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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.
| 251. |
On a two lane road, car A is travelling with a speed of `36 km h^(-1)`. Two cars B and C approach car A in opposite directions with a speed of ` 54 km h^(-1)` each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does, What minimum acceleration of car B is required to avoid an accident? |
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Answer» Velocity of car A, `v_A` = 36 km/h = 10 m/s Velocity of car B, `v_B` = 54 km/h = 15 m/s Velocity of car C, `v_C` = 54 km/h = 15 m/s Relative velocity of car B with respect to car A, `v_(BA) = v_B – v_A` ltbrge 15 – 10 = 5 m/s Relative velocity of car C with respect to car A, `v_(CA) = v_C – (– v_A)` = 15 + 10 = 25 m/s At a certain instance, both cars B and C are at the same distance from car A i.e.,s = 1 km = 1000 m Time taken (t) by car C to cover 1000 m =`1000/25=40s` Hence, to avoid an accident, car B must cover the same distance in a maximum of 40 s. From second equation of motion, minimum acceleration (a) produced by car B can be obtained as: `s=ut+1/2at^2` `1000=5xx40+1/2xxaxx(40)^2` `a=1600/1600=1 m//s^2` |
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| 252. |
The relation between time t and displacement x is `t = alpha x^2 + beta x,` where `alpha and beta` are constants. The retardation is |
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Answer» `t=alphax^(2)+betax` Differentiating w.r.t. x, we get `(dt)/(dx)=alpha.2x+beta.x` `(1)/v = 2alphax+beta` `v^(-1)=2alphax+beta` Differentiating w.r.t. t, we get `(-1)v^(-2)(dv)/(dt)=2alpha(dx)/(dt)` `-a/v^(2)=2alphav` `a=-2alphav^(3)` `a=-2alphav^(3)` OR `t=alphax^(2)+betax` Differentiating w.r.t. t, we get `1= alpha.2x(dx)/(dt)+beta(dx)/(dt)` `=(2alpha x+beta)v` Differentiating w.r.t. t, we get `0=(2alphax+beta)(dv)/(dt)+(2alpha(dx)/(dt)+0)v` `=(2alphax+beta)a+2alphav^(2)` `a= -(2alphav^(2))/((2alphax+beta))= - 2alphav^(3)` OR `t=alphax^(2)+betax` `alphax^(2)+betax-t=0` `x = (-beta+-(beta^(2)+4alphat)^(1//2))/(2alpha)` `=-(beta)/(2alpha)+(1)/(2alpha)(beta^(2)+4alphat)^(1//2)` `v=(dx)/(dt)=(1)/(2alpha).(1)/(2)(beta^(2)+4alphat)^(1//2-1)(4alpha)` `=(beta^(2)+4alphat)^(-1//2)` `a=(dv)/(dt)=-(1)/(2)(beta^(2) +4alphat)^(-1//2-1)(4alpha)` `=-2alpha(beta^(2)+4alphat)^(-3//2)` `=-2alphap(beta^(2)+4alphat)^(-1//2)]^(3)` `=-2alphav^(3)` |
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| 253. |
If the time and displacement of the particle along the positive x-axis are related as `t=(x^(2)-1)^(1//2)`, find the acceleration in terms of `x`. |
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Answer» `t=(x^(2)-1)^(1//2)` `(dt)/(dx)=1/2(x^(2)-1)^(1//2)(2x)=x/((x^(2)-1)^(1//2))` `v=((x^(2)-1)^(1//2))/x=(1-1/x^(2))^(1//2)` `(dv)/(dx)=1/2(1-1/x^(2))^(-1//2)(2/x^(3))` `a=v(dv)/(dx)=1/x^(3)` OR `t=(x^(2)-1)^(1//2)impliesx=(1+t^(2))^(1//2)` `v=(dx)/(dt)=1/2(1+t^(2))^(-1//2)(2t)` `=t/((1+t^(2))^(1//2))=1/((1/t^(2)+1)^(1//2))=(1+1/t^(2))^(-1//2)` `a=(dv)/(dt)=-1/2(1+1/t^(2))^(-3//2)(-2/t^(3))` `=1/(t^(3)(1+1/t^(2))^(3//2))=1/((t^(2)+1)^(3))` `=1/x^(3)` OR `t=(x^(2) -1)^(1//2)` `t^(2)=(x^(2)-1)` Differentiating w.r.t. t, we get `2t=2x(dx)/(dt)=2xv` `t=xv` Differentiating w.r.t. t we get `1=x(dv)/(dt)+v(dx)/(dt)=xa+v^(2)` (Product rule) `ax=1-v^(2)=1-(t^(2))/(x^(2))=(x^(2)-t^(2))/(x^(2))=(1)/(x^(2))` `a=(1)/(x^(3))` Though the above method is simple, but for this, you will require further knowledge of differentiation. e.g. if x and y are functions of `t` (not shown in expression), `y=x^(3)+3x^(2)` differentiating w.r.t. t, we get `(dy)/(dt)=3x^(2)(dx)/(dt)+3.2x(dx)/(dt)` Differentiating w.r.t. `t`, we get `2y(dy)/(dt)=3x^(2)(dx)/(dt)` We can also write `2y dy = 3x^(2) dx` |
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| 254. |
A cyclist moving a car circular track of raidus 40 cm completes half a revolution in 40 s. Its average velocity isA. zeroB. `4 pi ms^(-1)`C. `2 m s^(-1)`D. `8 pi m s^(-1)` |
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Answer» Correct Answer - C Average velocity = Displacement / Time taken `= (2R)/t = 2 xx 40/40 = 2 m s^(-1)` |
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| 255. |
A balloon starts rising with constant acceleration `2 m//s^(2)` from ground at `t=0s`. A stone is dropped at `t=5s`. S-t graph for the given situation is shown in figure answer the following Q. The maximum hight reached by the stone isA. 30 mB. 40 mC. 45 mD. 28 m |
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Answer» Correct Answer - A `s=ut+(1)/(2)at^(2),a=-g,v=0` |
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| 256. |
The velocity-time graph of a stone thrown vertically upward with an initial velocity of `30ms^(-1)` is shown in the figure. The velocity in the upward direction is taken as positive and that in the downward direction as negative. What is the maximum height to which the stone rises?A. 30 mB. 45 mC. 60 mD. 90 m |
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Answer» Correct Answer - B `h=(u^(2))/(2a)`, from graph `u=30 m//s`, `a=(30)/(3)=10 m//s^(2)` |
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| 257. |
A point initially at rest moves along x-axis. Its acceleration varies with time as `a = (6t + 5) m//s^(2)`. If it starts from origin, the distance covered in 2 s is:A. 20 mB. 18 mC. 16 mD. 25 m |
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Answer» Correct Answer - B `v = 3t^(2) + 5` `int_(0)^(x) dx = int_(0)^(2) (3t^(2) + 5) dt` |
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| 258. |
A point initially at rest moves along x-axis. Its acceleration varies with time as `a = (6t + 5) m//s^(2)`. If it starts from origin, the distance covered in 2 s is:A. 20 mB. 28 mC. 16 mD. 25 m |
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Answer» Correct Answer - B `(dv)/(dt) = (6t + 5)` `int_(0)^(v) dv = (6t^(2))/(2) + 5t` `v = 3 t^(2) + 5t` `(dx)/(dt) = 3 t^(2) + 5t` `x = t^(3) + 5 t^(2)` |
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| 259. |
Consider a particle moving along x-axis. Its distance from origin O is described by the co-ordinate x which varies with time. At a time `t_(1)`, the particle is at point P, where its co-ordinate is `x_(1)` and at time `t_(2)`, the particle is at point Q, where its co-ordinate is `x_(2)`. The displacement during the time interval from `t_(1)` to `t_(2)` is the vector from P to Q, the x-component of this vector is `(x_(2) - x_(1))` and all other components are zero. It is convenient to represent the quantity `x_(2) - x_(1)` the change in x by means of a notation `Delta`, thus `Delta x = x_(2) - x_(1)` and `Delta t = t_(2) - t_(1)`. The average velocity `bar(V) = (x_(2) - x_(1))/(t_(2) - t_(1)) = (Delta x)/(Delta t)` The resistive force suffered by a motor boat is proportional to `v^(2)`, where `v` is instantaneous velocity. The engine was shut down when the velocity of the boat was `v_(0)`. Find the average velocity at any time `t`.A. `(v_(0) + v)/(2)`B. `(v v_(0))/(2 (v_(0) + v))`C. `(v v_(0) log_(e) ((v_(0))/(v)))/((v_(0) - v))`D. `(2 v v_(0) log_(e) ((v_(0))/(v)))/((v_(0) + v))` |
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Answer» Correct Answer - C `F prop v^(2)` `F = -Kv^(2)` `m (dv)/(dt)` or `mv (dv)/(ds) = - kv^(2)` `rArr v (dv)/(dt) = - cv^(2)` `rArr int_(v_(0))^(v) (dv)/(v) = - C int_(0)^(s) ds` |
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| 260. |
Consider a particle moving along x-axis. Its distance from origin O is described by the co-ordinate x which varies with time. At a time `t_(1)`, the particle is at point P, where its co-ordinate is `x_(1)` and at time `t_(2)`, the particle is at point Q, where its co-ordinate is `x_(2)`. The displacement during the time interval from `t_(1)` to `t_(2)` is the vector from P to Q, the x-component of this vector is `(x_(2) - x_(1))` and all other components are zero. It is convenient to represent the quantity `x_(2) - x_(1)` the change in x by means of a notation `Delta`, thus `Delta x = x_(2) - x_(1)` and `Delta t = t_(2) - t_(1)`. The average velocity `bar(V) = (x_(2) - x_(1))/(t_(2) - t_(1)) = (Delta x)/(Delta t)` A particle moves according to the equation `x = t^(2) + 3t + 4`. The average velocity in the first 5 s is:A. 8 m/sB. 7.6 m/sC. 6.4 m/sD. 5.8 m/s |
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Answer» Correct Answer - A `x = t^(2) + 3t + 4` `x_(1) = 4` `x_(2) = 44` `v_(av) = ((x_(2) - x_(1))/(Delta T))` |
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| 261. |
A body is projected upwards with a velocity `u`. It passes through a certain point above the ground after `t_(1)`, Find the time after which the body passes through the same point during the journey.A. `((u)/(g)-t_(1)^(2))`B. `2((u)/(g)-t_(1))`C. `3((u^(2))/(g)-t_(1))`D. `3((u^(2))/(g^(2))-t_(1))` |
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Answer» Correct Answer - B Suppose v be the velocity attained by body after time `t_(1)` then `v=u-g t_(1)` …(i) Let the body reach the samepoint at time `t_(2)` now velocity will be downwards with same magnitude v, then `-v=u-g t_(2)` …(ii) `(i)-(iii)implies2v=gU(t_(2)-t_(1))` or `t_(2)-t_(1)=(2v)/(g)=(2)/(g)(u-g t_(1))=2((u)/(g)-t_(1))` |
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| 262. |
An object is projected upwards with a velocity of 4.9 m/s. It will strike the ground in approximatelyA. 2 sB. 1 sC. 1.5 sD. 0.5 s |
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Answer» Correct Answer - B `T = (2v_(0))/(g)` |
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| 263. |
If in the previous problem, particle `A` moves with constant acceleration `4 m//s^(2)` with initial speed `5 m//s` and `B` moves with uniform speed `12 m//s`, when and where the particle meet? |
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Answer» For `A:d =5t+1/2xx4t^(2)` `=5t+2t^(2)` (accelerated motion) …(i) For `B: 100-d=12t` (uniform motion) …(ii) Adding (i) and (ii), we get `100=2t^(2)+17t` `2t^(2)+17t-100=0` `t=(-17+-sqrt((17)^(2)+4(2)(-100)))/(2xx2)` `=(-17+-sqrt(289+800))/4` `=(-17+-33)/4=4 s, d=52m` |
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| 264. |
In the previous problem, if initial velocity is `v_(0)`, then velocity and displacement will beA. `v_(0)t+kt^(2), v_(0)+(kt^(3))/6`B. `v_(0)+(kt^(2))/2, v_(0)t+(kt^(3))/6`C. `v_(0)+kt^(2), v_(0)t+(kt^(3))/6`D. `v_(0)+(kt^(2))/2, v_(0)t+(kt^(3))/6` |
| Answer» Correct Answer - B | |
| 265. |
The splash of sound was heard `5.35 s` after dropping a stone into a well 122.5 m deep. Velocity of sound in air isA. `350 cm//s`B. `350 m//s`C. `392 cm//s`D. `0 cm//s` |
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Answer» Correct Answer - B `t=sqrt((2h)/(g))+(h)/(v_("sound"))` |
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| 266. |
A ball is falling freely from a certain height. When it reached `10 m` height from the ground its velocity is `v_(0)`. It collides with the horizontal ground and loses `50 %` of its energy and rises back to height of `10 m`. The value of velocity `v_(0)` isA. 7 m/sB. 10 m/sC. 14 m/sD. 16 m/s |
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Answer» Correct Answer - C The velocity of the body on reaching the ground is `V = sqrt(u^(2) + 2gh) = sqrt(V_(0)^(2) + 196)` Given that, `(1)/(2) (KE) = PE` `(1)/(2) ((1)/(2) mV^(2)) = mgh` `V^(2) = 4 gh` `V_(0)^(2) + 196 = 4 xx 9.8 xx 10` `V_(0)^(2) = 392 - 196 = 196` `V_(0) = 14 m//s` |
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| 267. |
A cone falling with a speed `v_(0)` strikes and penetrates the block of a packing material. The acceleration of the cone after impact is `a=g-cx^(2)`. Where c is a positive constant and x is the penetration distance. If maximum penetration depth is `x_(m)` then c equalsA. `(2gx_(m)+v_(0)^(2))/(x_(m)^(2))`B. `(2gx_(m)-v_(0)^(2))/(x_(m)^(2))`C. `(6gx_(m)-3v_(0)^(2))/(2x_(m)^(3))`D. `(6gx_(m)+3v_(0)^(2))/(2x_(m)^(3))` |
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Answer» Correct Answer - D `v(dv)/(dx)=g-cx^(2)` `impliesint_(v_(0))^(0)vdv=gint_(0)^(x)dx-cint_(0)^(x)x^(2)dx` `implies-(v_(0)^(2))/(2)=gx-(cx^(3))/(3)` `implies(x^(3)c)/(3)=gx+(v_(0)^(2))/(2)impliesc=(3g)/(x^(2))+(3v_(0)^(2))/(2x^(3))` |
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| 268. |
If the velocity of the particle is given by `v=sqrt(x)` and initially particel was at `x=4m` then which of the following are correct.A. at `t=2` sec, the position of the particle is `x=9 m`B. particle acceleration at `t=2` sec is `1 m//s^(2)`C. particle acceleration is `½ ms^(2)` through out the motionD. particle will never go in negative direction from its starting position. |
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Answer» Correct Answer - A::C::D `v=sqrt(x)` `(dx)/(dt)=sqrt(x)impliesint(dx)/x^(1//2)=int dt implies 2sqrt(x)=t+c` But at `t=0,x=4,impliesc=4` `impliesx=((t+4)^(2))/(4)impliesx=((6)^(2))/(4)=(36)/(4)=9` `a=v(dv)/(dx)=sqrt(x)xx(1)/(2sqrt(x))= ½ m//s^(2)`. |
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| 269. |
A swimmer crosses a flowing stream of width `omega` to and fro in time `t_(1)`. The time taken to cover the same distance up and down the stream is `t_(2)`. If `t_(3)` is the time the swimmer would take to swim a distance `2omega` in still water, thenA. `T_(1) = T_(2).T_(3)`B. `T_(1)^(2) = T_(2).T_(3)`C. `T_(2)^(2) = T_(1).T_(3)`D. `T_(2)^(3) = T_(1).T_(2)` |
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Answer» Correct Answer - B Let `v` and `u` be speeds of swimmer and river respectively. `T_(1) = (2b)/(sqrt(v^(2) - u^(2))) , T_(2) = (b)/(v - u) + (b)/(v + u) = (2bv)/(v^(2) - u^(2)) , T_(3) = (2b)/(v)` `:. T_(1)^(2) = t_(2) . T_(3)` |
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| 270. |
A parachutist drops first freely from a plane for `10 s` and then his parachute opens out. Now he descends with a net retardation of `2.5 ms^(-2)` If he bail out of the plane at a height of `2495 m` and `g=10 ms^(-2)`, his velocity on reaching the ground will be`.A. `5ms^(-1)`B. `10ms^(-1)`C. `15ms^(-1)`D. `20ms^(-1)` |
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Answer» Correct Answer - A The velocity v acquired by the parachutist after 10 s. `v=u+g t=9+10xx10=100ms^(-1)` Then, `s_(1)=ut+(1)/(2)g t^(2)=0+(1)/(2)xx10xx10^(2)=500m` The distance travelled by the parachutist under retardation is `s_(2)=2495-500=1995m` Let `v_(g)` be the velocity on reaching the ground then `v_(g)^(2)-v^(2)=2as_(2)` or `v_(g)^(2)-(100)^(2)=2xx(-2.5)xx1995` or `v_(g)=5ms^(-1)` |
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| 271. |
A body released from a great height, falls freely towards the earth. Another body is released from the same height exactly one second later. Then the separation between two bodies, two seconds after the release fo second body is. |
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Answer» According to given problem 2 nd body falls for 2 s so that `h_(2) = (1)/(2) g (2)^(2)`.....(i) while 1 st has fallen for `2 + 1 = 3 s`, so `h_(1) = (1)/(2) g (3)^(2)` ....(ii) `:.` Separation between two bodies 2 s after the release of 2nd body, `d = h_(1) - h_(2)` `= (1)/(2) g (3^(2) - 2^(2))` `= 4.9 xx 5` `=24.5 m` |
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| 272. |
A street car moves rectilinearly from station A to the next station B (from rest to rest) with an acceleration varying according to the law `f = a-bx,` where a and b are constants and x is the distance from station A. The distance between the two stations and the maximum velocity areA. `(2a)/(b)`B. `(a)/(b)`C. `(a)/(2b)`D. `2a - b` |
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Answer» Correct Answer - A Acc. `= (a - bx)` `v (dv)/(dx) = a - bx` `int v dv = int (a - bx) dx` `(v^(2))/(2) = ax - (bx^(2))/(2)` `v^(2) = 2 ax - bx^(2)` When `v = 0`, `0 = 2 ax - bx^(2)` `x = (2a)/(b)` |
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| 273. |
It takes one minute for a passenger standing on an escalator to reach the top. If the escalator does not move it takes him `3` minute to walk up. How long will it take for the passenger to arrive at the top if he walks up the moving escalator ?A. 2 minuteB. 1.5 minuteC. 0.75 minuteD. 1.25 minute |
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Answer» Correct Answer - C When the person will move on moving escalator the time will be less than one minute |
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| 274. |
A man throws ball with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two ball are in the sky at any time (Given `g=10(m)/(2^2)`)A. At least `0.8 m//s`B. Any speed less than `19.6 m//s`C. Only with speed `19.6 m//s`D. More than `19.6 m//s` |
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Answer» Correct Answer - D If balls are thrown with speed `19.6 m//s`, it will return to man after `4 s`. Third ball will be thrown, when the first ball comes to man. To keep more than two balls, balls should be thrown with speed more than `19.6 m//s`. |
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| 275. |
A wooden block is dropped from the top of a cliff 100m high and simultaneously a bullet of mass 10 g is fired from the foot of the cliff upwards with a velocity of `100 m//s`. The bullet and wooden block will meet each other after a time:A. 10 sB. 0.5 sC. 1 sD. 7 s |
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Answer» Correct Answer - C For block `h = (1)/(2) g t^(2)` For bullet `100 - h 100 t - (1)/(2) g t^(2)` `t = 1 sec` |
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| 276. |
Four person K,L,M and N are initally at the corners of a square of side of length d. If every person starts moving, such that K always heads towards L, L heads towards M, M heads directly towards N and N heads towards K, then the four perons will meet afterA. `(d)/(v)`B. `(sqrt(2)d)/(v)`C. `(d)/(2v)`D. `(d)/(sqrt(2)v)` |
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Answer» Correct Answer - A `T=(a)/(2vsin^(2)((pi)/(n))),(n=4)` |
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| 277. |
Four persons K,L,M,N are initially at the four corners of a square of side d. Each person now moves with a uniform speed v in such a way that K always moves directly towards L, L directly towards M, M directly towards N, and N directly towards K. The four persons will meet at a time............. . |
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Answer» It is obvious from considerations of symmetry that at any moment of time the persons will be at the corners of a square whose side gradually decreases (Fig.4.53) and so they will finally meet at the centre of the square O. The speed of each person along the line joining his initial position and O will be `v cos 45^(@) = v//sqrt2`. And as each person has to walk a distance `d cos 45^(@) = d//sqrt2` to reach the centre, the four persons will meet at the centre of the square O after time `t (d//sqrt2)/(v//sqrt2) = (d)/(v)` |
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| 278. |
A man standing on the edge of a cliff throws a stone straight up with initial speed (u) and then throws another stone straight down with same initial speed and from the same position. Find the ratio of the speeds. The stones would have attained when they hit ground at the base of the cliff. |
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Answer» As the stone thrown vertically up will come back to the point of projection with same speed, both stones will move downward with same initial velocity, so both will hit the ground with velocity `v^(2) = u^(2) + 2gh`, i.e., `v = sqrt((u^(2) + 2gh))` So the ratio of speeds attained when they hit the ground is `1 : 1`. [However, the stone projected up will take `(2u//g)` time more to react the ground than the stone projected downwards] |
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| 279. |
For the one dimensional motion, described by `x=t-sint`A. `x(t) gt 0` for all t gt 0B. `nu(t) gt 0` for all `t gt 0`C. `a(t) gt 0` for all `t gt 0`D. all of these |
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Answer» Correct Answer - A `x = t - sint, v = (dx)/(st) = 1 - cost, a = (dv)/(dt) = sint` `therefore x(t) gt 0` for all values of `t gt 0` and v(t) can be zero for one value of t, a(t) can zero for one value of t. |
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| 280. |
For the one dimensional motion, described by `x=t-sint`A. `x(t) gt 0` for all `t gt 0`B. `v(t) gt 0` for all `t gt 0`C. `a(t) gt 0` for all `t gt 0`D. `v(t)` lies between `0` and `2` |
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Answer» Correct Answer - A::D Given, `x = t - sint` velocity `v = (dx)/(dt) = (d)/(dt) [t - sin t]` `= 1- "cost"` Acceleration `a = (dv)/(dt) = (d)/(dt) [1-"cost"] = sint` As acceleration `a gt 0` for all `t gt 0` Hence, `x(t) gt 0` for all `t gt 0` Velocity `v = 1 - cos t` when, `cos t = 1, "velocity" v = 0` `v_(max) = 1 - (cos t)_(min) = 1 - (-1) = 2` `v_(min) = 1 - (cos t)_(max) = 1 - 1 = 0` Hence, v lies between 0 and 2 Acceleration `a = (dv)/(dt) = - sint` When , `t = 0, x = 0 , x = +1 , a = 0` When, `t = (pi)/(2), x = 1, v = 0, a = - 1` When `t = pi, x = 0, x = -1, a = 1` When `t = 2pi, x = 0, x = 0, a = 0` |
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| 281. |
The displacement-time graph of a moving particle with constant acceleration is shown in. The velocity-time is given by .A. B. C. D. |
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Answer» Correct Answer - A At `t=0` slope of the x-t graph is zero, hence, velocity is zero at `t=0` As time increases slope increases I negative direction, hence, velocity increases in negative direction at point `I` slope changes suddenly from negative to positive value: hence, velocity changes suddenly from negative to positive and then velocity starts decreasing and becomes zero at 2 optionn (a) represents all these clearly. |
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| 282. |
The velocity time plot for a particle moving on straight line is shown in the figure.A. The particle has a constant accelerationB. The particle has never turned aroundC. The particle has a zero displacementD. The average speed in the interval 0 to 10 s is the same as the average speed in the interval 10 s to 30 s |
| Answer» Correct Answer - A | |
| 283. |
The velocity time plot for a particle moving on straight line is shown in the figure.A. The particle has a constant accelerationB. The particle has never tuned aroundC. The particle has zero displacementD. The average speed in the interval 0 to 10 s is the same as the average speed in the interval 10 s to 20 s. |
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Answer» Correct Answer - A::D Since the graph is astraight line, its slope is constant it means acceleration of the pariticle is constant. Velocity of the particle changes from positive to negative at `t=10`s so particle changes direction at this time. The particle has zero displacement up to 20 s, but not for the entire motion. The average speed in the interval of 0 to 10 s is the same as the average speed in the internal of 10 s to 20 s because distance covered in both time interval is same. |
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| 284. |
Free fall on an object in vacuum is a case of motion withA. uniform velocityB. uniform accelerationC. variable accelerationD. uniform speed |
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Answer» Correct Answer - B Free fall of an object in vacuum is a case of motion with uniform acceleration. |
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| 285. |
Two inclined planes (AB) and (BC) are placed as shown in Fig. 2 (ABC).3 A particle is projected from the foot of the plane of angle ` alpha` along its line witn a velocity just sufficient to carry it to the top after which the particle slides down the other inclined plane. Fing the total time it will take to reach the pont (C ). |
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Answer» Correct Answer - `t = sqrt(((2h)/(g)))[(1)/(sin alpha) + (1)/(sin beta)]` For a body sliding down an inclined plane, `a = g sin beta` Using, `x = ut + (1)/(2) at^(2)` `l = 0 + (1)/(2) (g sin beta) t_(2)^(2)` `(h)/(sin beta) = (1)/(2) (g sin beta) t_(2)^(2), t_(2) sqrt((2h)/(g)) (1)/(sin beta)` Similarly, `t_(1) = sqrt((2h)/(g) (1)/(sin alpha)` `:. t = t_(1) + t_(2) = sqrt((2h)/(g)) [(1)/(sin alpha) + (1)/(sin beta)]` |
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| 286. |
A lawyer seeks your advice in one of his cases. The question is whether a driver was exceeding a `45 km//hr` speed limit before he made an emergency stop, brakes locked and wheels sliding. The length of skid marks on the road was 5 m. The policeman, assuming that maximum deceleration of the car would not exceed the acceleration of a freely falling body `(10 m//s^(2))`, arrested the driver for speeding What is your opinion ? |
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Answer» Correct Answer - The policeman was wrong in arresting the driver as the speed of car was `= 10 m//s = 36 km hr (lt 45 km//hr)` when breaks were applied As stopping distance `s = (u^(2)//2a)`, `u = sqrt(2as) = sqrt(2 xx 10 xx 5) = 10 m//s` |
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| 287. |
When the velocity is constant, can average velocity over any time interval differ from instantaneous velocity at any instant ? If so, give an example, if not, explain why? |
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Answer» No, by definition of velocity and average velocity we have `vec(v)=` slope of tangent to `vec(s)//t` curve at a given instant of time and `vec(v)_(av) =` slope of chord of `vec(s)//t` curve for a given interval of time. Now for constant velocity, the slope of `vec(s)//t` curve will be constant and so it will a straight line making a const, angle with time axis. Now as in case of straight line, Slope of tangent = Slope of chord, i.e., `vec(v) = vec(v)_(av)` i.e., if velocity is contant average velocity over any interval of time is always equal to velocity at any instant of time. |
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| 288. |
Consider the motion of the tip of the minute hand of a clock. In one hourA. a & b are correctB. 1,b & c are correctC. a & d are correctD. b,c & d are correct |
| Answer» Correct Answer - C | |
| 289. |
A ball is dropped from the top of a building 100 m high. At the same instant another ball is thrown upwards with a velocity of `40ms^(-1)` form the bottom of the building. The two balls will meet after.A. 5 sB. 2.5 sC. 2 sD. 3 s |
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Answer» Correct Answer - B `t=(d_(r))/(v_(r))=(h)/(u_(1)+u_(2))[because a_(r)=0],u_(1)=0` |
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| 290. |
A man starts chasing his dog 10 second after the latter runs along a straight track at a unifrom acceleration of `0.5 m//s^(2)`. The track is 2 km long after which it bends aways into the field. What will be the minimum constant speed of the man so that he may catch the dog before the bend in the track ?A. 1.50 km/hrB. 40 m/sC. 90 km/hrD. 20 m/s |
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Answer» Correct Answer - C `2.0 = (1)/(2) xx a xx t^(2)` Where `a = (1)/(2) m//s^(2)` For man `2.0 = v xx t` |
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| 291. |
The distance covered by a particle varies with as `x=k/b(1-e^(-bt))`. The speed of particle at time `t` isA. `ke^(-bt)`B. `kb e^(-bt)`C. `(k/b^(2))e^(-bt)`D. `(k/b)e^(-bt)` |
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Answer» Correct Answer - A `x=k/b(1-e^(-bt))` `v=(dx)/(dt)=k/b{0-e^(-bt)(-b)}` `=k e^(-bt)` |
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| 292. |
The position of and object moving along x-axis is gi en by `x=a +bt^(2)`, wher `a=8.5 m` and b=2.5 ms^(-2) and (t) is measured in ceconds. What is the velcoity `at t=1= 0`s and `t=2.0 s?` What is the average velocity between `t=2.0s` and `t=4.0 s`? |
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Answer» In notation of differential calculus, the velocity is `v = (dx)/(dt) = (d)/(dt) (a + bt^(2)) = 2b t = 5.0 ms^(-1)` At `t = 0s , v = 0 m s^(-1)` and at `t = 2.0 s, v = 10 ms^(-1)` Average velocity `= (x(4.0) - x (2.0))/(4.0 - 2.0)` `= (a + 16 b - a 4 b)/(2.0) = 6.0 xx b` `= 6.0 xx 2.5 = 15 ms^(-1)` |
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| 293. |
The position `x` of a particle varies with time `t` as `x=at^(2)-bt^(3)`. The acceleration at time `t` of the particle will be equal to zero, where (t) is equal to .`A. `alpha/beta`B. `(2alpha)/(3beta)`C. `alpha/(3beta)`D. zero |
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Answer» Correct Answer - C `x=alphat^(2)-betat^(3)` `v=(dx)/(dt)=2alphat-3betat^(2)` `a=(dv)/(dt)=2alpha-6betat` `a=0=2alpha-6betatimpliest=alpha/(3beta)` |
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| 294. |
The position-time relation of a particle moving along the x-axis is given by `x=a-bt+ct^(2)` where a, b and c are positive numbers. The velocity-time graph of the particle isA. B. C. D. |
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Answer» Correct Answer - C `x=a-bt+ct^(2)` `v=(dx)/(dt)=-b+2 ct` v-t graph `y=-b+2 cx`, straight line `y=C+mx` Slope `m=2c`, +ve Intercept, `c=-b`, -ve |
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| 295. |
When a body is thrown up in a lift with a velocity `u` relative to the lift, the time of flight is found to be `t`. The acceleration with which the lift is moving up isA. `(u - g t)/(t)`B. `(2u - g t)/(t)`C. `(u + g t)/(t)`D. `(2u + g t)/(t)` |
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Answer» Correct Answer - B `A_("rel") = (g + a)` Since `t = (2u)/(A_("rel"))` `rArr a = ((2u - g t)/(t))` |
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| 296. |
A ball is bouncing elastically with a speed 1 m/s between walls of a railway compartment of size 10 m in a direction perpendicular to walls. The train is moving at a constant velocity of 10 m/s parallel to the direction of motion of the ball. As seen from the ground choose the correct optionA. the direction of motion of the ball changes every 10 secondsB. speed of ball changes every 10 secondsC. average speed of ball over any 20 seconds interval is fixedD. the acceleration of the ball is the same as from the train. |
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Answer» Correct Answer - B::C::D As speed is changing after travelling 10 m and speed is `1 m//s` hence, time duration of the changing speed is 10. Since, the train is moving with constant velocity hence, it will act as inertial frame of rederence as that of earth and wacceleration will be same in both frames. |
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| 297. |
The velocity-time graph of a particle in one-dimensional motion is shown in Fig.: (a) Which of the following formulae are correct for describing the motion of the particle over the time-interval `t_(1)` to `t_(2)`: (a) `x(t_(2)) = x(t_(1)) + v (t_(1)) (t_(2) - t_(1)) + (1//2) a (t_(2) - t_(1))^(2)` (b) `v(t_(2)) = v(t_(1)) + a (t_(2) - t_(1))` (c) `v_("average") = (x (t_(2)) - x(t_(1)))//(t_(2) - t_(1))` (d) `a_("average") = (v(t_(2)) - v(t_(1)))//(t_(2) - t_(1))` (e) `x(t_(2)) = x(t_(1)) + v_("average") (t_(2) - t_(1)) + (1//2) a_("average") (t_(2) - t_(1))^(2)` (f) `x(t_(2)) - x(t_(1))` = area under the `v - t` curve bounded by the t-axis and the dotted line shown. |
| Answer» The correct formulae describing the motion of the particle are (c), (d) and, (f) The given graph has a non-uniform slope. Hence, the formulae given in (a), (b), and (e) cannot describe the motion of the particle. Only relations given in (c), (d), and (f) are correct equations of motion. | |
| 298. |
A branch of physics dealing with motion without considering its causes is known asA. staticsB. dynamicsC. kinematicsD. hydrodynamics |
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Answer» Correct Answer - C (c ) Kinematics is that branch of mechanics that deals with the study of motion without going into the causes of motion. |
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| 299. |
The motion of a body is given by the equation `dv//dt=6-3v`, where v is in m//s. If the body was at rest at `t=0` (i) the terminal speed is `2 m//s` (ii) the magnitude of the initial acceleration is `6 m//s^(2)` (iii) The speed varies with time as `v=2(1-e^(-3t)) m//s` (iv) The speed is `1 m//s`, when the acceleration is half initial value |
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Answer» (a) `(dv)/(dt)=6-3v=3(2-v)` `int_(0)^(v)(dv)/((2-v))=3int_(0)^(t) dt` `(| log_(e)(2-v)|_(0)^(v))/-1=3|t|_(0)^(t)` `log_(e)(2-v)-log_(e)v=-3t` `log_(e)((2-v)/v)=-3t` `(2-v)/v=e^(-3t)` `2-v=2e^(-3t)` `v=2(1-e^(-3t))` (b) Terminal velocity, i.e. at `t=oo`, `v=2(1-e^(-3t))=2(1-e^(-oo))` , `=2(1-1/e^(oo))=2(1-1/oo)` `=2(1-0)=2m//s` |
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| 300. |
A particle is moving in a straight line with initial velocity `u` and uniform acceleration `f`. If the sum of the distances travelled in `t^(th) and (t + 1)^(th)` seconds is `100 cm`, then its velocity after `t` seconds, in `cm//s`, is.A. 20B. 30C. 80D. 50 |
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Answer» Correct Answer - D `s_(t)=u+(t)/(2)(2t-1),s_(t+1)=u+(t)/(2)(2t+1)` `s_(t)+s_(t+1)=100,v=u+ft` |
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