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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.
| 1051. |
Two particles `1` and `2` are allowed to descend on the two frictionless chord `OA` and `OB` of a vertical circle, at the same instant from point `O`. The ratio of the velocities of the particles `1` and `2` respectively, when they reach on the circumference will be (OB is the diameter).A. `1/4`B. `1/2`C. 1D. `1/(2sqrt(2))` |
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Answer» Correct Answer - B |
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| 1052. |
An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle `alpha` with the vertical, the maximum possible value of `alpha` is given by A. `cot alpha=3`B. `tan alpha=3`C. `sec alpha=3`D. `cosec alpha = 3` |
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Answer» Correct Answer - A |
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| 1053. |
If the magnitudes of two vectors are 2 and 3 and magnitude of their scalar product is `2sqrt3` what is the angle between the vectors ? |
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Answer» `A=2B=3 " "barAbarB=2sqrt3` `costheta(barAbarB)/(AB)` `costheta=(vecA.vecB)/(AB)=(2sqrt3)/(2xx3)=(1)/(sqrt3)` `theta=cos^(-1)((1)/(sqrt3))` |
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| 1054. |
Find the angle between the force `vecF(5hati+4hatj-3hatk)`units and displacement `d=(3hatk+4hatj+5hatk)` unit. Also find the projection of `becF` on d |
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Answer» `vecFvecd=F_(x)d_(x)+F_(y)d_(y)+F_(z)d_(z)` `vecF.vecF=F^(2)=F_(x)^(2)+F_(z)^(2)" "vecd.vecd=d_(x)^(2)+d_(y)^(2)+d_(z)^(2)` `costheta(vecFvecd)/(Fd)` Projecto of `vecF on vecd=Fcostheta` `vecF.vecd=F_(x)d_(x)+F_(y)d_(y)+F_(z)d_(z)=5(3)+4(4)+(-3)(5)=16` unit `vecF.vecF=F^(2)+F_(x)^(2)+F_(y)^(2)+F_(z)^(2)=5^(2)+4^(2)+(-3)^(2)=50` unit `vecd.vecd=d^(2)=d_(x)^(2)+d_(y)^(2)+d_(z)^(2)=3^(2)+4^(2)+5^(2)=50` unit `cos theta=(vecF.vecd)/(Fd)=(16)/(sqrt50sqrt50)=0.32` `theta=cos^(-1)(0.32)` Projection of `vecFon vecd=F cos theta=sqrt50xx0.32=2.26` |
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| 1055. |
Find the angle between force `vecF=(5hati+5hatj+5hatk)` unit and displacement `vecd=(3hati+4hatj-3hatk)` unit Also find the projection of `vecF on vecd.` |
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Answer» `vecF.vecd=F_(x)d_(x)+F_(y)d_(y)+F_(z)d_(z)` `=5(3)+4(4)+5(-3)=16` unit `vecF.vecF=F^(2)=F_(x)^(2)+F_(x)^(2)+F_(z)^(2)=5^(2)+4^(2)+5^(2)=66` unit `vecF.vecd=Fd cos theta` `thereforecos theta=(vecF.vecd)/(Fd)=(16)/(sqrt(66)sqrt(34))=(16)/(47.37)=0.34` `therefore0=cos^(-1)(0.34)` Projection of `vecFon vecd=Fcostheta=sqrt(66)xx0.34=2.76` |
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| 1056. |
An engine exerts a force `F=(20hati-3hatj+5hatk)` N and moves with velocity `v=(6hati+20hatj-3hatk) ms^(-1)`. The power of the engine (in watt) isA. 45B. 75C. 20D. 10 |
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Answer» Correct Answer - A (a) Power of the engine, `P=F.v=(20hati-3hatj+5hatk).(6hati+20hatj-3hatk)` =120-60-15=45 W |
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| 1057. |
A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force. (Consider `g=10m//s^2`).A. ` 22N`B. `4 N`C. `16 N`D. `20 N` |
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Answer» Correct Answer - D Work done by hand `=` max. P.E. of the ball `Fxxs=mgh` `F=(mgh)/(s)=(0.2xx10xx2)/(0.2)=20N` |
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| 1058. |
Work done by a conservative force in bringing a body from infinity to A is 60 J and to B is 20J. What is the difference in potential energy between point A and B, i.e. `U_(B)-U_(A)`. |
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Answer» Correct Answer - D `U_(A) =-60J` `U_(B) =-20J` ` :. U_(B)-U_(A) =40J` |
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| 1059. |
At two positons kinetic energy and potential energy of a particle are `K_(1) =10J: U_(1) =-20J, K_(2) = 20 J, U_(2)=-10 J`. In moving from 1 to 2 .A. work done by conservative forces is positive.B. work done by conservative forces is negative.C. work done by all the forces is positive .D. work done by all the forces is negative. |
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Answer» Correct Answer - B::C Work done by conservative frtces `=U_(i) -U_(f)=-20+10=-10 J` Work done by all the forces, `K_(f)-K_(i)=20-10 =10 J` |
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| 1060. |
Potential energy of a body in position A is-40J. Work done by conservative force in moving the body from A to B is`-20J`. Find potential energy of the body in position B. |
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Answer» Correct Answer - B Work done by a conservtive force is given by `W_(ArarrB)=-DeltaU=-(U_(B)-U_A)=U_(A)-U_(B)` `rArr` `U_(B)=U_(A)-W_(ArarrB)=(-40)-(-20)=-20J`. |
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| 1061. |
The potential energy of a conservative force field is given by `U=ax^(2)-bx` where, a and b are positive constants. Find the equilibrium position and discuss whether the equilibrium is stable, unstable or neutral. |
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Answer» `U=alphax^2-betax` `(dU)/(dx)=2alphax-beta` For equilibrium,`F=-(dU)/(dx)=0` `2alphax=betaimpliesx=(beta)/(2alpha)` `(d^2U)/(dx^2)=2alpha` (positive), i.e., `U` is minimum. At `x=(beta)/(2alpha)`, equilibrium is stable. |
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| 1062. |
The potential energy function for a particle executing simple harmonic motion is given by `V(x)=(1)/(2)kx^(2)`, where k is the force constant of the oscillatore. For `k=(1)/(2)Nm^(-1)`, show that a particle of total energy 1 joule moving under this potential must turn back when it reaches `x=+-2m.` |
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Answer» At any instant, the total energy of an oscillator is the sum of K.E. and P.E. i.e. `E=K.E.+P.E.=(1)/(2)m u^(2)+(1)/(2)kx^(2)` The particle turns back at the instant, when its velocity becomes zero i.e. `u=0` . `:. E=0+(1)/(2)kx^(2)` As `E=1` joule and `k=(1)/(2)N//m` `:. 1=(1)/(2)xx(1)/(2)x^(2)` or `x^(2)=4, x=+-2m` |
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| 1063. |
Point out the correct alternative `:` (a) When a conservative force does positive work on a body , the potential energy of the body increases `//` decreases `//` remains unaltered. (b) Work done by a body against friction always result in a loss of its kinetic `//` potential energy. (c) The rate of change of total momentum of a many particle system is proportional to the external force `//` sum of the internal forces of the system. (d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy `//` total linear momentum `//` total energy of the system ot two bodies. |
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Answer» (a) Potential energy of the body decrease. The conservative force does positive work on a body, when it displaces the body in the direction of force. The obdy, therefore, approaches the centre of force, decreasing x. Hence, P.E. decreases. (b) Work is done by a body against friction at the expense of its kinetic energy. Hence K.E. of the body decreases. (c) Internal forces cannot change the total or net momentum of a system. Hence the rate of change of total momentum of many particle system system is proportional to the external force on the system. (d) In an inelastic collision of two bodies, the quantities which do not change after the collision are total linear momentum and total energy of the systme of two bodies. The total K.E. changes as some energy appears is other forms. |
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| 1064. |
A body of mass 2 kg makes an elastic collision with another body at rest and contiues to move in the original direction with a speed equal to one thrid of its original speed. Find the mass of the second body. |
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Answer» Here, `m_(1)=2kg, u_(1)=upsilon,m_(2)=? u_(2)=0,` `upsilon_(1)=upsilon//2` As `upsilon_(1)=((m_(1)-m_(2))u_(1)+2m_(2)u_(2))/((m_(1)+m_(2)))` `:. (upsilon)/(2)=((2-m_(2))upsilon+2m_(2)xx0)/(2+m_(2))` or `2+m_(2)=2(2-m_(2))=4-2m_(2)` `3m_(2)=4-2=2` `m_(2)=(2)/(3)=0.67kg` |
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| 1065. |
A bag of mass `M` hangs by a long massless rope. A bullet of mass in, moving horizontally with velocity `u`, is caught in the bag. Then for the combined (bag `+` bullet) system, just after collisionA. Momentum is `(mvM)/(M+m)`B. kinetic energy is `(mv^(2))/(2)`C. Momentum is `(mv(M+m))/(M)`D. kinetic energy is `(m^(2)v^(2))/(2(M+m))` |
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Answer» Correct Answer - D |
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| 1066. |
When a `1.0kg` mass hangs attached to a spring of length `50 cm,` the spring stretches by `2 cm.` The mass is pulled down until the length of the spring becomes `60 cm.` What is the amount of elastic energy stored in the spring in this condition. if `g = 10 m//s^(2).`A. 1.5 jouleB. 2.0 jouleC. 2.5 jouleD. 3.0 joule |
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Answer» Correct Answer - C |
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| 1067. |
A bullet leaving the muzzle of a rifle barrel with a velocity v penetrates a plank and loses one-fifth of its velocity. It then strikes second plank, which it just penetrates through. Find the ratio of the thickness of the planks, supposing the average resistance to the penetration is same in both the cases. |
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Answer» Let R be the resistance force offered by the planks, `t_1` be the thickness of first plank, and `t_2` be the thickness of second plank. For first plank: Loss in KE=work against resistance `1/2mv^2-1/2m(4/5v)^2=Rtimplies1/2mv^2(9/25)=Rt_1` (i) For second plank: `1/2m(4/5v)^2-0=Rt_2implies1/2mv^2(16/25)=Rt_2` (ii) Dividing Eq. (i) by Eq. (ii), we get `t_1/t_2=9/16` |
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| 1068. |
A ball hits a floor and rebounds after an inelastic collision. In this caseA. the total energy of the ball and the earth remains the sameB. the total momentum of the ball and the earth is conservedC. the momentum of the ball just after collision is same as that just before the collisionD. the mechanical energy of the ball remains the same during the collision. |
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Answer» Correct Answer - A::B A the collision is inelastic, body loses some energy, so that KE of ball does not remain the same. However, total energy and total momentum ball and earth remain the same. Choice `(a)` and `(b)` are correct. |
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| 1069. |
A ball dropped from a height of 2 m rebounds to a height of 1.5 m after hitting the ground. Then the percentage of energy lost isA. 25B. 30C. 50D. 100 |
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Answer» Correct Answer - A `U_1=mg(2)`,`U_2=mg(1.5)` `(U_2)/(U_1)=(1.5)/(2)=(3)/(4)` `%` change `=((U_2)/(U_1)-1)xx100=-2.5%`, i.e, `25%` loss |
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| 1070. |
A plastice ball is dropped from a height of 1m and rebounds several times from the floor. IF1.3 s elaspes from from the moment it is dropped to the second impact with the floor, what is the coefficient of restitution? |
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Answer» Here, `h=1m, u=0.` Velocity with which the ball strikes the floor for the first time, `v_(1)=sqrt(2gh)=sqrt(2xx9.8xx1)=4.427m//s` Time taken by the ball to strike the floor for the first time, `t_(1)=sqrt((2h)/(g))=sqrt((2xx1)/(9.8))=0.452s` Time taken by the ball between first and second impact with the floor is `t_(2)=1.3-t_(1)=1.3-0.452=0.848s. ` Time taken by the ball for upward journey after first imapct impact `=(t_(2))/(2)=(0.848)/(2)=0.424s` `:.` upward velocity after first impact `v_(2)=gxx(t_(2))/(2)=9.8xx0424=4.155m//s` `:` Coefficient of restitution. `e=(v_(2))/(v_(1))=(4.155)/(4.427)=0.939` |
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| 1071. |
20 J work is required to stretch a spring through 0.1 m. Find the force constant of the spring. If the spring is further stretched through 0.1 m. Calculate work done. |
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Answer» P.E. of spring when stretched through a distance 0·1 m, U = W.D. = 1/2 Kx2 = 20 J or K = 4000 N/m when spring is further stretched through 0·1 m, then P.E. will be : U' = 1/2k(0.2)2 = 80 J ∴ W.D. = U′ – U = 80 – 20 = 60 J. |
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| 1072. |
A body is moving along a circular path. How much work is done by the centripetal force ? |
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Answer» The object moves in a circular path. In this case the centripetal force acts on the object at right angles to the direction of motion of object. So, the work done by the object in circular path is zero. W = FS cos 90º = 0 |
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| 1073. |
A body constrained to move in z direction is subjected to a force given by `vecF=(3hatj-10hatj+5hatk)N.` What is the work done by this force in moving the body through a distance of 5 m along z-axis?A. 15 JB. `-15J`C. `-50J`D. `25J` |
| Answer» Correct Answer - D | |
| 1074. |
An agent applies force of constant magnitude `F_(0)` always in the tangential direction as shown in the figure`-3.77`. Find the speed of the bob when string becomes horizontal, assuming that it is at rest at its lowest point: A. `sqrt((l)/(m)(piF_(0)-2mg))`B. `sqrt(lg)`C. `sqrt((l)/(m)(piF_(0))-4mg)`D. `sqrt((l)/(m)F_(0))` |
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Answer» Correct Answer - A |
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| 1075. |
A small block of mass `M` move on a frictionless surface of an inclimed from as down is figure . The engle of the inclime suddenly change from `60^(@)` to `30^(@)` at point `B` . The block is initally at rest at `A` Assume the collsion between the block and the incline are totally inclassic `(g = 10m//s^(2)`) The speed of the block at point `C` immediately before is leaves the second incline isA. `sqrt(120) m//s`B. `sqrt(105) m//s`C. `sqrt(90) m//s`D. `sqrt(75) m//s` |
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Answer» Correct Answer - B In `delta BCE, tan 30^(@) = (BE)/(CE) rArr (1)/(sqrt(3) = (BE)/(3sqrt(3) rArr BE = 3m ` Appling mechanical energy conservation Machanical energy at `B = ` Machinical energy is `C` `(1)/(2) M(sqrt(45)^(2) + M xx 10 xx 3 = (1)/(2) Mv_(C)^(2)` `45 + 60 = v_(C)^(2) :. v_(2) = sqrt(105)m//s ` |
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| 1076. |
A `3.00 kg` object has a velocity `(6.00 hati - 2.00 hatj) m//s`. What is the net work done on the object if its velocity changes to `(8.00 hati - 4.00 hatj) m//s`?A. `64.5 J`B. `64.5 J`C. `64.5 J`D. `64.5 J` |
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Answer» Correct Answer - D `vec v_(i) = (6.00 hat i - 1.00 hatj) m//s^(2)` `v_(i) = sqrt(v_(I x)^(2) + v_(I y)^(2)) = sqrt(37.0) m//s` `K_(1) = (1)/(2) mv_(1)^(2) = (1)/(2) (3.00 kg) (37.0 m^(2)//s^(2)) = 55.5 J` `vec v_(f) = 8.00 hati + 4.00 hatj` `v_(f)^(2) = vec v_(f) . vec v_(f) = 64.0 + 16.0 = 80.0 m^(2)//s^(2)` `Delta K = K_(f) - K_(i) = (1)/(2) m (v)_(f)^(2) - v_(i)^(2)` `= (3.00)/(2) (80.0) - 55.5 = 64.5 J` |
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| 1077. |
When a rubber bandis streched by a distance `x` , if exerts resuring foprce of magnitube `F = ax + bx^(2)` where`a` and `b` are constant . The work in streached the unstreched rubber - band by `L` isA. `aL^(2) + bL^(2)`B. ` (1)/(2)(aL^(2) + bL^(3) )`C. `(aL^(2))/(2) + (aL^(3))/(3)`D. `(1)/(2) (9aL^(2))/(2) + (bL^(3))/(3)))` |
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Answer» Correct Answer - C (c ) Work done in stretching the rubber - band by a distance `dx` is ` dW = F dx = (ax + bx^(2) )dx` Integrating both sides , `W = int_(0)^(L) axdx + int _(0)^(L) dx^(2) dx = (aL^(2))/(2) + (bL^(3))/(3)` |
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| 1078. |
Two bodies moving towards each other collide and move away in opposite directions. There is some rise in temperature of bodies because a part of the kinetic energy is converted intoA) Heat energyB) Electrical energyC) Nuclear energyD) Mechanical energy |
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Answer» A) Heat energy Explanations: There is rise in temperature so; mechanical energy is converted into heat energy. |
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| 1079. |
Two bodies moving towards each other collide and move away in opposite directions. There is some rise in temperature of bodies because a part of the kinetic energy is converted intoA) Hear energyB) Electrical energyC) Nuclear energyD) Mechanical energy |
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Answer» A) Heat energy Explanations: There is rise in temperature so; mechanical energy is converted into heat energy. |
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| 1080. |
Two balls of equal masses moving with velocities 10 m/s and -7 m/s respectively collide elastically. Their velocities after collision will be (a) 3 ms-1 and 17 ms-1 (b) -7 ms-1 and 10 ms-1(c) 10 ms-1 and -7 ms-1 (d) 3 ms-1 and -70 ms-1 |
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Answer» Correct answer is (b) -7 ms-1 and 10 ms-1 |
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| 1081. |
Find expression for final velocities when two bodies of same mass m have head on collision when moving in opposite direction with equal speeds. Assuming the collision to be elastic. |
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Answer» Head on collision : When a body moving in a straight line collides with another body moving in the same straight line, the collision is called head-on collision. Here \(v_1=\frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2}\) = \(\frac{(m-m)v+2m(-v)}{m+m}\) = \(\frac{0-2mv}{2m}\) i.e., v1 = -v Similarly, \(v_2=\frac{(m_2-m_1)u_2+2m_1u_1}{m_1+m_2}\) = \(\frac{(m-m)(-v)+2mv}{m+m}\) = \(\frac{0+2mv}{2m}\) i.e. v2 = v After collision the bodies bounce back with same initial speed, i.e., v. |
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| 1082. |
Throwinig mud on a wall is an example of perfectly inelastic collision. Comment. |
| Answer» True. When mud is thrown on a wall. It sticks to the wall. KE of mud reduces to zero leadint to non-conservation of KE. There is absolutely no tendency of mud to separate from wall. So, the collision is perfectly inelastic. | |
| 1083. |
Two particles of masses 0.5 kg and 0.25kg moving with velocity 4.0 m`//`s and -3.0m`//` collide head on in a perfectly inelastic collision. Find the velocity of the composite particle after collision and KE lost in the collision. |
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Answer» Here, `m_(1)=0.5kg, m_(2)=0.25kg` `u_(1)=4.0m//s, u_(2)=-3.0m//s` If v is velosity of composite particle, then from the principla of conservation of linear momentum, `(m_(1)+m_(2))v=m_(1)u_(1)+m_(2)u_(2)` `(0.5+0.25)v=0.5xx4+0.25(-3.0)` `v=(2xx0.75)/(0.75)=(1.25)/(0.75)=1.67m//s` KE lost `=(1)/(2)m_(1)u_(1)^(2)+(1)/(2)m_(2)u_(2)^(2)-(1)/(2)(m_(1)+m_(2))v^(2)` `=(1)/(2)xx0.5xx4^(2)+(1)/(2)xx0.25xx(-3.0)^(2)` `-(1)/(2)(0.5+0.25)((5)/(3))^(2)` `=4+(9)/(8)-(25)/(24)=(98)/(28)=4.1J` |
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| 1084. |
A `10kg` ball and a `20kg` ball approach each other with velocities `20m//s` and `10m//s` respectively. What are their velocities after collision, if the collision is perfectly elastic ? |
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Answer» Here, `m_(1)=10kg, m_(2)=20kg` `u_(1)=20m//s,u_(2)=-10m//s, upsilon_(1)=?, upsilon_(2)=?` As the collision is perfectly elastic , `upsilon_(1)=((m_(1)-m_(2))u_(1))/(m_(1)+m_(2))+(2m_(2)u_(2))/(m_(1)+m_(2))` `=((10-20)20)/(10+20)+(2xx20(-10))/(10+20)` `upsilon_(1)=(-200)/(30)-(400)/(30)=(-600)/(30)=-20m//s` and `upsilon_(2)=(2m_(1)u_(1))/(m_(1)+m_(2))+((m_(2)-m_(1))u_(2))/(m_(1)+m_(2))` `=(2xx10xx20)/(10+20)+((20-10)(-10))/(20+10)=(300)/(30)=10m//s` |
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| 1085. |
Two masses 10kg and 20 kg are connected by a massless spring. A force of 200 N acts on 20 kg mass. At the instant when the 10 kg mass has an acceleration 12 ms-2, what will be the energy stored in the spring? k = 2400 M/m |
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Answer» We know that F = ma F = 10 × 12 = 120 N But F = k x = 2400 x = 120 n ∴ x = \(\frac{1}{20}\) ∴ energy stored in spring E = \(\frac{1}{2}\) kx2 = \(\frac{1}{2}\)× 2400 × \((\frac{1}{2})^2 = 3J\). |
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| 1086. |
A rocket accelerates straight up by ejecting gas downwards. In a small time interval ∆t, it ejects a gas of mass ∆m at a relative speed u. Calculate KE of the entire system at t + ∆t and t and show that the device that ejects gas work = (1/2) ∆ mu2 in this time interval (neglect gravity). |
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Answer» Let mass of rocket at any time t = M ∆M = mass of gas ejected in time interval ∆t. \((KE)_{t+Δt}=\frac{1}{2}(m-Δm)(v+Δv)^2+\frac{1}{2}Δm(v-u)^2\) For rocket For gas \(\frac{1}{2}mv^2+MvΔv-Δmvu+\frac{1}{2}Δmu^2\) Initial (KE)i = \(\frac{1}{2}mv^2\) \((KE)_{t+Δt}-(KE)t=(MΔv-Δmu)v+\frac{1}{2}Δmu^2\) By Newton’s third law, Reaction force on rocket (upward) = Action force by burnt gases (downward) \(\frac{Mdv}{dt}=\frac{dm}{dt}|u|\) (∵ F = mu) or M∆v = ∆mu ⇒ M∆v − u∆m = 0 substitute this value in (i) K = \(\frac{1}{2}\)∆mu2 |
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| 1087. |
A rocket accelerates straight up by ejecting gas downwards. In a small time interval `Deltat`, it ejects a gas of mass `Delta m` at a relative speed `u` . Calculate KE of the entire system at `t+Deltat` and `t` and show that the device that ejects gas does work `=((1)/(2))Delta m . u^(2)` in this time interval (neglect gavity). |
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Answer» LET M be the mass of rocket at any time t and `v_(1)` the velocity of rocket at the same time t. Let `Deltam`=mass of gas ejected in time interval `Deltat.` Relative speed of gas ejected=u. Consider at time `t+Deltat` `(KE)_(t)+Deltat="KE of rocket+KE of gas"` `=(1)/(2)(M-Deltam)(v+Deltav)^(2)+(1)/(2)Deltam(v-u)^(2)` `=(1)/(2)Mv^(2)+MvDeltav-Deltamvu+(1)/(2)Delta"mu"^(2)` `(KE)_(t)="KE of the rocket at time"t=(1)/(2)Mv^(2)` `DeltaK=(KE)_(t)+Deltat-(KE)_(t)` `=(MDeltav=Delta"mu")v+(1)/(2)"mu"^(2)` Hence, `M(dt)/(dt)=(dm)/(dt)|u|` `DeltaK=(1)/(2)"mu"^(2)` Now by work energy theoram=`DeltaK=DeltaW` `Rightarrow DeltaW=(1)/(2)Delta"mu"^(2)` |
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| 1088. |
A ball is placed on a compressed spring. When the spring is released, the ball is observed to fly away. 1. What form of energy does the compressed spring possess? 2. Why does the ball fly away? |
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Answer» (1) The compressed spring possesses potential energy. (2) The potential energy of the spring on releasing changes to kinetic energy. It is the kinetic energy which makes the ball to fly away. |
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| 1089. |
A bob is pulled side way so that string becomes parallel to horizontal and released. Length of the pendulum is 2 m. If due to air resistance loss of energy is 10%, what is the speed with which the bob arrived at the lowest point. |
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Answer» Given, Potential energy of pendulum is converted into kinetic energy and lost energy in air. So, PE = KE + lost energy......................(1) According to question, lost energy = 10% of PE =0.1PE Eqn (1) becomes PE = KE + 0.1PE => 0.9PE = KE =>0.9 mgh = 0.5 mv => 0.9 gh = 0.5 v => v = 1.8gh v = speed at the lowest point. h = length of pendulum = 2m g = 9.81 m/s Plug in the values of g and h v = 5.94 m/s |
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| 1090. |
A pendulum is swinging freely. The bob of pendulum has: (a) maximum K.E. at its extreme positions (b) minimum K.E. at its mean position (c) maximum K.E. at its mean position (d) both (b) and (c) |
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Answer» (c) maximum K.E. at its mean position. |
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| 1091. |
A pendulum is oscillating freely. Its bob has : (a) only kinetic energy (b) maximum kinetic energy at extreme position (c) maximum potential energy at its mean position (d) a constant energy which is the sum of potential and kinetic energy |
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Answer» (d) a constant energy which is the sum of potential and kinetic energy |
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| 1092. |
A man of 60 kg carries a stone of 20 kg to a height of 30 m. Calculate the work done by the man. |
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Answer» Given; mass of the man M = 60 kg; mass of the stone m = 20kg; h = 30m; g = 10 ms-2 |
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| 1093. |
A ball of mass m is dropped from height ‘h ’. (a) Potential energy of the ball at ground level is mgh. (b) Potential energy of the ball at height h is mgh. (c) kinetic energy of the ball at ground level is mgh (d) both (b) and (c) |
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Answer» (b) Potential energy of the ball at height h is mgh. |
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| 1094. |
A bullet gets embedded in a wooden block. Where does its KE go? |
| Answer» KE of bullet gets converted into heat and sound energy. | |
| 1095. |
A bullet of mass m hits a block of mass M. The transfer of energy is maximum whenA. `m gt gt M `B. `M gt gt m`C. `M =2m`D. `M=m` |
| Answer» Correct Answer - D | |
| 1096. |
A bullet of mass 200 g hits a block of thickness 6 cm with a speed of `100ms^(-1).` It emerges with `10%` of its initial KE. Find its emergent speed |
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Answer» `K_(i)"initial"KE=1/2mu^(2)" "m=200/1000kg" "u=100ms^(-1)" ""Final"KE=0.1K_(i)=1/2mv^(2)` `K_(i)` initial `KE=1/2m u ^(2)=1/2xx(200)/(1000)xx100xx100=1000J` `K_(f)="final" KE=0.1xxK_(i)=0.1xx1000=100J` `1/2mv^(2)=100` `v=sqrt(((2xx100)/(200))/(1000))=10sqrt10=31.6ms^(-1)` Speed is redued by `68%` (not98%`). |
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| 1097. |
If two masses `m_(1)` and `m_(2)` collide, the ratio of change in their respective velocities is proportional toA. `(m_(1))/(m_(2))`B. `sqrt((m_(1))/(m_(2)))`C. `(m_(2))/(m_(1))`D. `sqrt((m_(2))/(m_(1)))` |
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Answer» Correct Answer - C Let `u_(1), u_(2)` be velocities of `m_(1), m_(2)`, before collision, and `upsilon_(1), upsilon_(2)` be their respective velocities after collision. In every collision, final momentum `=` initial momentum `m_(1)upsilon_(1)+m_(2)upsilon_(2)= m_(1)u_(1)+m_(2)u_(2)` `:. m_(1)(upsilon_(1)-u_(1))=m_(2)(u_(2)-upsilon_(2))` or `(upsilon_(1)-u_(1))/(u_(2)-upsilon_(2))=(m_(2))/(m_(1))` |
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| 1098. |
A wooden block of mass `10 gm` is dropped from the top of a tower `100m` high. Simultaneously, a bullet of mass `10 gm` is fired from the foot of the tower vertically upwards with a velocity of `100m//sec`, figure. If the bullet is embedded in it, how high will it rise above the tower before it starts falling ? (Consider `g=10m//sec^(2))` A. `80m`B. `85 m`C. `75 m`D. `10 m` |
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Answer» Correct Answer - C Ignoring `g` compared to initial velocity, the bullet and block will meet after `t=(h)/(u)=(100)/(100)=1sec` Durinig this time, distance travelled by the block `s_(1)=(1)/(2)g t^(2)=(1)/(2)xx10xx1^(2)=5m` Distance travelled by the bullet, `s_(2)=100-s_(1)=100-5=95m` velocity of bullet just before collision, `u_(1)=100-10xx1=90 m//s,` upwards velocity of block just before collision, `u_(2)=10xx1=10 m//s, ` downwards According to law of conservation of linear momentum , `m_(1)u_(1)+m_(2)u_(2)=(m_(1)+m_(2))V` `0.01(-10+90)=0.02V` `V=40 m//s` Maximum height risen by the block `=(V^(2))/(2g)=(40xx40)/(2xx10)=80m` Height reached above the top of the tower `=80-s_(1)=80-5` `=75m` |
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| 1099. |
The momentum of an object is doubled. How does it’s. K.E. change? |
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Answer» K.E. becomes four times since K.E. = P2/2m. |
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| 1100. |
A ball of mass 1 kg moves inside a smooth fixed spherical shell of radius 1 m with an initial velocity `v=5 m//s` from the bottom. What is the total force acting on the particle at point B A. 10 NB. 25 NC. `5 sqrt(5)` ND. 5 N |
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Answer» Correct Answer - C |
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