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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.
| 201. |
An object falls from a height h (h<<R), the speed of the object when it reaches the ground is –(a) \(\frac{1}{2} gt^2\)(b) \(\sqrt{gt}\)(c) gh(d) \(\sqrt{2gh}\) |
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Answer» Correct answer is (d) \(\sqrt{2gh}\) |
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| 202. |
If P = mV then the direction of P along-(a) m (b) v (c) both (a) and (b) (d) neither m nor v |
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Answer» Correct answer is (b) v |
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| 203. |
The scalar product \(\vec A\). \(\vec B\) is equal to-(a) \(\vec A\, + \, \vec B\)(b) \(\vec A . \vec B\)(c) AB sin θ(d) (\(\vec A\) x \(\vec B\)) |
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Answer» Correct answer is (b) \(\vec A . \vec B\) |
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| 204. |
The direction of \(\vec A\)x \(\vec B\) is given by- (a) right hand screw rule (b) right hand thumb rule (c) both (a) and (b) (d) neither (a) and (b) |
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Answer» (c) both (a) and (b) |
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| 205. |
\(\vec A \times \vec B\) is equal to - (a) \(\vec B \times \vec A\)(b) \(\vec A\) + \(\vec B\)(c) - \(\vec B\) x \(\vec A\)(d) \(\vec A\) - \(\vec B\) |
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Answer» (c) \(- \vec B \times \vec A\) |
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| 206. |
For free falling body, its initial velocity is – (a) 0 (b) 1 (c) ∞ (d) none |
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Answer» Correct answer is (a) 0 |
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| 207. |
\(\hat i \times \hat i\) is -(a) 0(b) i(c) ∞(d) \(\hat j\) |
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Answer» Correct answer is (a) 0 |
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| 208. |
The scalar product \(\vec A\).\(\vec B\) is equal to- (a) \(\vec A\) + \(\vec B\) (b) AB sin θ (c) AB cos θ (d) \(\vec A\) + \(\vec B\) |
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Answer» Correct answer is (c) AB cos θ |
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| 209. |
\(\vec A \times \vec B\) is - (a) AB cos θ (b) AB sin θ (c) AB tan θ (d) AB sec θ |
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Answer» Correct answer is (b) AB sin θ |
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| 210. |
If vector P– vector Q then which of the following is incorrect. – (a) \(\hat P = \hat Q\)(b)\(|\hat P| = |\hat Q|\)(c) P\(\hat Q\) = \(Q \hat A\)(d) \(\hat P \hat Q = PQ\) |
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Answer» Correct answer is (d) \(\hat P \hat Q = PQ\) |
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| 211. |
The product of a vector with itself is equal to – (a) 0 (b) 1 (c) ∞ (d) A2 |
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Answer» Correct answer is (a) 0 |
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| 212. |
The magnitude of the vector is – (a) A2 (b) √A2 (c) √A (d) \(\sqrt [3]{A}\) |
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Answer» Correct answer is (b) \(\sqrt{A^2}\) |
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| 213. |
The magnitude of a vector can not be- (a) positive (b) negative (c) zero (d) 90 |
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Answer» Correct answer is (b) negative |
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| 214. |
The momentum of a particle is \(\vec P\) = cos θ + sin θ . The angle between momentum and the force acting on a body is –(a) 0° (b) 45° (c) 90° (d) 180° |
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Answer» Correct answer is (c) 90° |
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| 215. |
The vector product of two non-zero vectors will be minimum when O is equal to –(a) 0° (b) 180° (c) both (a) and (b) (d) neither (a) nor (b) |
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Answer» Correct answer is (c) both (a) and (b) |
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| 216. |
A and B are two vectors, if A and B are perpendicular to each other then –(a) \(\vec A \times \vec B\) = 0(b) \(\vec A \times \vec B\) = 1(c) \(\vec A \,\vec B\) = 0(d) \(\vec A \,\vec B\) = \(\vec A \,\vec B\) |
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Answer» (c) \(\vec A \, \vec B\) =0 |
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| 217. |
The radius vector is 2\(\hat i\) +\(\hat j\) + \(\hat k\) while linear momentum is 2\(\hat i\) +3\(\hat j\) + \(\hat k\) Then the angular momentum is(a) \(-2 \hat i + 4 \hat k\) (b) \(4\hat i - 8 \hat k\) (c) \(2 \hat k - 4 \hat j + 2\hat k\) (d) \(4 \hat i - 8\hat j\) |
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Answer» Correct answer is (a) \(-2 \hat i + 4 \hat k\) |
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| 218. |
Unit vector is – (a) having magnitude one but no direction (b) \(A \widehat A\)(c) \(\frac{\widehat A}{A}\)(d) |A| |
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Answer» Correct answer is (c) \(\frac{\widehat A}{A}\) |
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| 219. |
The angle between two vectors -3\(\hat i\) + 6\(\hat k\) and 2\(\hat i\) + 3\(\hat k\) + is –(a) 0° (b) 45° (c) 60° (d) 90° |
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Answer» Correct answer is (d) 90° |
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| 220. |
If \(|\vec P \times \vec Q|\) =\(|\vec P . \vec Q|\) then angle between \(\vec P\)and \(\vec Q\) then angle between P and Q will be -(a) 0° (b) 30° c) 45° (d) 60° |
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Answer» (c) 45° \(|\vec P \times \vec Q|\) = \(|\vec P . \vec Q|\) Expand the terms PQ sinθ = PQ cos θ tan θ = 1 θ = 45° |
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| 221. |
The angle between parallel vectors is – (a) 0°(b) 90° (c) 180° (d) 0° (or) 180° |
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Answer» Correct answer is (a) 0° |
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| 222. |
If \(|\vec P + \vec Q|\) = \(|\vec P|\) - \(|\vec P|\), then the angle between the vectors \(\vec P\) and \(\vec Q\)(a) 0° (b) 90°(c) 180° (d) 360° |
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Answer» (c) 180° \(|\vec P + \vec Q|\) = \(|\vec P|\) \(|\vec P|\) Square on both side, and the resultant becomes P2 + Q2 + 2PQ cos θ = P2 + Q2 – 2PQ . cos θ = -1 θ = 180° |
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| 223. |
The magnitude of distance is always- (a) positive (b) negative (c) either positive (or) negative (d) zero |
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Answer» Correct answer is (a) positive |
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| 224. |
Give an example of a physical wuantity (i) which has neigher unit nor direction (ii) has a dircetion but bot a vector (iii) can be either a vector or a scalar. |
| Answer» (i) Spectfic gravity (ii) pressure or current (iii) angular displacement. | |
| 225. |
Define a scalar. Give examples. |
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Answer» Scalar is a property which can be described only by magnitude. Example – mass, distance, speed. |
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| 226. |
Write a short note on vector product between two vectors. |
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Answer» The vector product or cross product of two vectors is defined as another vector having a magnitude equal to the product of the magnitudes of two vectors and the sine of the angle between them. The direction of the product vector is perpendicular to the plane containing the two vectors, in accordance with the right hand screw rule or right hand thumb rule. Thus, if \(\vec A\) and \(\vec B\) are two vectors, then their vector product is written as \(\vec A\)x \(\vec B\) which is a vector C defined by \(\vec c\) = \(\vec A\)x \(\vec B\) =(AB sin 0) \(\hat n\) The direction \(\hat n\) of \(\vec A\)x \(\vec B\), i.e., \(\vec c\) is perpendicular to the plane containing the vectors \(\vec A\) and \(\vec B\). |
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| 227. |
Referring `a-s` diagram in , find the velocity after particle travel `120 m` from starting. Assume `v_(0)=0`. . |
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Answer» `a=-(s)/(20)+200` `underset(0)overset(v)(int)v dv =underset(0)overset(250)(int)(-(s)/(20)+20)ds rArr (v^(2))/(2)=[-(s^(2)/(40)+20)s]_(0)^(250)` `(v^(2))/(2)=-((250)^(2))/(40)+20 xx250` `rArr v^(2)= 6875 rArr v=25 sqrt11 ms^(-1)`. |
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| 228. |
Statement-I : A projectile is thrown with an initial velocity of `(ahat(i)+bhat(j)) m//s`. If range of projectile is maximum then `a=b`. Statement-II : In projectile motion, angle of projection is equal to `45^(@)` for maximum range condition.A. Statement-I is true, Statement-II is true, Statement-II is correct explanation for Statement-IB. Statement-I is true, Statement-II is true, Statement-II is NOT a correct explanation for Statement-IC. Statement-I is true, Statement-II is falseD. Statement-I is false, Statement-II is true. |
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Answer» Correct Answer - A For max. range `((u^(2) sin 2theta)/g)`, the projection angle `(theta)` should be `45^(@)` So initial velocity `ai+bjrArr tan 45^(@)=b/arArr a=b` |
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| 229. |
Vector is having – (a) only magnitude (b) only direction (c) bot magnitude and direction (d) either magnitude or direction |
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Answer» (c) both magnitude and direction |
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| 230. |
Write a short note on the scalar product between two vectors. |
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Answer» The scalar product (or dot product) of two vectors is defined as the product of the magnitudes of both the vectors and the cosine of the angle between them. Thus if there are two vectors \(\vec A\) and \(\vec B\) having an angle 0 between them, then their scalar product is defined as \(\vec A\). \(\vec B\)= AB cos0. Here, AB and are magnitudes of \(\vec A\) and \(\vec B\). |
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| 231. |
A passenger reaches the platform and finds that the second least boggy of the train is passing him. The second last boggy takes `3 s` to pass the passenger, and the last boggy takes `2 s` to pass him. Find the time by which the passnger late for the departure of the trains? Assume that the train accelerates at constant rate and all the boggies are of equal length. |
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Answer» Correct Answer - `3.5 s` `(n-2)l=(1)/(2) at^(2)` `(n-1)l=(1)/(2)a(t+3)^(2)` `nl=(1)/(2)a (t+5)^(2)` `l=(1)/(2)a [(1+3)^(2)-t^(2)]=(1)/(2) a[(t+5)^(2)-(t+3)^(2)]` `t^(2)+9+6t-t^(2)=t^(2)+25+10 t-t^(2)-9-6t` `9+6t =16 +4t` `2t=7` `t=3.5 sec`. |
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| 232. |
Referring to `a-s` diagram as shown in , findthe velocity of the particle when the particle when the spaarticle justcovers `20 m`, `(v_(0)=sqrt50 ms^(-1)`. . |
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Answer» Correct Answer - `sqrt(350) m s^(-1)` `a =- (s)/(4) +10 rArr underset(v_(0)) overset(v) (int) (dv)/(ds) =-(s)/(4) +10` ` rArr underset(v_(0))overset(v)(int) v dv =underset(0)overset(20) (int) (-(s)/((ds)+10)ds` `rArr (v^(2)-v_(0)^(2))/(2) =[-s^(2)/(8) +10s]_(0)^(20)` ` rArr (v^(2) -50)/(2) =[-(s^(2))/(8)+10s]_(0)^(20)`. |
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| 233. |
Which of the following is not a scalar? (a) Volume(b) angular momentum (c) Relative density (d) time |
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Answer» (b) angular momentum |
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| 234. |
If the magnitude of two vectors are 3 and 4 and their scalar product is 6, find angle between them and also find \(|\vec A \times \vec B|\) |
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Answer» \(\vec A.\vec B\) = AB cos θ \(|\vec A \times \vec B|\) = AB sin θ or 6 = (3 x 4) cos θ = 3 x 4 x 60° or θ = 60° 3 x 4 x \(\frac{\sqrt3}{2}\) = 6\(\sqrt3\) |
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| 235. |
Example for scalar is – (a) distance (b) displacement (c) velocity (d) angular momentum |
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Answer» (a) distance |
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| 236. |
Give an example for scalar product of two vectors. |
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Answer» The work done by a force \(\vec F\) to move an object through a small displacement \(\vec {dr}\) then Work done W = \(\vec F\). \(\vec {dr}\) (or) W = F dr cos θ |
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| 237. |
Which one of the following physical quantities cannot be represented by a scalar?(a) Mass (b) length (c) momentum (d) magnitude of acceleration |
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Answer» Correct answer is (c) momentum |
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| 238. |
State, for each of the following physical quantities, if it is a scalar or a verctor. Volume, mass speed acceleration, density, number of moles, velocity, angular frequencey, displacemnt, angular velocity. |
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Answer» Scalars. Volume, mass, sped, density, number of moles, anular frequency. Vectors. Accelration, velocity displacement, angular velocity. |
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| 239. |
A particle move in `x-y` plane such that its position vector varies with time as `vec r=(2 sin 3t)hat j+2 (1-cos 3 t) hat j`. Find the equation of the trajectory of the particle. |
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Answer» Comparing `vec r=(2 sin 3t)hat j+ 2(1-cos 3t) hat j` With `vec r =xhat I + hat j`, we have `x =2 sin `3t` and `y =2(1-cost)`. This gives `sin3t=(x)/(2)` and `cos 3t =1-(y)/(2)`. Eliminating `t` by squaring and adding the above terms, we have `(x^(2)/(4)+(1-(y^(2))/(2))=1`. |
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| 240. |
Position vector `vec(r)` of a particle varies with time t accordin to the law `vec(r)=(1/2 t^(2))hat(i)-(4/3t^(1.5))hat(j)+(2t)hat(k)`, where r is in meters and t is in seconds. (a) Find suitable expression for its velocity and acceleration as function of time (b) Find magnitude of its displacement and distance traveled in the time interval `t=0` to `4 s`. |
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Answer» (a) Velocity `vec(v)` is defined as the first derivative vector with respect to time. `vec(v)=(dvec(r))/(dt)=t hat(i)-2sqrt(t)hat(j)+2hat(k) m//s` Acceleration `vec(a)` is defined as the first derivative of velocity vector with respect to time. `vec(a)=(dvec(v))/(dt)=hat(i)-1/sqrt(t)hat(j) m//s^(2)` (b) Displacement `Deltavec(r)` is defined as the change in place of position vector. `Deltavec(r)=8hat(i)-32/3hat(j)+8hat(k) m` Magnitude of displacement `Deltar=sqrt(8^(2)+(32/3)^(2)+8^(2))=1.55 m` Distance `Deltas` is defined as the path length and can be calculated by integrating speed over the concerned time interval. `Deltas=underset(0)overset(4)(int)vdt=underset(0)overset(4)(int) sqrt(t^(2)+4t+4)dt=underset(0)overset(4)(int)(t+2)dt=16 m` |
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| 241. |
If a particle takes `t` second less and acquire a velocity of `vms^(-1)` more in falling through the same disance on two planets where the accelerations due to gravity are `2g` and `8g` respectively, then `v=xgt`. Find value of `x` |
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Answer» Correct Answer - 4 |
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| 242. |
The velocity of a particle moving on the `x-` axis is gienv by `v=x^(2)+x(` for `xgt0)` where `v` is in `m//s` and `x` is in `m`. Find its acceleration in `m//s^(2)` when passing through the point `x=2m` |
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Answer» Correct Answer - D |
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| 243. |
A particle move so that its position verctor varies with time as `vec r=A cos omega t hat i + A sin omega t hat j`. Find the a. initial velocity of the particle, b. angle between the position vector and velocity of the particle at any time, and c. speed at any instant. |
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Answer» a. Position at time `t` `vec r=A cos omega t hat i+ A sin omega that j` Instantaneous velocity, `vec v=(dvec r)/(dt)` We have `vec v=A(d)/(dt)(cos omega t)hat i+A (d)/(dt)(sin omega t)hat j` `=-A omega sin 0 hat i+A omega cos t hat j` At `t=0,v=-A omega sin 0 hat i+A omega cos 0 hat i=A omega hat j` b. For calculating the angle between two vectors. The angle `theta` between `vec r` and `vec v` can be given as `the=cos^(-1)(vec r.vec v)/(|vec r||vec v|)` Where `vec r.vec v=(A cos omega t hat i+A sin omega hat j).(-A omega sin omega t hat i+A omega cos omega t hat j)` `=omega A^(2)(-cos omega t sin omega t+sin omega t cos omega t)=0` Hence, `theta=cos t sin omega t+ sin omega t=0` Hence, `theta=cos^(-1) 0=pi //2` That means `vec v_|_ vec r`. c. Speed at any time is the magnitude of instantaneous velocity, i.e., `v=|vec v|=sqrt((-A omega sin omega t)^2+(A omega cos omega t)^2) = A omega`. |
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| 244. |
A body is moving with uniform velocity of `8 ms^-1`. When the body just crossed another body, the second one starts and moves with uniform acceleration of `4 ms^-2`. The time after which two bodies meet will be :A. `2s`B. `4 s`C. `6 s`D. `8 s` |
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Answer» Correct Answer - B b. Let they meet after time `t`, then the diatance travelled by both in time `t` should be same `s=8t=(1)/(2) 4t^(2) rArr t =4 s`. |
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| 245. |
Particle A is moving with a constant velocity of `V_(A) = 50 ms^(–1)` in positive x direction. It crossed the origin at time `t = 10 s`. Another particle B started at `t = 0` from the origin and moved with a uniform acceleration of `a_(B) = 2 ms^( –2)` in positive x direction. (a) For how long was A ahead of B during the subsequent journey? (b) Draw the position (x) time (t) graph for the two particles and mark the interval for which A was ahead of B. |
| Answer» Correct Answer - (a) `10 sqrt(5) s` | |
| 246. |
A car `A` moves with velocit `20 m s^(-1)` and car `B` with velocity `15 m s^(-1)` as shown is. Find the relativety B` w.r.t. `A` w.r.t. `B`. . |
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Answer» Let us assume that right direction is positive. We are given: `v_(A)=20 m s^(-1), v_(B)=15 ms^(-1)`, Relative velocity of `B` w.r.t `A`: `v_(B//A)=v_(B)-v_(A)=15-20=-5 m s^(-1)` (Negative sign indecates that this relative velocity is in the left direction.) Relative velocity of `A` w.r.t `B`: `v_(A//B)=v_(A)-v_(B)=20-15=5 m s^(-1)` (Positive sign indicates that this relative velocity is in the right derction.) Hence, the separation between`A` and `B` is dereasing with time, here the velocity of approach of `A` w.r.t `B` is `5 m s^(-1)`. |
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| 247. |
A person walks up a stalled 15 m long escalator in 90 s. When standing on the same escalator, now moving, the person is carried up in 60 s. How much time would it take that person to walk up the moving escalator? Does the answer depend on the length of the escalator? |
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Answer» Correct Answer - C `v_1`=speed of persion `v_2`=speed pf escalator `v_1=l/(t_1)` and `v_2=l/(t_2)` `:. t=l/(v_1+v_2)=l/((l/t_1)+(l/t_2))` `=(t_1t_2)/(t_1+t_2)=(90xx60)/(90+60)=36s` |
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| 248. |
The acceleration of particle varies with time as shown. (a) Find an expression for velocity in terms of t. (b) Calculate the displacement of the particle in the interval from `t = 2 s` to `t = 4 s.` Assume that `v = 0` at `t = 0.` |
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Answer» Correct Answer - A::B (a) `a=2t-2` (from the graph) Now, `int_0^v dv=int_0^t adt=int_0^t (2t-2)dt` `:. V=t^2-2t` (b) `s=int_2^4 vdt=int_2^4 (t^2-2t)dt=6.67m` |
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| 249. |
A car `A` moves with velocity `15 m s^(-1)` and `B` with velocity `20 m s^(-1)` are moving in opposite directions as shown in . Find the relative velocity of `B` w.r.t. `A` and w.r.t. `B`. . |
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Answer» Here also let us assume right directionis positive, then left direction will be negative. Given: `v_(A)=-15 ms^(-1), v_(B)=20 m s^(-1)` Relative velocity of `B` w.r.t `A`: `v_(A//B)=v_(A)-v_(B)=-15-20=-35 ms^(-1)` (Negative sign indicates that this relative velocity is in back ward direction.) Hence the separation between `A` and `B` is increasing with time, hence velocity of separation is `35 m s^(-1)`. |
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| 250. |
An elevator without a ceiling is ascending up with an acceleration of `5 ms^-2.` A boy on the elevator shoots a ball in vertical upward direction from a height of 2 m above the floor of elevator. At this instant the elevator is moving up with a velocity of `10 ms^-1` and floor of the elevator is at a height of 50 m from the ground. The initial speed of the ball is `15 ms^-1` with respect to the elevator. Consider the duration for which the ball strikes the floor of elevator in answering following questions. (`g=10 ms^-2`) 4. The maximum separation between the floor of elevator and the ball during its flight would beA. 12 mB. 15 mC. 9.5 mD. 7.5 m |
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Answer» Correct Answer - C At maximum separation, their velocities are same `:. 25--10t=10+5t` or `t=1s` Maximum separation =`2+S_2-S_1` `=2+[25xx1-5xx(1)^2]-[10xx1+2.5(1)^2]` `=9.5m` |
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