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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.
| 101. |
A body of mass `10 kg` moving with a velocity of `5 ms^(-1)` hits a body of `1 gm` at rest. The velocity of the second body after collision. Assuming it to be perfectly elastic isA. `10 ms^(-1)`B. `5 ms^(-1)`C. `15 ms^(-1)`D. `0.10 ms^(-1)` |
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Answer» Correct Answer - A `v_(2) = ((2m_(1))/(m_(1)+m_(2)))u_(1)` |
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| 102. |
A marble going at a speed of `2 ms^(-1)` hits another marble of equal mass at rest. If the collision is perfectly elastic, then the velocity of the first marble after collision isA. `4 ms^(-1)`B. `0 ms^(-1)`C. `2 ms^(-1)`D. `3 ms^(-1)` |
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Answer» Correct Answer - B `vecv_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))vecu_(1)+((2m_(2))/(m_(1)+m_(2))) vecu_(2)`. |
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| 103. |
Consider the collision depicted in Figure, to be between two billiard balls with equal masses `m_(1)=m_(2)`. The first ball is called the cue and the second ball is called the target. The billiard player wants to sink the target ball in a corner pocket, which is at an angle `theta_(2)=phi=37^(@)`. Assume that the collision is elastic and that friction and rotational motion are not important. Obtain `theta_(1)=theta` .A. `37^@`B. `90^@`C. `45^@`D. `53^@` |
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Answer» Correct Answer - D `theta_(1) + theta_(2) = 90^@`. |
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| 104. |
A motor car of `m` travels with a uniform speed `v` on a convex bridge of radius `r`. When the car is at the middle point of the bridge, then the force exterted by the car on the bridge isA. `mg`B. `mg + (mv^(2))/( r)`C. `mg - (mv^(2))/( r)`D. `mg pm (mv^(2))/( r)` |
| Answer» Correct Answer - C | |
| 105. |
A car is moving up with uniform speed along a fly over bridge which is part of a vertical circle. The true statement from the following isA. Normal reaction on the car gradually decreases and becomes minimum at highest position of bridgeB. Normal reaction on the car gradually increases and becomes maximum at highest positionC. Normal reaction on car does not changeD. Normal reaction on the car gradually decreases and becomes zero at highest position |
| Answer» Correct Answer - B | |
| 106. |
A bottle of soda water is rotated in a vertical circle with the neck held in hand. The air bubbles are collectedA. near the neckB. near the bottomC. at the middleD. uniformly in the bottle |
| Answer» Correct Answer - A | |
| 107. |
A small block is given a velocity `v` from point `A`. Given `x = 3R,R = 20 m` and `g = 9.8 m//s^(2)`. If the block strikes the point `A` after it leaves the smooth circular track in vertical plane, the value of `v` is `7x`, find `v` ? . |
| Answer» Correct Answer - `5` | |
| 108. |
An object or mass `(m)` is located at the origin of a vertical plane. The body is projected at an angle `theta` with velocity `u`. The mean power developed by the gravitational force during the interval of time till it reaches maxmum heightA. `mgu sin theta`B. `(mgu sin theta)/(2)`C. `(mgu sin theta)/(3)`D. `(mgu sin theta)/(4)` |
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Answer» Correct Answer - B avg. Power `= (mgh)/(t_(1//2)) = (mgu sin theta)/(2)`. |
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| 109. |
A particle is projected along the inner surface of a smooth vertical circle of radius `R`, its velocity at the lowest point being `(1//5)(sqrt(95 gR))`. If the particle leaves the circle at an angular distance `cos^(-1)(x//5)` from the highest point, the value of `x` is. |
| Answer» Correct Answer - `3` | |
| 110. |
A ring of mass m can slide over a smooth vertical rod. The ring is connected to a spring of force constant `K=(4mg)/R` where 2R is the natural length of the spring. The other end spring is fixed to the ground at a horizontal distance 2R from the base of the rod. the mass is released at a height of 1.5 R from ground. A. work done by the spring will be `(mgR)/2`B. work done by the spring will be `9 mgR`C. the velocity of the ring when it reaches the ground will be `sqrt(gR)`D. the velocity of the ring when it reaches the ground will be `2sqrt(gR)` |
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Answer» Correct Answer - D `W_("Spring")=1/2 kx^(2)` `=1/2xx(4 mg)/R[sqrt(4R^(2)+(9R^(2))/4)-2R]^(2)` `=(mgR)/2` `W_("all forces")=DeltaKE` `(mgR)/2+mg.(3R)/2=1/2 mv^(2)` `V=2sqrt(gR)` |
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| 111. |
A ring of mass `m` can slide over a smooth vertical rod as shown in figure. The ring is connected to a spring of force constant `k = 4 mg//R`, where `2R` is the natural length of the spring . The other end of spring is fixed to the ground at a horizontal distance `2R` from base of the rod . If the mass is released at a height `1.5 J` then the velocity of the ring as it reaches the ground is A. `(mgR)/(2),2 sqrt(gR)`B. `mgR, 2 sqrt(gR)`C. `(mgR)/(2), sqrt(2 gR)`D. `(mgR)/(2),sqrt(gR)` |
| Answer» Correct Answer - A | |
| 112. |
A particle of mass `0.5 kg` is displaced from position `vec r_(1)(2,3,1)` to`vec r_(2)(4,3,2)` by applying a force of magnitude `30 N` which is acting along `(hati + hatj + hatk)`. The work done by the force isA. `10 sqrt(3) J`B. `30 sqrt(3)`C. 30 JD. 40 J |
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Answer» Correct Answer - B `F_(x) = F cos alpha, F_(y) = F cos beta, F_(z) = F cos gamma, W = F.S`. |
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| 113. |
A bullet fired into a trunk of a tree loses `1//4` of its kinetic energy in travelling a distance of 5 cm`. Before stopping it travels s further distance ofA. 150 cmB. 1.5 cmC. 1.25 cmD. 15 cm |
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Answer» Correct Answer - D `W = overlineF. overlineS = DeltaKE , (S_(1))/(S_(2)) = (Delta KE_(1))/(DeltaKE_(2))`. |
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| 114. |
Kinetic energy of a particle moving in a straight line varies with time `t` as `K = 4t^(2)`. The force acting on the particleA. is constantB. is increasingC. is decreasingD. first increase and then decreases |
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Answer» Correct Answer - A `(1)/(2) mv^(2) = 4t^(2)` , `:. v = sqrt((8)/(m))t` comparing with `v=at,a = "constant"` i.e., the force acting on the particle is constant. |
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| 115. |
Forces acting on a particle moving in a straight line varies with the velocity of the particle as `F = (alpha)/(upsilon)` where `alpha` is constant. The work done by this force in time interval `Delta t` is :A. `alpha Delta t`B. `(1)/(2) alpha Delta t`C. `2 alpha Delta t`D. `alpha^(2) Delta t` |
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Answer» Correct Answer - A `F=(alpha)/(upsilon) rArr m(d upsilon)/(dt)=(alpha)/(upsilon)rArr int m upsilon d upsilon = int alphadt` `((m upsilon^(2))/(2)) = alpha t , Delta KE = alpha Delta r =` work done. |
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| 116. |
A force `F` acting on a particle varies with the position `x` as shown in the graph. Find the work done by the force in displacing the particle from `x =-a` to `x = +2a`. .A. `(3ab)/(2)`B. `(4ab)/(2)`C. `(2)/(3ab)`D. `(2)/(4ab)` |
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Answer» Correct Answer - A `W` = area under `F-S` curve. |
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| 117. |
A force `vecF = 2 x hati + 2 hatj + 3z^(2)hatk N` is acting on a particle .Find the work done by this force in displacing the body from `(1, 2, 3) m` to `(3,6,1)m` |
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Answer» Work done `underset(x_(1)) overset(x_(2))(int) F_(x) dx + underset(y_(1)) overset(y_(2)) (int) F_(y) dy + underset(z_(1)) overset(z_(2)) (int) F_(z) dz` `W = underset(1) overset(3)(int) 2xdx + underset(2) overset(6)(int) 2dy + underset(3) overset(1)(int) 3z^(2) dz` `W =2[(x^(2))/(2)]_(1)^(3) + 2[y]_(2)^(6) + 3[(z^(3))/(3)]_(3)^(1) = -10 J`. |
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| 118. |
A ball hits the floor and rebounds after an inelastic collision. In this caseA. the momentum of the ball just after the collision is same as that just before the collisionB. The mechanical energy of the ball remains the same on the collisionC. The total momentum of the ball and the earth is conservedD. the total kinetic energy of the ball and the earth is conserved. |
| Answer» Correct Answer - C | |
| 119. |
The work done an a particle of mass `m` by a force `K[(x)/((x^(2) + y^(2))^(3//2)) hati +(y)/((x^(2) + y^(2^(3//2))) hatj)] (K being a constant of appropriate dimensions), when the partical is taken from the point `(a,0)` to the point `(0,a)` along a circular path of radius a about the origin in x - y plane isA. `(2 k pi)/(a)`B. `(k pi)/(a)`C. `(k pi)/(2a)`D. zero |
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Answer» Correct Answer - D `W = int vec F. d vec r` =`k underset(r_(A)) overset(r_(B))int[(x hat i)/((x^(2) +y^(2))^(3//2))+(y hat j)/((x^(2) +y^(2))^(3//2))](d xhat i+ dy hatj)` =`k underset(r_(A)) overset(r_(B))int (xdx)/((x^(2)+y^(2)))+(ydy)/((x^(2) +y^(2))^(3//2))` =`k underset(r_(A)) overset(r_(B)) int (1)/((x^(2)+y^(2))^(3//2))[d((x^(2))/(2))+d((y^(2))/(2))]` =`k underset(r_(A)) overset(r_(B)) int (1)/((x^(2)+y^(2))^(3//2)) (x^(2)+y^(2))` =`k underset(r_(A)) overset(r_(B)) int (1)/(2r^(3))d(r^(2)) = k underset(r_(A)) overset(r_(B)) int (2rdr)/(2r^(3)) = k underset(r_(A)) overset(r_(B)) int (dr)/(r^(2))` =`k[-(1)/(r)]_(r_(A))^(r_(B)) =k[(1)/(r_(A))-(1)/(r_(B))]` But `r_(A) =a `and `r_(B) =a` , `:. W = 0`. |
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| 120. |
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))/(2)+(bL^(3))/(3)`B. `(1)/(2)((aL^(2))/(2)+(bL^(3))/(3))`C. `(bL^(2))/(2)-(aL^(3))/(3)`D. `(1)/(2)((bL^(2))/(2)-(aL^(3))/(3))` |
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Answer» Correct Answer - A Restoring force on rubber-band, `F = ax + bx^(2)` Work done in stretching the rubber-band by a small amount `dx`. `dW = Fdx` Net work done in stretching the rubber-band by `L` is `W = int dW = underset(0)overset(L) int Fdx = underset(0)overset(L) int (ax + bx^(2)) dx` `rArr W =[a(x^(2))/(2) + b(x^(3))/(3)]_(0)^(L) = (aL^(2))/(2) + (bL^(2))/(3)`. |
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| 121. |
A simple pendulum having bob of maas m is suspended from the ceiling of a car used in a stunt film shooting. The car moves up along an inclined cliff at a speed v and makes a jump to leavwe the cliff and lands at some the top of the cliff. The tension in the string when the car is in air isA. `mg`B. `mg - (mv^(2))/(R )`C. `mg + (mv^(2))/( R)`D. zero |
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Answer» Correct Answer - D Acceleration of car is "g" so `g_(eff) = vec g-vec a`. |
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| 122. |
A wooden block of mass `10 gm` is dropped from the top of a cliff `100 m` high. Simultaneously a bullet of same mass is fired from the foot of the cliff vertically upwards with a velocity of `100 ms^(-1)`. If the bullet after collision gets embedded in the block, the common velocity of the bullet and the block immediately after collision is `(g=10 ms^(-2))`.A. `40 ms^(-1)` downwardB. `40 ms^(-1)` upwardC. `80 ms^(-1)` upwardD. zero |
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Answer» Correct Answer - B Time after which collision takes place`t = (h)/(u)` before collsion, initial velocity of the wooden block `u_(1) = u-g t` `rArr m_(1)u_(1) -m_(2)u_(2) = (m_(1) + m_(2)) v`. |
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| 123. |
An object is moving along a straight line path from P to Q under the action of a force `vec(F)=(4 hati-3hatj+hat2k)N`. If the co-ordinate of P and Q in meters are (3,2,-1) and (2,-1,4) respectively. Then the work done by the force isA. `-15 J`B. `+15 J`C. `1015 J`D. `(4hati-3hatj+2hatk)` |
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Answer» Correct Answer - B `vec(PQ)=(2-3)hati+(-1-2) hatj(4-(-1))hatk` `vec(F).vec(PQ)=-4+9+10=15 J` |
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| 124. |
A `10 kg` block is pulled along a frictionless surface in the form of an arc of a circle of radius `10 m`. The applied force is `200 N`. Find the work done by (a) applied force and (b) gravitational force in displacing through an angle `60^@`. . |
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Answer» Work done by applied force `W = Fr sin theta` `W = 200 xx 10 xx sin 60^@ = 200 xx 10 xx (sqrt(3))/(2) = 1732 J` work done by gravitational force `W = -mgr(1 - cos theta)` `W = -10 xx 9.8 xx 10(1 - cos 60^@)` `W = -98 xx 10(1 -(1)/(2)) = - 490 J`. |
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| 125. |
A lawn roller is pulled along a horizontal surface through a distance of `20 m` by a rope with a force of `200 N`. If the rope makes an angle of `60^@` with the vertical while pulling, the amount of work done by pulling force isA. 4000 JB. 1000 JC. `2000 sqrt(3) J`D. 2000 J |
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Answer» Correct Answer - C `W = vec F. vec S = FS cos theta` |
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| 126. |
The average work done by a human heat while it beats once is `0.5 J`. Calculate the power used by heat if it beats `72` times in a minute.A. 0.6 wB. 0.8 wC. 6 wD. 8 w |
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Answer» Correct Answer - A Given, average work done by a human Heart per beat `= 0.5 J` Total work done during `72` beats =`72 xx 0.5 J = 36 J` `power = ("work done")/("Time") = (36 J)/(60 s) = 0.6 W`. |
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| 127. |
Two identical ball bearings in contact with each other and resting on a frictionless table are hit heat-on by another ball bearing of the same mass moving initially with a speed `V` as shown in figure. If the collision is elastic, which of the following (figure) is a possible result after collision ?A. B. C. D. |
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Answer» Correct Answer - B When two bodies of equal mass collides elastically, their velocities are interchanged. When ball `1` collides with `ball-2`, then velocity of `ball-1, v_(1)` becomes zero and velocity of `ball-2,v_(2)` becomes `v`, i.e., similarly. `v_(1) = 0 rArr v_(2) = v` when ball `2` collides will ball `3 v_(2) = 0, v_(3) = v`. |
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| 128. |
The length of a ballistic pendulum is `1 m` and mass of its block is `1.9 kg`. A bullet of mass `0.1 kg` strikes the block in horizontal direction with a velocity `100 ms^(-1)` and got embedded in the block. After collision the combined mass swings away from lowest point. The tension in the strings when it makes an angle `60^@` with vertical is `(g=10 ms^(-2))`.A. 20 NB. 30 NC. 40 ND. 50 N |
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Answer» Correct Answer - C `m u=(m+M) v`, If `v_(theta)` is Velocity at an angle `theta`. `v^(2)-v_(dot)^(2) = 2gl(1-cos^(●))`, find `v_(theta)` then `T_(theta) = (mv_(theta)^(2))/(r) +mg cos theta`. |
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| 129. |
The length of ballistic pendulum is `1 m` and mass of its block is `0.98 kg`. A bullet of mass `20` gram strikes the block along horizontal direction and gets embedded in the block. If block + bullet completes vertical circle of radius `1m`, then the striking velocity of bullet isA. `280 m//s`B. `350 m//s`C. `420 m//s`D. `490 m//s` |
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Answer» Correct Answer - B According to law of conservation of linear momentum `m u = (M+m)v` `u=((M+m)sqrt(5gr))/(m)`. |
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| 130. |
The block of mass `M` moving on the frictionless horizontal surface collides with the spring constant `k` and compresses it by length `L` . The maximum momention of the block after collision isA. `(ML^(2))/(K)`B. zeroC. `(KL^(2))/(2M)`D. `sqrt(MKL)` |
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Answer» Correct Answer - D When block of mass `M` collides with the spring, its kinetic energy gets converted into elastic potential energy of the spring. From the law of conservation of energy, `(1)/(2) Mv^(2) = (1)/(2) KL^(2) : v = sqrt((K)/(M)) L` Where `v` is the velocity of block by which it collides with spring. So, its maximum momentum, `P = Mv = M sqrt((K)/(M)) L = sqrt(MK) L` After collision, the block will rebound with same linear momentum. |
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| 131. |
A stone of mass "m" initially at rest and dropped from a height "h" strikes the surface of the earth with a velocity "v". If the gravitational force acting on the stone is `W`, then which of the following identities is correct ?A. `mv - mh = 0`B. `.^(1)//_(2) mv^(2) - Wh^(2) = 0`C. `.^(1)//_(2) mv^(2) - Wh = 0`D. `.^(1)//_(2) mv^(2) - mh = 0` |
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Answer» Correct Answer - C gain in `K.E` = Loss of `P.E` |
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| 132. |
A `2kg` block slides on a horizontal floor with the a speed of `4m//s` it strikes a uncompressed spring , and compresses it till the block is motionless . The kinetic friction force is compresses is `15 N` and spring constant is `10000 N//m` . The spring byA. `2.5`B. `11.0`C. `8.5`D. `5.5` |
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Answer» Correct Answer - D Let the spring be compressed by `x`. Loss in `KE` of block = Gain in `PE` of spring + Work done against friction `rArr (1)/(2) xx 2 xx4^(2) =(1)/(2) 1000 x^(2) + 15 x` On solving, we get `x = 5.5 cm`. |
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| 133. |
A `2kg` block slides on a horizontal floor with the a speed of `4m//s` it strikes a uncompressed spring , and compresses it till the block is motionless . The kinetic friction force is compresses is `15 N` and spring constant is `10000 N//m` . The spring by |
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Answer» `KE = (1)/(2) mv^(2) = W_("friction") + (1)/(2) Kx^(2)` `rArr (1)/(2) xx 2xx 4^(2) = 15 x + (1)/(2) xx 10000 xx x^(2)` `rArr 5000 x^(2) + 15 x - 16 = 0` `rArr x = 0.055 m` or `x = 5.5 cm`. |
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| 134. |
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. `(3 mg)/(2k)`B. `(2 mg)/k`C. `(3mg)/(4k)`D. `(mg)/(4k)` |
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Answer» Correct Answer - A `mg(h+(3mg)/K)=1/2K((3mg)/K)^(2)` |
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| 135. |
Which of the diagrams corectly shows the change in kinetic energy of an iron sphere falling freely in a lake having sufficient depth to impart it a terminal velocity ?A. .B. .C. .D. . |
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Answer» Correct 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)`. |
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| 136. |
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. `(1)/(sqrt(2))`B. `2`C. `(3)/(sqrt(2))`D. `sqrt(2)` |
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Answer» Correct Answer - C Total mechanical energy of particle, `E_(T) = 2J` When kinetic energy is maximum, the potential energy should be minimum. The potential of the particle is given by `V (x) = (x^(4))/(4) -(x^(2))/(2)` or `(dV)/(dx) =(4x^(3))/(4)-(2x)/(2) = x^(3) - x = x (x^(2) -1)` For `V` to be minimum, `(dV)/(dx) = 0` `:. x(x^(2) -1) = 0`, or `x = 0 pm 1` At `x = 0, V (x) = 0` At `x = "pm" 1, V (x) = - (1)/(4) J` `:. ("Kinetic energy")_(max) = E_(T) - V_(min)`. or `("Kinetic energy")_(max) = 2 2-(-(1)/(4)) = (9)/(4) J` or `(1)/(2) mv_(m)^(2) = (9)/(4)` or `v_(m)^(2) = (9xx2)/(mxx4)` or `v_(m)^(2) = (9xx2)/(1xx4) = (9)/(2)` `rArr v_(m) = (3)/(sqrt(2)) m//s`. |
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| 137. |
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. `1//sqrt(2)`B. `2`C. `3//sqrt(2)`D. `sqrt(2)` |
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Answer» Correct Answer - C `U =(x^(4))/(4)-(x^(2))/(2),(dU)/(dx) =x^(3) -x = 0 rArrx =0, x = pm 1` `(d^(2)U)/(dx^(2)) = 3x^(2) -1(d^(2)U)/(dx^(2)) = +ve` for `x =pm1` `U(pm 1)=-(1)/(4)` `K_(max) =Ulmin=T.E = 2J` `K_(max) =(9)/(4)` `K_(max) =(1)/(2) mv^(2) , v = (3)/(sqrt(2))`. |
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| 138. |
A particle move in a straight line with retardation proportional to its displacement its loss of kinectic energy for any displacement `x` is proportional toA. xB. `e^(x)`C. `x^(2)`D. `log_(e) x` |
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Answer» Correct Answer - C According to problem Retardation `prop` displacement or `(dv)/(dt) = kx` or `v(dv)/(dt) = kx` or `vdv = kxdx` or `underset(v_(1)) overset(v_(2)) int vdv = underset(0) overset(x) int kdx` or `(v_(2)^(2))/(2) -(v_(1)^(2))/(2) =(kx^(2))/(2)` or `(1)/(m) ((mv_(2)^(2))/(2) -(mv_(1)^(2))/(2)) = (kx^(2))/(2)` or `(k_(2)-k_(1)) = (mk)/(2) x^(2)`. Thus, loss of kinetic enery is proportional to `x^(2)`. |
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| 139. |
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 magnitudeB. 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 When the man is squatting on the ground he is tilted somewhat, hence he also has to balance frictional force besides his weight in this case. `R =` reactional force = friction + mg `rArr R gt mg` When the man gets straight up in that case friction `= 0` `rArr "Reactional force" = mg`. |
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| 140. |
A block of mass `0.18 kg` is attached to a spring of force-constant `2 N//m`. The coefficient of friction between the block and the floor is `0.1` Initially the block is at rest and the spring is un-stretched. An impulse is given to the block as shown in the figure. The block slides a distance of `0.06 m` and comes to rest for the first time. The initial velocity of the block in m//s is `V = N//10`. Then `N` is : . |
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Answer» Correct Answer - `4` Decrease in mechanical energy = work done against friction `(1)/(2) m upsilon^(2) - (1)/(2) kx^(2) = (mu mg ) x` `v = sqrt((2 mugx+k)/(m))` Putting `m = 0.18 kg, x = 0.06 m, k = 2Nm^(-1)`, `mu = 01` we get `upsilon = 0.4 m//s = (4)/(10) m//s` `:. N = 4`. |
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| 141. |
A bullet is fired into a wooden block. If the bullet gets embedded in wooden block, thenA. momentum alone is conservedB. kinetic momentum and kinetic energy are conservedC. both momentum and kinetic energy are conservedD. neither momentum nor kinetic energy are conserved |
| Answer» Correct Answer - A | |
| 142. |
During collision, which of the following statement is wrong ?A. there is a change in momentum of individual bodiesB. the change in total momentum of the system of colliding particle is zeroC. the change in total energy is zeroD. law of conservation of momentum is not valid. |
| Answer» Correct Answer - D | |
| 143. |
A sphere of mass `m` moving with constant velocity `u`, collides with another stationary sphere of same mass. If `e` is the coefficient of restitution, the ratio of the final velocities of the first and second sphere isA. `(1+e)/(1-e)`B. `(1-e)/(1+e)`C. `( e)/(1-e)`D. `(1+e)/( e)` |
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Answer» Correct Answer - B `v_(1)=((m_(1)-em_(2))/(m_(1)+ m_(2))) u_(1)+0` `v_(2) =0+ (m_(1)(1+e))/(m_(1)+m_(2)) u_(1)`. |
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| 144. |
In two separate collisions, the coefficient of restitutions `e_(1)` and `e_(2)` are in the ratio `3 : 1`. In the first collision the relative velocity of approach is twice the relative velocity of separation. Then, the ratio between relativevelocity of approach and relative velocity of separation in the second collision isA. `1 :6`B. `2 : 3`C. `3 : 2`D. `6 : 1` |
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Answer» Correct Answer - D `e_(1):e_(2) = 3:1 , Delta U_(1) = 2Delta V_(1) rArr e_(1) = (Delta V_(1))/(Delta U_(1)) =(1)/(2)` `(Delta V_(1))/(Delta U_(1)) :(Delta V_(2))/(Delta U_(2)) =3:1`. |
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| 145. |
A constant power `P` is applied to a particle of mass `m`. The distance traveled by the particle when its velocity increases from `v_(1)` to `v_(2)` is (neglect friction):A. `(3P)/m(v_(2)^(2)-v_(1)^(2))`B. `m/(3P)(v_(2)^(3)-v_(1)^(3))`C. `m/(3P)(v_(2)^(2)-v_(1)^(2))`D. `m/(3P)(v_(2)-v_(1))` |
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Answer» Correct Answer - B acceleration `a=P/(mv)` `v(dv)/(dx)=P/(mv)` `int_(0)^(a)dx=m/P int_(v_(1))^(v_(2))v^(2)dv` `S=m/(3p)(v_(2)^(2)-v_(1)^(2))` |
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| 146. |
A constant power `P` is applied to a particle of mass `m`. The distance traveled by the particle when its velocity increases from `v_(1)` to `v_(2)` is (neglect friction):A. `(3P)/(m) (v_(2)^(2)-v_(1)^(2))`B. `(m)/(3P)(v_(2)-v_(1))`C. `(m)/(3P)(v_(2)^(3)-v_(1)^(3))`D. `(m)/(3P)(v_(2)^(2)-v_(1)^(2))` |
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Answer» Correct Answer - C `P = mav =m(dv)/(ds) v^(2) , (P)/(m) underset(0) overset(s) int ds = underset(v_(1)) overset(v_(2)) int v^(2) dv`. |
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| 147. |
A body of mass `5 kg` makes an elastic collision with another body at rest and continues to move in the original direction after the collision with a velocity equal to `1//10^(th)` of its original velocity. The mass of the second body isA. `4.09 kg`B. `0.5 kg`C. `5 kg`D. `5.09 kg` |
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Answer» Correct Answer - A `v_(1) = ((m_(1) -m_(2))/(m_(1) + m_(2))) u_(1)`. |
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| 148. |
A ball at rest is dropped freely from a height of `20 m`. It loses `30 %` of its energy on striking the ground and bounces back. The height to which it bounces back isA. 14 mB. 12 mC. 9 mD. 6 m |
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Answer» Correct Answer - A percentage loss of `PE = (h_(1) - h_(2))/(h_(1)) xx 100 %`. |
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| 149. |
A rock of mass `m` is dropped to the ground from a height `h`. A second rock, with mass `2m`, is dropped from the same height. When the second rock strikes the ground, what is its kinetic energy? (a) Twice that of the first rock, (b) four times that of the first rock, (c) same as that of the first rock, (d) half as much as that of the first rock, (e) impossible to determine.A. twice that of the first rockB. four times that of the first rockC. the same as that of the first rockD. half that of the first rock |
| Answer» Correct Answer - A | |
| 150. |
Which of the following forces is called a conservative force ?A. Frictional forceB. Air resistanceC. Electrostatic forceD. Viscous force |
| Answer» Correct Answer - C | |