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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.
| 351. |
Two vectors `vec(A) and vec(B)` lie in plane, another vector `vec(C )` lies outside this plane, then the resultant of these three vectors i.e., `vec(A)+vec(B)+vec(C )`A. cannot be zeroB. can be zeroC. Lies in the plane of `vec(A)` or `vec(B)`D. Lies in a plane different from that of any of the three vectors |
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Answer» Correct Answer - A::D The resultant of three vectors is zero only if they can form a triangle. But three vectors lying in diffeent planes cannot from a triangle. |
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| 352. |
A particle is simultaneously acted by two forces equal to `4 N` and `3N`. The net force on the particle isA. `7 N`B. `5 N`C. `1 N`D. Between `1 N and `7 N` |
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Answer» Correct Answer - D If two vectors `vec(A)` and `vec(B)` are given then the resultant `R_(max)= A+B= 7N and R_(min)= 4-3= 1 N` i.e., net force on the particle is between `1 N` and `7 N`. |
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| 353. |
Two vectors `vec(A) and vec(B)` lie in plane, another vector `vec(C )` lies outside this plane, then the resultant of these three vectors i.e., `vec(A)+vec(B)+vec(C )`A. Can be zeroB. cannnot be zeroC. Lies in the plane containing `vec(A)+vec(B)`D. Lies in the plane containing `vec(C )` |
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Answer» Correct Answer - B If `vec(C )` lies outside the plane them resultant force can not zero. |
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| 354. |
Which pair of the following forces will never give resultant force of `2 N`?A. `2 N and 2 N`B. `1 N and 1 N`C. `1 N and 3 N`D. `1 N and 4 N` |
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Answer» Correct Answer - D If two vectors A and B are given then range of their resultant can be written as `(A-B) le Rle(A+B)` i.e., `R_(max)= A+B and R_(min)= A-B` If `B=1` and `A= 4` then their resultant will lies in between `3N` and `5N`. It can never been `2N` If therse three vector are represented by three sides of triangle then they from equilateral triangle. |
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| 355. |
If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vector, the angle between these Vector isA. `180^(@)`B. `0^(@)`C. `90^(@)`D. `45^(@)` |
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Answer» Correct Answer - C `|vec(A)+vec(B)|=|vec(A)-vec(B)|= theta= 90^(@)` `(A)^(2)+(B)^(2)+2(A)(B) cos theta=(A)^(2)+(B)^(2)-2(A)(B) cos theta` `2 cos theta=0 implies theta= 90^(@)` |
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| 356. |
Three points with position vectors `vec(a), vec(b), vec(c ) ` will be collinear if there exist scalars x, y, z such thatA. x a + y b = z cB. xa + yb + zc = 0C. xa + yb + zc = 0D. xa + yb = c |
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Answer» Correct Answer - C Let A, B and C be the points with position vectors a, b and c respectively. These points will be collinear, if `AB = lambda AC` `implies b-a=lambda(c-a)` `implies (lambda-1)a+b+(-lambda)c=0` `implies xa+yb+zc=0` where, `x = lambda - 1, y = 1, z = - lambda` `implies xa + yb + zc = 0` such that x + y + z = 0 |
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| 357. |
Find the area of the parallelogram whose diagonals are represented by the vectors `vec(d)_(1)=(2 hat(i) - hat(j)+ hat(k)) and vec(d)_(2) = (3 hat(i) + 4 hat(j) - hat(k)).` |
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Answer» Given that `vec(d)_(1)=(2hat(i) - hat(j) + hat(k)) and vec(d)_(2) =(3 hat(i) + 4 hat(j) - hat(k)).` `" Vector area of the || gm is "1/2(vec(d)_(1) xxvec(d)_(2)).` Now,`(vec(d)_(1) xx vec(d)_(2)) = |(hat(i), hat(j), hat(k)),(2,-1,1),(3,4,-1)|` `=(1-4)hat(i) - (-2-3) hat(j) + (8+3) hat(k)` `= (-3hat(i) + 5 hat(j)+ 11 hat(k)).` Required area `=1/2|vec(d) xx vec(d)_(2)|` `= 1/2 sqrt((-3)^(2)+ 5^(2)+ (11)^(2)) "sq units "`. `= 1/2 sqrt(155) " sq units ".` |
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| 358. |
If `vec(A)` and `vec(B)` are two vectors, which of the following is not correct?A. `vec(A)+vec(B)= vec(B)+vec(A)`B. `vec(A).vec(B)= vec(B).vec(A)`C. `vec(A)xxvec(B)= vec(B)xxvec(A)`D. `vec(A)-vec(B)= -(vec(B)-vec(A))` |
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Answer» Correct Answer - C Vector product of two vectors in not commulative. i.e., `vec(A)xxvec(B)!=vec(B)xxvec(A)` `vec(A)xxvec(B)= -vec(B)xxvec(A)` |
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| 359. |
Unit vector `hat(P)` and `hat(Q)` are inclined at an angle `theta`. Prove that `|hat(P)-hat(Q)|= 2 sin(theta//2)`.A. `(2 sin)(theta)/2`B. `(2 cos)(theta)/2`C. `(2 tan)(theta)/2`D. `tan theta` |
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Answer» Correct Answer - A `|hat(A)-hat(B)|^(2)= (hat(A)-hat(B)).(hat(A)-hat(B))` `=hat(A).hat(A)-hat(A).hat(B)-hat(B).hat(A)+hat(B).hat(B)` `= 1-hat(A).hat(B)-hat(A).hat(B)+1` `=2-2 cos theta= 2(1-cos theta)` `=2((2sin)^(2)(theta)/2)= 4 (sin^(2))(theta)/2` `hat(A)-hat(B)| = (2sin^(2))(theta)/2` |
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| 360. |
If for two vectors `vec(A)` and `vec(B), vec(A)xxvec(B)=0`, the vectorsA. Are perpendicular to each otherB. Are parallel to each otherC. Act at an angle of `60^(@)`D. Act at an angle of `30^(@)` |
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Answer» Correct Answer - B `vec(A)xxvec(B)= 0 :. Sin theta= 0 implies theta= 0^(@)` Two vectors will be parallel to each other. |
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| 361. |
The angle made by the vecotr `vec(A)= hat(i)+hat(j)` with x-axis isA. `90^(@)`B. `45^(@)`C. `22.5^(@)`D. `30^(@)` |
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Answer» Correct Answer - B `vec(A)= hat(i)+hat(j)implies|A|= sqrt(1^(2)+1^(2))= sqrt(2)` `cos alpha= (A_(x))/(|A|)= 1/(sqrt(2))= cos 45^(@)` |
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| 362. |
A person aiming to reach the exactly opposite point on the bank of a stream is swimming with a speed of `0.5(m)/(s)` at an angle of `120^@` with the direction of flow of water. The speed of water in the stream isA. `1m//s`B. `0.5m//s`C. `0.25m//s`D. `0.433m//s` |
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Answer» Correct Answer - C |
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| 363. |
Work done by a force F on a body is W = F .s, where s is the displacement of body. Given that under a force `F = (2 hat I +3 hat j +4 hat k)` N a body is displaced from position vector `r_1 = (2 hat I +3 hat j + hat k)` m to the position vector `r_2 = (hat i +hat j+ hat k)` m. Find the work done by this force. |
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Answer» The body is displaced from `r_1 or r_2`. Therefore, displacement of the body is `s = r_2 - r_1 = (hat I +hat j + hat k)-(2hati+3hatj+ hat k) = (-hati- 2 hat j)m` Now, work done by the force is `W = F.s ` `=(2hati+3hatj+4hatk).(-hat I - 2 hatj)` `=(2)(-1) + (3)(-2) + (4)(0) = -8 J` |
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| 364. |
`vecA=2hati+hatj`,`B=3hatj-hatk` and `vecC=6hati-2hatk`. value of `vecA-2vecB+3vecC` would beA. `20hati+5hatj+4hatk`B. `20hati-5hatj-4hatk`C. `4hati+5hatj+20hatk`D. `5hati+4hatj+10hatk` |
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Answer» Correct Answer - B (b)` A=2hati+hatj,B=3hatj-hatkand c=6hati-2hatk,` `A-2B+3C=(2hati+hatj)-2(3hatj-hatk)+3(6hati-2hatk)` `=2hati+hatj+2hatk+18hati-6hatk` `=20hati-5hatj-4hatk` |
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| 365. |
For the resultant of the two vectors to be maximum, what must be the angle between themA. `0^(@)`B. `60^(@)`C. `90^(@)`D. `180^(@)` |
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Answer» Correct Answer - A |
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| 366. |
`vec(P)+vec(Q)` is a unit vector along x-axis. If `vec(P)= hat(i)-hat(j)+hat(k)`, then what is `vec(Q)`?A. `hati+hatj-hatk`B. `hatj-hatk`C. `hati+hatj+hatk`D. `hatj+hatk` |
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Answer» Correct Answer - B (b) Given ,`P=hati-hatj+hatk` and `P+Q=hatiimpliesQ=hati-hati+hatj+hatk=hatj-hatk` |
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| 367. |
The vector component of vector `vec(A) = 3 hat(i) + 4 hat(j) + 5 hat(k)` along vector `vec(B) = hat(i) + hat(j) + hat(k)` isA. `2 hat (i) + 2 hat(j) + 2 hat(k)`B. `3 hat (i) + 3 hat(j) + 3 hat(k)`C. `4 hat (i) + 4 hat(j) + 4 hat(k)`D. `5 hat (i) + 5 hat(j) + 5 hat(k)` |
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Answer» Correct Answer - C Vector component of `vec(A) along vec(B)` is `{|vec(A)| cos theta|} hat(B) = ( vec(A) . hat(B)) hat(B)` `hat(B) = (vec(B))/(|vec(B)|) = ( i + j + k)/(sqrt( 1 + 1 +1)) = ( i + j + k)/(sqrt(3))` `vec(A) . hat(B) = ( 3 i + 4 j + 5 k).((i + j + k)/(sqrt(3)))` `= ( 3 + 4 + 5)/(sqrt(3)) = (12)/(sqrt(3))` `(vec(A) . hat(B)) hat(B) = ((12)/(sqrt(3))) ( i + j + k)/(sqrt(3)) = 4 ( i + j + k)` |
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| 368. |
What are the minimum number of froces (all mumerically equal) whose vector sun can be zero ?A. TwoB. ThreeC. FourD. Any |
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Answer» Correct Answer - B (b) Minium three forces of unequal magniyude are required to make vector sum rqual to zero. |
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| 369. |
What is the angle between `vecP` and the resultant of `(vecP+vecQ)` and `(vecP-vecQ)`A. ZeroB. `tan^(-1)(P//Q)`C. `tan^(-1)(Q//P)`D. `tan^(-1)(P-Q)//(P+Q)` |
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Answer» Correct Answer - A |
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| 370. |
The expression `(1/(sqrt(2))hat(i)+1/(sqrt(2))hat(j))` is aA. unit vectorB. unll vectorC. vector of magnitude`sqrt(2)`D. Scalar |
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Answer» Correct Answer - A (a) Let `A=(1) /(sqrt(2))hati+(1)/(sqrt(2))hatj` `|A|=sqrt(((1)/(sqrt(2)))+((1)/(sqrt(2)))^(2))=sqrt((2)/(2))=1` As the magnitude o fgiven vetor is 1. `therfore ` It is a unit vector.` |
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| 371. |
Suppose `veca` is a vector of magnitude 4.5 unit due north. What is the vector (a) `3veca` (b) `-4 veca`? |
| Answer» Correct Answer - A::B::C::D | |
| 372. |
The expression `(1/(sqrt(2))hat(i)+1/(sqrt(2))hat(j))` is aA. Unit vectorB. Null vectorC. Vector of magnitude sqrt(2)`D. Scalar |
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Answer» Correct Answer - A `vec(P)= 1/sqrt(2)hat(i)+1/sqrt(2)hatj :. |vec(P)|= sqrt((1/(sqrt(2)))^(2)+(1/sqrt(2))^(2))=1` Hence, it is a unit vector |
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| 373. |
The unit vector along `hat(i)+hat(j)` isA. `hat(k)`B. `hat(i)+hat(j)`C. `(hat(i)+hat(j))/(sqrt(2))`D. `(hat(i)+hat(j))/(2)` |
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Answer» Correct Answer - C `vec(R )=(vec(R ))/(|vec(R)|)=(hat(i)+hat(j))/(sqrt(1^(2)+1^(2)))=1/(sqrt(2))hat(i)+1/(sqrt(2))hat(j)` |
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| 374. |
The magnitude of a given vector with end points`(4, -4,0)` and `(-2, -2,0)` must beA. 6B. `5sqrt(2)`C. 4D. `2sqrt(10)` |
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Answer» Correct Answer - D `vec(r)=vec(r)_(2)-vec(r)_(1)=(-2hat(i)-2hat(j)+0hat(k))-(4hat(i)-4hat(j)+0hat(k))` `impliesvec(r)= -6hat(i)+2hat(j)+0hat(k)` `:. |vec(r)|= sqrt((-6)^(2)+(2)^(2)+0^(2))` `=sqrt(36+4)=sqrt(40)= 2sqrt(10)` |
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| 375. |
Two vectors A and B have magnitudes 2 units and 4 units respectively. Find A. B is angle between these two vectors is (a) `0^(@)` (b) `60^(@)` (c) `90^(@)` (d) `120^(@)` . |
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Answer» Correct Answer - A::B::C::D `A.B = AB cos theta` |
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| 376. |
The resultant of `vecP` and `vecQ` is `vec R`. If `vecQ` is doubled, `vecR` is doubled, when `vecQ` is reversed, `vecR` is again doubled, find P : Q : R, |
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Answer» Let `theta` be the angle between `vec(P)` and `vec(Q)`.Then `R^(2)=|vec(p)+vec(Q)|^(2)=P^(2)+Q^(2)+2PQ cos theta`…..(i) If `vec(Q)` is doubled, `vec(R)` is doubled. That means ,the magnitude of resultant of `2vec(Q)` and `vec(P)` is `(2R)^(2)=P^(2)+(2Q)^(2)+2P(PQ)cos theta` This Yields `4R^(2)=P^(2)+4Q^(2)+4PQcos theta`....(ii) When `vec(Q)` is reversed,`vec(R)` is doubled. Hence, the magnitude of resultant of `vec(P)` and `(-vec(Q))`is `2R`. Then `(2R)^(2)=P^(2)+Q^(2)+2PQcos(180^(@)-theta)` This yield `4R^(2)=P^(2)+Q^(2)-2PQcos theta`....(iii) (ii)-(i) yields `3Q^(2)+2PQcos theta=3R^(2)`....(iv) (i)+(iii) yield `P^(2)+Q^(2)=(5R^(2))/(2)`...(v) (iii)+(iv) yield `P^(2)+4Q^(2)=7R^(2)`....(vi) Solving (v) and (vi), we obtain `Q=sqrt(3/2)R`and `P=R`. Hence,P:Q:R=`sqrt(2):sqrt(3):sqrt(2)`. |
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| 377. |
The resultant of `vecP` and `vecQ` is perpendicular to `vecP`. What is the angle between `vecP` and `vecQ`A. `cos^(-1)(P//Q)`B. `cos^(-1)(-P//Q)`C. `sin^(-1)(P//Q)`D. `sin^(-1)(-P//Q)` |
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Answer» Correct Answer - B |
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| 378. |
If the resultant of the two forces has a magnitude smaller than the magnitude of larger force ,then two forces must beA. Different both in magnitude and directionB. Mutually perpendicular to one directionC. Posses extremely small magnitudeD. Point in opposite directions |
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Answer» Correct Answer - D |
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| 379. |
Area of the parallelogram formed by vectors `vecA = hati + 2hatj+ 4 hatk` and `vecB= 3 hati - 2hatj` is :A. `4sqrt(17) "unit"`B. `2sqrt(17) "unit"`C. `17sqrt(2) "unit"`D. `17sqrt(3) "unit"` |
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Answer» Correct Answer - A `vectau =vecr xx vecF` |
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| 380. |
If the sum and difference of two vectors are at right angles, show that the vectors are equal in magnitude.A. perpendicular to each otherB. parallel to each otherC. of same magnitudeD. of unequal magnitude |
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Answer» Correct Answer - B Possible when `theta = 0^(@)` |
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| 381. |
The sum of two vectors A and B is at right angles to their difference. ThenA. A=BB. A=2BC. B=2AD. A and B have the same direction |
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Answer» Correct Answer - A `(A+B) _|_ (A-B)` `:. (A +B).(A-B) = 0` `:. A^2 + BA cos theta - AB cos theta - B^2 = 0` or `A =B` . |
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| 382. |
If A and B are two vectors such that |A+B|=2|A-B|. The angle between vectors A and B isA. `45^(@)`B. `60^(@)`C. `30^(@)`D. data insuffcient |
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Answer» Correct Answer - D (d ) We have `,|A+B|=2|A-B|` `therefore sqrt(A^(2)+B^(2)+2AB costheta)=2sqrt(A^(2)+B^(2)-2AB costheta)` From this equation we cannot find `theta` |
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| 383. |
check Wether the vector `((hati)/sqrt (2)+(hati)/sqrt(2))` is a unit vector or not . |
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Answer» Aunit vector is a vector with magnitude 1. The magnitude of given vector is `|(hati)/(sqrt(2)) +(hati)/(sqrt(2))|=sqrt (((1)/(sqrt(2)))^(2)+((1)/(sqrt(2)))^(2))=sqrt ((1)/(2)+(1)/(2))=sqrt(1)=1` As magnitude of given vector is 1. `therefore `It is a unit vector. |
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| 384. |
Find the unit vector of `4hati-3hatj+hatk.` |
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Answer» We know that , for unit vector `A=A_(x)hati+A_(y)hatj+A_(z)hatk` `=4hati-3hatj+hatk` `therefore A_(x)=4,A_(y)=-3,A_(z)=1` `|A|=sqrt(A_(x)^(2)+A_(y)^(2)+A_(x)^(2))=sqrt((4)^(2)+(-3)^(2)+(1)^(2))=sqrt(26)` `therefore "Unit vector", hatA=(A)/(|A|)=(hat4-3hatj+hatk)/sqrt(26)` |
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| 385. |
The position vector of a particle is given by `vecr=(3t^(2)hati+4t^(2)hatj+7hatk)m` at a given time `t`.The net displacement of the particle after `10 s` isA. `500m`B. `300m`C. `150m`D. `100m` |
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Answer» Correct Answer - A |
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| 386. |
If `|vec(A)+vec(B)|=|vec(A)|=|vec(B)|` then the angle between `vec(A)` and `vec(B)`isA. `90^(@)`B. `120^(@)`C. `0^(@)`D. `60^(@)` |
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Answer» Correct Answer - C Resultant of two vector `vec(A)` and `vec(B)` can be given by `vec(R )= vec(A)+vec(B)` `|vec(R )|=vec(A)+vec(B)|= sqrt(A^(2)+B^(2)+2AB cos theta)` If `theta= 0^(@)` then `|vec(R )|=A+B= |vec(A)|+|vec(B)|` |
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| 387. |
If a vector has an x-component of -250units and a y-component of 40.0 units, then the magnitude and direction of this vector isA. `5sqrt(89)` units, `sin^(-1)((5)/sqrt(89))` with +ve-x-axisB. `5sqrt(89)` units, `cos^(-1)((5)/sqrt(89))` with -ve-x-axisC. 45 units, `cos^(-1)((-5)/(9))` with x-axisD. 45 units, `sin^(-1)((-5)/(9))` with x-axis |
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Answer» Correct Answer - B `A=sqrt(X^(2)+Y^(2)), Tan theta=(y)/(x)` |
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| 388. |
A body, acted upon by a force of 50N, is displaced through a distance 10m in direction making an angle of `60^(@)` with the force. Find the work done by the force. |
| Answer» `W=vec(F).vec(S)=Fscos theta=50xx10xxcos 60^(@)=50xx10xx1/2=250J` | |
| 389. |
A force is inclined at `30^(@)` to the horizontal. If its rectangular component in the horizontal direction is 50N, find the magnitude of the force and its vertical component. |
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Answer» Correct Answer - `57.74 N, 28.87 N` |
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| 390. |
A force is inclined at `60^(@)` to the horozontal. If its rectangular component in the horizontal direction is `50 N`, then magnitude of the verticle components of force is approximatelyA. `25 N`B. `84 N`C. `87 N`D. `90 N` |
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Answer» Correct Answer - C `R_(x)= 50 = R cos 60^(@) impliesR= 100 N` `R_(y)= R sin 60^(@)= (100 sqrt(3))/(2)= 50 sqrt(3) N` `=86.6 N` ~~ 87 N` |
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| 391. |
A force of `5 N` acts on a particle along a direction making an angle of `60^(@)` with verticle. Its verticel components isA. `10N`B. `3N`C. `4N`D. `2.5N` |
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Answer» Correct Answer - D |
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| 392. |
दर्शाइए कि बिंदु `A(1, 2, 7), B(2, 6, 3)` और `C(3, 10, -1)` संरेख हैं |A. collinearB. coplanarC. non-collinearD. None of these |
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Answer» Correct Answer - A The given points are A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1). Now, `AB=PV" of "B - PV " of "A` `vec(AB)=(2hati+6hatj+3hatk)-(1hati+2hatj+7hatk)` `=hati+4hatj-4hatk` `implies |AB|=sqrt((1)^(2)+(4)^(2)+(-4)^(2))` `=sqrt(1+16+16)=sqrt(33)` `BC=PV" of "C-PV" of "B` `=(3hati+10hatj-1hatk)-(2hati+6hatj+3hatk)` `=hati+4hatj-4hatk` `implies |BC|=sqrt((1)^(2)+(4)^(2)+(-4)^(2))` `=sqrt(1+16+16)=sqrt(33)` `AC=PV" of "C-PV" of "A` `vec(AC)=(3hati+10hatj-1hatk)-(1hati+2hatj+7hatk)` `=2hati+8hatj-8hatk` `implies |AC|=sqrt(2^(2)+8^(2)+(-8)^(2))=sqrt(4+64+64)` `=sqrt(132)=2sqrt(33)=sqrt(33)+sqrt(33)` `therefore |AC|=|AB|+|BC|` Hence, the given points A, B and C are collinear. |
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| 393. |
The vectors `2 hati + 3 hatj , 5 hati + 6hatj and 8 hatj + lambda hatj ` have their initial points at (1, 1). The value of `lambda` so that the vectors terminate on one straight line, is |
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Answer» Since the vectors `2hati+3hatj and 5hati+6hatj` have (1, 1) as the initial point, their terminal points are (3, 4) and (6, 7), respectively. The equation of the line joining these two points is `x-y+1=0`. The terminal point of `8hati+lamdahatj` is `(9, lamda + 1)`. Since the vectors terminate on the same straight line, `(9, lamda+1)` lies on `x-y+1=0`. Therefore, `" "9-lamda-1+1=0` or `" "lamda=9` |
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| 394. |
The position vectors of `P` and `Q` are respectively `a` and `b`. If `R` is a point on `PQ` such that `PR=5PQ`, then the position vector of `R` isA. 5b - 4aB. 5b + 4aC. 4b - 5aD. 4b + 5a |
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Answer» Correct Answer - A Given, PR = 5 PQ It means R divides PQ externally in the ratio 5 : 4. `therefore` Position vector of `R=(5b-4a)/(5-4)=5b-4a` |
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| 395. |
Find a unit vector `vecc` if `-hati+hatj-hatk` bisects the angle between vectors `vecc` and `3hati+4hatj`. |
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Answer» Let `vecc=xhati+yhatj+zhatk`, where `x^(2)+y^(2)+z^(2)=1`. Unit vector along `3hati+4hatj` is `(3hati+4hatj)/(5)`. The bisector of these two is `-hati+hatj-hatk` (given). Therefore, `" "-hati+hatj-hatk=lamda(xhati+yhatj+zhatk+(3hati+4hatj)/(5))` `" " -hati+hatj-hatk=(1)/(5) lamda [(5x+3)hati+(5y+4)hatj +5zhatk]` `" "(lamda)/(5)(5x+3)=-1, (lamda)/(5)(5y+4)=1, (lamda)/(5)5z=-1` `" "x=-(5+3lamda)/(5lamda), y=(5-4lamda)/(5lamda),z=-(1)/(lamda)` Putting these values in (i), i.e., `x^(2)+y^(2)+z^(2)=1`, we get `" "(5+3lamda)^(2)+(5-4lamda)^(2)+25=25lamda^(2)` `" "25lamda^(2)-10lamda+75=25lamda^(2)` `" "lamda=(15)/(2)` `therefore" "vecc=(1)/(15)(-11hatik+10hatj-2hatk)` |
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| 396. |
The non-zero vectors are `vec a,vec b and vec c` are related by `vec a= 8vec b and vec c = -7vec b`. Then the angle between `vec a and vec c` isA. `pi`B. `0`C. `(pi)/(4)`D. `(pi)/(2)` |
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Answer» Correct Answer - A Since, a = 8b and c = - 7b `therefore` a is parallel to b and c is anti-parallel to b. `implies `a and c are anti-parallel `implies` Angle between a and c is `pi`. |
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| 397. |
`0.4hat(i)+0.8hat(j)+chat(k)` represents a unit vector, when c is :-A. `-0.2`B. `sqrt(0.2)`C. `sqrt(0.8)`D. 0 |
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Answer» Correct Answer - A |
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| 398. |
Let `hata and hatb` be two unit vectors. If the vectors `vecc=hata+2hatb and vecd=5hata-4hatb` are perpendicular to each other then the angle between `hata and hatb` is (A) `pi/2` (B) `pi/3` (C) `pi/4` (D) `pi/6`A. `(pi)/(6)`B. `(pi)/(2)`C. `(pi)/(3)`D. `(pi)/(4)` |
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Answer» Correct Answer - C Given, c and d are perpendicular to each other, `therefore c*d = 0` `rArr (hata + 2hatb)*(5 hata - 4 hatb) = 0 ` `rArr 5 hata * hata - 4 hat a *hat b + 10 hat b * hata - 8 hatb * hatb = 0 ` ` rArr 6 hata * hatb = 3 ` `rArr hata *hatb = (1)/(2)` So, the angle between `hata and hatb` is `(pi)/(3)`. |
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| 399. |
The vector `veca+vecb` bisects the angle between the vectors `hata and hatb` if (A) `|veca|+|vecb|=0` (B) angle between `veca and vecb` is zero (C) `|veca|=|vecb|=0` (D) none of these |
| Answer» We know that vector `veca+vecb` is along the diagonal of the parallelogram whose adjacent sides are vectors `veca and vecb`. Now if `veca+vecb` bisects the angle between vectors `veca and vecb`, then the parallelogram must be a rhombus , hence, `|veca|=|vecb|`. | |
| 400. |
Which of the following is a vectorA. PressureB. Surface tensionC. Moment of inertiaD. None of these |
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Answer» Correct Answer - D |
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