh-W)` in first case will be less or `K_1ltK_2`. |
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| 372. |
A block is released from the top of a smooth inclined plane of inclination `theta` as shown in figure. Let `upsilon` be the speed of the particle after travelling a distance s down the plane. Then which of the following will remain constant ? A. `v^(2)+2gs sintheta`B. `v^(2)-2gs sintheta`C. `v- sqrt(2gs) sintheta`D. `v+ sqrt(2gs) sintheta` |
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Answer» Correct Answer - B |
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| 373. |
A particle of mass `m_(1)` is fastened to one end of a string and one of `m_(2)` to the middle point, the other end of the string being fastened to a fixed point on a smooth horizontal table The particles are then projected, so that the two portions of the string are always in the same straight line and describes horizontal circles find the ratio of tensions in the two parts of the string A. `m_(1)/(m_(1)+m_(2))`B. `(m_(1)+m_(2))/m_(1)`C. `(2m_(1)+m_(2))/(2m_(1))`D. `(2m_(1))/(m_(1)+m_(2))` |
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Answer» Correct Answer - C |
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| 374. |
A partical is realeased from the top of two inclined rought surface of height `h` each. The angle of inclination of the two planes are `30^(@)` and `60^(@)` respectively. All other factors (e.g. coefficient of friction , mass of the block etc) are same in both the cases. Let `K_(1) and K_(2)` be the kinetic energy of the partical at the bottom of the plane in two cases. ThenA. `K_(1)=K_(2)`B. `K_(1) gt K_(2)`C. `K_(1) lt K_(2)`D. Data insufficient |
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Answer» Correct Answer - C |
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| 375. |
A stone at the end of `1m` long string is whirled in a vertical circle at a constant speed of `4ms^-1`. The tension in the string is `6N` when the stone isA. (a) `10ms^-1`B. (b) `5sqrt3ms^-1`C. (c) `10sqrt3ms^-1`D. (d) `20ms^-1` |
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Answer» Correct Answer - A `m=1kgimpliesL=10/3m` `(T_(max))/(T_(min))=4` (i) and `V_L=sqrt(V_H^2+4gL)` Tension at highest point, `T_(min)=(mv_H^2)/(K)-mg` Tension at lowest point, `T_(max)=mv_L^2//L+mg=(m(v_H^2+4gL))/(L)+mg` Now Eq. (i) can be written as `m[(v_H^2+4gL)/(L)+g]=4xxm((v_H^2)/(L)-g)` `implies v_H=sqrt(3gL)=10ms^-1` |
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| 376. |
A string with one end fixed on a rigid wall, passing over a fixed frictionless pulley at a distance of `2m` from the wall, has a point mass M of `2kg` attached to it at a distance of `1m` from the wall. A mass `m` of `0.5kg` is attached to the free end. The system is initially held at rest so that the stirng is horizontal between wall and pulley and vertical beyond the pulley as shown in figure. What will be the speed with which point mass M will hit the wall when the system is released? `(g=10ms^-2)` |
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Answer» Correct Answer - C |
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| 377. |
A constant force `vecF=(3hati+2hatj+2hatk)` N acts on a particle displacing it from a position `vec(r_1)=(-hati+hatj-2hatk)m` to a new position `vecr_2=(hati-hatj+3hatk)m`. Find the work done by the force. |
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Answer» The displacement vectors, `vecs=vecr_2-vecr_1` `vecs=(1+1)hati+(-1-1)hatj+(3+2)hatk=2hati-2hatj+5hatk` From `W=vecF.vecs`, we have `W=(3hati+2hatj+2hatk)*(2hati-2hatj+5hatk)=6-4+10=12J` |
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| 378. |
Find the power of a person who can chew `30g` of ice in one minute. Latent heat of ice `=80cal//gm`, `J=4.2`joule`//`cal. |
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Answer» Mass of ice chewe, `m=30g` time taken,`=1min=60s.` Latent heat of ice, `L=80cal//g` Heat produced `=mL=30xx80=2400cal`. `=2400xx4.2J.` Power `=("work")/("time")=(2400xx4.2)/(60)"watt"=168 "watt"` |
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| 379. |
A machine which is `75%` efficient uses 12 J of energy in lifting up a 2 kg mass through a cerfain distance. The mass is then allowed to fall through that distance. The velocity at the end of its fall is (in m/s)A. `sqrt(24)`B. `sqrt(18)`C. 6D. `sqrt9` |
| Answer» Correct Answer - D | |
| 380. |
A running man has half the KE that a body of half his mass has. The man speeds up by `1.0ms^(-1)` and then has the same energy as the boy. What were the original speeds of the man and the boy?A. `sqrt(2)m//s`B. `(sqrt(2)-1)m//s`C. `(1)/((sqrt(2)-1))m//s`D. `(1)/(sqrt(2))m//s` |
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Answer» Correct Answer - C |
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| 381. |
A running man has half the KE that a body of half his mass has. The man speeds up by `1.0ms^(-1)` and then has the same energy as the boy. What were the original speeds of the man and the boy?A. `(sqrt2+1),(sqrt2-1)`B. `(sqrt2+1),2(sqrt2+1)`C. `sqrt2,sqrt2`D. `(sqrt2+1),2(sqrt2-1)` |
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Answer» Correct Answer - B (b) According to the question, `(1)/(2) mv_(1)^(2)=(1)/(2)[(1)/(2)xx(m)/(2)xxv_(2)^(2)]` where, `v_(1)="speed of man and "v_(2)`=speed of boy Now, `(1)/(2)m(v_(1)+1)^(2)=(1)/(2)xx(m)/(2)xxv_(2)^(2)` Solving these two equations we get, `v_(1)=(sqrt2+1)ms^(-1)` and `v_(2)=2(sqrt2+1)ms^(-1)` |
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| 382. |
A running man has half the KE that a body of half his mass has. The man speeds up by `1.0ms^(-1)` and then has the same energy as the boy. What were the original speeds of the man and the boy?A. `2.4, 4.8 ms^(-1)`B. `2.4, 3.4 ms^(-1)`C. `3.4, 4.8 ms^(-1)`D. `3.4, 6.8 ms^(-1)` |
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Answer» Let M = mass of man , m= mass of boy V = speed of man , v = speed of boy Given `(1)/(2) MV^(2) = (1)/(2) ((1)/(2)mv^(2))`, As `m = (M)/(2)` So `(1)/(2) MV^(2) = (1)/(2) ((1)/(2) xx (M)/(2) mv^(2))` Hence `v^(2) = 4V^(2)` or `v = 2V` When the man speeds up by `1 m//s`, then we get `(1)/(2) M (V + 1)^(2) = (1)/(2)mv^(2) = (1)/(2) (M)/(2) (4V^(2))` or `(V + 1)^(2) = 2V^(2)` or `V^(2) - 2V - 1 = 0` Solving we get `V = 2.4 ms^(-1)` and` v = 2V = 4.8 ms^(-1)` |
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| 383. |
Assertion: In a elastic collision `(e = 1)` between two bodies, conservation of kinetic energy holds true. i.e., `(K_(1) + K_(2))_(i) = (K_(1) + K_(2))_(f)`. Reason: Conservation of momentum holds true i.e., `(P_(1) + P_(2))_(i) = (P_(1) + P_(2))_(f)`.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason are false. |
| Answer» In elastic collisions also conservation of momentum holds true. | |
| 384. |
Assertion: In an elasticcollision of two billard balls, the total `KE` is conservation during the short times of collision of the balls`(i.e., when they are in constant). Reason: Energy spend against friction does not follow the law of conservation of energy.A. If both assertion and reason are true and reason is a true explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion is and reason are false. |
| Answer» Since the balls are in constant in the deformed state, their total energy is in the form of potential energy and kinetic energy, so kinetic energy is less than the total energy. The energy spend against friction is dissipated in the form of heat which is not available for doing work. | |
| 385. |
A ball of mass `m` moving with velocity `vecu = u_(x) hati + u_(y) hatj` hits a vertical wall of infinite mass as shown in the figure. The ball slips up along the wall for the duration of collision and there is friction between the ball and the wall. Neglect the effect of gravity. Pick up the correct alternative. A. The wall provides the ball with net impulse along the negative x-axis for the duration of collision.B. The collision change only the x-component of velocity of thr ballC. The collision change only the y-component of velocity of thr ballD. The impulse provides by friction force to the ball for the duration of collision cannot be neglected in comparison to impulse provided by normal reaction |
| Answer» The wall exterts an impulsive normal reaction `N` on the ball along negative x-direction. Hence wall also exerts a friction force `mu N` on the ball in negative y-direction. Hence both `x and y` components of velocities are changed. Hence `d` is only correct choose. | |
| 386. |
A ball hits the floor and rebounds after an inelastic collision. In this caseA. The momentum of the ball just after the collision is the same as that just before the collisionB. The mechanical energy of the ball remains the same in the collisionC. The total momentum of the ball and the earth is conservedD. The total energy of the ball and the earth is conserved |
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Answer» Correct Answer - C |
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| 387. |
Assertion:The principal of conservation of energy is valid for inelastic collision. Reason: The principal of conservation of energy holds good in both elastic and inelastic collision. In case of inelastic collision kinetic energy before and after collision is not same.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason are false. |
| Answer» In case of inelastic collision , `KE` is not conserved but conservation of energy holds good always. | |
| 388. |
Assertion: In a perfectly inelastic collision in the absence of external forces , the kinetic energy is never conserved. Reason: The objects deformed and stick together in this type of collision.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If assertion and reason are false. |
| Answer» In a particaly inelastic collision the object in the obsence of external forces, get deformed and stick together hence the kinetic energy is never conserved. | |
| 389. |
A block is mass `m` is dropped from the fourth of an office building and hits the sidewalk below at speed `v` . From what floor should the block the dropped to double that impact speed?A. the either floorB. the thenth floorC. the twelfth floorD. the sixteenth floor |
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Answer» Correct Answer - D As the back falls freely, only the conservative gravitation force acts on it. Therefore, mechainical energy is conservate, or `KE_(f) = KE_(f) + PE_( C)` Assuming that the block is releasedfrom rest `(KE_(i) = 0)` and taking `y = 0` at ground level `(PE_(f) = 0)` we have that `KE_(f) = PE_(i)` or `(1)/(2) mv_(1)^(2)` = mgy and `y_(i) = (v_(f)^(2))/(2g)` Thus, to double the final speed, it is necessary to increase the initial height by a factor of four. |
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| 390. |
A bullet having a speed of `100 m//see` crashes through a plank of wood. After passing through a plank , its speed is `80 m//s`. Another bullet of the same mass and size , but traveling at `80 m//s` is fired at the plank . The speed of the second bullet after traveling through the plank is (Assume that resistance of the plank is independent of the speed of the bullet) :A. `10 sqrt(7) ms^(-1)`B. `20 sqrt(7) ms^(-1)`C. `30 sqrt(7) ms^(-1)`D. `20 sqrt(7) ms^(-1)` |
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Answer» Suppose `F` be the resistance force offered by the plank . Let thickness of plank be `x`. For first case: `Fx = (1)/(2) m [(100)^(2) - (80)^(2)]` (i) For second case, let `V` be the final velocity of bullet, then `Fx = (1)/(2) m [(80)^(2) - V^(2)]` (ii) From (i) and (ii) `80^(2) - V^(2) = 100^(2) - 80^(2)` or `V = 20 sqrt(7) m//s` |
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| 391. |
A bomb of mass `3.0 kg` explodes in air into two pieces of masses `2.0 kg` and `1.0 kg`. The smaller mass goes at a speed of `80 m//s`. The total energy imparted to the two fragments is :A. `1.07 kJ`B. `2.14 kJ`C. `2.4 kJ`D. `4.8 kJ` |
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Answer» Both fragments will posses the equal linear momentum `m_(1) v_(1) = m_(2) v_(2) implies 1 xx 80 = 2 xx v_(2) implies v_(2) = 40 m//s` `:.` Total energy of system ` = (1)/(2) m_(1) v_(1) + (1)/(2) m_(2) v_(2)^(2)` `= (1)/(2) xx 1 xx (80)^(2) + (1)/(2) xx 2 xx (40)^(2)` `= 4800 J = 4.8 kJ` |
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| 392. |
A body is allowed to fall on the ground from a height `h_(1)`. If it is rebound to a height `h_(2)` then the coefficient of restitution is:A. `(h_(2))/(h_(1))`B. `sqrt((h_(2))/(h_(1)))`C. `(h_(1))/(h_(2))`D. `sqrt((h_(1))/(h_(2)))` |
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Answer» Correct Answer - A `u_(1) = - sqrt(2 gh). V_(2) = sqrt(2 gh)`. Since `u_(2) = v_(2) = 0` `e = (-v_(1))/(u_(1)) = sqrt((h_(2))/(h_(1)))` |
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| 393. |
A ball hits a floor and rebounds after an inelastic collision. In this caseA. The momentum of the ball just after the collision is the same as that justbefore the collisionB. The mechanical energy of the ball remain the same collisionC. The total momentum of the ball and the earth is conservedD. The total energy of the ball and the earth is conserved |
| Answer» By the conservation of momentum in the absence of external force total momentum of the system (ball + earth) remain constant. | |
| 394. |
A body of mass `m` issuspended from a massless spring of natural length `l` It stretches the spring through a verticle distance `y`.The potential energy of the stretched spring isA. `mg(l +y)`B. `(1)/(2)mg(l +y)`C. `(1)/(2)mg y`D. `Mgy` |
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Answer» Correct Answer - C `ky = mg or k = (mg)/(y)` `U = (1)/(2)((mg)/(y))y^(2) = (1)/(2)mg y` |
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| 395. |
A aprtical moves on a rough horizontal ground with same initial velocity say `v_(0)`. If `(3//4)th` of its kinetic energy is lost in friction in time `t_(0)` , then coefficient of friction between the partical and the ground is:A. `(v_(0))/(2 gl_(0))`B. `(v_(0))/(4 gl_(0))`C. `(3 v_(0))/(4 gl_(0))`D. `(v_(0))/(gl_(0))` |
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Answer» `3//4^(th)` energy of lost, i.e., `1//4^(th)` kinetic energy is left . Hence, its velocity becomes `v_(0)//2` under a retardation of mmg in time `t_(0)` `(v_(0))/(2) = v_(0) - mu g t_(0)` or `mu = (v_(0))/(2g t_(0))` |
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| 396. |
The force constant of a weightless spring is `16 N m^(-1)`. A body of mass `1.0 kg` suspended from it is pulled down through `5 cm` and then released. The maximum energy of the sysytem (spring + body) will beA. `2 xx 10^(2) J`B. `4 xx 10^(2) J`C. `8 xx 10^(2) J`D. `16 xx 10^(2) J` |
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Answer» Correct Answer - A Maximum kinetic energy = maximum elastic potential energy . `= (1)/(2) xx 16 xx ((5)/(100))^(2)J` `= (8 xx 25)/(100 xx 100) J = 2 xx 10^(2)J` |
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| 397. |
A ball strickes a horizontal floor at `45^(@). 25%` of its kinetic energy is lost in collision. Find the coefficient of restitution.A. `(1)/(2)`B. `(1)/(sqrt2)`C. `(1)/(2 sqrt2)`D. `(1)/(4)` |
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Answer» Correct Answer - C Fraction of `KE` lost in collision `Delta K % = Delta K % = ((1)/(2) mu^(2) - (1)/(2) mv^(2))/((1)/(2) m u^(2)) = 1 - ((v)/(u))^(2) = (1)/(4)` (given) `v = y sqrt((3)/(4)` The ball strikes at `45^(@)` will hot change while component of velocity normal to wall will change. `v_(x) = u cos 45^(@) = (u)/(sqrt(2)` `v_(y) = eu cos 45^(@) = (eu)/(sqrt(2)` `v = sqrt(v_(x)^(2) + v_(y)^(2)) = [((u)/(sqrt2))^(2) + ((eu)/(sqrt2))^(2)])^(1)/(2)` `implies v = u [(1)/(2) + (e^(2))/(2)]^(1)/(2)` Solving (i) and (ii), we get `e = (1)/(sqrt2)`. |
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| 398. |
A block mass `m = 2 kg` is moving with velocity`v_(0)` towards a massless unstretched spring of the force constant `k = 10 N//m`. Coefficient of friction between the block and the ground is `mu = 0.2`. Find the maximum value of `v_(0)` so that after pressing the spring the block does not return back but stops there permanently. A. `2 sqrt((2)/(5)) m//s`B. `sqrt((1)/(5)) m//s`C. `4 sqrt((1)/(5)) m//s`D. `4 sqrt((2)/(5)) m//s` |
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Answer» Correct Answer - B Block will stop there if `kx = mu mg implies x = (mu mg)/(k)` ` implies x = (0.2 xx 2 xx 10)/(10) = (2)/(5) m` Now using energy conservation `(mv^(2))/(2) = (kx^(2))/(2) + mu mg (x+1) implies v = 4 sqrt((2)/(5)) m//s` |
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| 399. |
The pulleys and strings shown in the figure are smooth and of negligible mass. For the system to remain in equilibrium, the angle `theta` should be |
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Answer» Correct Answer - (a) `U=-2 mgL +mgd((2-cos theta))/(sin theta))` (b) `theta=60^(@)` (c) stable |
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| 400. |
A man weighing `60kg` `f` suports a body of `20kg f ` on his head. Calculate work done by him in moving a distance of `15m` up an incline of `1` in 10 . Take `g=9.8m//s^(2)` |
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Answer» Here, `m=60kg+20kg=80kg` `W=?s=15m, sintheta=(1)/(10),g=9.8ms^(2)` `W=(mgsintheta)s=80xx9.8xx(1)/(10)xx15` |
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