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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.
| 401. |
A particle moves in a circle of radius `20 cm`. Its linear speed is given by `v = 2t` where `t` is in seconds and `v` in `m s^-1`. ThenA. The radial acceleration at `t = 2 s` is `80 m s^-2`.B. Tangential acceleration at `t = 2 s` is `2 m s^-2`.C. Net acceleration at `t = 2 s` is greater than `80 m s^-2`.D. Tangential acceleration remains constant in magnitude. |
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Answer» Correct Answer - A::B::C::D (a.,b.,c.,d.) `v = 2t, a_c = (v^2)/( r) = ((2 t)^2)/(0.2) = 20 t^2 = 20 xx 2^2 = 80 ms^-2` `a_t = (dv)/(dt) = 2 ms^-2` Net acceleration `a = sqrt(a_c^2 + a_t^2) gt 80 ms^-2`. |
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| 402. |
A particle moves in the xy plane and at time t is at the point `(t^(2), t^(3)-2t)`. Then :-A. At `t=2//3 s`, directions of velocity and acceleration are perpendicularB. At `t=0`, directions of velocity and acceleration are perpendicularC. At `t=sqrt(2/3) s`, particle is moving parallel to x-axisD. Acceleration of the particle when it is at point `(4, 4)` is `2hat(i)+24 hat(j)` |
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Answer» Correct Answer - A::B::C::D `vec(r)=t^(2)hat(i)+(t^(3)-2t)hat(j)`, `vec(v)=(dvec(r))/(dt)=2that(i)+(3t^(2)-2)hat(j)` `vec(a)=(d^(2)vec(r))/(dt^(2))=2hat(i)+6that(j)` `vec(a).vec(v)=4t+18t^(3)-12t=0 ("For " bot)` `:. T=+- 2//3, 0`. For parallel to x-axis `rArr (dy)/(dx)=0 rArr (dy)/(dx)=(3t^(2)-2)/(2)` `:. at t=sqrt(2/3)` sec it becomes zero so (c) `vec(a)(4, 4)=2hat(i)+6xx2hat(j)=2hat(i)+12 hat(j)` |
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| 403. |
The positions corrdinate of particle that is confined to move along a straight line is given by `x = 2t^(2) -24t +6` where `x` is measured form a convenient origin and `t` is in second. Determine the distance travelled by the particle during the interval from `t = 1sec` to `t = 4sec`. |
| Answer» Correct Answer - `74m` | |
| 404. |
(a) Earth can be thought of as a sphere of radius ` 64 00 km`. Any object (or a person ) is performing circula motion around the axis os earth due to earth`s rotation (period 1 day ). What is acceleration o object on the surface of th earth 9at equator ) towards its centre ? What is its latitude ` theta` ? How does these accelerations compare with `g=9.8 m//s^2 `? (b) Earth also moves in circular orbit around sum every year withon orbital radius of ` 1.5 xx 10 ^(11) m`. What is the acceleration of earth ( or any object on the surface of the earth ) towards the centre of the sum ? How dies thsi acceleration comparte with ` g=9.8 ms^2 `? |
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Answer» (a) Here, ` R =6400 km =6.4 xx 10^6 m, T= 24 xx 60 xx 60 s` At latitude ` theta (=0^2)`, the value of (R )remains the same, hence no change in the value of ` a _c (= 0.034 m//s^2)` Now, ` a_c/g = (0.034)/(9.8) = 1/(288) ` which is much smaller than `1` ` (b) R= 1.5 xx 10^(11) m, T= 365 xx 24 xx 60 s = 3.15 xx 10^7 s` `a_c = (R4 pi^2)/T^2 = ((1.5 xx 10^(11) ) xx 4 xx (3. 142)^(2)) /((3.15 xx 10^7)^2) =5.97 xx 10^3 m//s^2` ` a_c /g = (5.97 xx 10^(-3))/(9.8) = 1/(1642)`. |
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| 405. |
Which of the following is true.A. path of particle moving along a straight line with respect to the observer moving along another straight line must be straight line.B. path of the particle depends upon the coordinate axis system taken.C. path of the man who projects a particle, will be parabolic with respect to the particle.D. none of these |
| Answer» Correct Answer - D | |
| 406. |
Particle is projected along a rough horizontal table, its path must be : (assuming only gravity and contanct force due to the table acting).A. straight lineB. circularC. parabolicD. elliptical |
| Answer» Correct Answer - A | |
| 407. |
During a accelerated motion of a particleA. average velocity of the particle is always less than its final velocityB. average velociyt of the particle is always greater than its final velocityC. average velocity of the particle may be zero alsoD. average velocity of the particle is half its final velocity |
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Answer» Correct Answer - C |
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| 408. |
If position time graph of particle is sincurve as shown, what will be its velocity-time graph. A. B. C. D. |
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Answer» Correct Answer - C |
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| 409. |
Average velocity of a particle moving in a straight line, with constant acceleration a and initial velocity `u` in first `t` seconds is.A. `u+1/2 at`B. `u+at`C. `(u+at)/2`D. `u/2` |
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Answer» Correct Answer - A |
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| 410. |
Cleck up the onlycorrect statements in the following :A. A body having a constant velocty still can have varying speed.B. A body having a constant speed can have varying velocity.C. A body having constant speed can have an acceleration.D. If body having accleration are in the same direction, then distance is equal to displacement. |
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Answer» Correct Answer - B::C::D A body having aconstant speed can have a varying velocitydue tochange in the direction of velocity. Thus a body having constant speed can in the same direction, If velocityand acceleration are in the same direction, then distance is equal to displacement, becouse then is no change in direction of motion. The body will contiuousl travel in one direction only. |
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| 411. |
Anoop (A) hits a ball along the ground with a speed u in a direction which makes an angle `30^@` with the line joining him and the fielder Babul (B). Babul runs to intercept the ball With a speed `(2u)/3.` At What angle theta should he run to intercept the ball ? A. `sin^-1[sqrt3/2]`B. `sin^-1[2/3]`C. `sin^-1[3/4]`D. `sin^-1[4/5]` |
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Answer» Correct Answer - C Relative velocity of A with respect to B should be along AB or absolute velocity components perpendicular to AB should be same. `:. (2u)/3 sin theta=u sin30^@.` `:. theta=sin^-1(3/4)` |
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| 412. |
Plot the acceleration-time graph of the welocity-time graph given in. .A. .B. .C. .D. . |
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Answer» Correct Answer - A For `0` to `5 s,` acceleration is positive, for `5` to `15 s` acceleration is negative, for `15` to `20 s` acceleration is positive. |
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| 413. |
A ball is thrown upwards with a speed of `40m//s`. When the speed becomes half of the initial speed, gravity is switched off for next 2 second. After that gravity is again switched on but magnitude gravity is doubled. The total distance travelled by the ball from `t=0` to the time when the ball reaches the maximum heighth is `55beta`. Find the value of `beta`. |
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Answer» Correct Answer - 2 |
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| 414. |
A particle `A` moves with velocity `(2hati-3hatj)m//s` from a point `(4,5m)m`. At the same instant a particle `B`, moving in the same plane with velocity` (4hati+hatj)m//s` passes through a point `C(0,-3)m`. Find the `x`-coordinate (in `m`) of the point where the particles collide. |
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Answer» Correct Answer - 8 |
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| 415. |
Anoop is moving due east with a velocity of `1 m//s` and Dhyani is moving due west with a velocity of `2 m//s.` what is the velocity of Anoop with respect to Dhyani? |
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Answer» Correct Answer - A It is a one dimensional motion. So, let us choose the east direction as positive and the west as negative. Now, given that `v_A`=velocity of Anoop=`1m//s` and `v_D`=velocity of Dhyani=`-2 m//s` Thus, `v_(AD)` =velocity of Anoop with respect to Dhyani `=v_A-v_D=1-(-2)=3 m//s` Hence, velocity of Anoop with respect to Dhyani is `3 m//s`due east. |
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| 416. |
A ball of mass `m` is attached to the lower end of a light vertical spring of force constant `K`. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal `(` unstretched `)` length, and comed to rest again after descending through a distance `x`.A. `x=mg//k`B. `x=2m g//k`C. The ball will have no acceleration at the position where it has descended through x/2.D. The ball will have an upward acceleration equal to g at its lowermost position. |
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Answer» Correct Answer - B::C::D |
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| 417. |
If ` vec A, vec B and vec C` are mutually perpendicular vectors, then find the value of ` vec A. vec (B + vec C)`. |
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Answer» ` vec A . (vec B+ vec C) = vec A. vec B+ vec A. vec C` =AB cos 90^(@) =0` . |
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| 418. |
Is ` hat i- hatj` a unit vectror ? Exolain. |
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Answer» ` (hat i-hat j)` not a unit vector. If ` vec R=hati -hatj`, then `R =sqrt ((1(^(2) + (-1)^(2)) =sqrt 2 ` and angle `beta` which ` ` vec R` makes with X-axis, is given by, ` cos beta = 1//sqrt 2 = cos 45^(@)`. |
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| 419. |
If the velocity is \(\vec V\) – 2\(\hat i\) +t2\(\hat j\) – 9 \(\vec k\) , then the magnitude of acceleration at t = 0.5 s is(a) 1 m s-2 (b) 1 m (c) zero (d) -1 m s s-2 |
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Answer» Correct answer is (a) 1 m s-2 |
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| 420. |
If a particle has negative velocity and negative acceleration, its speed (a) increases (b) decreases (c) remains same (d) zero |
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Answer» Correct answer is (a) increases |
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| 421. |
A projectile is thorwn with a velocity of `50ms^(-1)` at an angle of `53^(@)` with the horizontal The equation of the trajectroy is given byA. `180y = 240x - x^(2)`B. `180y = x^(2) -240x`C. `180y = 135x - x^(2)`D. `180y = x^(2) - 135x` |
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Answer» Correct Answer - A `u_(x) = u cos theta = 30m//s, u_(y) = u sin theta = 40m//s`, `tan theta = (4)/(3)` `x = u cos theta t …(1)` `y = usin theta t - (1)/(2) "gt"^(2) ….(2)` From (1) & (2) `y = x tan theta - (1)/(2) (gx^(2))/((ucos theta)^(2))` `y = (4)/(3) x- (gx^(2))/(2xx(30)^(2))rArr y = (4)/(3) x- (gx^(2))/(1800)` `rArr y = (4x)/(3) - (x^(2))/(180)` `180y = 240 x - x^(2)` |
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| 422. |
All the particles thrown with same initial velocity would strike the ground. .A. with same speed.B. simultaneouslyC. time would be least for the particle thrown with velocity `v` downward i.e., particle `1`.D. time would be maximum for the particle `2`. |
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Answer» Correct Answer - A::C::D (a.,c.,d.) `(KE + PE)_f = (KE + PE)_i` in all situations. Hence, `KE_f` is also equal as `PE_f = 0`. Hence, all the particles collide with the same speed. `-h = vt_1 -(1)/(2) "gt"_1^2` [for first particle] …(i) `-h = -vt_2 - (1)/(2) "gt"_1^2` [for second particle] ...(ii) From Eq. (i) and Eq. (ii), `t_2 gt t_1` `t_2` = maximum, `t_1` = minimum i.e., options ( c) and (d) are correct. |
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| 423. |
A person drops a stone from a building of height 20 m. At the same instant the front end of a truck passes below the building moving with constant acceleration of `1 m//s^(2)` and velocity of `2 m//s` at that instant. Length of the truck if the stone just misses to hit its rear part is :-A. `6 m`B. `4 m`C. `5 m`D. `2 m` |
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Answer» Correct Answer - A Time of fall of stone `=sqrt((2xx20)/10)=2` sec Horizontal displacement of truck in 2 sec `rArr S=2xx2+1/2xx1xx4` Length of truck `=6m` |
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| 424. |
A particle starts moving in +ve x direction with initial velocity of `10 ms^(-1)` with a uniform acceleration of magnitude `2 ms^(-2)` but directed in -ve x direction. What is the distance traversed by the particle in 12 seconds:A. `-24 m`B. `24 m`C. `70 m`D. `74 m` |
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Answer» Correct Answer - D |
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| 425. |
A body is moving with velocity `30 m//s` towards east. After `10 s` its velocity becomes `40 m//s` towards north. The average acceleration of the body is.A. ` 1 m//s^2`B. ` 7 m//s^2`C. ` sqrt 7 m//s^2`D. `5 m//s^@` |
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Answer» Correct Answer - B Velocity towaeds East,` vec v_1 = 30 hat I m//s` Velocity towaeds Noth, ` vec v_2 = 40 hat j m//s` Change in velocity, ` Delta vec v= vec r- vec v-1 = ( 40 hat j - 30 hat i)` ` |Delta vec v|= | 40 hat j - 30 aht i| = sqrt ( 40^2 + (- 30)^2) = 50 m//s` Average acceleration , ` a_(av) = (| Delta vec v|/(Deltat) = ( 50 m//s)/( 10 s) = 5 m//s^2`. |
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| 426. |
Acceleration vs time graph is shown in the figure for a particle moving along a straight line. The particle is initially at rest. Find the time instant(s) when the particle comes to rest? A. `t=0, 1, 2, 3, 4`B. `t=0, 2, 4`C. `t=1, 3`D. None of these |
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Answer» Correct Answer - B |
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| 427. |
A toy car is moving on a closed track whose curved portions are semicircules of radius 1 m. The adjacent graph describes the variation of speed of the car with distance moved by it (starting from point P). The time `t` required for the car to complete one lap is equal to 6K second. Find K. (take `piln 2~~2`) A. 4B. 8C. 12D. 16 |
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Answer» Correct Answer - B |
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| 428. |
A lift is coming from `8th` floor and is just about to reach `4th` floor. Taking ground floor as origin and positive direction upwards for all quantities, which one of the following is correct ?A. (a) ` x lt 0 , v lt 0 , a gt 0`B. (b0 ` x gt 0, lt 0, v lt 0 , a lt 0`C. (c ) ` x gt , v lt 0 , a gt 0`1D. (d0 ` x gt 0, v gt 0, a lt 0` |
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Answer» Correct Answer - A As lift is coming from ` 8th ` to ` 4th` floor, the value of (x) becomes less hence hegatve, i.e. ` xlt 0`. Velocity is downwards (i.e negative). So , ` v lt0`. Before reaching ` 4th` floor lift is retarded, i.e. acceleration is upwards. Hence, ` agt 0`. |
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| 429. |
A jet of water issunig from a hole mear the bottom of water tank is an…………………………. . |
| Answer» example of rojectile | |
| 430. |
In a two dimensional motion, instantaneous speed `v_(0)` is a positive constant. Then which of the following are neccessarily true?A. (a) The acceration of the particle is zeroB. (b) The acceleration of the particle is boundedC. (c ) The acceleration of the particle is necessarily in the plane of motonD. (d) The particle must abe undegoing a unifrom circular motion |
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Answer» Correct Answer - C In two dimensional motion if instantaneous speed is a positive constant, then the acceleration of the particle is necessarily in the plane of motion. |
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| 431. |
A cylclist is riding with a speed of `27 km h^-1`. As he approaches a circular turn on the road of radius `80 m`, he applies brakes and reduces his speed at the constant rate of `0.5 ms^-2`. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn ? |
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Answer» Correct Answer - `tan^-1(5//7)` with radius. `v = 27 xx (5)/(18) = (15)/(2) ms^-1, a_c = ((15//2)2)/(80) = 0.7 ms^-2` `a_t = 0.5 ms^-2` Net acceleration : `a = sqrt(a_c^2 + a_t^2) = 0.86 ms^-2` Direction `tan theta = (a_t)/(a_c) = (0.5)/(0.7)` `theta = tan^-1(5//7)` with radius. |
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| 432. |
The number of ocscillations performed by the bob at the end of 10 s is______A. `4^(1//2)`B. 5C. 6D. `2 ^(1//2)` |
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Answer» Correct Answer - B As the bob is performing `1` oscillation in 2s. The number of oscillations completed in 10s = 10/2 =5` oscillations. |
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| 433. |
Choose the correct statement:A. Unit of acceleration is `ms ^(-2)`B. Speed = distance/ time .C. if a body moves with unifrom elocity, its acceleration is zero.D. All the above |
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Answer» Correct Answer - D Unit of acceleration is ` ms ^(-2)` (2) speed = `("distance")/("time")` (3) As the body is moving with unifrom (constant) velocity ,i.e, there is no change in the velocity of body , and hence, acceletation of the body = zero |
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| 434. |
The number of ocscillations performed by the bob at the end of 10 s is______A. ` 4^(1//2)`B. 5C. 6D. `12^(1//2)` |
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Answer» Correct Answer - B As the time period of the oscillation, (T) = 2s ` T= 2pisqrt(l/g)` `T^(2) = 4pi^(2)l/g` ` l = (gT^(2))/(4pi^(2)) = (10xx4)/(4 xx 10) = 1 m` |
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| 435. |
The frequency of oscillation is ______Hz.A. 0.5B. 2C. 50D. 100 |
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Answer» Correct Answer - A The number of turns wound over the rod = 200 . The diameter of the rod , (d) = 10 cm R = 5cm = 0.5 m The time taken of wind, (t) = 2s The length of the wire wound in one second = `("distance")/("time")` ` = (200 xx 2pir)/2= (200 xx 2pixx0.05)/2 = 10 pi m` |
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| 436. |
A body begins to move with an initial velocity of `2 m s^(-1)` and continues to move at a constant acceleration a.Ten seconds later a second body begins to move from the same point with an initial velocity of `12 m s^(-1)` in the same direction with the same acceleration.What is the maximum acceleration at which it would be possible for the second body to overtake the first ? |
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Answer» Correct Answer - `1 ms^(-2)` |
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| 437. |
Acceleration-time graph of a particle moving in a straight line is as shown in figure. At time `t = 0,` velocity of the particle is zero. Find (a) average acceleration in a time interval from `t = 6 s` to `t = 12 s,` (b) velocity of the particle at `t = 14 s.` |
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Answer» Correct Answer - A::B (a) `a_(av)=(v_f-v_i)/t=(v_12-v_6)/(12-6)` `=((-10)-(20))/6=-5m//s^2` (b)`Deltav=v_f-v_i`=net area of a-t graph `:. v_(14)-v_0=40+30+40-20=90` But `v_0=0` `:. v_(14)=90m//s` |
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| 438. |
Instantaneous velocity of a particle moving in `+x` direction is given as `v = (3)/(x^(2) + 2)`. At `t = 0`, particle starts from origin. Find the average velocity of the particle between the two points `p (x = 2)` and `Q (x = 4)` of its motion path. |
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Answer» Correct Answer - [0.264 m/s] |
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| 439. |
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 are |
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Answer» Correct Answer - `[2 b//c]` |
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| 440. |
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. `x=2a/b,v_(max)=a/(sqrtb)`B. `x=a/2b,v_(max)=a/b`C. `x=a/2b,v_(max)=b/(sqrta)`D. `x=a/b,v_(max)=(sqrta)/b` |
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Answer» Correct Answer - A `f=v.(dv)/(dt)=a-bx` or `int_0^v vdv=int_0^x(a-bx)dx` `:. v=sqrt(2ax-bx^2)…..(i)` At other station, `v=0` `rArr x=(2a)/b` Further acceleration will change its direction when, `f=0 or a-bx=0 or x=a/b` At this x, velocity is maximum. Using Eq. (i), `v_(max)=sqrt(2a(a/b)-b(a/b)^2)=a/sqrtb` |
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| 441. |
A bicycle moves on a horizontal road with some acceleration. The forces of friction between the road and the front and rear wheels are `F_(1)` and `F_(2)` respectively.A. Both `F_(1)` and `F_(2)` act in the forward directionB. Both `F_(1)` and `F_(2)` act in the reverse directionC. `F_(1)` acts in the forward direction, `F_(2)` acts in the reverse direction.D. `F_(2)` acts in the forward direction, `F_(1)` acts in the reverse direction. |
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Answer» Correct Answer - D |
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| 442. |
Velocity and acceleration of a particle are `v=(2 hati-4 hatj) m/s` and `a=(-2 hati+4 hatj) m/s^2` Which type of motion is this? |
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Answer» Correct Answer - A::C::D v and a both are constant vectors. Further, these two vectors are antiparallel. |
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| 443. |
The displacement `x` of particle moving in one dimension, under the action of a constant force is related to the time `t` by the equation ` t = sqrt(x) +3` where `x is in meters and t in seconds` . Find (i) The displacement of the particle when its velocity is zero , and (ii) The work done by the force in the first ` 6 seconds`. |
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Answer» Correct Answer - [0] |
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| 444. |
A machine gun is mounted on the top of a tower of height h=100 m. At what angle should the gun be inclined to cover a maximum range of firing on the ground below ?The muzzle speed of the bullet is `u=150 m s^(-1)` and `g=10 m s^(-2)` . |
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Answer» Correct Answer - `[46.3^(@)]` |
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| 445. |
Two bodies were thrown simultaneously from the same point, one, straight up, and the other, at an angle of `theta=60^@` to the horizontal. The initial velocity of each body is equal to `v_0=25m//s`. Neglecting the air drag, find the distance between the bodies `t=1.70s` later. |
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Answer» Correct Answer - [22 m] |
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| 446. |
A ball is thrown from a point in level with velocity `u` and at a horizontal distance `r` from the top os a tower of height `h`. How must the speed and angle of the projection of the ball be related to `r` in order that the ball may just go grazing the top edge of the tower ? |
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Answer» Correct Answer - `[r g = u^(2) sin 2 theta, (u cos theta)/(g) {(u^(2) theta + 2 gh)^(1//2) - u sin theta}]` |
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| 447. |
The velocity time curve of a moving point is shown in Fig. Find the retardation of the particle for the porion `CD`. . |
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Answer» The slope of the velocity-time graph represents acceleration or positive, it represents accelecration and if the slope is negatitve, positive, it represents acceleation and if the slope is negative, it represents retardation. The section `CD` of the graphe retesebts retardation and magnitude of retardtion is `|vec a|=(Change in veloity)/(Time taken )=(60)/((70-40))=2 m s^(-2)`. |
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| 448. |
A bus starts form rest with an acceleration of `5 m s^(-2)` A man who is on a motorcyle, 24 m behind the bust, overtakes, the bus in 2s. If the motorcyclist moves with uniform velocity, find his velocity |
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Answer» (i) What is the initial velocity and acceleration of the bus ? Find the distance travelled by the bus in 2s, by using the formula, `s=ut+(1)/(2)at^(2)" "(1)` Is the distacne travelled by the motocrycle equal to `(s+24)m? " "(2)` Find the time taken by motorycle to overtake the bus. Find the velocity of the of the motorcycle from the formula, distance/time (ii) Uniform velocity of the motorcyclist `=17 m s^(-1)` |
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| 449. |
Find the average acceleration in first `20 s`. (Hint: Area under `a-t` fraph is equal to the change in velocity). . |
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Answer» Let us first find the charge in velocity Change in velcity is the area under the acceleration=time graph. For first `20 s`, Area `=Delta v (1)/(2) xx10 xx20+10xx20 =300 ms^(-1)` Average acceleration `=(Delta v)/(Delta t) =(300)/(20) =15 ms^(-2)`. |
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| 450. |
A body with constant acceleration always moves along a straight line. A body with constant magnitude of acceleration may not speed up.A. Statement (I) is true, Statement (II) is true , statement (II) is the correct explanation for Statement (I).B. Statement (I) is true, Statement (II) is true , statement (II) is not the correct explanation for Statement (I).C. Statement (I) is true, Statement (II) is false.D. Statement (i) is false, Statement (II) is true. |
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Answer» Correct Answer - D (d) Projectile motion is a motion with constant acceleration but it is not a straight line motion. A body with constant magnitude of acceleration may not speed up, this is possible in uniform circular motion. |
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